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SontlOtt : C. J. CLAY AND SONS, 











P. G. TAIT, M.A., SEC. R.S.E., 



rrayciv devdov 




[All Rights reserved.] 



Sambrilige : 



P. xxiv, 1. 18, for Vpp = read Vpp = 0. 

for Vpp y read Vpp = y. 

P. 232, 1. 15, for To=l read Tp = l. 

Ox A a en 




TN the present edition this work has been very greatly en- 
-*- larged ; to the extent, in fact, of more than one-third. Had I 
not determined to keep the book in moderate compass it might 
easily have been doubled in size. A good deal of re-arrangement 
also has been thought advisable, especially with reference to the 
elementary uses of q( )q~\ and of V. Prominent among the 
additions is an entire Chapter, On the Analytical Aspect of Quater 
nions, which I owe to the unsolicited kindness of Prof. Cayley. 

As will be seen by the reader of the former Preface (reprinted 
below) the point of view which I have, from the first, adopted 
presents Quaternions as a Calculus uniquely adapted to Euclidian 
space, and therefore specially useful in several of the most im 
portant branches of Physical Science. After giving the necessary 
geometrical and other preliminaries, I have endeavoured to de- 
velope it entirely from this point of view ; and, though one can 
scarcely avoid meeting with elegant and often valuable novelties 
to whatever branch of science he applies such a method, my chief 
contributions are still those contained in the fifth and the two 
last Chapters. When, twenty years ago, I published my paper 
on the application of V to Greens and other Allied Theorems, I 
was under the impression that something similar must have been 
contemplated, perhaps even mentally worked out, by Hamilton as 
the subject matter of the (unwritten but promised) concluding 
section of his Elements. It now appears from his Life (Vol. ill. 
p. 194) that such was not the case, and thus that I was not in any 
way anticipated in this application (from rny point of view by far 
the most important yet made) of the Calculus. But a bias in 
such a special direction of course led to an incomplete because one 
sided presentation of the subject. Hence the peculiar importance 
of the contribution from an Analyst like Prof. Cayley. 

T.Q.I. 781534 b 


It is disappointing to find how little progress has recently been 
made with the development of Quaternions. One cause, which 
has been specially active in France, is that workers at the subject 
have been more intent on modifying the notation, or the mode of 
presentation of the fundamental principles, than on extending the 
applications of the Calculus. The earliest offender of this class 
was the late M. Hoiiel who, while availing himself of my permis 
sion to reproduce, in his Theorie des Quantite s Complexes, large parts 
of this volume, made his pages absolutely repulsive by introducing 
fancied improvements in the notation. I should not now have 
referred to this matter (about which I had remonstrated with M. 
Hoiiel) but for a remark made by his friend, M. Laisant, which 
peremptorily calls for an answer. He says: "M. Tait...trouve 
que M. Hoiiel a altere 1 ceuvre du maitre. Perfectionner n est .pas 
detruire." This appears to be a parody of the saying attributed 
to Louis XIV.: " Pardonner n est pas oublier": but M. Laisant 
should have recollected the more important maxim " Le mieux est 
Pennemi du bien." A line of Shakespeare might help him: 

"...with taper-light 

To seek the beauteous eye of heaven to garnish, 
Is wasteful and ridiculous excess." 

Even Prof. Willard Gibbs must be ranked as one of the retarders 
of Quaternion progress, in virtue of his pamphlet on Vector 
Analysis; a sort of hermaphrodite monster, compounded of the 
notations of Hamilton and of Grassmann. 

Apropos of Grassmann, I may advert for a moment to some 
comparatively recent German statements as to his anticipations 
&c. of Quaternions. I have given in the last edition of the Encyc. 
Brit. (Art. QI}ATEKNIONS ; to which I refer the reader) all that is 
necessary to shew the absolute baselessness of these statements. 
The essential points are as follows. Hamilton not only published 
his theory complete, the year before the first (and extremely 
imperfect) sketch of the Ausdehnungslehre appeared; but had 
given ten years before, in his protracted study of Sets, the very 
processes of external and internal multiplication (corresponding to 
the Vector and Scalar parts of a product of two vectors) which 
have been put forward as specially the property of Grassmann. 
The scrupulous care with which Hamilton drew up his account of 


the work of previous writers (Lectures, Preface) is minutely detailed 
in his correspondence with De Morgan (Hamilton s Life, Vol. in.). 

Another cause of the slow head-way recently made by Qua 
ternions is undoubtedly to be ascribed to failure in catching the 
"spirit" of the method: especially as regards the utter absence 
of artifice, and the perfect naturalness of every step. To try to 
patch up a quaternion investigation by having recourse to quasi- 
Cartesian processes is fatal to progress. A quaternion student loses 
his self-respect, so to speak, when he thus violates the principles 
of his Order. Tannhauser has his representatives in Science as 
well as in Chivalry ! One most insidious and dangerous form of 
temptation to this dabbling in the unclean thing is pointed out in 
500 below. All who work at the subject should keep before 
them Hamilton s warning words (Lectures, 513): 

"I regard it as an inelegance and imperfection in this calculus, 
or rather in the state to which it has hitherto [1853] been un 
folded, whenever it becomes, or seems to become, necessary to have 

recourse to the resources of ordinary algebra, for the SOLUTION 


As soon as my occupation with teaching and with experimental 
work perforce ceases to engross the greater part of my time, I 
hope to attempt, at least, the full quaternion development of 
several of the theories briefly sketched in the last chapter of 
this book ; provided, of course, that no one have done it in the 
meantime. From occasional glimpses, hitherto undeveloped, I feel 
myself warranted in asserting that, immense as are the simplifi 
cations introduced by the use of quaternions in the elementary 
parts of such subjects as Hydrokinetics and Electrodynamics, they 
are absolutely as nothing compared with those which are to be 
effected in the higher and (from the ordinary point of view) vastly 
more complex branches of these fascinating subjects. Complexity 
is no feature of quaternions themselves, and in presence of their 
attack (when properly directed) it vanishes from the subject 
also : provided, of course, that what we now call complexity 
depends only upon those space-relations (really simple if rightly 
approached) which we are in the habit of making all but incom 
prehensible, by surrounding them with our elaborate scaffolding of 
non-natural coordinates. 

viii PREFACE. 

Mr Wilkinson has again kindly assisted me in the revision of 
the proofs; and they have also been read and annotated by Dr 
Plarr, the able French Translator of the second edition. Thanks 
to their valuable aid, I may confidently predict that the present 
edition will be found comparatively accurate. 

With regard to the future of Quaternions, I will merely quote 
a few words of a letter I received long ago from Hamilton : 

" Could anything be simpler or more satisfactory ? Don t you 
feel, as well as think, that we are on a right track, and shall be 
thanked hereafter ? Never mind when." 

The special form of thanks which would have been most 
grateful to a man like Hamilton is to be shewn by practical 
developments of his magnificent Idea. The award of this form of 
thanks will, I hope, not be long delayed. 

P. G. TA1T. 


(Only the more important are noticed, and they are indicated by the 
sectional numbers.) 

Chap. I. 31 (k), (m), 40, 43. 

II. 51, 89. 

III. 105, 108, 116, 119122, 1334. 

IV. 140 (8) (12), 145149. 

V. 174, 187, 1934, 196, 199. 

VI. The whole. 

VIII. 247, 250, 250*. 

IX. 285, 286, 287. 

X. 326, 336. 

XI. 3578, 382, 3846. 

XII. 390, 393403, 407, 458, 473 (a) (Z), 480, 489, 493, 499, 500, 
503, 508511, 51213. 

There are large additions to the number of Examples, some in fact 
to nearly every Chapter. Several of these are of considerable importance ; 
as they contain, or suggest, processes and results not given in the text. 


To the first edition of this work, published in 1867, the 
following was prefixed : 

THE present work was commenced in 1859, while I was a 
Professor of Mathematics, and far more ready at Quaternion 
analysis than I can now pretend to be. Had it been then 
completed I should have had means of testing its teaching 
capabilities, and of improving it, before publication, where found 
deficient in that respect. 

The duties of another Chair, and Sir W. Hamilton s wish that 
my volume should not appear till after the publication of his 
Elements, interrupted my already extensive preparations. I had 
worked out nearly all the examples of Analytical Geometry in 
Todhunter s Collection, and I had made various physical applica 
tions of the Calculus, especially to Crystallography, to Geometrical 
Optics, and to the Induction of Currents, in addition to those on 
Kinematics, Electrodynamics, Fresnel s Wave Surface, &c., which 
are reprinted in the present work from the Quarterly Mathematical 
Journal and the Proceedings of the Royal Society of Edinburgh. 

Sir W. Hamilton, when I saw him but a few days before his 
death, urged me to prepare my work as soon as possible, his being 
almost ready for publication. He then expressed, more strongly 
perhaps than he had ever done before, his profound conviction of 
the importance of Quaternions to the progress of physical science ; 
and his desire that a really elementary treatise on the subject 
should soon be published. 

I regret that I have so imperfectly fulfilled this last request 


of my revered friend. When it was made I was already engaged, 
along with Sir W. Thomson, in the laborious work of preparing 
a large Treatise on Natural Philosophy. The present volume has 
thus been written under very disadvantageous circumstances, 
especially as I have not found time to work up the mass of 
materials which I had originally collected for it, but which I 
had not put into a fit state for publication. I hope, however, 
that I have to some extent succeeded in producing a thoroughly 
elementary work, intelligible to any ordinary student ; and that 
the numerous examples I have given, though not specially 
chosen so as to display the full merits of Quaternions, will 
yet sufficiently shew their admirable simplicity and naturalness 
to induce the reader to attack the Lectures and the Elements; 
where he will find, in profusion, stores of valuable results, and 
of elegant yet powerful analytical investigations, such as are 
contained in the writings of but a very few of the greatest 
mathematicians. For a succinct account of the steps by which 
Hamilton was led to the invention of Quaternions, and for 
other interesting information regarding that remarkable genius, 
I may refer to a slight sketch of his life and works in the 
North British Review for September 1866. 

It will be found that I have not servilely followed even so 
great a master, although dealing with a subject which is entirely 
his own. I cannot, of course, tell in every case what I have 
gathered from his published papers, or from his voluminous 
correspondence, and what I may have made out for myself. 
Some theorems and processes which I have given, though wholly 
my own, in the sense of having been made out for myself before 
the publication of the Elements, I have since found there. Others 
also may be, for I have not yet read that tremendous volume 
completely, since much of it bears on developments unconnected 
with Physics. But I have endeavoured throughout to point out 
to the reader all the more important parts of the work which I 
know to be wholly due to Hamilton. A great part, indeed, may 
be said to be obvious to any one who has mastered the pre 
liminaries ; still I think that, in the two last Chapters especially, 
a good deal of original matter will be found. 


The volume is essentially a working one, and, particularly in 
the later Chapters, is rather a collection of examples than a 
detailed treatise on a mathematical method. I have constantly 
aimed at avoiding too great extension ; and in pursuance of 
this object have omitted many valuable elementary portions 
of the subject. One of these, the treatment of Quaternion 
logarithms and exponentials, I greatly regret not having given. 
But if I had printed all that seemed to me of use or interest to 
the student, I might easily have rivalled the bulk of one of 
Hamilton s volumes. The beginner is recommended merely to 
read the first five Chapters, then to work at Chapters VI, 
VII, VIII* (to which numerous easy Examples are appended). 
After this he may work at the first five, with their (more 
difficult) Examples ; and the remainder of the book should 
then present no difficulty. 

Keeping always in view, as the great end of every mathe 
matical method, the physical applications, I have endeavoured 
to treat the subject as much as possible from a geometrical 
instead of an analytical point of view. Of course, if we premise 
the properties of i, j, k merely, it is possible to construct from 
them the whole system -f-; just as we deal with the imaginary 
of Algebra, or, to take a closer analogy, just as Hamilton 
himself dealt with Couples, Triads, and Sets. This may be 
interesting to the pure analyst, but it is repulsive to the 
physical student, who should be led to look upon i, j, k, from 
the very first as geometric realities, not as algebraic imagi- 

The most striking peculiarity of the Calculus is that mul 
tiplication is not generally commutative, i.e. that qr is in general 
different from rg, r and q being quaternions. Still it is to 
be remarked that something similar is true, in the ordinary 
coordinate methods, of operators and functions : and therefore 

* [In this edition these Chapters are numbered VII, VIII, IX, respectively 
Aug. 1889.] 

t This has been done by Hamilton himself, as one among many methods he has 
employed; and it is also the foundation of a memoir by M. Allegret, entitled Essai 
sur le Calcul des Quaternions (Paris, 1862). 


the student is not wholly unprepared to meet it. No one is 
puzzled by the fact that log. cos. x is not equal to cos. log. x, 

or that A -~ is not equal to ^-\A/. Sometimes, indeed, this 
V dx dx l 

rule is most absurdly violated, for it is usual to take cos 2 # as 
equal to (cos#) 2 , while cos" 1 ^ is not equal to (cos a?)" 1 . No such 
incongruities appear in Quaternions ; but what is true of operators 
and functions in other methods, that they are not generally com 
mutative, is in Quaternions true in the multiplication of (vector) 

It will be observed by those who are acquainted with the 
Calculus that I have, in many cases, not given the shortest or 
simplest proof of an important proposition. This has been done 
with the view of including, in moderate compass, as great a 
variety of methods as possible. With the same object I have 
endeavoured to supply, by means of the Examples appended 
to each Chapter, hints (which will not be lost to the intelli 
gent student) of farther developments of the Calculus. Many 
of these are due to Hamilton, who, in spite of his great origi 
nality, was one of the most excellent examiners any University 
can boast of. 

It must always be remembered that Cartesian methods are 
mere particular cases of Quaternions, where most of the distinctive 
features have disappeared ; and that when, in the treatment of 
any particular question, scalars have to be adopted, the Quaternion 
solution becomes identical with the Cartesian one. Nothing there 
fore is ever lost, though much is generally gained, by employing 
Quaternions in preference to ordinary methods. In fact, even 
when Quaternions degrade to scalars, they give the solution 
of the most general statement of the problem they are applied 
to, quite independent of any limitations as to choice of particular 
coordinate axes. 

There is one very desirable object which such a work as this 
may possibly fulfil. The University of Cambridge, while seeking 
to supply a real want (the deficiency of subjects of examination for 
mathematical honours, and the consequent frequent introduction 
of the wildest extravagance in the shape of data for " Problems "), 


is in danger of making too much of such elegant trifles as Trilinear 
Coordinates, while gigantic systems like Invariants (which, by the 
way, are as easily introduced into Quaternions as into Cartesian 
methods) are quite beyond the amount of mathematics which 
even the best students can master in three years reading. 
One grand step to the supply of this want is, of course, the 
introduction into the scheme of examination of such branches 
of mathematical physics as the Theories of Heat and Electricity. 
But it appears to me that the study of a mathematical method 
like Quaternions, which, while of immense power and compre 
hensiveness, is of extraordinary simplicity, and yet requires 
constant thought in its applications, would also be of great 
benefit. With it there can be no "shut your eyes, and write 
down your equations," for mere mechanical dexterity of analysis 
is certain to lead at once to error on account of the novelty of 
the processes employed. 

The Table of Contents has been drawn up so as to give the 
student a short and simple summary of the chief fundamental 
formulae of the Calculus itself, and is therefore confined to an 
analysis of the first five [and the two last] chapters. 

In conclusion, I have only to say that I shall be much obliged 
to any one, student or teacher, who will point out portions of the 
work where a difficulty has been found; along with any inaccuracies 
which may be detected. As I have had no assistance in the revision 
of the proof-sheets, and have composed the work at irregular in 
tervals, and while otherwise laboriously occupied, I fear it may 
contain many slips and even errors. Should it reach another 
edition there is no doubt that it will be improved in many 
important particulars/ 

To this I have now to add that I have been equally surprised 
and delighted by so speedy a demand for a second edition and the 
more especially as I have had many pleasing proofs that the 
work has had considerable circulation in America. There seems 
now at last to be a reasonable hope that Hamilton s grand 
invention will soon find its way into the working world of science, 
to which it is certain to render enormous services, and not be laid 
T. Q. I. c 


aside to be unearthed some centuries hence by some grubbing 

It can hardly be expected that one whose time is mainly en 
grossed by physical science, should devote much attention to the 
purely analytical and geometrical applications of a subject like this; 
and I am conscious that in many parts of the earlier chapters I 
have not fully exhibited the simplicity of Quaternions. I hope, 
however, that the corrections and extensions now made, especially 
in the later chapters, will render the work more useful for my chief 
object, the Physical Applications of Quaternions, than it could 
have been in its first crude form. 

I have to thank various correspondents, some anonymous, for 
suggestions as well as for the detection of misprints and slips of 
the pen. The only absolute error which has been pointed out to 
me is a comparatively slight one which had escaped my own notice : 
a very grave blunder, which I have now corrected, seems not to 
have been detected by any of my correspondents, so that I cannot 
be quite confident that others may not exist. 

I regret that I have not been able to spare time enough to 
rewrite the work ; and that, in consequence of this, and of the 
large additions which have been made (especially to the later 
chapters), the whole will now present even a more miscellaneously 
jumbled appearance than at first. 

It is well to remember, however, that it is quite possible to 
make a book too easy reading, in the sense that the student may 
read it through several times without feeling those difficulties 
which (except perhaps in the case of some rare genius) must 
attend the acquisition of really useful knowledge. It is better to 
have a rough climb (even cutting one s own steps here and there) 
than to ascend the dreary monotony of a marble staircase or a 
well-made ladder. Royal roads to knowledge reach only the 
particular locality aimed at and there are no views by the way. 
It is not on them that pioneers are trained for the exploration of 
unknown regions. 

But I am happy to say that the possible repulsiveness of my 
early chapters cannot long be advanced as a reason for not attack 
ing this fascinating subject. A still more elementary work than 


the present will soon appear, mainly from the pen of my colleague 
Professor KELLAND. In it I give an investigation of the properties 
of the linear and vector function, based directly upon the Kine 
matics of Homogeneous Strain, and therefore so different in method 
from that employed in this work that it may prove of interest to 
even the advanced student. 

Since the appearance of the first edition I have managed (at 
least partially) to effect the application of Quaternions to line, 
surface, and volume integrals, such as occur in Hydrokinetics, 
Electricity, and Potentials generally. I was first attracted to 
the study of Quaternions by their promise of usefulness in 
such applications, and, though I have not yet advanced far in 
this new track, I have got far enough to see that it is certain 
in time to be of incalculable value to physical science. I have 
given towards the end of the work all that is necessary to put 
the student on this track, which will, I hope, soon be followed to 
some purpose. 

One remark more is necessary. I have employed, as the 
positive direction of rotation, that of the earth about its axis, or 
about the sun, as seen in our northern latitudes, i.e. that opposite 
to the direction of motion of the hands of a watch. In Sir W. 
Hamilton s great works the opposite is employed. The student 
will find no difficulty in passing from the one to the other ; but, 
without previous warning, he is liable to be much perplexed. 

With regard to notation, I have retained as nearly as possible 
that of Hamilton, and where new notation was necessary I 
have tried to make it as simple and, as little incongruous with 
Hamilton s as possible. This is a part of the work in which great 
care is absolutely necessary; for, as the subject gains development, 
fresh notation is inevitably required ; and our object must be to 
make each step such as to defer as long as possible the revolution 
which must ultimately come. 

Many abbreviations are possible, and sometimes very useful in 
private work ; but, as a rule, they are unsuited for print. Every 
analyst, like every short-hand writer, has his own special con 
tractions ; but, when he comes to publish his results, he ought 
invariably to put such devices aside. If all did not use a common 


mode of public expression, but each were to print as he is in the 
habit of writing for his own use, the confusion would be utterly 

Finally, I must express my great obligations to my friend 
M. M. U. WILKINSON of Trinity College, Cambridge, for the care 
with which he has read my proofs, and for many valuable sug 

P. G. TAIT. 

October 1873. 



Sketch of the attempts made to represent geometrically the imaginary of 

algebra. 113. 

De Moivre s Theorem interpreted in plane rotation. 7, 8. 
Curious speculation of Servois. 11. 

Elementary geometrical ideas connected with relative position. 14. 
Definition of a VECTOR. It may be employed to denote translation. Definition 

of currency. 16. 
Expression of a vector by one symbol, containing implicitly three distinct 

numbers. Extension of the signification of the symbol = . 18. 
The sign + defined in accordance with the interpretation of a vector as 

representing translation. 19. 

Definition of - . It simply reverses the currency of a vector. 20. 
Triangles and polygons of vectors, analogous to those of forces and of simul 

taneous velocities. 21. 
When two vectors are parallel we have 

a = xp. 22. 

Any vector whatever may be expressed in terms of three distinct vectors, which 
are not coplanar, by the formula 

which exhibits the three numbers on which the vector depends. 23. 
Any vector in the same plane with a and may be written 

p = xa + y(3. 24. 
The equation Tff p, 

between two vectors, is equivalent to three distinct equations among 

numbers. 25. 

The Commutative and Associative Laws hold in the combination of vectors by 
the signs + and - . 27. 


The equation p = xfi, 

where p is a variable, and j8 a fixed, vector, represents a line drawn through 
the origin parallel to |8. 

is the equation of a line drawn through the extremity of a and parallel 
to |8. 28. 

represents the plane through the origin parallel to a and /3, while 

denotes a parallel plane through the point 7. 29. 
The condition that p, a, /3 may terminate in the same line is 

pp + qa + rp = Q, 
subject to the identical relation 

Similarly pp + q a + r p + S y = Q, 

with jj + 2 + r + s = 0, 

is the condition that the extremities of four vectors lie in one plane. 30. 
Examples with solutions. Conditions that a vector equation may represent a 
line, or a surface. 

The equation p = 0i 

represents a curve in space ; while 

is a cone, and 

is a cylinder, both passing through the curve. 31. 

Differentiation of a vector, when given as a function of one number. 32 37. 
If the equation of a curve be 


where s is the length of the arc, dp is a vector-tangent to the curve, and its 
length is ds. 38, 39. 
Examples with solutions. 4043. 



Here we begin to see what a quaternion is. When two vectors are parallel 
their quotient is a number. 45, 46. 

When they are perpendicular to one another, their quotient is a vector perpen 
dicular to their plane. 47, 64, 72. 

When they are neither parallel nor perpendicular the quotient in general 
involves four distinct numbers and is thus a QUATEKNION. 47. 

A quaternion q regarded as the operator which turns one vector into another. 
It is thus decomposable into two factors, whose order is indifferent, the 
stretching factor or TENSOR, and the turning factor or VERSOR. These are 
denoted by Tq, and Uq. 48. 


The equation p = qa 


gives =<? or 8a~ l = q, "but not in general 


a- 1 8 = q. 49. 

q or /Sa" 1 depends only on the relative lengths, and directions, of /3 and a. 
Reciprocal of a quaternion defined, 

8 . I a 

q = - gives or q~ 1 = - , 
a q 8 

Definition of the Conjugate of a quaternion, 

and qKq=Kq. q = (Tq)v. 52. 

Eepresentation of versors by arcs on the unit-sphere. 53. 
Versor multiplication illustrated by the composition of arcs. The process 

proved to be not generally commutative. 54. 
Proof that K (qr) = Kr . Kq. 55. 

Proof of the Associative Law of Multiplication 

p.qr = pq.r. 5760. 
[Digression on Spherical Conies. 59*.] 
Quaternion addition and subtraction are commutative. 61. 
Quaternion multiplication and division are distributive. 62. 
Integral powers of a versor are subject to the Index Law. 63. 
Composition of quadrantal versors in planes at right angles to each other. 

Calling them i, j, k, we have 

ijk=-l. 6471. 
A unit-vector, when employed as a factor, may be considered as a quadrantal 

versor whose plane is perpendicular to the vector. Hence the equations 

just written are true of any set of rectangular unit-vectors i, j, k. 72. 
The product, and also the quotient, of two vectors at right angles to each other 

is a third perpendicular to both. Hence 
Ka = - a, 

and (Ta) 2 = aKa= -a 2 . 73. 

Every versor may be expressed as a power of some unit-vector. 74. 
Every quaternion may be expressed as a power of a vector. 75. 
The Index Law is true of quaternion multiplication and division. 76. 
Quaternion considered as the sum of a SCALAR and a VECTOR. 


Proof that SKq = Sq, VKq = - Vq, 2Kq = K2q . 79. 
Quadrinomial expression for a quaternion 

An equation between quaternions is equivalent to four equations between 
numbers (or scalars). 80. 


Second proof of the distributive law of multiplication. 81 83. 

Algebraic determination of the constituents of the product and quotient of two 

vectors. 83, 84. 

Second proof of the associative law of multiplication. 85. 
Proof of the formulae Sap = Spa, 

S .qr = S .rq, 
S . qrs = S . rsq = S . sqr, 

2S.ap...<t> X )_ 

the upper sign belonging to the scalar if the number of factors is even. 
Proof of the formulae 

F. aVpy = ySap - pSya, 
V.apy = aSpy - pSya + ySap, 
V.apy= V.ypa, 
V. VapVyd = aS . 1878 - pS . 07$, 
= 8S . apy - yS . a/35, 
dS . apy = aS . pyd + pS . yad + yS . apd, 

= VapSyd + VpySad + VyaSpd. 9092. 

Hamilton s proof that the product of two parallel vectors must be a scalar, and 
that of perpendicular vectors, a vector; if quaternions are to deal with 
space indifferently in all directions. 93. 




If 6 be the angle between two vectors, a and p, we have 




a Ta 

Applications to plane trigonometry. 97. 
The condition Sap = 

shews that a is perpendicular to /3, while 

Fa/3 = 0, 

shews that a and p are parallel. 
The expression S . apy 

is the volume of the parallelepiped three of whose conterminous edges are 
o, /3, 7. Hence 

S.apy = 
shews that a, p, y are coplanar. 


Expression of S . apy as a determinant. 98102. 
Proof that (Tq) 2 = (Sq)* + ( TVq)*, 

and T(qr) = TqTr. 103. 

Simple propositions in plane trigonometry. 104. 
Proof that - a/3a~ l is the vector reflected ray, when /3 is the incident ray and a 

normal to the reflecting surface. 105. 
Interpretation of apy when it is a vector. 106. 
Examples of variety in simple transformations. 107. 

The relation among the distances, two and two, of five points in space. 108. 
De Moivre s Theorem, and Plane Trigonometry. 109 111. 
Introduction to spherical trigonometry. 112 116. 

Representation, graphic, and by quaternions, of the spherical excess. 117, 118. 
Interpretation of the Operator 

( ) T 1 
in connection with rotation. Astronomical examples. 119 122. 

Loci represented by different equations points, lines, surfaces, and volumes. 


Proof that r~ l (?V)* V~ l = U (rq + KrKq). 127. 

Proof of the transformation 


Convenient abbreviations of notation. 133, 134. 



Definition of a differential, 

where dq is any quaternion whatever. 

We may write dFq =f (q, dq), 

where / is linear and homogeneous in dq ; but we cannot generally write 

dFq = f(q)dq. 135138. 
Definition of the differential of a function of more quaternions than one. 

d (qr) = qdr + dq.r, but not generally d (qr) = qdr + rdq. 139. 
Proofs of fundamental differential expressions : 

Successive differentiation; Taylor s theorem. 141. 142. 


If the equation of a surface be 

F(p) = C, 
the differential may be written 

Sfdp = 0, 

where v is a vector normal to the surface. 144. 
Definition of Hamilton s Operator 

. d . d , d 

V = i~ +? - +k . 
dx J dy dz 

Its effects on simple scalar and vector functions of position. Its square 
the negative of Laplace s Operator. Expressions for the condensation and 
rotation due to a displacement. Application to fluxes, and to normals to 
surfaces. Precautions necessary in its use. 145 149. 

EXAMPLES TO CHAPTER IV. ...... 106, 107 


FIRST DEGREE ....... 108 141 

The most general equation of the first degree in an unknown quaternion q, 
may be written 

SF. aqb + S.cq = d, 

where a, &, c, d are given quaternions. Elimination of Sq, and reduction 
to the vector equation 

0/> = S.ctS/3/> = 7. 150,151. 

General proof that 3 /> is expressible as a linear function of /o, 0/>, and 2 p. 

Value of for an ellipsoid, employed to illustrate the general theory. 

Hamilton s solution of 0/ 9 = 7- 

If we write Sa^>p = S/o0V, 

the functions and are said to be conjugate, and 

W10- 1 VX/JL = V<f> X0 V*. 

Proof that ra, whose value may be written as 
S . <> 

is the same for all values of X, /A, v. 156158. 
Proof that if m g = m- m^g + w 2 # 2 - s , 

, _ S (\0 /U,0V + Xytt0 j> + X0 /X I/ ) 

_ S (XM0V + X/uv + X0 V) 
and flic, - - pj - , 

S . \fJLV 

(which, like m, are Invariants,) 
then m g (0 - g)~ l V\^ = (W0- 1 -gx + 9 

Also that X = m 2-<> 

whence the final form of solution 

W0- 1 = m 1 - Wo0 + 2 . 159, 160. 


Examples. 161173. 
The fundamental cubic 

3 - ni 2 2 + TO 1 - m= (0 - ft) (0 - <7 2 ) (0 - 3 ) = 0. 

When is its own conjugate, the roots of the cubic are real; and the 


or (0-<7)p = 0, 

is satisfied by a set of three real and mutually perpendicular vectors. 
Geometrical interpretation of these results. 174 178. 
Proof of the transformation of the self-conjugate linear and vector function 

<f>p=fp + hV. (i + ek) p(i- ek) 
where (0-^)i = 0, 

Another transformation is 

0p = aaVap + bpSpp. 179181. 
Other properties of 0. Proof that 

Sp(<t>-g)~ l p = 0, and Sp (<j>-h)- l p= 
represent the same surface if 

Proof that when is not self-conjugate 

<t>p = <t> p+Vep. 182184. 
Proof that, if q = a0ct + /30/3 + 707, 

where a, , 7 are any rectangular unit-vectors whatever, we have 

Sq = - m 2 , Vq = e, 
where Vcp = J (0 - ) p. 

This quaternion can be expressed in the important form 

gr-V0/3. 185. 

A non-conjugate linear and vector function of a vector differs from a self- 

conjugate one solely by a term of the form 

Fep. 186. 
Graphic determination of the conditions that there may be three real vector 

solutions of 

Fp0p = 0. 187. 
Degrees of indeterminateness of the solution of a quaternion equation 

Examples. 188 191. 
The linear function of a quaternion is given by a symbolical biquadratic. 

Particular forms of linear equations. Differential of the nth root of a quaternion. 

A quaternion equation of the rath degree in general involves a scalar equation 

of degree ra 4 . 197. 
Solution of the equation q~ = qa + b. 198. 

EXAMPLES TO CHAPTER V. . ...... 142 145 





AND PLANE . 160174 





EXAMPLES TO CHAPTER IX ...... . . . 224229 


EXAMPLES TO CHAPTER X ........ 270 278 


A. Kinematics of a Point. 354366. 

If p = $t be the vector of a moving point in terms of the time, p is the vector 
velocity, and p the vector acceleration. 

<r p = (p (t) is the equation of the Hodograph. 
p = iip + v 2 p" gives the normal and tangential accelerations. 
VppsxQ if acceleration directed to a point, whence Vpp = y. 
Examples. Planetary acceleration. Here the equation is 
.. nUp 

p= w 

giving Vpp = y ; whence the hodograph is 

p = ey- l -/j.Up.y- 1 , 
and the orbit is the section of 

ftTp = Se(y*-e-i-p) 
by the plane Syp = Q. 

Cotes Spirals, Epitrochoids, &c. 354366. 

B. Kinematics of a Rigid System. 367375. 

Rotation of a rigid system. Composition of rotations. If the position of a 
system at time t is derived from the initial position by q( )q~ l , the 
instantaneous axis is 

Rodrigues coordinates. 367 375. 


C. Kinematics of a Deformable System. 376 386. 

Homogeneous strain. Criterion of pure strain. Separation of the rotational 
from the pure part. Extraction of the square root of a strain. A strain 
is equivalent to a pure strain /^/^ ^ followed by the rotation 0/^/0 0. 
Simple Shear. 376383. 

Displacements of systems of points. Consequent condensation and rotation. 
Preliminary about the use of V in physical questions. For displacement 
<r, the strain function is 

(pr = T- SrV . a. 384 386. 

D. Axes of Inertia. 387. 
Moment of inertia. Binet s Theorem. 387. 



A. Statics of a Rigid System. 389 403. 

Condition of equilibrium of a rigid system is SS . /35ct = 0, where /3 is a vector 
force, a its point of application. Hence the usual six equations in the 
form 2/3 = 0, SFa/3 = 0. Central axis. Minding s Theorem, Ac. 389 

B. Kinetics of a Rigid System. 404 425. 

For the motion of a rigid system 


whence the usual forms. Theorems of Poinsot and Sylvester. The 

2q = q<t>~ l (q- l yq), 

where 7 is given in terms of t and q if forces act, but is otherwise constant, 
contains the whole theory of the motion of a rigid body with one point 
fixed. Eeduction to the ordinary form 

dt _ dw _ dx _ dy _ dz 

~%~W~~X = ~Y = ~Z 

Here, if no forces act, W, X, Y, Z are homogeneous functions of the third 
degree in w, x, y, z. 404425. 

C. Special Kinetic Problems. 426430. 

Precession and Nutation. General equation of motion of simple pendulum. 
Foucault s pendulum. 426430. 

D. Geometrical and Physical Optics. 431 452. 

Problem on reflecting surfaces. 431. 

Fresnel s Theory of Double Eefraction. Various forms of the equation of 

Fresnel s Wave-surface; 
S.p(0-p 2 )-ip = -l, T(p- 2 -0- 1 )-*p = 0, l = -pp ~=F(TS)V\pVw. 

The conical cusps and circles of contact. Lines of vibration, &c. 



E. Electrodynamics. 453472. 

Electrodynamics. The vector action of a closed circuit on an element of 
current c^ is proportional to Varfi where 

_ [Vada_ [dUa 
-J T -J 

the integration extending round the circuit. It can also be expressed 
as - Vft, where ft is the spherical angle subtended by the circuit. This 
is a many-valued function. Special case of a circular current. Mutual 
action of two closed circuits, and of solenoids. Mutual action of magnets. 
Potential of a closed circuit. Magnetic curves. 453 472. 

F. General Expressions for the Action between Linear Elements. 473. 

Assuming Ampere s data .1, II, III, what is the most general expression for 
the mutual action between two elements? Particular cases, determined 
by a fourth assumption. Solution of the problem when I and II, alone, 
are assumed. Special cases, including v. Helmholtz form. 473 (a)... (I). 

G. Application of V to certain Physical Analogies. 474 478. 

The effect of a current-element on a magnetic particle is analogous to dis 
placement produced by external forces in an elastic solid, while that 
of a small closed circuit (or magnet) is analogous to the corresponding 
vector rotation. 

//. Elementary Properties of V. 479481. 
SdpV=-d=SdffV <r 

where <r = i +.717 + fcf and V<r = i~ + j - + A- ; 
so that, if d<r = (f)dp, 

J. Applications of V to Line, Surface and Volume Integrals. 

Proof of the fundamental theorem for comparing an integral over a closed 
surface with one through its content 

Hence Green s Theorem. Examples in potentials and in conduction of 
heat. Limitations and ambiguities. Determination of the displacement 
in a fluid when the consequent rotation is given. 482 494. 
Similar theorem for double and single integrals 

fSvdp =ffS . UvVffds. 495497. 
The fundamental forms, of which all the others are simple consequences, are 

ffSVuds=ffUvuds. 498, 499. 
The first is a particular case of the second. 500. 

Expression for a surface in terms of an integral through the enclosed volume. 


K. Application of the V Integrals to Magnetic &c. Problems. 

Volume and surface distributions due to a given magnetic field. Solenoidal 
and Lamellar distributions. 502. Magnetic Induction and Vector 
potential. 503. Ampere s Directrice. 504. Gravitation potential of 
homogeneous solid in terms of a surface-integral. 505. 

L. Application of V to the Stress- Function. 507 511. 

When there are no molecular couples the stress-function is self-conjugate. 

507. Properties of this function which depend upon the equilibrium 

of any definite portion of the solid, as a whole. 508. Expression for 

the stress-function, in terms of displacement, when the solid is isotropic : 

W = - n (SwV . <r + VSaxr) - (c - $n) wSVtr. 

Examples. 509. Work due to displacement in any elastic solid. Green s 
21 elastic coefficients. 511. 

M. The Hydrokinetic Equations. 512, 513. 

Equation of continuity, for displacement a : 

dt dt 

Equation for rate of change of momentum in unit volume: 

*- v * 512 - 


Term introduced by Viscosity: 

ocV 2 <r + -|VSV<r. (Miscellaneous Examples, 36.) 
Helmholtz s Transformation for Vortex- motion: 

Thomson s Transformation for Circulation: 

- jj= \Q - it; 2 T , where/- - ( 

dt L Ja J 

N. Use of V in connection with Taylor s Theorem. 514 517. 
Proof that 

Applications and consequences. Separation of symbols of operation and their 
treatment as quantities. 

0. Applications of V in connection with the Calculus of Variations. 


If A=fQTd P 

we have dA=- [QSUdp8p] +f{ 8QTdp + S.dp(Q Udp) } , 
whence, if A is a maximum or minimum, 

~(Q P )-VQ = 0. 

Applications to Varying Action, Brachistochrones, Catenaries, &c. 518 526. 
Thomson s Theorem that there is one, and but one, solution of the equation 

S.V(e 2 VH) = 4irr. 527. 




1. FOR at least two centuries the geometrical representation 
of the negative and imaginary algebraic quantities, 1 and ,J 1 
has been a favourite subject of speculation with mathematicians. 
The essence of almost all of the proposed processes consists in 
employing such expressions to indicate the DIRECTION, not the 
length, of lines. 

2. Thus it was long ago seen that if positive quantities were 
measured off in one direction along a fixed line, a useful and lawful 
convention enabled us to express negative quantities of the same 
kind by simply laying them off on the same line in the opposite 
direction. This convention is an essential part of the Cartesian 
method, and is constantly employed in Analytical Geometry and 

Applied Mathematics. 

3. Wallis, towards the end of the seventeenth century, proposed 

to represent the impossible roots of a quadratic equation by going 
out of the line on which, if real, they would have been laid off. 
This construction is equivalent to the consideration of J 1 as a 
directed unit-line perpendicular to that on which real quantities 
are measured. 

4. In the usual notation of Analytical Geometry of two 
dimensions, when rectangular axes are employed, this amounts 
to reckoning each unit of length along Oy as + J 1, and on 
Oy as J 1 ; while on Ox each unit is + 1, and on Ox it is 1. 

T. Q. I. 1 


If we look at these four lines in circular order, i.e. in the order of 
positive rotation (that of the northern hemisphere of the earth 
about its axis, or opposite to that of the hands of a watch), they 

1, N/-1, -1, -V-l. 

In this series each expression is derived from that which precedes 
it by multiplication by the factor J 1. Hence we may consider 
J 1 as an operator, analogous to a handle perpendicular to the 
plane of xy, whose effect on any line in that plane is to make it 
rotate (positively) about the origin through an angle of 90. 

5. Iu such a system, (which seems to have been first developed, 
in 1805, by Buee) a point in the plane of reference is defined by a 
single im&giiiai7 expression. Thus a + b J 1 may be considered 
as a single quantity, denoting the point, P, whose coordinates are 
a and b. Or, it may be used as an expression for the line OP 
joining that point with the origin. In the latter sense, the ex 
pression a + b V 1 implicitly contains the direction, as well as the 
length, of this line ; since, as we see at once, the direction is 
inclined at an angle tan" 1 b/a to the axis of oc, and the length is 
Jo? + b z . Thus, say we have 

the line OP considered as that by which we pass from one 

J A 

extremity, 0, to the other, P. In this sense it is called a VECTOR. 
Considering, in the plane, any other vector, 

OQ = a f + Vj^l- 
the addition of these two lines obviously gives 

OR = a + a + (b + b ) J^l; 

and we see that the sum is the diagonal of the parallelogram on 
OP, OQ. This is the law of the composition of simultaneous 
velocities ; and it contains, of course, the law of subtraction of one 
directed line from another. 

6. Operating on the first of these symbols by the factor J 1, 
it becomes - b -f a J - 1 ; and now, of course, denotes the point 
whose x and y coordinates are - b and a ; or the line joining this 
point with the origin. The length is still J a* + V, but the angle 
the line makes with the axis of x is tan 1 (- a/b) ; which is 
evidently greater by ?r/2 than before the operation. 


7. De Moivre s Theorem tends to lead us still further in the 
same direction. In fact, it is easy to see that if we use, instead 
of J 1, the more general factor cos a + J 1 sin a, its effect on 
any line is to turn it through the (positive) angle a. in the plane 
of x, y. [Of course the former factor, J 1, is merely the par 
ticular case of this, when a = 7r/2.~| 

Thus (cos a + J 1 sin a) (a + b J 1 ) 

= a cos a b sin a + J 1 (a sin a -f 6 cos a), 

by direct multiplication. The reader will at once see that the new 
form indicates that a rotation through an angle a has taken place, 
if he compares it with the common formulae for turning the co 
ordinate axes through a given angle. Or, in a less simple manner, 

Length = J (a cos a b sin a) 2 + (a sin a + b cos a) 2 
= J a 2 + b* as before. 

Inclination to axis of x 


, tan a + - 

_, a sin a 4- b cos a. a 

= tan = : = tan . 

a cos a b sin a n 6 , 

1 tan a 


= a + tan" 1 b/a. 

8. We see now, as it were, why it happens that 

(cos a + v 1 sin a) m = cos ma. + J 1 sin ma. 

In fact, the first operator produces m successive rotations in the 
same direction, each through the angle a ; the second, a single 
rotation through the angle ma. 

9. It may be interesting, at this stage, to anticipate so far as to 
remark that in the theory of Quaternions the analogue of 

cos 6 + J ^1 sin 6 

is cos -f tff sin 0, 

where w 2 = - 1. 

Here, however, & is not the algebraic J 1, but is any directed 
unit-line whatever in space. 

10. In the present century Argand, Warren, Mourey, and 
others, extended the results of Wallis and Buee. They attempted 



to express as a line the product of two lines each represented by a 
symbol such asa+6/s/ 1. To a certain extent they succeeded, 
but all their results remained confined to two dimensions. 

The product, II, of two such lines was defined as the fourth 
proportional to unity and the two lines, thus 

1 :a + bJ^I::a + V l J~^l : II, 
or n = (aa - bV) + (a b + If a} J^l. 

The length of II is obviously the product of the lengths of the 
factor lines ; and its direction makes an angle with the axis of x 
which is the sum of those made by the factor lines. From this 
result the quotient of two such lines follows immediately. 

11. A very curious speculation, due to Servois and published 
in 1813 in Gergonne s Annales, is one of the very few, so far as has 
been discovered, in which a well-founded guess at a possible mode 
of extension to three dimensions is contained. Endeavouring to 
extend to space the form a + 6 J 1 for the plane, he is guided by 
analogy to write for a directed unit-line in space the form 

p cos a + q cos ft + r cos 7, 

where a, ft, 7 are its inclinations to the three axes. He perceives 
easily that p, q, r must be non-reals : but, he asks, " seraient-elles 
imaginaires re ductibles a la forme generale A+Bjl?" The 
i,j t k of the Quaternion Calculus furnish an answer to this question. 
(See Chap. II.) But it may be remarked that, in applying the 
idea to lines in a plane, a vector OP will no longer be represented 
(as in 5) by 

but by OP =pa + qb. 

And if, similarly, OQ = pa + qb , 

the addition of these two lines gives for OR (which retains its 
previous signification) 

12. Beyond this, few attempts were made, or at least recorded, 
in earlier times, to extend the principle to space of three dimen 
sions; and, though many such had been made before 1843, none, 
with the single exception of Hamilton s, have resulted in simple, 
practical methods ; all, however ingenious, seeming to lead almost 
at once to processes and results of fearful complexity. 


For a lucid, complete, and most impartial statement of the 
claims of his predecessors in this field we refer to the Preface to 
Hamilton s Lectures on Quaternions. He there shews how his long 
protracted investigations of Sets culminated in this unique system 
of tridimensional-space geometry. 

13. It was reserved for Hamilton to discover the use and 
properties of a class of symbols which, though all in a certain sense 
square roots of 1, may be considered as real unit lines, tied down 
to no particular direction in space ; the expression for a vector is, 
or may be taken to be, 

but such vector is considered in connection with an extraspatial 
magnitude w, and we have thus the notion of a QUATERNION 

w + p. 

This is the fundamental notion in the singularly elegant, and 
enormously powerful, Calculus of Quaternions. 

While the schemes for using the algebraic J 1 to indicate 
direction make one direction in space expressible by real numbers, 
the remainder being imaginaries of some kind, and thus lead to 
expressions which are heterogeneous ; Hamilton s system makes all 
directions in space equally imaginary, or rather equally real, there 
by ensuring to his Calculus the power of dealing with space 
indifferently in all directions. 

In fact, as we shall see, the Quaternion method is independent 
of axes or any supposed directions in space, and takes its reference 
lines solely from the problem it is applied to. 

14. But, for the purpose of elementary exposition, it is best 
to begin by assimilating it as closely as we can to the ordinary 
Cartesian methods of Geometry of Three Dimensions, with which 
the student is supposed to be, to some extent at least, acquainted. 
Such assistance, it will be found, can (as a rule) soon be dispensed 
with; and Hamilton regarded any apparent necessity for an oc 
casional recurrence to it, in higher applications, as an indication 
of imperfect development in the proper methods of the new 

We commence, therefore, with some very elementary geometrical 
ideas, relating to the theory of vectors in space. It will subsequently 
appear how we are thus led to the notion of a Quaternion. 


15. Suppose we have two points A and B in space, and sup 
pose A given, on how many numbers does -B s relative position 
depend ? 

If we refer to Cartesian coordinates (rectangular or not) we find 
that the data required are the excesses of B s three coordinates 
over those of A. Hence three numbers are required. 

Or we may take polar coordinates. To define the moon s 
position with respect to the earth we must have its Geocentric 
Latitude and Longitude, or its Right Ascension and Declination, 
and, in addition, its distance or radius-vector. Three again. 

16. Here it is to be carefully noticed that nothing has been 
said of the actual coordinates of either A or B, or of the earth 
and moon, in space ; it is only the relative coordinates that are 

Hence any expression, as AB, denoting a line considered with 
reference to direction and currency as well as length, (whatever 
may be its actual position in space) contains implicitly three 
numbers, and all lines parallel and equal to AB, and concurrent 
with it, depend in the same way upon the same three. Hence, all 
lines which are equal, parallel, and concurrent, may be represented 
by a common symbol, and that symbol contains three distinct numbers. 
In this sense a line is called a VECTOR, since by it we pass from 
the one extremity, A, to the other, B , and it may thus be 
considered as an instrument which carries A to B: so that a 
vector may be employed to indicate a definite translation in space. 

[The term " currency " has been suggested by Cayley for use 
instead of the somewhat vague suggestion sometimes taken to 
be involved in the word "direction." Thus parallel lines have 
the same direction, though they may have similar or opposite 
currencies. The definition of a vector essentially includes its 

17. We may here remark, once for all, that in establishing a 
new Calculus, we are at liberty to give any definitions whatever 
of our symbols, provided that no two of these interfere with, or 
contradict, each other, and in doing so in Quaternions simplicity 
and (so to speak) naturalness were the inventor s aim. 

18. Let AB be represented by a, we know that a involves 
three separate numbers, and that these depend solely upon the 
position of B relatively to A. Now if CD be equal in length to AB 


and if these lines be parallel, and have the same currency, we may 
evidently write 

where it will be seen that the sign of equality between vectors 
contains implicitly equality in length, parallelism in direction, 
and concurrency. So far we have extended the meaning of an 
algebraical symbol. And it is to be noticed that an equation 
between vectors, as 

a = /3, 

contains three distinct equations between mere numbers. 

19. We must now define + (and the meaning of will follow) 
in the new Calculus. Let A, B, G be any three points, and (with 
the above meaning of = ) let 

If we define + (in accordance with the idea ( 16) that a vector 
represents a translation) by the equation 


we contradict nothing that precedes, but we at once introduce the 
idea that vectors are to be compounded, in direction and magnitude, 
like simultaneous velocities. A reason for this may be seen in 
another way if we remember that by adding the (algebraic) differ 
ences of the Cartesian coordinates of B and A, to those of the 
coordinates of G and B, we get those of the coordinates of C and 
A. Hence these coordinates enter linearly into the expression for 
a vector. (See, again, 5.) 

20. But we also see that if C and A coincide (and G may be 
any point) 


for no vector is then required to carry A to G. Hence the above 
relation may be written, in this case, 

AB + BA = 0, 

or, introducing, and by the same act defining, the symbol , 

BA=-AB t 

Hence, the symbol , applied to a vector, simply shews that its 
currency is to be reversed. 


And this is consistent with all that precedes ; for instance, 

and AB = AC-BC, 

or =AC + CB, 

are evidently but different expressions of the same truth. 

21. In any triangle, ABC, we have, of course, 

and, in any closed polygon, whether plane or gauche, 
AB+BC+ ...... + YZ + ZA = Q. 

In the case of the polygon we have also 

AB + BC+ ...... + YZ = AZ. 

These are the well-known propositions regarding composition 
of velocities, which, by Newton s second law of motion, give us 
the geometrical laws of composition of forces acting at one point. 

22. If we compound any number of parallel vectors, the result / 
is obviously a numerical multiple of any one of them. 
Thus, if A, B, G are in one straight line, 

where x is a number, positive when B lies between A and C, other 
wise negative : but such that its numerical value, independent 
of sign, is the ratio of the length of BC to that of AB. This is 
at once evident if AB and BC be commensurable ; and is easily 
extended to incommensurables by the usual reductio ad absurdum. 

23. An important, but almost obvious, proposition is that any 
vector may be resolved, and in one way only, into three components 
parallel respectively to any three given vectors, no two of which are 
parallel, and which are not parallel to one plane. 

Let OA, OB, OC be the three fixed c 
vectors, OP any other vector. From P draw 
PQ parallel to CO, meeting the plane BO A 
in Q. [There must be a definite point Q, 
else PQ, and therefore CO, would be parallel 
to BOA, a case specially excepted.] From Q 
draw QR parallel to BO, meeting OA in R. 

Then we have OP=OR + RQ + QP ( 21), 

and these components are respectively parallel to the three given 


vectors. By 22 we may express OR as a numerical multiple 
of OA, RQ of OB, and QP of OC. Hence we have, generally, for 
any vector in terms of three fixed non-coplanar vectors, a, 0, 7, 

OP = p = XQL + yft + 7, 

which exhibits, in one form, the three numbers on which a vector 
depends ( 16). Here x, y, z are perfectly definite, and can have 
but single values. 

24. Similarly any vector, as OQ, in the same plane with OA 
and OB, can be resolved (in one way only) into components OR, 
RQ, parallel respectively to A and OB ; so long, at least, as these 
two vectors are not parallel to each other. 

25. There is particular advantage, in certain cases, in em 
ploying a series of three mutually perpendicular unit-vectors as 
lines of reference. This system Hamilton denotes by i,j, k. 

Any other vector is then expressible as 
p xi + yj + zk. 

Since i, j, k are unit-vectors, x, y, z are here the lengths of 
conterminous edges of a rectangular parallelepiped of which p 
is the vector-diagonal ; so that the length of p is, in this case, 

Let w = f i + ijj + J& 

be any other vector, then (by the proposition of 23) the vector 

equation p = & 

obyiously involves the following three equations among numbers, 

a?=f, y = n, z = 

Suppose i to be drawn eastwards, j northwards, and k upwards, 
this is equivalent merely to saying that if two points coincide, they 
are equally to the east (or west) of any third point, equally to the 
north (or south) of it, and equally elevated above (or depressed below) 
its level. 

26. It is to be carefully noticed that it is only when a, /3, 7 
are not coplanar that a vector equation such as 

p = *&, 

or XQL + y/3 -f zy = f a + rj/3 + 7, 

necessitates the three numerical equations 

= fc V = V. * = 


For, if a, /3, 7 be coplanar ( 24), a condition of the following form 
must hold 

7 = aa. + b/3. 

Hence p = (# + za) a 

and the equation p = in 

now requires only the two numerical conditions 

x + za = 

27. TAe Commutative and Associative Laws hold in the com 
bination of vectors by the signs + and . It is obvious that, if we 
prove this for the sign + , it will be equally proved for , because 
before a vector ( 20) merely indicates that it is to be reversed 
before being considered positive. 

Let A, B, C, D be, in order, the corners of a parallelogram ; we 
have, obviously, 

And = ^ = . 

Hence the commutative law is true for the addition of any two 
vectors, and is therefore generally true. 

Again, whatever four points are represented by A, B, G, D, we 


or substituting their values for AD, BD, AC respectively, in these 
three expressions, 

And thus the truth of the associative law is evident. 
28. The equation 


where p is the vector connecting a variable point with the origin, 
/3 a definite vector, and x an indefinite number, represents the 
straight line drawn from the origin parallel to @ ( 22). 

The straight line drawn from A, where OA = a, arid parallel 
to /3, has the equation 

P = + */8 ................................. (1). 

In words, we may pass directly from to P by the vector OP or 
p ; or we may pass first to A, by means of OA or a, and then to P 
along a vector parallel to /3 ( 16). 


Equation (1) is one of the many useful forms into which 
Quaternions enable us to throw the general equation of a straight 
line in space. As we have seen ( 25) it is equivalent to three 
numerical equations ; but, as these involve the indefinite quantity 
x, they are virtually equivalent to but two, as in ordinary Geometry 
of Three Dimensions. 

29. A good illustration of this remark is furnished by the fact 
that the equation 


which contains two indefinite quantities, is virtually equivalent to: 
only one numerical equation. And it is easy to see that it re 
presents the plane in which the lines a. and j3 lie ; or the surface 
which is formed by drawing, through every point of OA, a line 
parallel to OB. In fact, the equation, as written, is simply 24 
in symbols. 

And it is evident that the equation 


P 7 + 2/ a + # 

is the equation of the plane passing through the extremity of 7, 
and parallel to a and ft. 

It will now be obvious to the reader that the equation 

where a t , a 2 , &c. are given vectors, and p lt p 9 , &c. numerical quan 
tities, represents a straight line if p t , p 9 , &c. be linear functions of 
one indeterminate number ; and a plane, if they be linear expres 
sions containing two indeterminate numbers. Later ( 31 ()), 
this theorem will be much extended. 
Again, the equation 

refers to any point whatever in space, provided a, ft, 7 are not 
coplanar. (Ante, 23.) 

30. The equation of the line joining any two points A and B, 
where A = a and OB = ft, is obviously 


These equations are of course identical, as may be seen by putting 
1 y for x. 

12 QUATE&NIONS. [31. 

The first may be written 

" tP 

subject to the condition p + q + r = Q identically. That is A 
homogeneous linear function of three vectors, equated to zero, 
expresses that the extremities of these vectors are in one straight 
line, if the sum of the coefficients be identically zero. 

Similarly, the equation of the plane containing the extremities 
A y B, C of the three non-coplanar vectors a, /3, 7 is 

where x and y are each indeterminate. 
This may be written 

pp + qa + r/3 + sy = 0, 
with the identical relation 

which is one form of the condition that four points may lie in one 

31. We have already the means of proving, in a very simple 
manner, numerous classes of propositions in plane and solid 
geometry. A very few examples, however, must suffice at this 
stage ; since we have hardly, as yet, crossed the threshold of the 
subject, and are dealing with mere linear equations connecting two 
or more vectors, and even with them we are restricted as yet to 
operations of mere addition. We will give these examples with a 
painful minuteness of detail, which the reader will soon find to be 
necessary only for a short time, if at all. 

(a) The diagonals of a parallelogram bisect each other. 
Let A BCD be the parallelogram, the point of intersection of 
its diagonals. Then 

AO + OB=AB=DC= DO + 00, 
which gives AO-OC = DO-OB. 

The two vectors here equated are parallel to the diagonals respect 
ively. Such an equation is, of course, absurd unless 

(1) The diagonals are parallel, in which case the figure 
is not a parallelogram ; 

(2) AO = OC, and DO = OB, the proposition. 


(6) To shew that a triangle can be constructed, whose sides 
are parallel, and equal, to the bisectors of the sides of any 

Let ABC be any triangle, Aa, Bb, Cc the bisectors of the 

Then Aa = AB + Tla = AB + BC, c 

55 - 

~Cc - - 

Hence ~A~a + ~Bb + ~Cc = f ( + J9C + ) = ; 

which (21) proves the proposition. 
Also Aa=AB+BC 

= AB-$(CA+AB) 

results which are sometimes useful. They may be easily verified 
by producing A a to twice its length and joining the extremity 
with B. 

(b 1 ) The bisectors of the sides of a triangle meet in a point, 
which trisects each of them. 

Taking A as origin, and putting a, /3, 7 for vectors parallel, and 
equal, to the sides taken in order BG, GA, AB; the equation of 
Bb is ( 28 (1)) 

That of Cc is, in the same way, 
p = -(l H 

At the point 0, where Bb and Cc intersect, 


Since 7 and /S are not parallel, this equation gives 

From these x = y = f . 

Hence ZO = J (7 - 0) = | Aa. (See Ex. (6).) 

This equation shews, being a vector one, that A a passes 
through 0, and that AO : Oa :: 2:1. 


(c) If 


04 = 0, 

OC = la + m/3, 

be three given co-planar vectors, c the intersection of AB, OC, and 
if the lines indicated in the figure be drawn, the points a v b v c l lie 
in a straight line. 

We see at once, by the process indicated in 30, that 


l + m 

Hence we easily find 



Oa = - 



m l 

l-l-2m l-21-m 

These give 

- (1 - I - 2m) ~0a l + (l-2l- m) ~Ob l - (m - I) Oc^ = 0. 

But - (1 - I - 2m) + (1 - 21 - m) - (m - I) = identically. 
This, by 30, proves the proposition. 

(d) Let OA = a, OB = /S, be any two vectors. If MP be a 
given line parallel to OB ; and OQ, BQ, be drawn parallel to AP, 
OP respectively ; the locus of Q is a straight line parallel to OA. 




Hence the equation of OQ is 

and that of BQ is p = ft + z (ea. + acj3). 

At Q we have, therefore, 

These give xy = e, and the equation of the locus of Q is 

p = e& + 2/X 

i.e. a straight line parallel to 0.4, drawn through N in OjB pro 
duced, so that 

COR. If BQ meet MP in q, Pq = ; and if AP meet NQ in 
p, Qp = a. 

Also, for the point R we have pR = AP, QR = Bq. 

Further, the locus of R is a hyperbola, of which MP and NQ 
are the asymptotes. See, in this connection, 31 (k) below. 

Hence, if from any two points, A and B, lines be drawn inter 
cepting a given length Pq on a given line Mq ; and if, from R their 
point of intersection, Rp be laid off= PA, and RQ = qB ; Q and p 
lie on a fixed straight line, and the length of Qp is constant. 

(e) To find the centre of inertia of any system of masses. 
If OA = a, OB = a, l , be the vector sides of any triangle, the 
vector from the vertex dividing the base AB in C so that 
BC : CA :: m : m l 
ma + m,GL 


For AB is a x a, and therefore AC is 

Hence 00 = OA + AC 

_ ma. + m 1 1 

m + ra t 

This expression shews how to find the centre of inertia of two 
masses ; m at the extremity of a, m l at that of a r Introduce m g 



at the extremity of 2 , then the vector of the centre of inertia of the 
three is, by a second application of the formula, 

(m -f 


From this it is clear that, for any number of masses, expressed 
generally by m at the extremity of the vector a, the vector of the 
centre of inertia is 

2 (ma) 

This may be written 2m (a - ft) = 0. 

Now a l - ft is the vector of m x with respect to the centre of inertia. 
Hence the theorem, If the vector of each element of a mass, drawn 
from the centre of inertia, be increased in length in proportion to the 
mass of the element, the sum of all these vectors is zero. 

(/) We see at once that 
the equation 

where t is an indeterminate 
number, and a, ft given vec 
tors, represents a parabola. 
The origin, 0, is a point on 
the curve, ft is parallel to 
the axis, i.e. is the diameter 
OB drawn from the origin, 
and a is OA the tangent at the origin. In the figure 

QP = at, OQ = ^-. 

The secant joining the points where t has the values t and t is 
represented by the equation 

*-*? (30) 

V } 

Write x for x (f - 1) [which may have any value], then put tf = t, 
and the equation of the tangent at the point (t) is 


In this put x = t, and we have 


p ~ "T 

or the intercept of the tangent on the diameter is equal in length 
to the abscissa of the point of contact, but has the opposite 

Otherwise : the tangent is parallel to the vector a + fit or 

at + /3f or f + at + ^f or OQ + OP. But TP=TO + OP, 

L - 

hence TO = OQ. 

(g) Since the equation of any tangent to the parabola is 


+ fit), 

let us find the tangents which can be drawn from a given point. 
Let the vector of the point be 

p=p + q/3 (24). 

Since the tangent is to pass through this point, we have, as con 
ditions to determine t and #, 



by equating respectively the coefficients of a and fi. 

Hence t=pJp* 2q. 

Thus, in general, m> tangents can be drawn from a given point. 
These coincide if p 2 = 2q ; 

that is, if the vector of the point from which they are to be drawn 

is p=pa + qP = pa + ^0, 

i.e. if the point lies on the parabola. They are imaginary if 
Zq >p z , that is, if the point be 

r being positive. Such a point is evidently within the curve, as at 
R, where OQ = ^/3, QP=pa, PR = rfi. 

T. Q. I. 2 

18 QUATERNIONS. [3 T (^)- 

(h) Calling the values of t for the two tangents found in (g) 
^ and t 2 respectively, it is obvious that the vector joining the 
points of contact is 

which is parallel to 4-/3 -- s > 

or, by the values of ^ and 2 in (g), 

a + p/3. 

Its direction, therefore, does not depend on q. In words, If pairs of 
tangents be drawn to a parabola from points of a diameter produced, 
the chords of contact are parallel to the tangent at the vertex of the 
diameter. This is also proved by a former result, for we must have 
OT for each tangent equal to QO. 

(i) The equation of the chord of contact, for the point whose 
vector is 

Rt 2 
is thus p = 0-^ + -- + y (a + pft). 

Suppose this to pass always through the point whose vector is 

p = act + 6/3. 
Then we must have ^ + y = a, \ 

or t^p + Jp*- 2pa + 26. 

Comparing this with the expression in (g), we have 

q = pa - b ; 

that is, the point from which the tangents are drawn has the vector 
p=pa + (pa-b) fi 
= - b/3 +p (a + aft), a straight line ( 28 (1)). 

The mere form of this expression contains the proof of the usual 
properties of the pole and polar in the parabola ; but, for the sake 
of the beginner, we adopt a simpler, though equally general, 

Suppose a = 0. This merely restricts the pole to the particular 


diameter to which we have referred the parabola. Then the pole 
is Q, where p = b/3; 

and the polar is the line TU, for which 


Hence the polar of any point is parallel to the tangent at the 
extremity of the diameter on which the point lies, and its inter 
section with that diameter is as far beyond the vertex as the pole 
is within, and vice versa. 

(j) As another example let us prove the following theorem. 
If a triangle be inscribed in a parabola, the three points in which 
the sides are met by tangents at the angles lie in a straight line. 

Since is any point of the curve, we may take it as one corner 
of the triangle. Let t and t t determine the others. Then, if 
w,, -sr 2 , OT 3 represent the vectors of the points of intersection of the 
tangents with the sides, we easily find 


tt l 

f -\- 1 
These values give 

j. 2 / 2 

-.- J ir-.=- 

v 2 t 2 

Also ^- - - "* - " 1 - = identically. 

t tt. 

Hence, by 30, the proposition is proved. 

(k) Other interesting examples of this method of treating 
curves will, of course, suggest themselves to the student. Thus 

p = a. cos t + @ sin t 

represents an ellipse, of which the given vectors a and ft are semi- 
conjugate diameters. If t represent time, the radius-vector of this 
ellipse traces out equal areas in equal times. [We may anticipate 
so far as to write the following : 

2 Area = TJVpdp = TVap . fdt ; 
which will be easily understood later.] 


20 QUATERNIONS. [31 (/). 


Again, p = at + or p = a tan x + (3 cot x 


evidently represents a hyperbola referred to its asymptotes. [If 
t represent time, the sectorial area traced out is proportional to 
log t t taken between proper limits.] 
Thus, also, the equation 

p = a. (t + sin t) + /3 cos t 

in which a and ft are of equal lengths, and at right angles to one 
another, represents a cycloid. The origin is at the middle point of 
the axis (2/9) of the curve. [It may be added that, if t represent 
time, this equation shews the motion of the tracing point, provided 
the generating circle rolls uniformly, revolving at the rate of a 
radian per second.] 

When the lengths of a, ft are not equal, this equation gives the 
cycloid distorted by elongation of its ordinates or abscissae : not a 
trochoid. The equation of a trochoid may be written 

p = a (et + sin f) + ft cos t, 

e being greater or less than 1 as the curve is prolate or curtate. 
The lengths of a and ft are still taken as equal. 

But, so far as we have yet gone with the explanation of the 
calculus, as we are not prepared to determine the lengths or in 
clinations of vectors, we can investigate only a very limited class of 
the properties of curves, represented by such equations as those 
above written. 

(1) We may now, in extension of the statement in 29, make 
the obvious remark that 

(where, as in 23, the number of vectors, a, can always be reduced 
to three, at most) is the equation of a curve in space, if the 
numbers p v p z , &c. are functions of one indeterminate. In such 
a case the equation is sometimes written 

P =<t>(t). 

But, if p v p 2 , &c. be functions of two indeterminates, the locus of 
the extremity of p is a surface; whose equation is sometimes written 

[It may not be superfluous to call the reader s attention to the 
fact that, in these equations, </> (t) or c/> (t, u) is necessarily a vector 
expression, since it is equated to a vector, p,] 


(m) Thus the equation 

p = a cos t + /3 sin t 4- yt ..................... (1) 

belongs to a helix, while 

p = a cos t + @ sin t + 7^ .................. (2) 

represents a cylinder whose generating lines are parallel to 7, and 
whose base is the ellipse 

p = a cos + /3 sin t. 

The helix above lies wholly on this cylinder. 
Contrast with (2) the equation 

p = u(acost + /3sin t + 7) ..................... (3) 

which represents a cone of the second degree : made up, in fact, 
of all lines drawn from the origin to the ellipse 

p = a cos t + sin t + 7. 
If, however, we write 

p = u (a cos t + @ sin 4- yt), 

we form the equation of the transcendental cone whose vertex is 
at the origin, and on which lies the helix (1). 
In general 

p = u<f> (t) 
is the cone whose vertex is the origin, and on which lies the curve 

while p = (f>(t) + UOL 

is a cylinder, with generating lines parallel to a, standing on the 
same curve as base. 

Again, p = pa + q/3 + ry 

with a condition of the form 

op 2 + bf + cr 2 = I 

belongs to a central surface of the second order, of which a, (3, 7 
are the directions of conjugate diameters. If a, b, c be all positive, 
the surface is an ellipsoid. 

32. In Example (/") above we performed an operation equi 
valent to the differentiation of a vector with reference to a single 
numerical variable of which it was given as an explicit function. 
As this process is of very great use, especially in quaternion investi 
gations connected with the motion of a particle or point ; and as it 
will afford us an opportunity of making a preliminary step towards 


overcoming the novel difficulties which arise in quaternion differen 
tiation ; we will devote a few sections to a more careful, though 
very elementary, exposition of it. 

33. It is a striking circumstance, when we consider the way 
in which Newton s original methods in the Differential Calculus 
have been decried, to find that Hamilton was obliged to employ 
them, and not the more modern forms, in order to overcome the 
characteristic difficulties of quaternion differentiation. Such a thing 
as a differential coefficient has absolutely no meaning in quaternions, 
except in those special cases in which we are dealing with degraded 
quaternions, such as numbers, Cartesian coordinates, &c. But a 
quaternion expression has always a differential, which is, simply, 
what Newton called a fluxion. 

As with the Laws of Motion, the basis of Dynamics, so with the 
foundations of the Differential Calculus ; we are gradually coming 
to the conclusion that Newton s system is the best after all. 

34. Suppose p to be the vector of a curve in space. Then, 
generally, p may be expressed as the sum of a number of terms, 
each of which is a multiple of a constant vector by a function of some 
one indeterminate ; or, as in 31 (1), if P be a point on the curve, 

OP= P = <HO. 

And, similarly, if Q be any other point on the curve, 
06 = ^ = ^ + 8^=0(0 = *^+^), 

where St is any number whatever. 

The vector-chord PQ is therefore, rigorously, 

35. It is obvious that, in the present case, because the vectors 
involved in c/> are constant, and their numerical multipliers alone vary, 
the expression * (t + $t) is, by Taylor s Theorem, equivalent to 

*, . 

*. . 

Hence, Sp = ^- ( + 

And we are thus entitled to write, when Bt has been made inde 
finitely small, 

T . . /&p\ dp d<f) (t) , , A 
Limit = - = -^ = </> (*) 




In such a case as this, then, we are permitted to differentiate, 
or to form the differential coefficient of, a vector, according to the 
ordinary rules of the Differential Calculus. But great additional 
insight into the process is gained by applying Newton s method. 

36. Let OP be 

and OQ 1 

Pl = (f> (t + dt), 

where dt is any number whatever. 

The number t may here be taken 
as representing time, i.e. we may 
suppose a point to move along the 
curve in such a way that the value 
of t for the vector of the point P of 
the curve denotes the interval which 
has elapsed (since a fixed epoch) when the moving point has 
reached the extremity of that vector. If, then, dt represent any 
interval, finite or not, we see that 

will be the vector of the point after the additional interval dt. 

But this, in general, gives us little or no information as to the 
velocity of the point at P. We shall get a better approximation 
by halving the interval dt, and finding Q 2 , where OQ 2 = $ (t + |cft), 
as the position of the moving point at that time. Here the vector 
virtually described in \dt is PQ 2 . To find,_on this supposition, 
the vector described in dt, we must double PQ 2 , and we find, as a 
second approximation to the vector which the moving point would 
have described in time dt, if it had moved for that period in the 
direction and with the velocity it had at P, 

= 2 

The next approximation gives 

- 8 

And so on, each step evidently leading us nearer the sought truth. 
Hence, to find the vector which would have been described in time 
dt had the circumstances of the motion at P remained undisturbed, 
we must find the value of 

dp = Tq = 

(t 4 i dt} - 4> (t)\ 

\ X / 


We have seen that in this particular case we may use Taylor s 
Theorem. We have, therefore, 

dp = ^ = 00 X {< (0 \ dt + f (t) A ^ + &C.J 

= (#> (t) dt. 
And, if we choose, we may now write 

s -*< 

37. But it is to be most particularly remarked that in the 
whole of this investigation no regard whatever has been paid to 
the magnitude of dt. The question which we have now answered 
may be put in the form A point describes a given curve in a given 
manner. At any point of its path its motion suddenly ceases to be 
accelerated. What space will it describe in a definite interval ? As 
Hamilton well observes, this is, for a planet or comet, the case 
of a celestial Atwood s machine. 

38. If we suppose the variable, in terms of which p is expressed, 
to be the arc, s, of the curve measured from some fixed point, we 
find as before 

dp = fi(s) ds. 

From the very nature of the question it is obvious that the length 
of dp must in this case be ds, so that </> () is necessarily a unit- 
vector. This remark is of importance, as we shall see later ; and 
it may therefore be useful to obtain afresh the above result without 
any reference to time or velocity. 

39. Following strictly the process of Newton s Vllth Lemma, 
let us describe on Pg 2 an arc similar to PQ 2 , and so on. Then 
obviously, as the subdivision of ds is carried farther, the new arc 
(whose length is always ds) more and more nearly (and without 
limit) coincides with the line which expresses the corresponding 
approximation to dp. 

40. As additional examples let us take some well-known 
plane curves; and first the hyperbola (31 

Here db-1 -)<& 


This shews that the tangent is parallel to the vector 


In words, if the vector (from the centre) of a point in a hyperbola 
be one diagonal of a parallelogram, two of whose sides coincide with 
the asymptotes, the other diagonal is parallel to the tangent at the 
point, and cuts off a constant area from the space between the 
asymptotes. (For the sides of this triangular area are t times the 
length of a, and I/t times the length of ft, respectively ; the angle 
between them being constant.) 

Next, take the cycloid, as in 31 (&), 

p = a (t + sin t)+ft cos t. 
We have 

dp = {(!+ cos t) ft sin t} dt. 
At the vertex 

t = 0, cos t = l, sin t = 0, and dp = 2adt. 
At a cusp 

t = TT, cos = 1, sin = 0, and dp = 0. 
This indicates that, at the cusp, the tracing point is (instan 
taneously) at rest. To find the direction of the tangent, and the 
form of the curve in the vicinity of the cusp, put t = TT + T, where 
powers of T above the second are omitted. We have 

dp = ftrdt + -J"- dt, 

so that, at the cusp, the tangent is parallel to ft. By making the 
same substitution in the expression for p, we find that the part of 
the curve near the cusp is a semicubical parabola, 

or, if the origin be shifted to the cusp (p = TTOL - ft\ 

41. Let us reverse the first of these questions, and seek the 
envelop of a line which cuts off from two fixed axes a triangle of 
constant area. 

If the axes be in the directions of a and ft, the intercepts may 

evidently be written at and ^-. Hence the equation of the line is 

//? \ 

p = a.t 4- x ( j at) . 


The condition of envelopment is, obviously, (see Chap. IX.) 

This gives = la - 


x f + a dt + - ctf 


Hence (1 as) dt tdx = 0, 

x , dx 
and - - dt + = 0. 

6 C 

From these, at once, x = J, since cfcc and cfa are indeterminate. 
Thus the equation of the envelop is 

the hyperbola as before ; a, /3 being portions of its asymptotes. 

42. It may assist the student to a thorough comprehension 
of the above process, if we put it in a slightly different form. 
Thus the equation of the enveloping line may be written 

which gives dp = = CLd {t(l - as)} + @d . 


Hence, as a is not parallel to /3, we must have 


and these are, when expanded, the equations we obtained in the 
preceding section. 

43. For farther illustration we give a solution not directly 
employing the differential calculus. The equations of any two of 
the enveloping lines are 

* Here we have opportunity for a remark (very simple indeed, but) of the 
utmost importance. We are not to equate separately to zero the coefficients of dt 
and dx; for we must remember that this equation is of the form 

where p and q are numbers ; and that, so long as a and /3 are actual and non-parallel 
vectors, the existence of such an equation requires ( 24) 


(R \ 
p = at + x ( - at ) , 

\i / 

t and t t being given, while x and x l are indeterminate. 

At the point of intersection of these lines we have ( 20), 

These give, by eliminating x v 

t n _ .\ 
i \L x) 

or x . 

Hence the vector of the point of intersection is 

_ aflt + ff 
P= tt + t 

and thus, for the ultimate intersections, where < - = 1, 


+ ^- ) as before. 

COR. If. instead of the ultimate intersections, we consider 
the intersections of pairs of these lines related by some law, we 
obtain useful results. Thus let 

OL + /3 


or the intersection lies in the diagonal of the parallelogram on a, 0. 
If ^ = mt, where m is constant, 


mtot + 

P = 

But we have also x = 


m-f 1 

Hence the locus of a point which divides in a given ratio a line 
cutting off a given area from two fixed axes, is a hyperbola of which 
these axes are the asymptotes. 


If we take either 


tt^t + tfj) = constant, or *- = constant, 
r + c, 

the locus is a parabola ; and so on. 

It will be excellent practice for the student, at this stage, to 
work out in detail a number of similar questions relating to the 
envelop of, or the locus of the intersection of selected pairs from, a 
series of lines drawn according to a given law. And the process 
may easily be extended to planes. Thus, for instance, we may 
form the general equation of planes which cut off constant tetra- 
hedra from the axes of coordinates. Their envelop is a surface of 
the third degree whose equation may be written 

p = CCOL + y/3 + 7 ; 
where ocyz = a 3 . 

Again, find the locus of the point of intersection of three of 
this group of planes, such that the first intercepts on /3 and 7, the 
second on 7 and a, the third on a and /3, lengths all equal to one 
another, &c. But we must not loiter with such simple matters as 

44. The reader who is fond of Anharmonic Ratios and Trans 
versals will find in the early chapters of Hamilton s Elements of 
Quaternions an admirable application of the composition of vectors 
to these subjects. The Theory of Geometrical Nets, in a plane, 
and in space, is there very fully developed; and the method is 
shewn to include, as particular cases, the corresponding processes of 
Grassmann s Ausdehnungslehre and Mobius Barycentrische Calcul. 
Some very curious investigations connected with curves and surfaces 
of the second and third degrees are also there founded upon the 
composition of vectors. 


1. The lines which join, towards the same parts, the extremities 
of two equal and parallel lines are themselves equal and parallel. 
(Euclid, I. xxxiii.) 

2. Find the vector of the middle point of the line which joins 
the middle points of the diagonals of any quadrilateral, plane or 


gauche, the vectors of the corners being given ; and so prove that 
this point is the mean point of the quadrilateral. 

If two opposite sides be divided proportionally, and two new 
quadrilaterals be formed by joining the points of division, the mean 
points of the three quadrilaterals lie in a straight line. 

Shew that the mean point may also be found by bisecting the 
line joining the middle points of a pair of opposite sides. 

3. Verify that the property of the coefficients of three vectors 
whose extremities are in a line ( 30) is not interfered with by 
altering the origin. 

4. If two triangles ABC, abc, be so situated in space that Aa, 
Bb, Cc meet in a point, the intersections of AB, ab, of BO, be, and 
of CA, ca, lie in a straight line. 

5. Prove the converse of 4, i. e. if lines be drawn, one in each 
of two planes, from any three points in the straight line in which 
these planes meet, the two triangles thus formed are sections of a 
common pyramid. 

6. If five quadrilaterals be formed by omitting in succession 
each of the sides of any pentagon, the lines bisecting the diagonals 
of these quadrilaterals meet in a point. (H. Fox Talbot.) 

7. Assuming, as in 7, that the operator 

cos 6 + J 1 sin 6 

turns any radius of a given circle through an angle 6 in the 
positive direction of rotation, without altering its length, deduce 
the ordinary formulae for cos (A -f B), cos (A B), sin (^L + B), and 
sin (A B), in terms of sines and cosines of A and B. 

8. If two tangents be drawn to a hyperbola, the line joining 
the centre with their point of intersection bisects the lines join 
ing the points where the tangents meet the asymptotes : and the 
secant through the points of contact bisects the intercepts on 
the asymptotes. 

9. Any two tangents, limited by the asymptotes, divide each 
other proportionally. 

10. If a chord of a hyperbola be one diagonal of a parallelogram 
whose sides are parallel to the asymptotes, the other diagonal passes 
through the centre. 


11. Given two points A and B, and a plane, C. Find the 
locus of P, such that if AP cut C in Q, and BP cut G in R, QR 
may be a given vector. 

12. Shew that p = # 2 a + ?/ 2 /3 + (a? + y)V 

is the equation of a cone of the second degree, and that its section 
by the plane 

_ pa. + q/3 + ry 
^ p + q + r 

is an ellipse which touches, at their middle points, the sides of the 
triangle of whose corners a, /3, 7 are the vectors. (Hamilton, 
Elements, p. 96.) 

13. The lines which divide, proportionally, the pairs of opposite 
sides of a gauche quadrilateral, are the generating lines of a hyper 
bolic paraboloid. (Ibid. p. 97.) 

14. Shew that p = x s a + y s /3 + z s y, 
where # + 2/ + 2 = 0, 

represents a cone of the third order, and that its section by the 

__ pa 4- q/3 + ry 
p + q + r 

is a cubic curve, of which the lines 

pCL + q/3 p 
p = - - , &c. 
p + q 

are the asymptotes and the three (real) tangents of inflection. Also 
that the mean point of the triangle formed by these lines is a 
conjugate point of the curve. Hence that the vector a + /3 + y is 
a conjugate ray of the cone. (Ibid. p. 96.) 



45. WE now come to the consideration of questions in which 
the Calculus of Quaternions differs entirely from any previous 
mathematical method ; and here we shall get an idea of what a 
Quaternion is, and whence it derives its name. These questions 
are fundamentally involved in the novel use of the symbols of 
multiplication and division. And the simplest introduction to 
the subject seems to be the consideration of the quotient, or ratio, 
of two vectors. 

46. If the given vectors be parallel to each other, we have 
already seen ( 22) that either may be expressed as a numerical 
multiple of the other; the multiplier being simply the ratio of 
their lengths, taken positively if they have similar currency, 
negatively if they run opposite ways. 

47. If they be not parallel, let OA and OB be drawn parallel 
and equal to them from any point ; and the question is reduced 
to finding the value of the ratio of two vectors drawn from the 
same point. Let us first find upon how many distinct numbers this 
ratio depends. 

We may suppose OA to be changed into OB by the following 
successive processes. 

1st. Increase or diminish the length of OA till it becomes 
equal to that of OB. For this only one number is required, viz. 
the ratio of the lengths of the two vectors. As Hamilton remarks, 
this is a positive, or rather a signless, number. 

2nd. Turn OA about 0, in the common plane of the two 
vectors, until its direction coincides with that of OB, and (remem- 


bering the effect of the first operation) we see that the two vectors 
now coincide or become identical. To specify this operation three 
numbers are required, viz. two angles (such as node and inclination 
in the case of a planet s orbit) to fix the plane in which the rotation 
takes place, and one angle for the amount of this rotation. 

Thus it appears that the ratio of two vectors, or the multiplier 
required to change one vector into another, in general depends upon 
four distinct numbers, whence the name QUATERNION. 

A quaternion q is thus defined as expressing a relation 

between two vectors a, /3. By what precedes, the vectors a, /3, 
which serve for the definition of a given quaternion, must be in a 
given plane, at a given inclination to each other, and with their 
lengths in a given ratio ; but it is to be noticed that they may be 
any two such vectors. [Inclination is understood to include sense, 
or currency, of rotation from a to /3.] 

The particular case of perpendicularity of the two vectors, where 
their quotient is a vector perpendicular to their plane, is fully 
considered below ; 64, 65, 72, &c. 

48. It is obvious that the operations just described may be 
performed, with the same result, in the opposite order, being per 
fectly independent of each other. Thus it appears that a quaternion, 
considered as the factor or agent which changes one definite vector 
into another, may itself be decomposed into two factors of which 
the order is immaterial. 

The stretching factor, or that which performs the first operation 
in 47, is called the TENSOR, and is denoted by prefixing T to the 
quaternion considered. 

The turning factor, or that corresponding to the second operation 
in 47, is called the VERSOR, and is denoted by the letter U pre 
fixed to the quaternion. 

49. Thus, if OA = a, OB = {3, and if q be the quaternion 
which changes a. to /3, we have 

which we may write in the form 

a ^ * 

if we agree to define that 




Here it is to be particularly noticed that we write q before a to 
signify that a is multiplied by (or operated on by) q, not q 
multiplied by ot. 

This remark is of extreme importance in quaternions, for, as we 
shall soon see, the Commutative Law does not generally apply to 
the factors of a product. 

We have also, by 47, 48, 

q = Tq.Uq=Uq. Tq, 

where, as before, Tq depends merely on the relative lengths of 
a and /3, and Uq depends solely on their directions. 

Thus, if t and /9 X be vectors of unit length parallel to a. and (3 

T = 

= U/3J Ua^ 

As will soon be shewn, when a is perpendicular to /3, i.e. when the 
versor of the quotient is quadrantal, it is a unit-vector. 

50. We must now carefully notice that the quaternion which 
is the quotient when (3 is divided by a in no way depends upon 
the absolute lengths, or directions, of these vectors. Its value 
will remain unchanged if we substitute for them any other pair 
of vectors which 

(1) have their lengths in the same ratio, 

(2) have their common plane the same or parallel, 
and (3) make the same angle with each other. 

Thus in the annexed figure 

if, and only if, 

-^w !J;=S> 

(2) plane AOB parallel to plane Afl^B v 

/o\ / A (~\~R / A C\ 2? 
\*J/ Z. xl L/-O == . 4TL.\J.-D . 

[Equality of angles is understood to include 
concurrency of rotation. Thus in the annexed 
figure the rotation about an axis drawn upwards 
from the plane is negative (or clock- wise) from / 
OA to OB, and also from O l A l to 0^.] 
T. Q. I. 


It thus appears that if 

f} = qaL, & = qy, 

the vectors a, /3, 7, S are parallel to one plane, and may be repre 
sented (in a highly extended sense) as proportional to one another, 

thus : 

j3 : a = 8 : 7. 

And it is clear from the previous part of this section that this 
may be written not only in the form 

a :=7 : 8 

but also in either of the following forms : 

7 : a = B : . 
a : 7 = : 3. 

While these proportions are true as equalities of ratios, they 
do not usually imply equalities of products. 

Thus, as the first of these was equivalent to the equation 

^= S =g, or /3or l = bf l = q , 
OL 7 

the following three imply separately, (see next section) 

a _ ?_-! 7_S a _/3 -i 

jg--* -0- 7~S~ 
or, if we please, 

afiT 1 = y&- 1 = q~\ 7~ l = S/3" 1 = r, a<f l = /3S 1 = r~ l ; 
where r is a new quaternion, which has not necessarily anything 
(except its plane), in common with q. 

But here great caution is requisite, for we are not entitled to 
conclude from these that 

ctS = /3y, &c. 

This point will be fully discussed at a later stage. Meanwhile 
we may merely state that from 

we are entitled to deduce a number of equivalents such as 

a/rT^ = 7, or a = 7S~ 1 /9, or yS 1 ^ = a ^y, &c. 

51. The Reciprocal of a quaternion <? is defined by the 



Hence if = 7 or 

we must have 5 ==s ~ =s *?~ 1 

For this gives -~. ft = q~ l . qy, 

and each member of the equation is evidently equal to a. 
Or thus : 

= q 

Operate by q~ l , 

? - /3 = . 
Operate on {3 l , 

Or, we may reason thus : since q changes OA to 07?, t?" 1 must 
change OB to O.A, and is therefore expressed by ^ ( 49). 

The tensor of the reciprocal of a quaternion is therefore the 
reciprocal of the tensor ; and the versor differs merely by the 
reversal of its representative angle. The versor, it must be 
remembered, gives the plane and angle of the turning it has 
nothing to do with the extension. 

[Remark. In 49 51, above, we had such expressions as 
- = /3oT l . We have also met with a" 1 /#. Cayley suggests that this 

also may be written in the ordinary fractional form by employing 
the following distinctive notation : 

(It might, perhaps, be even simpler to use the solidus as 
recommended by Stokes, along with an obviously correlative 
type : thus, 

I have found such notations occasionally convenient for private 
work, but I hesitate to introduce changes unless they are abso 
lutely required. See remarks on this point towards the end of the 
Preface to the Second Edition reprinted above.] 





52. The Conjugate of a quaternion q, written Kq, has the 
same tensor, plane, and angle, only the angle is taken the reverse 
way; or the versor of the conjugate is the reciprocal of the versor 
of the quaternion, or (what comes to the same thing) the versor of 
the reciprocal. 

Thus, if OA, OB, OA , lie in one plane, and if 
OA = OA, and Z A OB = Z BOA, we have 


- = q, and =. = coniugate oi q = Kq. 

By last section we see that 

Hence qKq = Kq . q = (Tq)\ 

This proposition is obvious, if we recollect that 
the tensors of q and Kq are equal, and that the 
J versors are such that either annuls the effect of the other ; while 
the order of their application is indifferent. The joint effect of 
these factors is therefore merely to multiply twice over by the 
common tensor. 

53. It is evident from the results of 50 that, if a. and ft be 
of equal length, they may be treated as of unit-length so far as 
their quaternion quotient is concerned. This quotient is therefore 
a versor (the tensor being unity) and may be represented indif 
ferently by any one of an infinite number of concurrent arcs of 
given length lying on the circumference of a circle, of which the 
two vectors are radii. This is of considerable importance in the 
proofs which follow. 

r\ 7> 

Thus the versor = may be represented 

in magnitude, plane, and currency of rota 
tion ( 50) by the arc AB, which may in 

this extended sense be written AB. 

And, similarly, the versor -=4 is repre- 

OA t 

sented by AJBj which is equal to (and concurrent with) AB if 

i.e. if the versors are equal, in the quaternion meaning of the 


54. By the aid of this process, when a versor is represented as 
an arc of a great circle on the unit-sphere, we can easily prove 
that quaternion multiplication is not generally commutative. 

^-^ (VR 
Thus let q be the versor AB or _ _ , 

where is the centre of the sphere. 

Take BC = AB, (which, it must be re 
membered, makes the points A, B, C, lie 
in one great circle), then q may also be 

represented by = . 

In the same way any other versor r may be represented by 

^_^ ^^ /\ 7) 

DB or BE and by = or = . 
3 OD OB 

[The line OB in the figure is definite, and is given by the 
intersection of the planes of the two versors.] 

Now rOD = OB, and qOB=OC. 

Hence qrOD = OC, 


or qr = T=T , and may therefore be represented by the arc DC of 

a great circle. 

But rq is easily seen to be represented by the arc AE. 
For qOA=OB, and rOB=OE 9 

whence rqOA OE. and rq = 


Thus the versors rq and qr, though represented by arcs of equal 
length, are not generally in the same plane and are therefore 
unequal: unless the planes of q and r coincide. 

Remark. We see that we have assumed, or defined, in the 
above proof, that q . m = qr . a. and r.qz = rq.oLm the special case 
when qcn, ra, q . rot, and r . qx are all vectors. 

55. Obviously CB is Kq, BD is Kr, and CD is K (qr). But 

CD = BI) . CB, as we see by applying both to OC. This gives us 
the very important theorem 

K (qr) = Kr . Kq, \/ 

i.e. the conjugate of the product of two versors is the product of their 


conjugates in inverted order. This will, of course, be extended to 
any number of factors as soon as we have proved the associative 
property of multiplication. ( 58 below.) 

56. The propositions just proved are, of course, true of quater 
nions as well as of versors; for the former involve only an additional 
numerical factor which has reference to the length merely, and not 
the direction, of a vector ( 48), and is therefore commutative with 
all other factors. 

57. Seeing thus that the commutative law does not in general 
hold in the multiplication of quaternions, let us enquire whether 
the Associative Law holds generally. That is if p, q, r be three 
quaternions, have we 

p . qr=pq. r ? 

This is, of course, obviously true if^>, q, r be numerical quantities, 
or even any of the imaginaries of algebra. But it cannot be con 
sidered as a truism for symbols which do not in general give 

pq = qp. 

We have assumed it, in definition, for the special case when r, 
qr, and pqr are all vectors. ( 54.) But we are not entitled to 
assume any more than is absolutely required to make our 
definitions complete. 

58. In the first place we remark that p, q, and r may be 
considered as versors only, and therefore represented by arcs of / 
great circles on the unit sphere, for their tensors may obviously 

( 48) be divided out from both sides, being commutative with the 

LetAB=p,EI) = CA = q, and FE = r. 

Join BC and produce the great circle till it meets EF in H, and 

make KH=FE = r, and HG = CB =pq ( 54). 

Join GK. Then JtG = HG . KH = pq . r. 


Join FD and produce it to meet AB in M. Make 
LM = FD, and MN = AB, 

and join NL. Then LN= MN .Lbl=p.qr. 

Hence to shew that p . qr = pq . r 

all that is requisite is to prove that LN, and KG, described as 
above, are equal arcs of the same great circle, since, by the figure, 
they have evidently similar currency. This is perhaps most easily 
effected by the help of the fundamental properties of the curves 
known as Spherical Conies. As they are not usually familiar to 
students, we make a slight digression for the purpose of proving 
these fundamental properties ; after Chasles, by whom and Magnus 
they were discovered. An independent proof of the associative 
principle will presently be indicated, and in Chapter VIII. we shall 
employ quaternions to give an independent proof of the theorems 
now to be established. 

59.* DEF. A spherical conic is the curve of intersection of a 
cone of the second degree with a sphere, the vertex of the cone being 
the centre of the sphere. 

LEMMA. If a cone have one series of circular sections, it has 
another series, and any two circles belonging to different series lie 
on a sphere. This is easily proved as follows. 

Describe a sphere, A, cutting the cone in one circular section, 
C, and in any other point whatever, and let the side OpP of the 
cone meet A in p, P ; P being a point in G. Then PO.Op is 
constant, and, therefore, since P lies in a plane, p lies on a sphere, 
a, passing through 0. Hence the locus, c, of p is a circle, being 
the intersection of the two spheres A and a. 

Let OqQ be any other side of the cone, q and Q being points in 
c, C respectively. Then the quadrilateral qQPp is inscribed in a 
circle (that in which its plane cuts the sphere A) and the exterior 


angle at p is equal to the interior angle at Q. If OL, OM be the 
lines in which the plane POQ cuts the cyclic planes (planes through 
parallel to the two series of circular sections) they are obviously 
parallel to pq, QP, respectively ; and therefore 

z LOp = z Opq = z OQP = z M OQ. 

Let any third side, OrR, of the cone be drawn, and let the 
plane OPR cut the cyclic planes in 01, Om respectively. Then, 

^IOL= Z qpr, 

and these angles are independent of the position of the points p 
and P, if Q and R be fixed points. 

In the annexed section of the above space-diagram by a sphere 
whose centre is 0, IL, Mm are the great circles which represent 
the cyclic planes, PQR is the spherical conic which represents the 
cone. The point P represents the line OpP, and so with the 
others. The propositions above may now be stated thus, 

Arc PL = arc MQ ; 

and, if Q and R be fixed, M m and IL are constant arcs whatever be 
the position of P. 

60. The application to 58 is now obvious. In the figure of 
that article we have 

= GB, LM 

Hence L, (7, G, D are points of a spherical conic whose cyclic 
planes are those of AB, FE. Hence also KG passes through L, 

and with LM intercepts on AB an arc equal to AB. That is, it 
passes through j\ r , or KG and LN are arcs of the same great circle : 
and they are equal, for G and L are points in the spherical 




Also, the associative principle holds for any number of 
quaternion factors. For, obviously, 

qr . st = qrs . t = &c., &c., 

since we may consider qr as a single quaternion, and the above 
proof applies directly. 

61. That quaternion addition, and therefore also subtraction, 
is commutative, it is easy to shew. 

For if the planes of two quaternions, 
q and r, intersect in the line OA, we 
may take any vector OA in that line, 
and at once find two others, OB and 
00, such that 

OB = qOA, 
and CO = rOA. 

And (q + r)OA**OB+OC=OC+OB=(r + q) OA, 

since vector addition is commutative ( 27). 

Here it is obvious that (q + r) OA, being the diagonal of the 
parallelogram on OB, OC, divides the angle between OB and OG 
in a ratio depending solely on the ratio of the lengths of these 
lines, i.e. on the ratio of the tensors of q and r. This will be useful 
to us in the proof of the distributive law, to which we proceed. 

62. Quaternion multiplication, and therefore division, is 
distributive. One simple proof of this depends on the possibility, 
shortly to be proved, of representing any quaternion as a linear 
function of three given rectangular unit- vectors. And when the 
proposition is thus established, the associative principle may readily 
be deduced from it. 

[But Hamilton seems not to have noticed that we may employ 
for its proof the properties of Spherical Conies already employed 



in demonstrating the truth of the associative principle. "For 
continuity we give an outline of the proof by this process. 

Let BA, CA represent the versors of q arid r, and be the great 
circle whose plane is that of p. 

Then, if we take as operand the vector OA, it is obvious that 
U (q + r) will be represented by some such arc as D A where 
B, D, C are in one great circle; for (q + r) OA is in the same plane 
as qO A and rOA, and the relative magnitude of the arcs BD and 
DC depends solely on the tensors of q and r. Produce BA, DA, 
CA to meet be in b, d, c respectively, and make 

Eb = BA t Fd = DA, Gc = CA. 

Also make bj3 = d& = cy=p. Then E, F, G, A lie on a spherical 
conic of which BC and be are the cyclic arcs. And, because 

b/3 = d8 = cry, (3E, SF, 7$, when produced, meet in a point H 
which is also on the spherical conic ( 59*). Let these arcs meet 
BC in J, L, K respectively. Then we have 


Also LJ 

and KL = CD. 

And, on comparing the portions of the figure bounded respectively 

by HKJ and by ACB we see that (when considered with reference 

to their effects as factors multiplying OH and OA respectively) 

pU(q 4- r) bears the same relation to pUq and pUr 
that U(q + r) bears to Uq and Ur. 
But T(q + r)U(q + r) = q + r=TqUq+TrUr. 

Hence T (q + r~) .pU(q + r) = Tq .pUq + Tr .pUr; 

or, since the tensors are mere numbers and commutative with all 
other factors, 

In a similar manner it may be proved that 
(q + r)p = qp + rp. 

And then it follows at once that 

(p + q) (r + s) = pr + ps + q r + qs, 
where, by 61, the order of the partial products is immaterial.] 



63. By similar processes to those of 53 we see that versors, 
and therefore also quaternions, are subject to the index-law 

at least so long as m and n are positive integers. 

The extension of this property to negative and fractional 
exponents must be deferred until we have defined a negative or 
fractional power of a quaternion. 

64. We now proceed to the special case of quadrantal versors, 
from whose properties it is easy to deduce all the foregoing 
results of this chapter. It was, in fact, these properties whose 
invention by Hamilton in 1843 led almost intuitively to the 
establishment of the Quaternion Calculus. We shall content 
ourselves at present with an assumption, which will be shewn 
to lead to consistent results ; but at the end of the chapter we 
shall shew that no other assumption is possible, following for this 
purpose a very curious quasi-metaphysical speculation of Hamilton. 

65. Suppose we have a system of three mutually perpendicular 
unit-vectors, drawn from one point, which we may call for shortness 
i, j, k. Suppose also that these are so situated that a positive 
(i.e. left-handed) rotation through a right angle about i as an axis 
brings j to coincide with k. Then it is obvious that positive 
quadrantal rotation about j will make k coincide with i; and, about 
k, will make i coincide with j. 

For defmiteness we may suppose i to be drawn eastwards, j 
northwards, and k upwards. Then it is obvious that a positive 
(left-handed) rotation about the eastward line (i) brings the north 
ward line (j) into a vertically upward position (k) ; and so of the 

\J 66. Now the operator which turns j into k is a quadrantal 
versor ( 53) ; and, as its axis is the vector i, we may call it i. 

Thus = i, or k = ?j (1). 

Similarly we may put - = j, or i = jk (2), 


and j=&, or J = H (3). 

[It may be here noticed, merely to shew the symmetry of the 
system we arc explaining, that if the three mutually perpendicular 




vectors i, j, k be made to revolve about a line equally inclined to 
all, so that i is brought to coincide with j, j will then coincide 
with k, and k with i : and the above equations will still hold good, 
only (1) will become (2), (2) will become (3), and (3) will become 


67. By the results of 50 we see that 

i.e. a southward unit- vector bears the same ratio to an upward 
unit-vector that the latter does to a northward one ; and therefore 
we have 

or -j =ik 



=j, or -k=ji. 

= K, or i = 



G8. By (4) and (1) we have 

j = ^k = i(ij) (by the assumption in 54) = i"j. 

Hence i 2 = - 1 (7). 

Arid in the same way, (5) and (2) give 

J 2 = -1 (8), 

and (6) and (3) & 2 = -l (9). 

Thus, as the directions of i, j, k are perfectly arbitrary, we see 
that the square of every quadrantal versor is negative unity. 

[Though the following proof is in principle exactly the same as 
the foregoing, it may perhaps be of use to the student, in shewing 
him precisely the nature as well as the simplicity of the step we 
have taken. 

Let ABA be a semicircle, whose centre 
is 0, and let OB be perpendicular to AOA . 


Then -=- , = q suppose, is a quadrantal 

versor, and is evidently equal to - ; 

50, 53. 

-^=. = = -=^ 




69. Having thus found that the squares of i, j, k are each 
equal to negative unity ; it only remains that we find the values of 
their products two and two. For, as we shall see, the result is such 
as to shew that the value of any other combination whatever of 
i,j, k (as factors of a product) may be deduced from the values of 
these squares and products. 

Now it is obvious that 

k i 

("Us VH>f 



r W 

T, fv 

(i.e. the versor which turns a westward unit-vector into an upward 

one will turn the upward into an eastward unit) ; 

or k=j(-i) = -ji* ........................ (10). 

Now let us operate on the two equal vectors in (10) by the 
same versor, i t and we have 

ik = i (ji) = iji. 
But by (4) and (3) 

ik = -J = -H. 

Comparing these equations, we have 

or, by 54 (end), 
and symmetry gives 

ij = k,\ 
jk i t \ 
hi =. } 


The meaning of these important equations is very simple ; and 
is, in fact, obvious from our construction in 54 for the multiplica 
tion of versors ; as we see by the annexed figure, where we must 
remember that i, j, k are quadrantal versors whose planes are at 
right angles, so that the figure represents 
a hemisphere divided into quadrantal 
triangles. [The arrow-heads indicate the 
direction of each vector arc.] 

Thus, to shew that ij = k, we have, 
being the centre of the sphere, N, E, 
8, W the north, east, south, and west, 
and ^the zenith (as in 65) ; 

60 JOW=OZ, 
whence ij OW = iOZ=OS = kOW. 

* The negative sign, being a mere numerical factor, is evidently commutative 
with j ; indeed we may, if necessary, easily assure ourselves of the fact that to turn 
the negative (or reverse) of a vector through a right (or indeed any) angle, is the 
same thing as to turn the vector through that angle and then reverse it. 



4fi QUATERNIONS. [70. 

70. But, by the same figure, 

whence jiON =jOZ = OE = - OW = - kON. 

71. From this it appears that 

and similarly kj = i, J- (12) 

a jj 

and thus, by comparing (11), 

jf c = -kj = i, ((11), (12)). 
ki = ik=j, 
These equations, along with 

?=f=V = -l ((7), (8), (9)), 

contain essentially the whole of Quaternions. But it is easy to see 
that, for the first group, we may substitute the single equation 

ijk=-l, (13) 

since from it, by the help of the values of the squares of i,j, k, all 
the other expressions may be deduced. We may consider it proved 
in this way, or deduce it afresh from the figure above, thus 

kON= OW, 
jkON= JOW=OZ, 
ijkON = 

72. One most important step remains to be made, to wit the 
assumption referred to in 64. We have treated i, j, k simply as 
quadrantal versors ; and i, j, k as unit-vectors at right angles to 
each other, and coinciding with the axes of rotation of these versors. 
But if we collate and compare the equations just proved we have 

f =-i. (7) .a 

li = -l, (9) 

(*> = 1; (11) 

Ki= k (i) 

{> = -* 02) 

b i = -k, (5) 

with the other similar groups symmetrically derived from them. 


Now the meanings we have assigned to i, j, k are quite inde 
pendent of, and not inconsistent with, those assigned to i, j, k. 
And it is superfluous to use two sets of characters when one will 
suffice. Hence it appears that i, j, k may be substituted for i, j, k ; 
in other words, a unit-vector when employed as a factor may be con 
sidered as a quadrantal versor whose plane is perpendicular to the 
vector. (Of course it follows that every vector can be treated as the 
product of a number and a quadrantal versor.) This is one of the 
main elements of the singular simplicity of the quaternion calculus. 

73. Thus the product, and therefore the quotient, of two perpen 
dicular vectors is a third vector perpendicular to both. 

Hence the reciprocal ( 51) of a vector is a vector which has 
the opposite direction to that of the vector, arid its length is the 
reciprocal of the length of the vector. 

The conjugate ( 52) of a vector is simply the vector reversed. 

Hence, by 52, if a be a vector 

(Ta) 2 - aKcL = a ( - a) = - a 2 . 

74. We may now see that every versor may be represented by 
a power of a unit-vector. 

For, if a be any vector perpendicular to i (which is any definite 

id, = /3, is a vector equal in length to a, but perpendicular to 
both i and a. ; ^ / 

i Z CL = - *, 

*a fa A _/7r 

i 4 a = ift = i*a = a. 

Thus, by successive applications of i, a. is turned round i as an axis 
through successive right angles. Hence it is natural to define i m as 
a versor which turns any vector perpendicular to i through m right 
angles in the positive direction of rotation about i as an axis. Here 
m may have any real value whatever, whole or fractional, for it is 
easily seen that analogy leads us to interpret a negative value of m 
as corresponding to rotation in the negative direction. 

75. From this again it follows that any quaternion may be 
expressed as a power of a vector. For the tensor and versor 
elements of the vector may be so chosen that, when raised to the 
same power, the one may be the tensor and the other the versor 
of the given quaternion. The vector must be, of course, perpen 
dicular to the plane of the quaternion. 



[ 7 6. 

76. And we now see, as an immediate result of the last two 
sections, that the index-law holds with regard to powers of a 
quaternion ( 63). 

77. So far as we have yet considered it, a quaternion has been 
regarded as the product of a tensor and a versor : we are now to 
consider it as a sum. The easiest method of so analysing it seems 
to be the following. 


Let - - represent any quaternion. 


BG perpendicular to OA, produced if neces 

Then, 19, OB = OC + CB. 

But, 22, OC=xOA, 
where a? is a number, whose sign is the same 
as that of the cosine of Z A OB. 

Also, 73, since CB is perpendicular to OA, 

A w^*- 

w H* 


> . 

where 7 is a vector perpendicular to OA and CB, i.e. to the plane 
of the quaternion; and, as the figure is drawn, directed towards the 




Thus a quaternion, in general, may be decomposed into the sum 
of two parts, one numerical, the other a vector. Hamilton calls 
them the SCALAR, and the VECTOR, and denotes them respectively 
by the letters $ and V prefixed to the expression for the 

78. Hence q = Sq+ Vq, and if in the above example 



then OB 

The equation above gives 


* The points are inserted to shew that S and V apply only to q, and not to qOA 


79. If, in the last figure, we produce BC to D, so as to double 
its length, and join OD, we have, by 52, 

so that OI)=OC+CD = SKq.OA + VKq.OA. 

Hence OC = SKq.OA, 

and CD = VKq.OA. 

Comparing this value of OC with that in last section, we find 

SKq = Sq, ................................. (1) 

or the scalar of the conjugate of a quaternion is equal to the scalar of 
the quaternion. 

Again, CD = CB by the figure, and the substitution of their 
values gives 

VKq = -Vq, ........................ (2) 

or the vector of the conjugate of a quaternion is the vector of the 
quaternion reversed. 

We may remark that the results of this section are simple con 
sequences of the fact that the symbols S, V, K are commutative*. 

Thus SKq = KSq = Sq, 

since the conjugate of a number is the number itself; and 

VKq = KVq = - Vq 73). 
Again, it is obvious that, 

and thence 

80. Since any vector whatever may be represented by 

cci + yj + zk 

where a?, y, z are numbers (or Scalars), and i, j, k may be any three 
non-coplanar vectors, 23, 25 though they are usually under 
stood as representing a rectangular system of unit-vectors and 

* It is curious to compare the properties of these quaternion symbols with those 
of the Elective Symbols of Logic, as given in BOOLE S wonderful treatise on the 
Laws of Thought; and to think that the same grand science of mathematical 
analysis, by processes remarkably similar to each other, reveals to us truths in the 
science of position far beyond the powers of the geometer, and truths of deductive 
reasoning to which unaided thought could never have led the logician. 

T. Q. T. 4 


since any scalar may be denoted by w ; we may write, for any 
quaternion q, the expression 

q = w + a?i + yj+zk( 78). 

Here we have the essential dependence on four distinct numbers, 
from which the quaternion derives its name, exhibited in the most 
simple form. 

Arid now we see at once that an equation such as 

? =? 

where q = w + otfi + y j + z k, 

involves, of course, the four equations 

w = w, x = x, y = y, z = z. 

81. We proceed to indicate another mode of proof of the dis 
tributive law of multiplication. 

We have already defined, or assumed ( 61), that 

7_ff + 7 
a a a 

or /3cr 1 + 7 a- 1 = (/3 + 7 )a- 1 , 

and have thus been able to understand what is meant by adding 
two quaternions. 

But, writing a for of 1 , we see that this involves the equality 

(13 + 7) a = @OL + 7 ; 

from which, by taking the conjugates of both sides, we derive 
oL (ff + y) = OL p + a y (5:>). 

And a combination of these results (putting {3 + y for a! in the 
latter, for instance) gives 

= 13/3 + 7/5 + J3y -f 77 by the former. 

Hence the distributive principle is true in the multiplication of 

It only remains to shew that it is true as to the scalar and 
vector parts of a quaternion, and then we shall easily attain the 
general proof. 

Now, if a be any scalar, a any vector, and q any quaternion, 

(a + a) q = aq + aq- 
For, if /3 be the vector in which the plane of q is intersected by 


a plane perpendicular to a, we can find other two vectors, 7 and S, 
one in each of these planes such that 

7 13 

*=/ *-? 

And, of course, a may be written -~ ; so that 

And the conjugate may be written 

?X + ) = ? + ? * ( 55). 
Hence, generally, 

(a + a) (6 + ) = a& + a/3 -f- 6a + a/3 ; 

or, breaking up a and 6 each into the sum of two scalars, and a, fi 
each into the sum of two vectors, 

(a, + a 2 + a, + * 2 ) (6, + & + , + s ) 

= (a, + a,) (6, + 6 8 ) + (a, -f a,) ( 

(by what precedes, all the factors on the right are distributive, so 
that we may easily put it in the form) 

+ K + ,) (6 2 + /3 a ) + (a 2 + 2 ) ( 

Putting 0, + ^=^, a 2 + 2 = ^, 

we have (^ + q) (r + s) =pr + ps + qr + ^5. 

82. Cayley suggests that the laws of quaternion multiplication 
may be derived more directly from those of vector multiplication, 
supposed to be already established. Thus, let a. be the unit vector 
perpendicular to the vector parts of q and of q . Then let 

p = q.*, a = -a.q, 
as is evidently permissible, and we have 

pa = q . OLOL = q ; a<r = act. . q q , 
so that q . q = px . acr = p . cr. 

The student may easily extend this process. 
For variety, we shall now for a time forsake the geometrical 
mode of proof we have hitherto adopted, and deduce some of our 



next steps from the analytical expression for a quaternion given in 
80, and the properties of a rectangular system of unit-vectors as 
in 71. 

We will commence by proving the result of 77 anew. 

83. Let a = xi + yj + zk, 

ft = x i + yj + zk. 

Then, because by 71 every product or quotient of i,j t k is reducible 
to one of them or to a number, we are entitled to assume 

q = - = v + & + vj + &, 

where o>, f, 77, f are numbers. This is the proposition of 80. 

[Of course, with this expression for a quaternion, there is no 
necessity for a formal proof of such equations as 

p + (q+r) = (p + q) + r, 

where the various sums are to be interpreted as in 61. 

All such things become obvious in view of the properties of 


84. But it may be interesting to find co, , 77, f in terms of 

x, y, z, x, y , z . 

We have ft qa, 

or x i + yj + z k = (a) + %i + rjj + %k) (xi + yj + zk) 

as we easily see by the expressions for the powers and products of 
i>j> k> gi ven i n ^1- But the student must pay particular attention 
to the order of the factors, else he is certain to make mistakes. 
This ( 80) resolves itself into the four equations 

= f x + vjy + z, 

z = coz + f y - TJX. 
The three last equations give 

j/ xx + yy + zz = w ( 2 + if + 2 ) 
which determines o>. 

Also we have, from the same three, by the help of the first, 

f .r + w + ty = : 


which, combined with the first, gives fy *M~~ 

yz zy zx xz xy yx 

and the common value of these three fractions is then easily seen 
to be 


x* + y 2 + z z 

It is easy enough to interpret these expressions by means of 
ordinary coordinate geometry : but a much simpler process will 
be furnished by quaternions themselves in the next chapter, and, in 
giving it, we shall refer back to this section. 

85. The associative law of multiplication is now to be proved 
by means of the distributive ( 81). We leave the proof to the 
student. He has merely to multiply together the factors 

w + xi + yj + zk, w + xi + yj + z k, and w" 4- x"i + y"j + z"k t 

as follows : 

First, multiply the third factor by the second, and then multiply 
the product by the first ; next, multiply the second factor by the 
first and employ the product to multiply the third: always re 
membering thatjthe multiplier in any product is placed before the 
multiplicand. He will find the scalar parts and the coefficients of 
i,j, k, in these products, respectively equal, each to each. 

86. With the same expressions for a, /3, as in section 83, we 

a/3 = (xi + yj 4 zk) (xi 4 y j + zk) 

= - (ocx +yy + zz) + (yz - zy ) i + (zx - xz )j + (xy - yx ) k. 
But we have also 

/3a = (xx 4 yy 4 zz) (yz zy) i (zx xz )j (xy yx) k. 
The only difference is in the sign of the vector parts. 

Hence Safi = S/3a, (1) 

Fa = -F/3a, (2) 

also a/3 + /3a = 2$a/3, (3) 

a/3-a = 2Fa#, (4) 

and, finally, by 79, a/3 = ^T./3a (5). 

87. If a = /3 we have of course ( 25) 

/y - /y 77 /j /y ^ 

* ^j y ~~y > z z , 


and the formulae of last section become 

which was anticipated in 73, where we proved the formula 

and also, to a certain extent, in 25. 

88. Now let q and r be any quaternions, then 

S.qr = S.(Sq+Vq) (Sr+Vr), 

= S . (8q Sr + Sr. Vq + Sq . Vr + VqVr), 
= SqSr + S. VqVr, 

since the two middle terms are vectors. 
Similarly, S.rq = SrSq + S . Vr Vq. 

Hence, since by (1) of 86 we have 

8. VqVr = S. VrVq, 
we see that S.qr = S.rq, (1) 

a formula of considerable importance. 

It may easily be extended to any number of quaternions, 
because, r being arbitrary, we may put for it rs. Thus we have 

S . qrs = S . rsq, 
= S . sqr 

by a second application of the process. In words, we have the 
theorem the scalar of the product of any number of given 
quaternions depends only upon the cyclical order in which they are 

89. An important case is that of three factors, each a vector. 
The formula then becomes 

But S.OL@y = SOL (S/3y 

= SOL Vfiy, since a8/3y is a vector, 
= -aF 7 /3, by (2) of 86, 
= - SOL (Syj3 + Vy/3) 
= -S. ay $. 

Hence the scalar of the product of three vectors changes sign when 
the cyclical order is altered. 


By the results of 55, 73, 79 we see that, for any number 
of vectors, we have 

K . ct/3y ... </>x = %</> V@<* 

(the positive sign belonging to the product of an even number of 
vectors) so that 

S . j3 . . . (f>x = S x<t> - &<* 

F. ... x = *F.x .... 
Thus we may generalize (3) and (4) of 86 into 


2V. a/3 ... <fo = a/3 ... <x + % c - , 

the upper sign still being used when the -number of factors is 

Other curious propositions connected with this will be given 
later (some, indeed, will be found in the Examples appended to 
this chapter), as we wish to develop the really fundamental 
formulae in as compact a form as possible. 

90. By (4) of 86, 

Hence 2V.otV/3y = V.a(8y-yl3) 

(by multiplying both by a, and taking the vector parts of each 


= F (a/3y + flay j3ay - ay {3) 

(by introducing the null term @ay pay). 

That is 

2 F . aV/3y = F. (a/3 + 0a) 7 - F (/3a 7 + /5Fa 7 + S*y./3+ Vay . ) 

(if we notice that F (F 7 . /3) = - F . /3Fa 7? by (2) of 86). 
Hence F . a F/fy = ySz{3 - jSSya .................. (1), 

a formula of constant occurrence. 

Adding aS{3y to both sides, we get another most valuable 

V.a0y = aSfa-l3Sya + y8oL0 ............ (2); 

and the form of this shews that we may interchange 7 and a 
without altering the right-hand member. This gives 

F . OLj3y = V . yj3a, 
a formula which may be greatly extended. (See 89, above.) 


Another simple mode of establishing (2) is as follows : 

K . a/37 = - 7/3a, 

.-. 2 V . a/3 7 - a/3 7 - K . a/3 7 (by 79 (2)) 
= a/37 -f 7/3a 
= a (Py + 7/3) - (7 4- 7) /3 + 7 (a/3 + /3a) 

91. We have also 

FFa/3FyS = - FFySFa/3 by (2) of 86 : 

- &S( 7 Fa/3 - ySS Fa/3 = SS . a/3 7 - yS . a/ 
= - /3SoL FyS + aS/3 1 78 = - /3S . a 7 S + a 

all of these being arrived at by the help of 90 (1) and of 89 ; 

and by treating alternately Fa/3 and FyS as simple vectors. 
Equating two of these values, we have 

$S . a/3 7 = aS . /3jS + 08 . 7S + 7$ . a/3S ......... (3), 

a very useful formula, expressing any vector whatever in terms 
of three given vectors. [This, of course, presupposes that a, /3, 7 
are not coplanar, 23. In fact, if they be coplanar, the factor 
8. a/37 vanishes, and thus (3) does not give an expression for 8. 
This will be shewn in 10 i below.] 

92. That such an expression as (3) is possible we knew already 
by 23. For variety we may seek another expression of a similar 
character, by a process which differs entirely from that employed 
in last section. 

a, ft, 7 being any three non-coplanar vectors, we may derive 
from them three others Fa/3, V(B^ y Vya. and, as these will not be 
coplanar, any other vector 8 may be expressed as the sum of the 
three, each multiplied by some scalar. It is required to find this 
expression for 8. 

Let 8 = # Fa/3 + 7/F/37 + zVy*- 

Then SyS = xS . yOL/3 = xS . a/3 7 , 

the terms in y and z going out, because 

7 F/3 7 = S . 7/37 = 80v* = 7 2 S/3 = 0, 
for 7 2 is ( 73) a number. 

Similarly S/3S = zS . /3 7 a = zS . a/3 7 , 

and $a = yS . ifty. 

Thus $8 . a/3 7 = Fa/3# 7 8 + F/3 7 /Sfa8 + Vy*S/3B ......... (4). 


93. We conclude the chapter by shewing (as promised in 64) 
that the assumption that the product of two parallel vectors is 
a number, and the product of two perpendicular vectors a third 
vector perpendicular to both, is not only useful and convenient, 
but absolutely inevitable, if our system is to deal indifferently with 
all directions in space. We abridge Hamilton s reasoning. 

Suppose that there is no direction in space pre-eminent, and 
that the product of two vectors is something which has quantity, 
so as to vary in amount if the factors are changed, and to have its 
sign changed if that of one of them is reversed ; if the vectors be 
parallel, their product cannot be, in whole or in part, a vector 
inclined to them, for there is nothing to determine the direction in 
which it must lie. It cannot be a vector parallel to them ; for by 
changing the signs of both factors the product is unchanged, 
whereas, as the whole system has been reversed, the product 
vector ought to have been reversed. Hence it must be a number. 
Again, the product of two perpendicular vectors cannot be wholly 
or partly a number, because on inverting one of them the sign of 
that number ought to change ; but inverting one of them is simply 
equivalent to a rotation through two right angles about the other, 
and (from the symmetry of space) ought to leave the number 
unchanged. Hence the product of two perpendicular vectors must 
be a vector, and a simple extension of the same reasoning shews 
that it must be perpendicular to each of the factors. It is easy to 
carry this farther, but enough has been said to shew the character 
of the reasoning. 


1. It is obvious from the properties of polar triangles that any 
mode of representing versors by the sides of a spherical triangle 
must have an equivalent statement in which they are represented 
by angles in the polar triangle. 

Shew directly that the product of two versors represented 
by two angles of a spherical triangle is a third versor represented 
by the supplement of the remaining angle of the triangle ; and 
determine the rule which connects the directions in which these 
angles are to be measured. 


2. Hence derive another proof that we have not generally 

pq = qp. 

3. Hence shew that the proof of the associative principle, 
57, may be made to depend upon the fact that if from any point 
of the sphere tangent arcs be drawn to a spherical conic, and also 
arcs to the foci, the inclination of either tangent arc to one of the 
focal arcs is equal to that of the other tangent arc to the other 
focal arc. 

4. Prove the formulae 

2$ . a/3y = a/37 - 7 /3a, 
2F.a/37 = a/37 + 7a. 

5. Shew that, whatever odd number of vectors be represented 
by a, j3, 7, &c., we have always 

F. a/3y$efr = V. rteSypa, &c. 

6. Shew that 

8 . Fa/3F/3 7 F 7 a = -(S. a/3 7 ) 2 , 

F. Vot{3V/3yVyoL=Voi{3(ry 2 SoL/3-S/3vSvoi) + ...... , 

and F ( Fa/3 F . Vj3y Fya) = ((3 Say - a/3 7 ) 8 . ay. 

7. If a, (3, 7 be any vectors at right angles to each other, shew 

(a 3 -f /3 3 -h y 3 ) S . a/3 7 = a 4 F/3 7 + /3 4 F>a + 7 4 Fa/3. 

(a" 1 " 1 + P n ~ l + 7 in - 1 ) 8 . a/fy = 2 " V/3y + /3 2w Fya + 7 2n Fa/3. 

8. If a, /3, 7 be non-coplanar vectors, find the relations among 
the six scalars, x, y, z and f, 77, f, which are implied in the 
equation XOL + yfi + 27 = f F/37 + 77 Fya + f FayS. 

9. If a, /3, 7 be any three non-coplanar vectors, express any 
fourth vector, S, as a linear function of each of the following sets of 
three derived vectors. 

F.yaft F.a^7, F. ya, 
and F. Fa/3F/3 7 F 7 a, F. F/3 7 Fya Fa/3, F. FyaFa/3F/3 7 . 

10. Eliminate p from the equations 

where a, /3, 7, 8 are vectors, and a, 6, c, d scalars. 

11. In any quadrilateral, plane or gauche, the sum of the 
squares of the diagonals is double the sum of the squares of the 
lines joining the middle points of opposite sides. 



94. AMONG the most useful characteristics of the Calculus of 
Quaternions, the ease of interpreting its formulae geometrically, 
and the extraordinary variety of transformations of which the 
simplest expressions are susceptible, deserve a prominent place. 
We devote this Chapter to some of the more simple of these, 
together with a few of somewhat more complex character but of 
constant occurrence in geometrical and physical investigations. 
Others will appear in every succeeding Chapter. It is here, 
perhaps, that the student is likely to feel most strongly the 
peculiar difficulties of the new Calculus. But on that very account 
he should endeavour to master them, for the variety of forms 
which any one formula may assume, though puzzling to the 
beginner, is of the utmost advantage to the advanced student, not 
alone as aiding him in the solution of complex questions, but 
as affording an invaluable mental discipline. 

95. If we refer again to the figure of 77 we see that 



Hence, if OA = a, OB = /3, and Z A OB = 0, we have 
OB=T0, OA = Ta, 

00 = Tfi cos 0, OB = T{3 sin 6. 

a /3 00 T/3 

Hence S - = TTT = m~ cos 6. 


cv -1 i mtrfi CB J (3 . A 

Similarlv TV - - 77-7 = 7f P sin 0. 

a OA To. 


Hence, if ?; be a unit-vector perpendicular to a and /3, and such 
that positive rotation about it, through the angle 6, turns a 
towards /3, or 



we have V - -~ sin 6 . 77. (See, again, 84.) 

96. In the same way, or by putting 

S- + V- 

a a. 

we may shew that 

Sa/3 = - TOL T8 cos 0, 

and Fa/3 - 2 7 a T sin 6 . 7; 

where rj = UVaft =U(- V/3a) =UV /3 . 

Thus the scalar of the product of two vectors is the continued 
product of their tensors and of the cosine of the supplement of the 
contained angle. 

The tensor of the vector of the product of two vectors is the con 
tinued product of their tensors and the sine of the contained angle ; 
and the versor of the same is a unit-vector perpendicular to both, 
and such that the rotation about it from the first vector (i. e. the 
multiplier) to the second is left-handed or positive. 

Hence also T VOL ft is double the area of the triangle two of whose 
sides are a, j3. 

97. (a) In any plane triangle ABC we have 

Hence AC 2 = 8. AC AC = S . A 

With the usual notation for a plane triangle the interpretation 
of this formula is 

1} Z = be cos A ab cos G, 
or 6 = a cos C + c cos A . 


(b) Again we have, obviously, 


= V.ABBC ) 

or cb sin A = ca sin B, 

sin A sin B sin G 

whence = > --= . 

a b c 

These are truths, but not truisms, as we might have been led 
to fancy from the excessive simplicity of the process employed. 

98. From 96 it follows that, if a and /3 be both actual (i. e. 
real and non-evanescent) vectors, the equation 

shews that cos 6 = 0, or that a. is perpendicular to (3. And, in fact, 
we know already that the product of two perpendicular vectors is 
a vector. 

Again : if Fa/3 = 0, 

we must have sin 6 = 0, or a is parallel to j3. We know already 
that the product of two parallel vectors is a scalar. 

Hence we see that 

is equivalent to a = 

where 7 is an undetermined vector; and that 

is equivalent to a = xj3, 

where x is an undetermined scalar. 

99. If we write, as in SS 83, 84, 


a = isc +jy +kz, 
j3 = ix +jy 4- kz , 
we have, at once, by 86, 

Sa/3 = xx yy zz 

(IT IT II II % % 

r r r r r r 

where r = Jx* 4- y* + z\ r = Jx * + y"* + z*. 

Tr n , (yz zy . zx xz . xy yx 

Also VOL 8 rr \- ~- i + /?+ 7 - A- 

rr rr rr 


These express in Cartesian coordinates the propositions we have 
just proved. In commencing the subject it may perhaps assist 
the student to see these more familiar forms for the quaternion 
expressions ; and he will doubtless be induced by their appearance 
to prosecute the subject, since he cannot fail even at this stage to 
see how much more simple the quaternion expressions are than 
those to which he has been accustomed. 

100. The expression S . a/3y 
may be written SV (aft) 7, 

because the quaternion a/3y may be broken up into 

of which the first term is a vector. 
But, by 96, 

8 V (a/3) y = TaT/3 sin 
Here Trj = 1, let < be the angle between 77 and 7, then finally 
S .a/3y = - TOL Tft Ty sin cos <j>. 

But as rj is perpendicular to a and {3, Ty cos <j> is the length of the 
perpendicular from the extremity of 7 upon the plane of a, ft. And 
as the product of the other three factors is ( 96) the area of the 
parallelogram two of whose sides are a, ft, we see that the mag 
nitude of S . a/37, independent of its sign, is the volume of the 
parallelepiped of which three coordinate edges are a, ft, y : or six 
times the volume of the pyramid which has a, ft, y for edges. 

101. Hence the equation 

8. a/37 = 0, 
if we suppose a, ft, y to be actual vectors, shews either that 

sin (9 = 0, 
or cos (/> = 0, 

i. e. two of the three vectors are parallel, or all three are parallel to 
one plane. 

This is consistent with previous results, for if 7 =pft we have 

S.a/3y = pS.a/3* = 0; 

and, if 7 be coplanar with a, ft, we have 7 =poi 4- qft, and 
S.aj3y = S.aij3 (pa + q/3) = 0. 


102. This property of the expression 8 . afiy prepares us to 
find that it is a determinant. And, in fact, if we take a, ft as in 
83, and in addition 7 = ix" +jy" + kz", 

we have at once 

8 . a/3y = - x" (yz f - zy ) - y" (zx f - xz ) - z" (xy f - yx ), 
x y z 
x y z 
x" y" z 

The determinant changes sign if we make any two rows change 
places. This is the proposition we met with before ( 89) in the 
form 8. afty = - S . pay = 8 . j3ya, &c. 

If we take three new vectors 

a 1 = ix +jx + kx", 

I 3 1 = iy +jy + %"> 

y^iz+jz + kz", 
we thus see that they are coplanar if a, /3, y are so. That is, if 

8 . afiy = 0, 
then S.afitf^Q. 

103. We have, by 52, 

(Tqf = qKq = (Sq + Vq) (Sq - Vq) ( 79), 
by algebra, 

If q = a/3, we have Kq = /3a, and the formula becomes 

*$.$* = 2 /3 2 = (Sap)* - ( Vapy. 
In Cartesian coordinates this is 

(^ + 2/ 2 + /)( ^2 + ^ 2+/2) 

= (xaf + yy + zzj + (yz - zyj + (zx f - xzj + (xy f - yxj. 
More generally we have 
(T(qr)) z = 

If we write q w + OL = W + IX +jy + kz, 

r = w -f P = w 4- ix +jy + kz \ 
this becomes 

(w 2 -f x* + y 1 + ^ 2 ) (w 2 + x 2 + y"* + / 2 ) 

= (low xx f yy zz)* + (wx + wx + yz zy J 
+ (wy + w y + zx xz ) z + (wz + w z + xy yx ^, 

a formula of algebra due to Euler. 

64 QUATERNIONS. [104. 

104. We have, of course, by multiplication, 

(a + ft) 2 = a 2 + a/3 + /3a + /3 2 = a 2 + 2Sa/3 + ft 2 ( 86 (3)). 
Translating into the usual notation of plane trigonometry, this 
becomes c 2 = a 2 2ab cos G + 6 2 , 

the common formula. 

Again, F . (a + ft) (a - /3) = - Fa/3 + F/3a = - 2 Fa/3 ( 86 (2)). 
Taking tensors of both sides we have the theorem, the paral 
lelogram whose sides are parallel and equal to the diagonals of a 
given parallelogram, has double its area ( 96). 

Also S (a + ft) (a - ft) = a 2 - ft 2 , 

and vanishes only when a 2 = ft 2 , or Ten = Tft ; that is, the diagonals 
of a parallelogram are at right angles to one another, when, and 
only when, it is a rhombus. 

Later it will be shewn that this contains a proof that the angle 
in a semicircle is a right angle. 

105. The expression p = afta~ l 

obviously denotes a vector whose tensor is equal to that of ft. 

But we have S . ftap = 0, 

so that p is in the plane of a, ft. 

Also we have Sap = Saft, 

so that ft and p make equal angles with a, evidently on opposite 
sides of it. Thus if a be the perpendicular to a reflecting surface 
and ft the path of an incident ray, p will be the path of the 
reflected ray. 

Another mode of obtaining these results is to expand the above 
expression, thus, 90 (2), 

p = 2 a" 1 Sa ft ft 

= 2- 1 a/3 - of 1 (Saft + Fa/3 ) 
- a 1 (Saft - Fa/3), 

so that in the figure of 77 we see that if A = a, and OB = ft, we 
have OD = p = afta~ l . 

Or, again, we may get the result at once by transforming the 
equation to ^ = K (a" 1 p) - K ? . 


106. For any three coplanar vectors the expression 

p = ct/3y 

is ( 101) a vector. It is interesting to determine what this vector 
is. The reader will easily see that if a circle be described about 
the triangle, two of whose sides are (in order) a and /3, and if from 
the extremity of /3 a line parallel to 7 be drawn, again cutting the 
circle, the vector joining the point of intersection with the origin 
of a. is the direction of the vector a/3y. For we may write it in the 

p = a /3*/3- 7 = - (T/9) Vy = - ( W 1 7, 
which shews that the versor [ -5] which turns yS into a direction 

parallel to a, turns 7 into a direction parallel to p. And this ex 
presses the long-known property of opposite angles of a quadri 
lateral inscribed in a circle. 

Hence if a, /3, y be the sides of a triangle taken in order, the 
tangents to the circumscribing circle at the angles of the triangle 
are parallel respectively to 

a/3y, ftya, and ya(3. 
Suppose two of these to be parallel, i. e. let 
afiy = xfiya = xay/3 ( 90), 
since the expression is a vector. Hence 

which requires either 

x=l, Vyj3 = or 711/3, 
a case not contemplated in the problem ; 
or a? = -l, S/3y=0, 

i. e. the triangle is right-angled. And geometry shews us at once 
that this is correct. 

Again, if the triangle be isosceles, the tangent at the vertex is 
parallel to the base. Here we have 

or x (a -f 7) = a (a + 7) 7 ; 

whence x 7* = a 2 , or Ty = Tea, as required. 

As an elegant extension of this proposition the reader may 
T. Q. I. 5 

66 QUATERNIONS. [107. 

prove that the vector of the continued product a/rtyS of the vector- 
sides of any quadrilateral inscribed in a sphere is parallel to the 
radius drawn to the corner (a, 8). [For, if 6 be the vector from 8, 
a to /3, 7, a/9e and 78 are (by what precedes) vectors touching the 
sphere at a, 8. And their product (whose vector part must be 
parallel to the radius at a, 8) is 

OL/3e . 78 = e 2 . a 

107. To exemplify the variety of possible transformations 
even of simple expressions, we will take cases which are of 
frequent occurrence in applications to geometry. 

Thus T(p+a)=:T(p-Qi), 

[which expresses that if 

Ol=a, 01 = -a, and OP = p, 
we have AP = A P, 

and thus that P is any point equidistant from two fixed points,] 
may be written (p + a) 2 = (p a) 2 , 

or p 2 + 2Sap + a 2 - p 9 - 2Sap + a 2 ( 104), 

whence Sap = 0. 

This may be changed to 

a/3 + pa. = 0, 
or dp + Kaip = 0, 


or finally, TVU?=l, 

all of which express properties of a plane. 
Again, Tp - TOL 

may be written T- = l, 

or finally, T. (p + a) (p - a) = 2TVap. 

All of these express properties of a sphere. They will be 
interpreted when we come to geometrical applications. 


108. To find the space relation among five points. 

A system of five points, so far as its internal relations are 
concerned, is fully given by the vectors from one to the other four. 
If three of these be called a, /?, y, the fourth, S, is necessarily 
expressible as xa. + yfi + zy. Hence the relation required must be 
independent of x, y, z. 

But SaiS = ax 2 + ySa/3 + 

Sy$ = xSya + ySy/3 -f z 
SB8 = tf = xS8 

The elimination of a?, y, z gives a determinant of the fourth order, 
which may be written 

SoLOi Sa/3 Say S 
S/3a S{3j3 S{3y 8/38 
Syot Sy(3 Syy SyS 

Now each term may be put in either of two forms, thus 

S/3y = 4 {P* + y*-(p- y) 2 } = - TpTy cos py. 

If the former be taken we have the expression connecting the 
distances, two and two, of five points in the form given by Muir 
(Proc. R. S. E. 1889) ; if we use the latter, the tensors divide out 
(some in rows, some in columns), and we have the relation among 
the cosines of the sides and diagonals of a spherical quadrilateral. 

We may easily shew (as an exercise in quaternion manipulation 
merely) that this is the only condition, by shewing that from it 
we can get the condition when any other of the points is taken as 
origin. Thus, let the origin be at a, the vectors are a, P a, 
y a, 8 a. But, by changing the signs of the first row, and first 
column, of the determinant above, and then adding their values 
term by term to the other rows and columns, it becomes 

S( -)(-) S( -a)(P~a) S( -a)( 7 -a) S( -a) (8-a) 
S(y-a)(-a) S(y- a)(/3 - a) 8 (y - a) (7 - a) 

which, when equated to zero, gives the same relation as before. 
[See Ex. 10 at the end of this Chapter.] 





An additional point, with e = x a. -f y ft + 2 % gives six additional 
equations like (1) ; i. e. 

Sae = a/a 2 + tfSa/3 + z Say, 

= x Sya 

+ yj + s 

from which corresponding conclusions may be drawn. 

Another mode of solving the problem at the head of this 
section is to write the identity 

where the ms are undetermined scalars, and the as are given 
vectors, while is any vector whatever. 

Now, provided that the number of given vectors exceeds four, we 
do not completely determine the ms by imposing the conditions 

2m = 0, 2ma = 0. 

Thus we may write the above identity, for each of five vectors 
successively, as 

2m (a a a ) 2 = 2ma 2 , 

2m (a a 2 ) 2 = 2ma 2 , 

2m (a - 5 ) 2 = 2ma 2 . 

Take, with these, 2m = 0, 

and we have six linear equations from which to eliminate the ms. 
The resulting determinant is 

_a* a 

a a a a 2 1 2ma 2 = 0. 

W. . tA- <-*_ X 

1 1 ..10 

This is equivalent to the form in which Cayley gave the 
relation among the mutual distances of five points. (Camb. Math. 
Journ. 1841.) 


109. We have seen iu 95 that a quaternion may be divided 
into its scalar and vector parts as follows : 

where is the angle between the directions of a and ft and e UV 

is the unit-vector perpendicular to the plane of a. and /3 so situated 
that positive (i. e. left-handed) rotation about it turns a towards ft 

Similarly we have ( 96) 

= TaTj3 (- cos + e sin 0), 
arid e having the same signification as before. 

110. Hence, considering the versor parts alone, we have 

U = cos + e sin 0. 

Similarly U -j. = cos (f> + e sin ; 

</> being the positive angle between the directions of 7 and ft and e 
the same vector as before, if a, ft 7 be coplanar. 

Also we have 

U ^ = cos (6 + <) + e sin (0 + 0). 
But we have always 

and therefore U.U= U; 

pa a 

or cos (</> + 6) + e sin (< + 0) = (cos </> -f e sin (/>) (cos + e sin 6) 

= cos (/> cos 6 sin </> sin 6 + e (sin </> cos + cos $ sin 0), 

from which we have at once the fundamental formulae for the 
cosine and sine of the sum of two arcs, by equating separately the 
scalar and vector parts of these quaternions. 

And we see, as an immediate consequence of the expressions 
above, that 

cos mO + e sin mO = (cos 6 -f e sin 0) m . 

if m be a positive whole number. For the left-hand side is a versor 

70 QUATERNIONS. [ill. 

which turns through the angle m6 at once, while the right-hand 
side is a versor which effects the same object by m successive turn 
ings each through an angle 6. See 8, 9. 

111. To extend this proposition to fractional indices we have 


only to write - for 0, when we obtain the results as in ordinary 


From De Moivre s Theorem, thus proved, we may of course 
deduce the rest of Analytical Trigonometry. And as we have 
already deduced, as interpretations of self-evident quaternion trans 
formations ( 97, 104), the fundamental formulae for the solution 
of plane triangles, we will now pass to the consideration of spherical 
trigonometry, a subject specially adapted for treatment by qua 
ternions ; but to which we cannot afford more than a very few 
sections. (More on this subject will be found in Chap. XI. in con 
nexion with the Kinematics of rotation.) The reader is referred to 
Hamilton s works for the treatment of this subject by quaternion 

112. Let a, /3, 7 be unit-vectors drawn from the centre to the 
corners A, B, C of a triangle on the unit-sphere. Then it is evident 
that, with the usual notation, we have ( 96), 

Sa/3 = cos c, S/3y = cos a, Sya. = cos b, 
TVQLJ3 = sin c, TV/3y = sin a, T Fya = sin b. 

Also [7 Fa/3, UV/3y, UVyct are evidently the vectors of the corners 
of the polar triangle. 

Hence S . UVajS UVj3y = cos B t &c., 

TV. U Fa/3 UVfa = sin B, &c. 
Now ( 90 (1)) we have 

Remembering that we have 

SVafiVPy = TVaQTVPyS . 
we see that the formula just written is equivalent to 
sin a sin c cos B = cos a cos c -f cos b, 
or cos b cos a cos c + sin a sin c cos J5. 


1 1 3. Again, V . Fa/3 F/3 7 = - /3a/3 7 , 
which gives 

TV. FaF/3 7 = TS . a/3 7 = TS . aF/3 7 = TS . /3F 7 a = TS . 7 Fa/3, 
or sin a sin c sin B = sin a sin p a = sin 6 sin p & = sin c sin ^) c ; 
where p a is the arc drawn from A perpendicular to BC, &c. 
Hence sin p a = sin c sin J9, 

sin a sin c . 

sn p = sn a sn . 

114. Combining the results of the last two sections, we have 
Vap . V/3y = sin a sin c cos B /3 sin a sin c sin B 
= sin a sin c (cos jB /3 sin B). 

Hence ^7 . Fa/3 F/3 7 = (cos 5 - /3 sin )) 

and U . Fy/3 F/3a = (cos B + ft sin B) } 

These are therefore versors which turn all vectors perpendicular to 
OB negatively or positively about OB through the angle B. 

[It will be shewn later ( 119) that, in the combination 
(cos B+/3smB)( ) (cos B - /3 sin B) t 

the system operated on is made to rotate, as if rigid, round the 
vector axis ft through an angle 2B.] 

As another instance, we have 
sin B 

tan B = 

cos B 

S.Fa/3F/3 7 

F.Fa/3F/3 7 
P S.Fa/3F/3 7 


The interpretation of each of these forms gives a different theorem 
in spherical trigonometry. 

115. Again, let us square the equal quantities 
F a/3 7 and a#/3 7 - 

7 2 QUATERNIONS. [ I I 6. 

supposing a, @, 7 to be any unit-vectors whatever. We have 

- ( F, a/3 7 ) 2 - 8*0y + V + fy 
But the left-hand member may be written as 

T 2 . afiy $ 2 . a/3<y, 

or 1 cos 2 tt cos 2 6 cos 2 c + 2 cos a cos 6 cos c 

= sin 2 a sm 2 j9 a = &c. 

= sin 2 a sin 2 6 sin 2 = &c., 
all of which are well-known formulae. 

116. Again, for any quaternion, 

so that, if n be a positive integer, 

f = (Sqy + n (Sq)- 1 Vq + W f= (Sg) M ( Vqf + . . . 
From this at once 

F. " = F ,S- - T*V + &c 

If q be a versor we have 

q = cos u + sin u t 
so that 

S.q n = (cos u) n 2 (cos u) n ~ 2 (sin w) 2 + . . . 

= cos nu ; 

r n . ^TTi . 7^2 1 

F . ^ n = 6 sin M w (cos uf 1 19*3 ^ C S U ^ * ^^ ^) 2 + 

= 6 sin nu ; 
as we might at once have concluded from 110. 

Such results may be multiplied indefinitely by any one who has 
mastered the elements of quaternions. 


117. A curious proposition, due to Hamilton, gives us a 
quaternion expression for the spherical excess in any triangle. 
The following proof, which is very nearly the same as one of his, 
though by no means the simplest that can be given, is chosen here 
because it incidentally gives a good deal of other information. 
We leave the quaternion proof as an exercise. 

Let the unit-vectors drawn from the centre of the sphere to 
A, B, C, respectively, be a, j3, 7. It is required to express, as an 
arc and as an angle on the sphere, the quaternion 

The figure represents an orthographic projection made on a 
plane perpendicular to 7. Hence C is the centre of the circle DEe. 
Let the great circle through A, B meet DEe in E, e, and let DE be 

a quadrant. Thus DE represents 7 ( 72). Also make EF = AB 
= /3a~\ Then, evidently, 

which gives the arcual representation required. 

Let DF cut Ee in G. Make Ca = EG, and join D, a, and a, F. 
Obviously, as D is the pole of Ee, Da is a quadrant ; and since 
EG = Ca, Ga EG, a quadrant also. Hence a is the pole of DG, 
and therefore the quaternion may be represented by the angle 

Make Cb = Ca, and draw the arcs Pa/3, P6a from P, the pole of 


A B. Comparing the triangles Ebi and ea/3, we see that Ecu = e/3. 
But, since P is the pole of AB, Ffia is a right angle: and therefore 
as Fa is a quadrant, so is P/3. Thus AB is the complement of Eot 
or fte, and therefore 

Join 6J. and produce it to c so that Ac = bA; join c, P, cutting 
AB in o. Also join c, 5, and 5, a. 

Since P is the pole of AB, the angles at o are right angles ; 
and therefore, by the equal triangles baA, coA, we have 

aA = Ao. 

But a/3 = 2AB, 

whence oB = 5/3, 

and therefore the triangles coB and Bafi are equal, and c, 5, a lie 
on the same great circle. 

Produce cA and cB to meet in H (on the opposite side of the 
sphere). H and c are diametrically opposite, and therefore cP, 
produced, passes through H. 

Now Pa = Pb = P^T, for they differ from quadrants by the 
equal arcs a/3, ba, oc. Hence these arcs divide the triangle Hab 
into three isosceles triangles. 

But Z PHb + Z PHa = Z aHb = Z 6ca. 

Also Z Pa& = TT - Z ca& - Z 

Z P&a = Z Pa& = TT - 
Adding, 2 Z Pa& = 2?r Z ca& Z c6a Z 6ca 

= TT (spherical excess of abc). 
But, as Z Pa/3 and Z Dae are right angles, we have 
angle of /Ba^y = Z FaD = Z /3ae = Z Pa& 

= - (spherical excess of a&c). 

[Numerous singular geometrical theorems, easily proved ab 
initio by quaternions, follow from this : e.g. The arc AB, which 
bisects two sides of a spherical triangle abc, intersects the base at 
the distance of a quadrant from its middle point. All spherical 
triangles, with a common side, and having their other sides 
bisected by the same great circle (i.e. having their vertices in a 


small circle parallel to this great circle) have equal areas, &c. 


118. Let Oa = a. , Ob = /3 , Oc = 7 , and we have 

^ - Ga . cA . Be 

= Ca.BA 

= EG.FE=FG. 

But FG is the complement of DF. Hence the angle of the 

is Aa/" iAe spherical excess of the triangle whose angular points are 
at the extremities of the unit-vectors a , /3 , 7 . 

[In seeking a purely quaternion proof of the preceding proposi 
tions, the student may commence by shewing that for any three 
unit-vectors we have 

The angle of the first of these quaternions can be easily assigned ; 
and the equation shews how to find that of ffa. 1 ^. 

Another easy method is to commence afresh by forming from 
the vectors of the corners of a spherical triangle three new vectors 
thus : 

Then the angle between the planes of a, @ and 7 , a ; or of ft, 7 
and a , j3 ; or of 7, a and /3 , 7 ; is obviously the spherical excess. 

But a still simpler method of proof is easily derived from the 
composition of rotations.] 

119. It may be well to introduce here, though it belongs 
rather to Kinematics than to Geometry, the interpretation of the 

9( )<?" 

By a rotation, about the axis of q, through double the angle of q ) 
the quaternion r becomes the quaternion qrq~ l . Its tensor and 
angle remain unchanged, its plane or axis alone varies. 


A glance at the figure is sufficient for 
the proof, if we note that of course 
T . qrq~ l = Tr, and therefore that we need 
consider the versor parts only. Let Q 
be the pole of q, n/ 

AB = q, AB = q~\ BC = r. 
Join C A, and make AC=C A. Join 
CB. c 

Then CB is qrq 1 , its arc CB is evidently equal in length to that 
of r, B C ; and its plane (making the same angle with B B that 
that of B C does) has evidently been made to revolve about Q, the 
pole of q, through double the angle of q. 

It is obvious, from the nature of the above proof, that this 
operation is distributive ; i. e. that 

q (r 4- s) q 1 = qrq~ l + qsq \ 

If r be a vector, = p, then qpq~ l (which is also a vector) is the 
result of a rotation through double the angle of q about the axis 
of q. Hence, as Hamilton has expressed it, if B represent a rigid 
system, or assemblage of vectors, 


is its new position after rotating through double the angle of q 
about the axis of q. 

120. To compound such rotations, we have 
r . qBq~ l . r 1 = rq . B . (rq)~ l . 

To cause rotation through an angle -fold the double of the angle 
of q we write cfBf*. 

To reverse the direction of this rotation write q^Bq*. 

To translate the body B without rotation, each point of it moving 
through the vector a, we write a + B. 

To produce rotation of the translated body about the same axis, 
and through the same angle, as before, 

Had we rotated first, and then translated, we should have had 

a + qBq~\ 


From the point of view of those who do not believe in the 
Moon s rotation, the former of these expressions ought to be 

qaq- 1 + B 
instead of 

qaq~ l + qSq~\ 

But to such men quaternions are unintelligible. 

121. The operator above explained finds, of course, some 
of its most direct applications in the ordinary questions of 
Astronomy, connected with the apparent diurnal rotation of the 
stars. If X be a unit-vector parallel to the polar axis, and h the 
hour angle from the meridian, the operator is 

/ h h\f \fh h\ 

(^cosg - X sin -J ^ J (^cos g + X sin ^J , 

or L~ l ( ) L, 

the inverse going first, because the apparent rotation is negative 

If the upward line be i, and the southward j, we have 
X = i sin I j cos I, 

where I is the latitude of the observer. The meridian equatorial 
unit vector is 

fjb = i cosl+j sin I; 

and X, /-t, k of course form a rectangular unit system. 
The meridian unit-vector of a heavenly body is 
8 = i cos (I d) + j sin (I d), 

= X sin d + jj, cos d, 
where d is its declination. 

Hence when its hour-angle is h t its vector is 

v = L- I SL. 

The vertical plane containing it intersects the horizon in 
so that 

/ A i \ 
tan (azimuth) = -7 .................. (1). 

[This may also be obtained directly from the last formula (1) 
of 114.] 

78 QU ATERNIONS. [ I 2 2 . 

To find its Amplitude, i.e. its azimuth at rising or setting, 
the hour-angle must be obtained from the condition 

& 8 = (2). 

These relations, with others immediately deducible from them, 
enable us (at once and for ever) to dispense with the hideous 
formulae of Spherical Trigonometry. 

122. To shew how readily they can be applied, let us 
translate the expressions above into the ordinary notation. This 
is effected at once by means of the expressions for X, //,, L, and 
S above, which give by inspection 

= x sin d + (fjb cos h k sin h) cos d, 
and we have from (1) and (2) of last section respectively 

. sin h cos d 

tan (azimuth) = j. T -. = , (1), 

cos I sin d sin / cos d cos h 

cos h + tan I tan d = (2). 

In Capt. Weir s ingenious Azimuth Diagram, these equations 
are represented graphically by the rectangular coordinates of a 
system of confocal conies : viz. 

x = sin h sec I 
y = cos h tan I 

The ellipses of this system depend upon / alone, the hyperbolas 
upon h. Since (1) can, by means of (3), be written as 

tan (azimuth) = = 

tan d y 

we see that the azimuth can be constructed at once by joining 
with the point 0, tan d, the intersection of the proper ellipse and 

Equation (2) puts these expressions for the coordinates in the 

x = sec lj\ tan 2 j tan 8 d \ 

y = tan 2 1 tan d j 

The elimination of d gives the ellipse as before, but that of I 
gives, instead of the hyperbolas, the circles 

x* + 7/ 2 y (tan d cot d) = 1. 

The radius is 

J (tan d 4- cot d) ; 

and the coordinates of the centre are 

0, J (tanrf cot d). 


123. A scalar equation in p, the vector of an undetermined 
point, is generally the equation of a surface; since we may use 
in it the expression p = oca., 

where x is an unknown scalar, and a any assumed unit-vector. 
The result is an equation to determine x. Thus one or more 
points are found on the vector XOL, whose coordinates satisfy the 
equation ; and the locus is a surface whose degree is determined 
by that of the equation which gives the values of x. 

But a vector equation in p, as we have seen, generally leads to 
three scalar equations, from which the three rectangular or other 
components of the sought vector are to be derived. Such a vector 
equation, then, usually belongs to a definite number of points in 
space. But in certain cases these may form a line, and even a 
surface, the vector equation losing as it were one or two of the 
three scalar equations to which it is usually equivalent. 

Thus while the equation cup ft 
gives at once p = a" 1 /?, 

which is the vector of a definite point (since by making p a vector 
we have evidently assumed 

Sa/3 = 0); 

the closely allied equation Vap = j3 
is easily seen to involve Sa{3 = 0, 
and to be satisfied by p = of */8 + xa, 

whatever be x. Hence the vector of any point whatever in the 
line drawn parallel to a from the extremity of a~ l (3 satisfies the 
given equation. [The difference between the results depends 
upon the fact that Sap is indeterminate in the second form, but 
definite (= 0) in the first.] 

124. Again, Vap .Vp/3 = ( Fa/3) 2 

is equivalent to but two scalar equations. For it shews that Vap 
and V/3p are parallel, i.e. p lies in the same plane as a and /3, and 
can therefore be written ( 24) 

p = xa. + yfl, 

where x and y are scalars as yet undetermined. 
We have now Vap = yVa/3, 


which, by the given equation, lead to 

ajy = l, or y = -, 


or finally p = xa.+ -ft; 


which ( 40) is the equation of a hyperbola whose asymptotes are 
in the directions of a and ft. 

125. Again, the equation 

though apparently equivalent to three scalar equations, is really 
equivalent to one only. In fact we see by 91 that it may be 

whence, if a be not zero, we have 

and thus ( 101) the only condition is that p is coplanar with a, ft. 
Hence the equation represents the plane in which a and ft lie. 

126. Some very curious results are obtained when we extend 
these processes of interpretation to functions of a quaternion 

q = w + p 
instead of functions of a mere vector p. 

A scalar equation containing such a quaternion, along with 
quaternion constants, gives, as in last section, the equation of a 
surface, if we assign a definite value to w. Hence for successive 
values of w, we have successive surfaces belonging to a system ; 
and thus when w is indeterminate the equation represents not a 
surface, as before, but a volume, in the sense that the vector of any 
point within that volume satisfies the equation. 

Thus the equation (Tqf = a 8 , 


or w p = a, 


represents, for any assigned value of w, not greater than a, a sphere 
whose radius is Jd 2 w\ Hence the equation is satisfied by the 
vector of any point whatever in the volume of a sphere of radius a, 
whose centre is origin. 


Again, by the same kind of investigation, 

where q = iu + p, is easily seen to represent the volume of a sphere 
of radius a described about the extremity of ft as centre. 

Also S(q*) = a 2 is the equation of infinite space less the space 
contained in a sphere of radius a about the origin. 

Similar consequences as to the interpretation of vector equa 
tions in quaternions may be readily deduced by the reader. 

127. The following transformation is enuntiated without proof 
by Hamilton (Lectures, p. 587, and Elements, p. 299). 

To prove it, let r~ l (r 2 q*) 2 q~ l = t, 

then Tt = 1, 

and therefore Kt = t~\ 

But ( ? Y) 4 = rfy, 

or r*q* = rtqrtq, 

or rq = tqrt. 

Hence KqKr = t~ l KrKqr\ 

or KrKq = tKqKrt. 

Thus we have U (rq KrKq) = tU(qr KqKr) t, 
or, if we put s = U (qr KqKr), 

Ks= tst. 

Hence sKs = (Ts) 2 =l = stst, 

which, if we take the positive sign, requires 


or t = s 1 = UKs, 

which is the required transformation. 

[It is to be noticed that there are other results which might 
have been arrived at by using the negative sign above ; some 
involving an arbitrary unit-vector, others involving the imaginary 
of ordinary algebra.] 

128. As a final example, we take a transformation of Hamil 
ton s, of great importance in the theory of surfaces of the second 

Transform the expression 

T. Q. I. 

82 QUATERNIONS. [128. 

in which a, /3, 7 arc any three mutually rectangular vectors, into 
the form 

v -? 

which involves only two vector-constants, i, K. 

[The student should remark here that t, K, two undetermined 
vectors, involve six disposable constants : and that a, ft, 7, being 
a rectangular system, involve also only six constants.] 

[T(ip + pK)} 2 = (ip + />*) (pi + KP) ( 52, 55) 


Hence (%>) + (S/3 P )* + (>S 7P ) 2 = /V^C 

K L 

But ^ 2 (Sapf + /T (/>) + 7 " 2 (^) 2 = p 2 (i 25, 73). 
Multiply by /3 2 and subtract, we get 

The left side breaks up into two real factors if /3 2 be intermediate 
in value to a 2 and 7 2 : and that the right side may do so the term 
in p 2 must vanish. This condition gives 

/3 2 = ,42X8 ; an d the identity becomes 

- l ) (P 

Hence we must have 


where jp is an undetermined scalar. 

To determine p, substitute in the expression for /3 2 , and we find 

= p* + ( a2 - r) 


Thus the transformation succeeds if 

P +- 

which gives p + - = + 2 A / 

p y * 2 - 7 2 


-6 2 ) /I 2 \ 

Tf = t -* J ^ - 7*) = 

*) = * ,/*, 

or (** - i*)" 1 = TaTy. 

Toc+Ty 1 Ta-Ty 

Again, ft = , - = - 

/ 2 -3. rt) 

and therefore 

Thus we have proved the possibility of the transformation, and 
determined the transforming vectors i, K. 

129. By differentiating the equation 
(Sap? + (Sto? + (Hyp? = 

we obtain, as will be seen in Chapter IV, the following, 
+ S/3 P S/3 P 

_ _ 

(K i ) 

where p also may be any vector whatever. 

This is another very important formula of transformation ; and 
it will be a good exercise for the student to prove its truth by 
processes analogous to those in last section. We may merely 
observe, what indeed is obvious, that by putting p = p it becomes 
the formula of last section. And we see that we may write, with 
the recent values of i and K in terms of a, /3 y, the identity 

i J 


84 QUATERNIONS. [ 1 30. 

130. In various quaternion investigations, especially in such 
as involve imaginary intersections of curves and surfaces, the old 
imaginary of algebra of course appears. But it is to be particularly 
noticed that this expression is analogous to a scalar and not to a 
vector, and that like real scalars it is commutative in multipli 
cation with all other factors. Thus it appears, by the same proof 
as in algebra, that any quaternion expression which contains this 
imaginary can always be broken up into the sum of two parts, one 
real, the other multiplied by the first power of V 1. Such an 
expression, viz. 

q = q + >f^lq", 

where q and q" are real quaternions, is called by Hamilton a 
BIQUATERNION. [The student should be warned that the term 
Biquaternion has since been employed by other writers in the 
sense sometimes of a "set" of 8 elements, analogous to the 
Quaternion 4 ; sometimes for an expression q + Oq" where 6 is not 
the algebraic imaginary. By them Hamilton s Biquaternion is 
called simply a quaternion with non-real constituents.] Some 
little care is requisite in the management of these expressions, but 
there is no new difficulty. The points to be observed are : first, 
that any biquaternion can be divided into a real and an imaginary 
part, the latter being the product of V - 1 by a real quaternion ; 
second, that this V - 1 is commutative with all other quantities in 
multiplication ; third, that if two biquaternions be equal, as 

g fV-lg W + V-l/ , 

we have, as in algebra, q = /, q" = r" ; 

so that an equation between biquaternions involves in general 

eight equations between scalars. Compare 80. 

131. We have obviously, since \l - 1 is a scalar, 

8 (q + V - 1 q") = Sq + V - 1 Sq", 
V (q + V ~1" q") = Vq + V 7 -T Vq". 
Hence ( 103) 

= (Sq + V - 1 Sq"+ Vq + V^l Vq") (Sq + V - 1 Sq" - Vq 

- V - 1 Vq") 

= (Sq> + V^l Sq"T ~(Vq + V - 1 Vq J, 
= (TqJ - (Tq J + 2 V^LSf . q Kq". 


The only remark which need be made on such formulae is this, that 
the tensor of a biquaternion may vanish while both of the component 
quaternions are finite. 

Thus, if Tq = Tq", 

and S.q Kq" = 0, 

the above formula gives 

The condition S . q Kq = 

may be written 


where a is any vector whatever. 

Hence Tq = Tq" = TKq" = ~ , 

and therefore 

Tq (Uq f - V^T U* . Uq ) = (1 - V^l Ua) q 
is the general form of a biquaternion whose tensor is zero. 

132. More generally we have, q, r, q, r being any four real 
and non-evanescent quaternions, 

(q + V^lg ) (r + V~lr ) = qr - q r + V^ 1 (qr + q r). 
That this product may vanish we must have 

qr = q r , 

and qr = - qr. 

Eliminating r we have qq ~ 1 qr = qr ) 
which gives (q ~ l q) 2 = ~ 1. 

i.e. q = qa. 

where a is some unit-vector. 

And the two equations now agree in giving 

r = ar , 

so that we have the biquaternion factors in the form 
q (OL + V^T) and - (a - V~^T) r ; 
and their product is 

-q(QL + V^T) (a - V -^1) r, 
which, of course, vanishes. 

86 QUATERNIONS. [133. 

[A somewhat simpler investigation of the same proposition 
may be obtained by writing the biquaternions as 

q (q" 1 q + V^ ) and (rr ~ l + V^l) r , 
or q (q" + V ^1) and (r" + V ^1) r, 

and shewing that 

q" r" = a, where To. 1.] 

From this it appears that if the product of two livectors 
p + crV^T and p +er V-l 

is zero, we must have 

o-- 1 p = -p f a - 1 =Ua, 

where a may be any vector whatever. But this result is still more 
easily obtained by means of a direct process. 

133. It may be well to observe here (as we intend to avail our 
selves of them in the succeeding Chapters) that certain abbreviated 
forms of expression may be used when they are not liable to confuse, 
or lead to error. Thus we may write 

T>q for (Tqf, 

just as we write cos 2 for (cos 0) 2 , 

although the true meanings of these expressions are 
T(Tq) and cos (cos 0). 

The former is justifiable, as T (Tq) = Tq, and therefore T q is not 
required to signify the second tensor (or tensor of the tensor) of q. 
But the trigonometrical usage is defensible only on the score of 
convenience, and is habitually violated by the employment of 
cos" 1 x in its natural and proper sense. 
Similarly we may write 

S 2 q for (Sq) z , &c, 

but it may be advisable not to use 


as the equivalent of either of those just written; inasmuch as it 
might be confounded with the (generally) different quantity 

although this is rarely written without the point or the brackets. 

The question of the use of points or brackets is one on which 
no very definite rules can be laid down. A beginner ought to use 


them freely, and he will soon learn by trial which of them are 
absolutely necessary to prevent ambiguity. 

In the present work this course has been adopted : the 
earlier examples in each part of the subject being treated with 
a free use of points and brackets, while in the later examples 
superfluous marks of the kind are gradually got rid of. 

It may be well to indicate some general principles which 
regulate the omission of these marks. Thus in 8 .a@ or V. a/3 
the point is obviously unnecessary : because So. = 0, and Va. = a, 
so that the S would annihilate the term if it applied to a alone, 
while in the same case the V would be superfluous. But in S.qr 
and V . qr, the point (or an equivalent) is indispensable, for Sq . r, 
and Vq.r are usually quite different from the first written 
quantities. In the case of K, and of d (used for scalar differen 
tiation), the omission of the point indicates that the operator acts 
only on the nearest factor : thus 

Kqr = (Kq) r = Kq . r, dqr = (dq) r=dq.r; 
while, if its action extend farther, we write 

K . qr = K (qr), d . qr = d (qr} } &c. 

In more complex cases we must be ruled by the general 
principle of dropping nothing which is essential. Thus, for 



may be written without ambiguity as 


but nothing more can be dropped without altering its value. 

Another peculiarity of notation, which will occasionally be 
required, shows which portions of a complex product are affected 
by an operator. Thus we write 

if V operates on a and also on r, but 

if it operates on r alone. See, in this connection, the last Example 
at the end of Chap. IV. below. 

134. The beginner may expect to be at first a little puzzled 
with this aspect of the notation ; but, as he learns more of the 
subject, he will soon see clearly the distinction between such an 
expression as 

88 QUATERNIONS. [134. 

where we may omit at pleasure either the point or the first V 
without altering the value, and the very different one 

Sa/3. F/3 7 , 
which admits of no such changes, without alteration of its value. 

All these simplifications of notation are, in fact, merely examples 
of the transformations of quaternion expressions to which part of 
this Chapter has been devoted. Thus, to take a very simple ex 
ample, we easily see that 

S. Fa/3 V/3y=S Fa/3 V/3y = .a/3F/3 7 - SaV./SVpy = - SOL V. ( F/3 7 )/3 
= SOL V. ( Vyfl/3 = S.a F( 7 /3)/3 = S. F( 7 /3)/9a = S F 7 /3 F/3a 

= S . Vy/3V/3ot, &c., &c. 

The above group does not nearly exhaust the list of even the simpler 
ways of expressing the given quantity. We recommend it to the 
careful study of the reader. He will find it advisable, at first, to 
use stops and brackets pretty freely; but will gradually learn to 
dispense with those which are not absolutely necessary to prevent 

There is, however, one additional point of notation to which 
the reader s attention should be most carefully directed. A very 
simple instance will suffice. Take the expressions 

.2 and &. 

7 OL ya. 

The first of these is 

/3 7 - 1 . 7 a- 1 = ,9a- 1 , 

and presents no difficulty. But the second, though at first sight 
it closely resembles the first, is in general totally different in 
value, being in fact equal to 

For the denominator must be treated as one quaternion. If, 
then, we write 

^ - ft 

7 a 
we have 

so that, as stated above, 

q = 
We see therefore that 




1. Investigate, by quaternions, the requisite formulae for 
changing from any one set of coordinate axes to another ; and 
derive from your general result, and also from special investiga 
tions, the usual expressions for the following cases : 

(a) Rectangular axes turned about z through any angle. 

(6) Rectangular axes turned into any new position by rota 
tion about a line equally inclined to the three. 

(c) Rectangular turned to oblique, one of the new axes 
lying in each of the former coordinate planes. 

2. Point out the distinction between 

a a 

and find the value of their difference. 
= 1, then U 
OL + ft Fa/3 

If ZW = 1, then U- -. 

a. \a 

Shew also that 

ai-ft 1 + tfa/3 

, a - B Fa 


provided a and ft be unit-vectors. If these conditions are not 
fulfilled, what are the true values ? 

3. Shew that, whatever quaternion r may be, the expression 

ar + rft, 

in which a and ft are any two unit- vectors, is reducible to the 

where I and m are scalars. 

4. If Tp = To. = Tft = 1, and S . a/3p = 0, shew by direct trans 
formations that 

S.U(p-oi} U(p-p)= 
Interpret this theorem geometrically. 

( JO 


5. If 8*0 = 0, Ta = T/3 = I, shew that 

W7T - m 

(1 + a" 1 ) = 2 cos " - a* = 2Sa * . a* {3. 

6. Put in its simplest form the equation 

pS. Fa/3F/3 7 F 7 a = ttFFyaFa/3 + bV. Fa/3F/3 7 + cFF/3 7 F 7 a ; 
and shew that a = S . (Byp, &c. 

7. Shew that any quaternion may in general, in one way only, 
be expressed as a homogeneous linear function of four given 
quaternions. Point out the nature of the exceptional cases. Also 
find the simplest form in which any quaternion may generally be 
expressed in terms of two given quaternions. 

8. Prove the following theorems, and exhibit them as proper 
ties of determinants : 

(a) S.(* + 0)(0 + y)(y + a ) = 2S.*0y, 

(b) S . Fa/3 F/3 7 Vya = -(8. a/3 7 ) 2 , 

(c) 8. V(a + /3) (0 + y) F(/3 + 7 ) (7 + a) V(y -f a) (a + 0) 

(d) S.V(Va(3V{3y)V( F/3 7 Fya) F ( F 7 a Fa/3) = - (S . a/3 7 ) 4 , 

(e) > 7 ) 4 , 

where 3 = F(F(a + 0) (0 + 7) F(^ + 7 ) ( 7 + a)), 

9. Prove the common formula for the product of two determi 
nants of the third order in the form 

Soft W, Sy/3, 
Say, S/3 7l S Wl 

10. Shew that, whatever be the eight vectors involved, 

W t >W! S/3 7l S/33, 

Vt tyft ^7% ^ 
SfSa, fifS^ S8 yi SSS, 

If the single term tfaaj be changed to Sai a,, the value of the 
determinant is 


State these as propositions in spherical trigonometry. 
Form the corresponding null determinant for any two groups 
of five quaternions : and give its geometrical interpretation. 

11. If, in 102, a, /3, 7 be three mutually perpendicular 
vectors, can anything be predicated as to c^, j3 v y l ? If a, /3, 7 be 
rectangular unit-vectors, what of 15 /3 t , y l ? 

12. If a, /3, 7, a , /3 , 7 be two sets of rectangular unit-vectors, 
shew that 

t &c., &c. 

13. The lines bisecting pairs of opposite sides of a quadri 
lateral (plane or gauche) are perpendicular to each other when the 
diagonals of the quadrilateral are equal. 

14. Shew that 

(a) S . q* = 2S*q - T 2 q, 

(b) S.q* = S?q-3SqrVq, 

(c) 2 /3V + S 2 . a/3y = V 2 . a@y, 


and interpret each as a formula in plane or spherical trigonometry. 

15. If q be an undetermined quaternion, what loci are repre 
sented by 

(a) ( ? O = -a 2 , 
(6) ( 3 a-y = o 4 , 
(c) S.(g-a) = o , 
where a is any given scalar and a. any given vector ? 

16. If q be any quaternion, shew that the equation 

<2 2 = ? 2 

is satisfied, not alone by Q = q, but also by 

Q = J-l(Sq.UVq-TVq). 

(Hamilton, Lectures, p. 673.) 

1 7. Wherein consists the difference between the two equations 


What is the full interpretation of each, a being a given, and p an 
undetermined, vector ? 

18. Find the full consequences of each of the following 
groups of equations, as regards both the unknown vector p and 
the given vectors a, /3, 7 : 

S*p =0, Sap =0, 

(") * fl n. (&) S.afip = 0, (c) S.aft, =0, 
8 to pm 8$p = 0; s!#yp = <X 

19. From 74, 110, shew that, if 6 be any unit-vector, and m 

mir mir 

any scalar, e = cos - + e sin . 

- 2i 

Hence shew that if a, ft, 7 be radii drawn to the corners of a tri 
angle on the unit-sphere, whose spherical excess is m right angles, 

Also that, if A, B, G be the angles of the triangle, we have 

2C 2 ZA 

7 7r /3 7r ct 7r = 1. 
20. Shew that for any three vectors a, /3, 7 we have 

= -2. 

(Hamilton, Elements, p. 388.) 

21. If a v 2 , a s , x be any four scalars, and p lf p 2 , p :i any three 
vectors, shew that 

(S . Pl p 2 p 3 f + (S, . a, Vp 2 p 3 Y + x* (2 V Pl p 2 }* - a? (2 . a, (p 2 - p,)) 8 

+ 2H (x 2 + S Pl p 2 + a^ 2 ) = 2H (x* -4- p z ] + 2Ha 2 
+ S {(x* + o-j 2 + /3 X 2 ) (( Vp 2 p s ) 2 + 2a 2 a 3 ( 2 + 8p z p 9 ) a? (p 3 p 3 } 2 )} ; 
where Ha 2 = a*a*a*. 

Verify this formula by a simple process in the particular case 

22. Eliminate p from the equations 

V.0pap = t ^7/0 = 0; 
and state the problem and its solution in a geometrical form. 


23. If p, q, r, s be four versors, such that 

qp = - S r = a, 

rq = -ps = /3, 
where a and /3 are unit-vectors ; shew that 

8(V. VsVqV. VrVp) = 0. 
Interpret this as a property of a spherical quadrilateral. 

24. Shew that, if pq, rs, pr, and qs be vectors, we have 


25. If a, /3, 7 be unit-vectors, 

- (3 (S 

26. If i, j, k, i , j , k y be two sets of rectangular unit-vectors, 
shew that 

8 . Vii Vjj Vkk = (SijJ - (Sji) 2 

and find the values of the vector of the same product. 

27. If a, /3, 7 be a rectangular unit-vector system, shew that, 
whatever be X, /A, v, 


are coplanar vectors. What is the connection between this and 

the result of the preceding example ? 



135. IN Chapter I. we have already considered as a special 
case the differentiation of a vector function of a scalar independent 
variable : and it is easy to see at once that a similar process is 
applicable to a quaternion function of a scalar independent variable. 
The differential, or differential coefficient, thus found, is in general 
another function of the same scalar variable ; and can therefore be 
differentiated anew by a second, third, &c. application of the same 
process. And precisely similar remarks apply to partial differentia 
tion of a quaternion function of any number of scalar independent 
variables. In fact, this process is identical with ordinary differ 

136. But when we come to differentiate a function of a vector, 
or of a quaternion, some caution is requisite ; there is, in general 
(except, of course, when the independent variable is a mere scalar), 
nothing which can be called a differential coefficient ; and in fact 
we require (as already hinted in 33) to employ a definition of a 
differential, somewhat different from the ordinary one but, coinciding 
with it when applied to functions of mere scalar variables. 

137. If T = F (q) be a function of a quaternion q, 

where n is a scalar which is ultimately to be made infinite, is 
defined to be the differential of r or Fq. 

Here dq may be any quaternion whatever, and the right-hand 
member may be written 


where f is a new function, depending on the form of F; homo 
geneous and of the first degree in dq ; but not, in general, capable 
of being put in the form 

138. To make more clear these last remarks, we ma,y observe 
that the function 


thus derived as the differential of F (q), is distributive with respect 
to dq. That is 

r and s being any quaternions. 

*) - Fq\ 

w/ ] 

And, as a particular case, it is obvious that if x be any scalar, 

t r). 

139. And if we define in the same way 

dF (q, r, s ) 

as being the value of 

dq dr ds 
n n n 

where q, r, s, ... dq, dr, ds, are any quaternions whatever; we 

shall obviously arrive at a result which may be written 

f(q, r, s,...dq, dr, ds, ), 

where / is homogeneous and linear in the system of quaternions 

dq, dr, ds, and distributive with respect to each of them. Thus, 

in differentiating any power, product, &c. of one or more quater 
nions, each factor is to be differentiated as if it alone were variable ; 
and the terms corresponding to these are to be added for the com 
plete differential. This differs from the ordinary process of scalar 
differentiation solely in the fact that, on account of the non-com 
mutative property of quaternion multiplication, each factor must in 
general be differentiated in situ. Thus 

d (qr) dq.r-\- qdr, but not generally = rdq + qdr. 

96 QUATERNIONS. [ T 4 O - 

140. As Examples we take chiefly those which lead to results 
which will be of constant use to us in succeeding Chapters. Some 
of the work will be given at full length as an exercise in quaternion 


The differential of the left-hand side is simply, since Tp is a 



That of f is C4(W-T-? 2 

r ^ 1 V r n f 

Hence Tp dTp = - Spdp, 

or dTp = - 

dTp Q dp 

or ^=3^. 

Tp p 

(2) Again, p = Tp Up 

dp = dTp. Up + TpdUp, 

dp dTp dUp 

whence = rn + -j r 

p Tp Up 

d Up Tr dp 

Hence -jj^ = V . 

Up p 

This may be transformed into F 2 or - /^- , &c. 

p lp 


qKdq + dqKq 1 

= -Cao w + -2 dqKdq 

\ n n 

= qKdq + dqKq, 

= qKdq + IT (qKdq) ( 55), 

= 28. qKdq = 2S. dqKq. 


Hence dTq = S . dqUKq = S. dqUq~ l = TqS^ , 

since Tq = TKq, and UKq = Uq~\ 

[If q = p, a vector, Kq = Kp = -p, and the formula becomes 

Again, dq = Tqd Uq + UqdTq, 

which gives dq = dTq dUq , 

q Tq Uq 

whence, as 


we have ydq = dUq_ 

q Uq 

= qdq + dq.q 

= ZS.qdq + 2Sq . Vdq + 2Sdq . Vq. 
If q be a vector, as p, Sq and Sdq vanish, and we have 
d(p*) = 2Spdp,SiS in (1). 

(5) Let q = r*. 
This gives dr* = dq. But 

dr = d (q 2 ) = qdq + dq . q. 
This, multiplied by q and into Kq, gives the two equations 

qdr = q*dq + qdq . g, 

and drKq = dg ?V + qdq . Kq. 

Adding, we have 

qdr + dr.Kq = (q* + Tq 2 + 2Sq .q)dq = 4<Sq . qdq ; 

whence dq, i.e. dr , is at once found in terms of dr. This process 
is given by Hamilton, Lectures, p. 628. See also 193 below, and 
No. 7 of the Miscellaneous Examples at the end of this work. 

(6) ?<f = l> 

. . dq~ l = q~ l dq . q 1 . 
If q is a vector, = p suppose, 

dq-+ = dp- l = -p- l dp.p- 1 

T. Q. I. 

9 8 QUATERNIONS. [ 1 4- 

P P 

dp ~ dp 
P P 1 P 

(7) q = Sq+Vq, 

But dq = $d# 4- 

Comparing, we have 

dSq = Sdq, dVq=Vdq. 
Since Kq = Sq- Vq, we find by a similar process 

(8) In the expression qaq~\ where a is any constant quaternion, 
q may be regarded as a mere versor, so that 

Thus S.dqKq = , 

and hence dg^ 1 = - (J^ 1 , 

as well as 

q~ l dq = - dq~*q, 

are vectors. But, if a = a + a, where a is a scalar, q^q 1 = a, i.e. 

constant, so that we are concerned only with d (q<*q~ l ). 
Hence d (qvq 1 ) = dq aq~ l - qctq" l dqq l t 

= dqq l . qaq~ l - qaq~ l . dqq~\ 
= 2V. d- a 1 = -2V. 

(9) With the restriction in (8) above we may write 

q = cos u + 6 sin u, 

where T0 = I; S0d0 = 0. 

Hence q 1 = cosu-0 sin u ; 

- q~ l dq = dq~ l q = { - (sin u+6 cos u) du - dO sin u} (cos u + sin ?< 

= -0 du-d6 sin w (cos w + ^ sin w) ; 
= ^ q~ l = 0du + d0 sin u (cos w - ^ sin u). 
Both forms are represented as linear functions of the rectangula 

system of vectors 

0, d-0 t OdB. 


If the plane of q be fixed, 6 is a constant unit vector, and 
dq q 1 = dq 1 q = Odu. 

(10) The equation (belonging to a family of spheres) 


gives 8 dp {(p + a)-e 2 (p-a)} = 0; 

or, by elimination of e, 

sdp{( P +ar-(p-*r}=o, 

whose geometrical interpretation gives a well-known theorem. 

If we confine our attention to a plane section through the 
vector a, viz. 

= U; or 

where /3||F7a||Fap ; 

we have 
dp\\ V. $ {(p + a) 1 - (p - a) 1 } or F. dp F/3 {(p + )- - (p - a) 1 } = 0. 

(11) Again, from 


(which is the equation of the family of tores produced by the 
rotation of a group of circles about their common chord) we have 

Also this gives VU. (p + a) (p - a) = j3 = Jl - e*U. VOL P . 
We obtain from the first of these, by differentiation, 

or 8 . &dp {(p + a) 1 - (p - a)" 1 } = 0. 

If we consider /3 to be constant, we limit ourselves to a meridian 
section of the surface (i.e. a circle) and the form of the equation 
shews that, as /3 is perpendicular to the plane of a, p (and, of 
course, dp), 

v.d p {(p+*r-(p-*r}=o- 

We leave to the reader the differentiation of the vector form of 
the equation above. 

These results are useful, not only as elementary proofs of 
geometrical theorems but, as hints on the integration of various 
simple forms. 


100 QUATERNIONS. [141. 

(12) As a final instance, take the equation 

P~YP = > 

where p stands for dp/ds, s being the arc of a curve. 

By 38, a is a unit vector, and the expression shews by its 
form that it belongs to a plane curve. Let (3 be a vector in its 
plane, and perpendicular to a. Operate by S . /3 and we get 

whose integral is 

P 2 

the tensor of /3 being the constant of integration. 

141. Successive differentiation of course presents no new 

Thus, we have seen that 

d (f) = dq.q + qdq. 
Differentiating again, we have 

and so on for higher orders. 

If q be a vector, as p, we have, 140 (1), 

Hence d 2 (p 2 ) = 2 (dpY + 2Spd*p, and so on. 
Similarly d*Up=- d (^ Vpd^j . 

l_ 2dT P _2Spdp 
Tp*~ ~ Tp s : Tp* 
and d . Vpdp = V . pd 2 p. 

Up /jr . X2 UpVptfp ZUpVpdpSpdp 
Hence d* Up = - ( Vpdpf - - -f- 

2 p - 2 VpdpSpdp}*. 

142. If the first differential of ^ be considered as a constant 
quaternion, we have, of course, 

tfq = 0, d 3 q = 0, &c. 
and the preceding formulaa become considerably simplified. 

* This may be farther simplified ; but it may be well to caution the student that 
we cannot, for such a purpose, write the above expression as 



Hamilton has shewn that in this case Taylors -Theorem admits 
of an easy extension to quaternions. That is, we may write 

f(q + xdq) =/(<?) + xdf (q) + <Ff(g) + ...... 

if d 2 q = ; subject, of course, to particular exceptions and limita 
tions as in the ordinary applications to functions of scalar variables. 
Thus, let 

f (q) = q 3 > an d we have 

d f(q) = q*dq + qdq.q + dq. q\ 

= 2dq . qdq + 2q (dq)* + 2 (dq) z q, 

and it is easy to verify by multiplication that we have rigorously 
(q + ocdqY = q* + x (q*dq + qdq .q + dq.q*) + 

x* (dq .qdq + q (dq)* + (dqfq) + ** (dq) 3 ; 

which is the value given by the application of the above form of 
Taylor s Theorem. 

As we shall not have occasion to employ this theorem, and as 
the demonstrations which have been found are all too laborious for 
an elementary treatise, we refer the reader to Hamilton s works, 
where he will find several of them. 

143. To differentiate a function of a function of a quaternion 
we proceed as with scalar variables, attending to the peculiarities 
already pointed out. 

144. A case of considerable importance in geometrical and 
physical applications of quaternions is the differentiation of a scalar 
function of p, the vector of any point in space. 

Let F(p) = C, 

where F is a scalar function and C an arbitrary constant, be the 
equation of a series of surfaces. Its differential, 

f(p, dp) = 0, 

is, of course, a scalar function : and, being homogeneous and linear 
in dp, 137, may be thus written, 

where v is a vector, in general a function of p. 

This vector, v, is shewn, by the last written equation, to have 
the direction of the normal to the given surface at the extremity 
of p. It is, in fact, perpendicular to every tangent line dp ; 
1 36, 98. 


145. This- leads us directly to one of the most remarkable 
operators peculiar to the Quaternion Calculus ; viz. 

__ . d . d j d 

V = i^ r + <)- r + k- r ..................... (1), 

dx J dy dz 

to whose elementary properties we will devote the remainder of 
the chapter. The above definition is that originally given by 
Hamilton, before the calculus had even partially thrown off its 
early Cartesian trammels. Since i, j, k stand for any system of 
rectangular unit vectors, while a?, y, z are Cartesian co-ordinates 
referred to these as axes, it is implied in (1) that V is an Invariant. 
This will presently be justified. Meanwhile it is easy to see that 
if p be the vector of any point in space, so that 

p = ix + jy + kz t 
we have V/> = -3 ................................. (2), 

"p ............ (4), 

of which the most important case is 

vJL a P .UP 

Tp (Tp)*~ Tp 9 " 
A second application gives 

v ^=-$- v ^= ............... (G) - 


so that VUp = -~ ........................... (7). 

By the definition (1) we see that 

\ a (d\* fd\*\ 

} +(3-) + b~ r ............... ( 8 ) 

dxj \dyj \dz) } 

the negative of what has been called Laplace s Operator. Thus 
(6) is merely a special case of Laplace s equation for the potential 
in free space. 

Again we see by (2), a being any constant vector, 

SaV . p - VSap = F. F Vp = 0, 
from which 

VTap + FaV. p = (SaV.p - otSVp) + (aSVp - VSap) = 0. 


[The student should note here that, in expanding the terms of 
the vector function on the left by the formula (1) of 90, the 
partial terms are written so that V is always to the left of (though 
not necessarily contiguous to) its subject, p.] 

146. By the help of these elementary results, of which (3) 
and (7) are specially noteworthy, we easily find the effect of V 
upon more complex functions. 

For instance, taking different modes of operating, we have with 

a = ia +jb + kc 

SaV . p = VSap = - V (ax + by + cz) = - (ia + jb + kc) = - ...(!), 
or thus VSoip = iSai + jSaj + kScnk a ; 

while - VaV .p = WoLp = -WpoL = -V (poi-SpOL) 

= 3a-a = 2a ...... , .............. (2), 

or V Vap = i Vai + j VOLJ + k Vcnk = 2 a. 

From the latter of these we have 

Vap _ 2a _ 



Tp* " Tp> Tp 5 Tp 5 Tp 5 

[where note that the first of these values is obtained thus, 

- 1 

, q , 

The order is of vital importance.] 
This, in its turn, gives 

- g 

Tf" ~W~ T? 

where 8 is. a symbol of variation. This is a result of great physical 
importance, especially in electro-dynamics. We may alter the 
right-hand member (by 145, (5)) so as to write the whole in the 

......... (4 ). 

And it is easy to see that 8 may be substituted for V in the left- 
hand member. [The reason for this may be traced in the result of 

As an addition to these examples, note that (by (2) of 148, 


104 QUATERNIONS. [147. 

which may be contrasted with (4) above. The altered position of 
the point produces a complete change in the meaning of the left- 
hand member. 

Finally, we see that 

V2Sp = -2a (5), 

a result which will be found useful in next Chapter. 

147. Still more important are the results obtained from the 
operator V when it is applied to 

and functions of this vector. (Here , 77, f are functions of x, y, z, 
so that a is a vector whose value is definite for each point of 

We have at once 


/ dz 

Those who are acquainted with mathematical physics will 
recognize at a glance the importance of this expression. For, if a 
denote the displacement (or the velocity) of a point originally 
situated at p, it is clear that 

\dx dy dz) 

represents the consequent condensation of the group of points (say 
particles of a fluid) originally in the neighbourhood of p, while 

- ...(4), 

dz) J \dz dx) \dx dyj 

represents double the (vector) axis of rotation of the same group. 

Other, and more purely quaternion, methods will be employed 
later, to deduce these results afresh, and to develop their applica 
tions. They are introduced here in their semi-Cartesian form 
merely to shew the importance of the operator V. 

148. Let us recur to the equation of 144, viz. 

F( P } = G .......................... (1). 

Ordinary complete differentiation gives 

dF . dF , dF 1 

dF = -j- dx 4- -T dy + -j- dx, 
dx dy dz 


or, what is obviously the same, 

dF=-SdpVF ........................ (2), 

which we may write, if we please, as 


Here the point is obviously unnecessary, but we shall soon 
come to cases in which it, or some equivalent, is indispensable. 
Thus it appears that the operator 

is equivalent to total differentiation as involved in the passage 
from p to p + dp. Hence, of course, as in 144 

dF (p)=0 = Svdp = - SdpVF, 

and thus (as dp may have any of an infinite number of values) 

v = -VF ........................... (3). 

If we pass from one surface of the series (1) to a consecutive 
one by the vector Sp, we have 

Hence v~ l SC is a value of Sp ; so that the tensor of v is, at every 
point, inversely as the normal distance between two consecutive 
surfaces of the series. 

Thus, if (1) be the equation of a series of equipotential surfaces, 
v, as given by (3), represents the vector force at the point p ; if 
(1) be a set of isothermals, v (multiplied by the conductivity, a 
scalar) is the flux of heat, &c. 

149. We may extend the result (2) of 148 to vector functions 
by multiplying both sides into a constant vector, a, and adding 
three such expressions together. Thus if 

cr = 0^+^ + 7^, 
we obtain at once 

d<r = -S(dpV)<r = -SdpV.(r ............... (4). 

But here the brackets, or the point, should (at first, at least) be 
employed ; otherwise we might confound the expression with 

which, as equating a vector to a scalar, is an absurdity (unless both 
sides vanish). See again, 148. 

Finally, from (2) and (4), we have for any quaternion 

dq = -SdpV.q ........................ (5). 

106 QUATERNIONS. [l49- 

The student s attention is particularly called to the simple 
processes we have adopted in obtaining (4) and (5) from (2) of 
148 ; because, in later chapters, other and more complex results 
obtained by the same processes will frequently be taken for 
granted ; especially when other operators than 8 (dpV) are em 
ployed. The precautions necessary in such matters are two-fold, 
(a) the operator must never be placed anywhere after the operand ; 
(6) its commutative or non- commutative character must be carefully 
kept in view. 


1. Shew that 

(a) d.SUq = S.UqV = -S TVUq, 

(b) d.VUq=V. Uq- l 

(c) d. 

(d) d.a. x = a? + l dx, 


(e) d z .Tq={S 2 .dqq- 1 -S. (dqq 1 )*} Tq = - Tq V* ^ . 

2. If Fp = 2.SoLpS/3p + gp* 
give dFp = Svdp, 

shew that v = 2 F . ap/3 + (g + 2a/3) p. 

3. Find the maximum and minimum values of Tp, when 

(a) (p-) 2 - -a 2 ; 

(b) (p-a? = - 

(c) p 2 - SapS/3p = - a 2 ; 

(d) p* - SoLpS/Bp = - a 2 ,) 

S 7P = V. } 

Point out the differences, geometrical and analytical, between 
(a), (c) on the one hand, and (6), (d) on the other. 
State each of the problems in words. 


4. With V as in 145, shew that 

8V (qaq~ l ) = 28 . Vq uf 1 = 28 . aq 1 Vq. 

where q is a function of p, and a any constant vector. 

5. Shew that, if a, /3, 7 be a constant rectangular unit-vector 

qaq~ l d . qa.q~ l + qftq~ l d . qj3q~ l + qyq~ l d . qfiq 1 = 4>dqq~\ 

6. Integrate the differential equations : 

(a) /3 + xSftp = 7, 

(6) q + aq = b, 

(c) 6 = VoiO. Proc. R. S. E. 1870. 

7. Shew that 

(a) fVadpSfip = V.a ( P S/3p + V . PjVpdp). 

(b) fdp Vap = (pVap-V. afVpdp) + 8 . afVpdp. 

(c) fV . V*dp V(3p = i (*S . PfVpdp + (38 . afVpdp - P 8 . #>), 

(d) fS . Vadp V(3p = J (Szp8j3p p 2 Sz/3 8 . afifVpdp). 

When these integrals are taken round a closed plane curve we 

jVpdp = 2Ay, 

where A is the area, and 7 a unit vector perpendicular to its plane. 
In this case 

JdpVap = A Vya + 2 A Syot, 

fV.VadpVpp = A (z8j3Y + 

8. State, in words, the geometrical theorems involved in the 
equations of 140, (10), (11), (12). 

9. Shew (by means of 91) that 

where V, V t , operate respectively on a-, <r l ; but, after the 
operations are performed, we put 

v = v * = *. 



150. WE have seen that the differentiation of any function 
whatever of a quaternion, q, leads to an equation of the form 


where /is linear and homogeneous in dq. To complete the process 
of differentiation, we must have the means of solving this equation, 
so as to be able to exhibit directly the value of dq. 

This general equation is not of so much practical importance 
as the particular case in which dq is a vector; and, besides, as we 
proceed to shew, the solution of the general question may easily be 
made to depend upon that of the particular case ; so that we shall 
commence with the latter. 

The most general expression for the function / is easily seen 
to be 

dr =f(q, dq) = 2 V. adqb + S . cdq, 

where a, b, and c may be any quaternion functions of q whatever, 
including even scalar constants. Every possible term of a linear 
and homogeneous function is reducible to this form, as the reader 
may see at once by writing down all the forms he can devise. 

Taking the scalars of both sides, we have 

Sdr = S.cdq = SdqSc + S.Vdq Vc. 
But we have also, by taking the vector parts, 

Vdr = 2F. adqb = Sdq . 2Fa& + 2F. a(Vdq) b. 

Eliminating Sdq between the equations for Sdr and Vdr it is 
obvious that a linear and vector expression in Vdq will remain. 
Such an expression, so far as it contains Vdq, may always be 
reduced to the form of a sum of terms of the type aS . /3 Vdq, by 
the help of formulae like those in 90, 91. Solving this, we have 
Vdq, and Sdq is then found from the preceding equation. 


151. The problem may now be stated thus. 
Find the value of p from the equation 

aS/3p + fl^ftp + ... = 2. aSfa = 7, 
where a, ft, a lt /3 lf ...y are given vectors. 

The most general form of the left-hand member requires but 
three distinct or independent terms. These, however, in con 
sequence of the form of the expression, involve scalar constants 
only; since the whole can obviously be reduced to terms of the 
forms AiSip, BiSjp, CjSjp, &c. and there are only nine such forms. 
In fact we may write the most general form either as 

iSaip +J8j3p + kSyp, 
or as afiip + faSjp + yfikp, 

according as we arrange it by the vector, or by the scalar, factors 
of the several terms. But the form 

is that which, as committing us to no special system of vectors of 
reference, is most convenient for a discussion of its properties. 
If we write, with Hamilton, 

4,p = 2.a80p ........................... (1), 

the given equation may be written 

<t>p = 7, 
or p = ^ry, 

and the object of our investigation is to find the value of the 
inverse function 0" 1 . 

It is important to remark that the definition (1) shews < to be 
distributive, so that 

A particular case of this is 

(f) (xp} = xcj)p, 
where x is a scalar. 

Also, by the statement above, it is clear that <, in its most 
general form, essentially involves nine independent scalars. 

152. We have seen that any vector whatever may be expressed 
linearly in terms of any three non-coplanar vectors. Hence, we 
should expect d priori that a vector such as </></>$/9, or <f> 3 p, for 
instance, should be capable of expression in terms of p, (ftp, and 

[This is, of course, on the supposition that p, <f)p, and ftp are 

110 QUATERNIONS. [152. 

not generally coplanar. But it may easily be seen to extend to 
that case also. For if these vectors be generally coplanar, so are <pp, 
2 p, and cj) s p, since they may be written cr, (fxr, and <V. And thus, 
of course, <f?p can be expressed as above. If in a particular case, 
we should have, for some definite vector p, $p = gp where g is a 
scalar, we shall obviously have <f> z p = g*p and (f> 3 p = g 3 p, so that the 
equation will still subsist. And a similar explanation holds for 
the particular case when, for some definite value of p, the three 
vectors p, <pp, (/> 2 p are coplanar. For then we have an equation of 
the form 

which gives <fip = A$p + Bfip 

= AB P +(A +B 2 )<f>p, 

so that </> 3 /j is in the same plane.] 
If, then, we write 

-<pp = xp + y<j>p + z<Fp ........................ (1), 

and bear in mind the distributive character of the operator </>, it is 
evident (if only ex absurdo) that x t y, z are quantities independent 
of the vector p. 

[The words above, "it is evident," have been objected to by 
more than one correspondent. But, on full consideration, I not 
only leave them where they are, but put them in Italics. For 
they are, of course, addressed to the reader only ; and it is to be 
presumed that, before he reaches them, he has mastered the 
contents of at least the more important previous sections which 
bear on this question, such as 23, 151. If, with these sections 
in his mind, and a homogeneous linear equation such as (1) before 
him, he does not see the " evidence," he has begun the study of 
Quaternions too soon. A formal demonstration, giving the values 
of x, y, z, will however be found in 156 9 below.] 

If any three vectors, as i, j, k, be substituted for p, they will in 
general enable us to assign the values of the three coefficients on 
the right side of the equation, and the solution of the problem of 
1 5 1 is complete. For by putting $T l p for p and transposing, the 
equation becomes 


that is, the unknown inverse function is expressed in terms of 
direct operations. Should x vanish, while y remains finite, we 
must substitute $~*p for p, and have 


and if x and y both vanish 

- z$- l p = p. 

[We may remark here that it is in general possible to determine 
x,y,z by putting one known vector for p in (1). The circumstances 
in which some particular vector does not suffice will be clear from 
the theory to be given below.] 

153. To illustrate this process by a simple example we shall 
take the very important case in which $ belongs to a central 
surface of the second order ; suppose an ellipsoid ; in which case it 
will be shewn (in Chap. IX.) that we may write 

(frp = - cfiSip - tfjSjp - c*kSkp, 

where i,j, k are parallel to the principal diameters, and the semi- 
lengths of these are I/a, 1/6, 1/c. 
Here we have 

<l>k = c 2 &, tfk = cVj, <pk = ck. 

Hence, putting separately i, j, k for p in the equation (1) of 
last section, we have 

- a 6 = x + 7/a 2 + za*, 

- 6 6 = x + yV + zb\ 
c 6 = x + yc* + 2C 4 . 

Hence a 2 , 6 2 , c 2 are the roots of the cubic 

which involves the conditions 

y = + c 
x = - a 2 b*c 2 . 
Thus, with the above value of (/>, we have 

(a 2 

154. Putting </>~V in place of p (which is any vector what 
ever) and changing the order of the terms, we have the desired 
inversion of the function (j> in the form 

aWc/TV - (a 2 6 2 + 6V + cV) a - (a 2 + 6 2 + c 2 ) fa + (/>V, 
where the inverse function is expressed in terms of the direct 

112 QUATERNIONS. [ T 5 5 

function. For this particular case the solution we have given 
is complete, and satisfactory; and it has the advantage of pre 
paring the reader to expect a similar form of solution in more 
complex cases. 

155. It may also be useful as a preparation for what follows, 
if we put the equation of 153 in the form 

= cf> (p) = <p p - (a 2 + 6 2 + c 2 ) <pp + (a 2 6 2 + 6V + cV) <j>p - a Wp 
= {<#> 3 - (a 2 + 6 2 + c 2 ) c/> 2 + (a 2 6 2 + 6V + cV) - a 2 6 2 c 2 ) p 
= {((/,- a 2 ) (</>-6 2 )(0-c 2 )} P ........................... (2). 

This last transformation is permitted because (151) (f> is com 
mutative with scalars like a 2 , i.e. (f> (a?p) = cffyp. The explanation 
of its meaning must, however, be deferred to a later section. 
( 177.) 

Here we remark that the equation 

V.p<f>p = 0, or <t>p = gp, 

where </> is as in 153, and g is some undetermined scalar, is 
satisfied, not merely by every vector of null-length, but by the 
definite system of three rectangular vectors Ai, Bj, Ck whatever 
be their tensors, the corresponding particular values of g being 
a 2 , V, c\ 

156. We now give Hamilton s admirable investigation. 

The most general form of a linear and vector function of a 
vector may of course be written as 

where q and r are any constant quaternions, either or both of which 
may degrade to a scalar or a vector. 

Hence, operating by S . cr where a is any vector whatever, 

S<r<l>p = S<T?,V.qpr = Sp2V.r(rq = Sp<l> (r ............ (1), 

if we agree to write </>V = lV .rcrq, 

and remember the proposition of 88. The functions <p and < 
are thus conjugate to one another, and on this property the whole 
investigation depends. 

157. Let X, fi be any two vectors, such that 

Operating by 8 . X and S. fi we have 

= 0, Sfi^p = 0. 



But, introducing the conjugate function </> , these become 

SpfiX = 0, Spffj, = 0, 
and give p in the form mp = V$ \<f> /z, 

where ra is a scalar which, as we shall presently see, is independent 
of \, p, and p. 

But our original assumption gives 

hence we have m^" 1 FXyu, = F^ Xc//^ (2), 

and the problem of inverting <fi is solved. 

It remains to find the value of the constant m, and to express 
the vector 

as a function of FX/-t. 

158. To find the value of m, we may operate on (2) by S. <f>v, 
where v is any vector not coplanar with X and /j,, and we get 

mS . <f> v<l)~ l FX/4 = mS . zx/x/T 1 FXyu, (by (1) of 156) 
= mS . X//-Z/ = S . tfr XfiiJbfiv, or 

wi = FT^: ~~ ~ (3). 

S. \fjiV 

[That this quantity is independent of the particular vectors 
X, /*, v is evident from the fact that if 

X = x\ + yfM + zv t fji = ^X + y^ + z^v, and v = % 2 \ + y^ + zp 

be any other three vectors (which is possible since X, //,, v are not 
coplanar), we have 

from which we deduce 

8 . <f> X <f> V<V = os y z 


* y* z * 

x y z 

*i 2/i ^ 

^2 2/2 ^2 

S.\flV t 

so that the numerator and denominator of the fraction which ex 
presses m are altered in the same ratio. Each of these quantities 
is in fact an Invariant, and the numerical multiplier is the same 
for both when we pass from any one set of three vectors to another. 
T. Q. I. 8 

114 QUATERNIONS. [l59- 

A still simpler proof is obtained at once by writing X + xp for X 
in (3), and noticing that neither numerator nor denominator is 

159. We have next to express 

as a function of FX//,. For this purpose let us change </> to </> g, 
where g is any scalar. It is evident that < becomes </> g, and 
our equation (2) becomes 

= (rac/T 1 -gx + f f] V*>p, suppose. 
In this equation (see (3) above) 

_ S.($-g)\(f-g)iL(<l> - g) v 


= m-m 1 g + m 2 g 2 -g 3 ............................ (4) 

is what m becomes when is changed into <f) g ; m l and m z being 
two new scalar constants whose values are 

S . 


S . 

If, in these expressions, we put X + osfi for X, we find that the terms 
in x vanish identically ; so that they also are invariants. 

Substituting for m gt and equating the coefficients of the 
various powers of g after operating on both sides by </> g, we 
have two identities and the following two equations, 

[The first determines ^, and shews that we were justified in 
treating V (0 X//. + Xc/> ^) as a linear and vector function of V . X/A. 
The result might have been also obtained thus, 

S . 

S . v% V\JJL = S . 

= mjSKfjiv S . 
= 8,v 


and all three (the utmost generality) are satisfied by 

% = w 2 -</>.] 

160. Eliminating ^ from these equations we find 

or mc/T 1 = m, ??? 2 (/> 4- < 2 ......................... (5), 

which contains the complete solution of linear and vector equa 

161. More to satisfy the student as to the validity of the 
above investigation, about whose logic he may at first feel some 
difficulties, than to obtain easy solutions, we take a few very 
simple examples to begin with : we treat them with all desirable 
prolixity, as useful practice in quaternion analysis ; and we append 
for comparison easy solutions obtained by methods specially 
adapted to each case. The advanced student need therefore pay 
but little attention to the next ten sections. 

162. Example I. 
Let <f>p= V.ap/3 = y. 

Then <>= = 0p. 

Hence m = \ S ( V. a\j3 V. ap/3 V. ai/). 

>O . i^lLV 

Now A, p, v are any three non-coplanar vectors ; and we may 
therefore put for them a, ft, 7 , if the latter be non-coplanar. 
With this proviso 

B ^ ( a / 32 F a 7/ 3 + a2 ^ ^ F af y + a2 8 7) 

Hence we have by (5) above 

. <F l v = a 2 /3 2 Sa/3 . p = - ^^ + F. 7 /3 + F. a (F. a 
which is one form of solution. 


116 QUATERNIONS. [ 1 6 3 . 

By expanding the vectors of products we may easily reduce it 
to the form 

*2/D2CV /O rs,2/Q2 i -, /O2 Ct n , Q^CfQ 

a p ootp . p = a p y + ap bay + pa >py, 

a" 1 Say + (3~ l S/3y - y 
p= -SaW~ - 

163. To verify this solution, we have 

V.apft = ^-g (ftSay + aSfty - V. ay/3) = y, 
which is the given equation. 

164. An easier mode of arriving at the same solution, in this 
simple case, is as follows : 

Operating by 8. a and 8. ft on the given equation 

we obtain afSffp = Say, 

ff8*p = S 

and therefore aS/3p = of 1 

But the given equation may be written 

aS/3p - pSa/3 + @Sap = y. 
Substituting and transposing we get 

pSa/3 = a- 1 Say + P^SjSy - y, 
which agrees with the result of 162. 

[Note that, at first sight, one might think that the value of p 
should have a term with an arbitrary scalar factor added. But the 
notation p is limited to a vector. Had the equation been written = y 
we should have had 

aq/3 = x + y, or q = xa 1 ^ 1 + a~ l y/3~\ 

But because q is to be a vector 

Sq = Q, or xSa/3 + S.ay/3 = ) 

and, with this value of x, the expression for q takes the form given 
above for p.] 

165. If a, ft, y be coplanar, the above mode of solution is 
applicable, but the result may be deduced much more simply. 

For ( 101) 8 . a/By = 0, and the equation then gives 8 . a/3p = 0, 
so that p also is coplanar with a, ft, y. 


Hence the equation may be written 

and at once 

p = 
and this, being a vector, may be written 

This formula is equivalent to that just given, but not equal to 
it term by term. [The student will find it a good exercise to 
prove directly that, if a, ft, y are coplanar, we have 

- 7) = a^ 

The conclusion that 

.a/3p = 

in this case, is not necessarily true if 

But then the original equation becomes 

aSpp+pSap = y, 

which is consistent with 

This equation gives 

2 /3 2 <? 2 fi\ - ^ a ^ ^ a @ i Q ^^ ^ ay ^ 

Say < 

by comparison of which with the given equation we find 

Sap and S/3p. 

The value of p remains therefore with an indeterminate vector 
part, parallel to aft ; i.e. it involves one arbitrary scalar. 

166. Example II. 

Let <f)p = V . aftp = y. 

Suppose a, ft, 7 not to be coplanar, and employ them as X, yu,, v 
to calculate the coefficients in the equation for </>~ 1 . We have 

S . o"</>/9 = S . <raftp = S . p V . o~a/3 = S . p<f) cr. 
Hence ftp = V . pa ft = V . ftap. 

We have now 


s-8(*./ + *.l3afl. 

Hence by (5) of 1 GO 

= (2 (S*/3) 2 + a 2 /? 2 ) 7 - 3Sa/3F. a/3 7 + F. a/3F. a/3 7 , 

which, by expanding the vectors of products, takes easily the 
simpler form 

. p = a 2 /3 2 7 - a/ 

167. To verify this, operate by F.a/3 on both sides, and we 

or F . a/3p = 7 . 

168. To solve the same equation without employing the 
general method, we may proceed as follows : 

ry = F. {3p = pSoL/3 + F. F 

Operating by 8. Fa/3 we have 

Divide this by Sa/3, and add it to the given equation. We thus 

Hence p = r 

a form of solution somewhat simpler than that before obtained. 

To shew that they agree, however, let us multiply by a 2 y# 2 $a/3, 
and we get 

. p = 


In this form we see at once that the right-hand side is a vector, 
since its scalar is evidently zero ( 89). Hence we may write 

a 2 /3 2 a/3 .p=V. /3a 7 a/3 - Fa/3S . a/3 7 . 
But by (3) of 91, 

- jS . a/3 Fa/3 + a . /3 ( Fa/3) 7 + /3S . F(a/3) a 7 + Fa/3 . a/3 7 = 0. 
Add this to the right-hand side, and we have 

a 2 /3 2 Sa/3 . p = 7 ((Sa(3) 2 - S . a/3 Fa/3) - a (Sot{3S/3y -S.0( Fa/3) 7) 

+ /3 (SajSSoLy + S . F(a) a 7 ). 

But (Sa/3) 2 - . a/3 Fa/3 = (Sa/3) 2 - ( Fa) 2 - a 2 /3 2 , 

tfa/3>S 7 - 8 . j3 ( Fa/3) 7 - SoL(3S/3y - S/3aS/3y + /3 2 a 7 = /3 2 /Sfa 7 , 
>Sfa/3>Sfa7 + S . F (a/3) a 7 = Sa/BSoLy + /S Y 

and the substitution of these values renders our equation identical 
with that of 166. 

[If a, /3, 7 be coplanar, the simplified forms of the expression 
for p lead to the equation 

801/3 . p-^y = 7 - a 1 Sa 7 + 2/3>Sfa- 1 /3- 1 ASfa 7 - /T S/ty, 
which, as before, we leave as an exercise to the student.] 

169. Example III. The solution of the equation 

Vep = 7 

leads to the vanishing of some of the quantities m. Before, how 
ever, treating it by the general method, we shall deduce its solu 
tion from that of 

F. OL/3p = 7 

already given. Our reason for so doing is that we thus have an 
opportunity of shewing the nature of some of the cases in which 
one or more of m, m^ m 2 vanish; and also of introducing an 
example of the use of vanishing fractions in quaternions. Far 
simpler solutions will be given in the following sections. 
The solution of the last- written equation is, 166, 

a 2 /3 2 Sa/3 , p = a 2 /3 2 7 - a/3 2 a 7 - {B^Spy + 2/3a/3a 7 . 
If we now put QL/3 = e + e 

where e is a scalar, the solution of the first-written equation will 
evidently be derived from that of the second by making e gradually 
tend to zero. 

120 QUATERNIONS. [ 1 7<D. 

We have, for this purpose, the following necessary transforma 
tions : 

a/3 2 Say + /3a 2 S/3y = a/3 . ft Say + ft a . aSfty, 

= (e + e) ftSay + (e-e) aSfty, 
= e (PSay + aSfty) + eV. 7 Fa/3, 
e (ffSoiy + a fifty) + eVye. 
Hence the solution becomes 

(e 2 - e 2 ) ep = (e 2 -e*)y-e ({BSay + otSjSy) -eVye + ZeftSay, 
= ( _ 6 2) ry + eV. y Va/3 -eVye, 
= (e 2 - e 2 ) y + e Vye + ye 2 - eSye, 
= e*y + eVye - eSy. 
Dividing by e, and then putting e = 0, we have 

Q / 
Now, by the form of the given equation, we see that 

Sye = 0. 

Hence the limit is indeterminate, and we may put for it x, where 
x is any scalar. Our solution is, therefore, 

or, as it may be written, since >Sfye = 0, 

p = e 1 (y + x). 

The verification is obvious for we have 

ep = y + x. 

170. This suggests a very simple mode of solution. For we 
see that the given equation leaves Sep indeterminate. Assume, 

Sep = x 

and add to the given equation. We obtain 

ep = x + 7, 

or p = e -1 (7 + x\ 

if, and only if, p satisfies the equation 

Vep = 7. 

171. To apply the general method, we may take e, 7 and 67 
(which is a vector) for X, //,, v. 


We find 


p = Vpe. 
m = 0, 




= 0, 

That is, p = - 

= e -1 7 + a?e, as before. 

Our warrant for putting a?e, as the equivalent of c/)" 2 is this : 

The equation 0V = 

may be written V . eFecr = ae 2 eSecr. 
Hence, unless a = 0, we have a \\ e = xe. 

[Of course it is well to avoid, when possible, the use of expressions 
such as 0~ 2 &c. but the student must be prepared to meet them ; 
and it is well that he should gain confidence in using them, by 
verifying that they lead to correct results in cases where other 
modes of solution are available.] 

172. Example IV. As a final example let us take the most 
general form of (/>, which, as has been shewn in 151, may be 
expressed as follows : 

0p = aSl3p + a^p + ajSfij) = 7. 
Here $p = fiSap + ftSfl^p + 2 2 p, 

and, consequently, taking a, a v a 2 , which are in this case non- 
coplanar vectors, for X, ^, v, we have 

8 . OLOL^ 





A = 

122 QU ATEBNIONS. [172. 

= - 8 . 

Hence the value of the determinant is 
- (SaaS . Ffl^a, Va^ + Sa^S . Va 2 a Va^ + Sa^aS . VOL^ Va^) 

= -S.a( Va^S . a A) {by 92 (4)} = - (S . aatf. 
The interpretation of this result in spherical trigonometry is 
very interesting. (See Ex. (9) p. 90.) 
By it we see that 


m < = s~i^ s - [a (/3/Saa + /Si<Sa a + /8 srv ) 

s c< 2 ) + &c.] 

+ . . .) 

. a ( F/3/3.S . Faa, Fa.o 

o . aa ct 

+ S.a l (V/3(3 l S.Vaa l Va 2 a+...) 

or, taking the terms by columns instead of by rows, 

y 4 
+ .... I 

= - 8 ( Fact, F/3& + Fa x a 2 F/9 A + 

or, grouping as before, 

= ^ ^ ^f [^ ( Vaafiw, + Va^Saa, + Fa^aa) + ...], 


*i a ) +-]( 92 (4)), 

And the solution is, therefore, 

. Fa otj F/3& 

[It will be excellent practice for the student to work out in 
detail the blank portions of the above investigation, and also to 
prove directly that the value of p we have just found satisfies the 
given equation.] 

173. But it is not necessary to go through such a long process 
to get the solution though it will be advantageous to the student 
to read it carefully for if we operate on the proposed equation by 
8 . a^g, 8 . 2 a, and 8 . aa t we get 

S . o^aa^/^/o = 8 . 2 7, 

8 . aLOL^SPyp = S . OLOL t y. 

From these, by 92 (4), we have at once 
pS . aot^S . /3/3 A = Vft^S . l7 + F/S^S . a^y 

The student will find it a useful exercise to prove that this is 
equivalent to the solution in 172. 

To verify the present solution we have 
(*SP P + afifo + o^S&p) S . awfi . 0J3& 

174. It is evident, from these examples, that for special cases 
we can usually find modes of solution of the linear and vector 
equation which are simpler in application than the general process 
of 160. The real value of that process however consists partly in 
its enabling us to express inverse functions of </>, such as(< g)~ l 
for instance, in terms of direct operations, a property which will be 
of great use to us later ; partly in its leading us to the fundamental 

(f> 3 w 2 2 + m^ m = 0, 

which is an immediate deduction from the equation of 160, and 
whose interpretation is of the utmost importance with reference to 
the axes of surfaces of the second order, principal axes of inertia, 

124 QUATERNIONS. [175. 

the analysis of strains in a distorted solid, and various similar 

We see, of course, that the existence of the cubic renders it a 
mere question of ordinary algebra to express any rational function 
whatever of (f> in a rational three-term form such as 

In fact it will be seen to follow, from the results of 177 below, 
that we have in general three independent scalar equations of the 

which determine the values of A, B, C without ambiguity. The 
result appears in a form closely resembling that known as 
Lagrange s interpolation formula. 

175. When the function <f> is its own conjugate, that is, when 

Sp<f)(T = 8(7(j)p 

for all values of p and cr, the vectors for which 

form in general a real and definite rectangular system. This, of 
course, may in particular cases degrade into one definite vector, and 
any pair of others perpendicular to it ; and cases may occur in 
which the equation is satisfied for every vector. 
To prove this, suppose the roots of 

( 159 (4)) to be real and different, then 

<M = #iPij 
0/> 2 = 9*P* \ 


where p v p 2 , p 3 are three definite vectors determined by the 
constants involved in <j). 

Hence, operating on the first by Sp 2 , and on the second by Sp v 
we have 

The first members of these equations are equal, because $ is its 
own conjugate. 

Thus (fc-&)%p, = <>; 

which, as g l and g t2 are by hypothesis different, requires 

<>> =o. 


Similarly Sp,p 3 = 0, Sp^ = 0. 

If two roots be equal, as g v g 3 , we still have, by the above proof, 
Sp^ = and Sp^ = 0. But there is nothing farther to determine 
ps and p 9 , which are therefore any vectors perpendicular to p lf 

If all three roots be equal, every real vector satisfies the equation 
(</>-<7)p = 0. 

176. Next as to the reality of the three directions in this 

Suppose g + h \l 1 to be a root, and let p + cr V 1 be the 
corresponding value of p, where g and h are real numbers, p and cr 
real vectors, and \/" 1 the old imaginary of algebra. 

Then <f>(p + (T V~l) = (g + h\/ - 1) (p + a V^l), 
and this divides itself, as in algebra, into the two equations 

(f)p = gp- ho; 
c/>cr = hp + go: 

Operating on these by 8 . cr, S.p respectively, and taking the 
difference of the results, remembering our condition as to the 
nature of c/> 

we have h (<r 2 + p 2 ) = 0. 

But, as cr and p are both real vectors, the sum of their squares 
cannot vanish, unless their tensors separately vanish. Hence h 
vanishes, and with it the impossible part of the root. 

The function c/> need not be self-conjugate, in order that the 
roots of 

ra, = 

may be all real. For we may take g lt # 2 , g 3 any real scalars, and 
a, /3 } 7 any three real, non-coplanar, vectors. Then if be such that 

S.a@y.<l>p= g^S . /3yp + g^S . yctp + g^S . afip, 
we have obviously 

((/> -gj * = 0, (< -&) /3 = 0, (</> - ffa ) 7 = 0. 

Here c/> is self-conjugate only if a, /3, 7 form a rectangular 

177. Thus though we have shewn that the equation 

g 3 m^g* 4- m^y m = 
has three real roots, in general different from one another, when c/> 

126 QUATERNIONS. [178. 

is self-conjugate, the converse is by no means true. This must be 
most carefully kept in mind. 

In all cases the cubic in may be written 

(*-sO(*-fc)(*-0 i ) = o (i), 

and in this form we can easily see its meaning, provided the values 
of g are real. For there are in every such case three real (and in 
general non-coplanar) vectors, p v p 2 , p s for which respectively 

(*-^)P! = O, -&)ft = 0, W>-<7 8 )p 8 = 0. 
Then, since any vector p may be expressed by the equation 

pS PiPiPs = Pi S PiPaP + P* S PsPiP +P* S - PiPzP ( 91 )> 

we see that when the complex operation, denoted by the left-hand 
member of the symbolic equation, (1), is performed on p, the 
first of the three factors makes the term in p l vanish, the second 
and third those in p^ and p s respectively. In other words, by the 
successive performance, upon a vector, of the operations <f>g v <j> g z , 
$ g a , it is deprived successively of its resolved parts in the direc 
tions of p v p v p 3 respectively ; and is thus necessarily reduced to 
zero, since p t , p 2 , p a are (because we have supposed g v g z , g 3 to be 
distinct) distinct and non-coplanar vectors. 

178. If we take p v p 2 , p 3 as rectangular unit- vectors, we have 

~P= Pl$PlP + 

whence ^ = - g lPl S Pl p - 

or, still more simply, putting i,j, k for p v p 2 , p 3 , we find that any 

self-conjugate function may be thus expressed 

fo^-ffiiSip-gJ&ip-gJcSkp (2), 

provided, of course, i, j, k be taken as the roots of the equation 

Vpj>p = 0. 

A rectangular unit-vector system requires three scalar quan 
tities, only, for its full specification. g v </ 2 , g 3 are other three. 
Thus any self-conjugate function involves only six independent 

179. A very important transformation of the self-conjugate 
linear and vector function is easily derived from this form. 

We have seen that it involves, besides those of the system i,j, k, 
three scalar constants only, viz. g v g^, g y Let us enquire, then, 
whether it can be reduced to the following form 

<l>p=fp + hV.(i + ek) P (i-ek) (3), 


which also involves but three scalar constants/, h, e, in addition to 
those of i, j y k, the roots of 

= 0. 

Substituting for p the equivalent 

p = iSip - jSjp kSkp, 

expanding, and equating coefficients of i,j, k in the two expressions 
(2) and (3) for <f>p, we find 

These give at once 

Hence, as we suppose the transformation to be real, and therefore 
e* to be positive, it is evident that g l <7 2 and # 2 g. A have the same 
sign ; so that we must choose as auxiliary vectors in the last term 
of <j)p those two of the rectangular directions i, j, k for which the 
coefficients g have respectively the greatest and least values. 
We have then 

and / 

180. We may, therefore, always determine definitely the 
vectors X, //,, and the scalar/, in the equation 

= + V. 

when (j) is self-conjugate, and the corresponding cubic has not equal 
roots ; subject to the single restriction that 


is known, but not the separate tensors of X and //,. This result is 
important in the theory of surfaces of the second order, and in that 
of Fresnel s Wave-Surface, and will be considered in Chapters IX. 
and XII. 

181. Another important transformation of <f> when self- 
conjugate is the following, 

<f)p = aaVap + b/SSffp, 
where a and b are scalars, and a and /3 unit-vectors. This, of 

128 QUATERNIONS. [182. 

course, involves six scalar constants, and belongs to the most 
general form 

<I>P=- 9lPl S PlP ~ 9*P$P*P - 93P* S P3P> 

where p l} p 2 , p 3 are the rectangular unit-vectors for which p and (f>p 
are parallel. We merely mention this form in passing, as it 
belongs to the focal transformation of the equation of surfaces of 
the second order, which will not be farther alluded to in this work. 
It will be a good exercise for the student to determine a, /3, a and 
6, in terms of g v g 2) g 3y and p v p 2 , p 3 . 

182. We cannot afford space for a detailed account of the 
singular properties of these vector functions, and will therefore 
content ourselves with the enuntiation and proof of one or two of 
the more important. 

In the equation m<f^V\i^= F^ X^V ( 157), 
substitute \ for <fi\ and /it for (jb /it, and we have 

Change </> to (/> g, and therefore (/> to (/> g, and m to m g , we have 

, V (* - g) ^ (f - 0)~V = (* - 9) V\n ; 
a formula which will be found to be of considerable use. 

183. Again, by 159, 

Similarly S.pft- h)~ l p = Sp^p - S PXP + hp 2 . 

wi. rv / , x^, m,, ci /i T \-i 7 \ f 9 wiSpd) p] 

-jt8.p(4>-grpj*8.p(4>-hrp=te- h) Y - -^-^| 

That is, the functions 

^S.ptf-grp, and ^S.pQ-hrp 

are identical, i.e. when equated to constants represent the same series 
of surfaces, not merely when 


but also, whatever be g and h, if they be scalar functions of p which 
satisfy the equation 

mS . p<f>~ l p = 


This is a generalization, due to Hamilton, of a singular result ob 
tained by the author*. 

184. It is easy to extend these results ; but, for the benefit of 
beginners, we may somewhat simplify them. Let us confine our 
attention to cones, with equations such as 

- (?)-> = 0,1 

These are equivalent to mSp^p gSpXP+9*P* = > 

mSp^~ l p - hSpxp + h p* = 0. 

m (1 - x) Sp^p -(g- hx) Spxp + (g* - Wx] p* = 0, 
whatever scalar be represented by x. 

That is, the two equations (1) represent the same surface if this 
identity be satisfied. As particular cases let 

(1) x = \, in which case 

(2) g hx=0, in which case 

or mSp~ l ^~ l p -gh=0. 

(3) ^ = ^, giving 

or -m(h + g) Sp^p + ghSpxp = 0. 

185. In various investigations we meet with the quaternion 
= < + 0/8 + 707 ..................... (1), 

where a, 0, 7 are three unit-vectors at right angles to each other. 
It admits of being put in a very simple form, which is occasionally 
of considerable importance. 

We have, obviously, by the properties of a rectangular unit- 

q = /#70a + 70/3 + a/307. 
As we have also 

S./8 7 = -l (71 (13)), 

* Note on the Cartesian equation of the Wave -Surface. Quarterly Math. Journal, 
Oct. 1859. 

T. Q. I. 9 

130 QUATERNIONS. [l86. 

a glance at the formulae of 159 shews that 

Sq = -m a , 

at least if <j> be self-conjugate. Even if it be not, still (as will be 
shewn in 186) the term by which it differs from a self-conjugate 
function is of such a form that it disappears in Sq. 
We have also, by 90 (2), 
Vq = OL 

. yea + 7$ . ae/3 ( 186) 
= - (aae + /3S/36 + jSye) = e. 

[We may note in passing that the quaternion (1) admits of 
being expressed in the remarkable forms 


, __ d ~ d d 

where ( 145) V = a-,- + /5 -y- + 7 ^- , 

cfo? cZy ^ 

and p = a^ + ^7 + 72. 

We will recur to this towards the end of the work.] 

Many similar singular properties of <j> in connection with a rect 
angular system might easily be given ; for instance, 

) = mV. 

which the reader may easily verify by a process similar to that just 
given, or (more directly) by the help of 157 (2). A few others 
will be found among the Examples appended to this Chapter. 

186. To conclude, we may remark that, as in many of the 
immediately preceding investigations we have supposed <f> to be 
self-conjugate, a very simple step enables us to pass from this to 
the non-conjugate form. 

For, if < be conjugate to </>, we have 

Spficr = 
and also Spfya = 

Adding, we have 

so that the function (< + < ) is self-conjugate. 

Again, Spfp = Spflp, 

which gives Sp ($ <}> ) p = 0. 

Hence (< $ ) p= Yep, 


where, if < be not self-conjugate, e is some real vector, and 

Thus every non-conjugate linear and vector function differs from 
a conjugate function solely ly a term of the form 


The geometric signification of this will be found in the Chapter on 

The vector e involves, of course, three scalar constants. Hence 
( 151, 178) the linear and vector function involves, in general, nine. 

187. Before leaving this part of the subject, it may be well 
to say a word or two as to the conditions for three real vector 
solutions of the equation 

Vp$p = 0. 

This question is very fully treated in Hamilton s Elements, and 
also by Plarr in the Trans. K S. E. For variety we adopt a semi- 
graphic method*, based on the result of last section. By that 
result we see that the equation to be solved may be written as 
<f)p = &p + Vep = xp ........................ (1) 

where w is a given self-conjugate function, e a given vector, and x 
an unknown scalar. 

Let a v 2 , a 3 and g v g 2 , g 3 (the latter taken in descending order 
of magnitude), be the vector and scalar constants of OT, so that 
( 177) 

(-&)*! = (), &C. 

We have obviously, by operating on (1) with S . o^ &c., three 
equations of the form 

fy{( S P 1 -flOa 1 -Fe 1 )=0 ..................... (2). 

Eliminating p (whose tensor is not involved) we have 
S K0> -*K- V**J {(g 2 - x) , - FeJ {(?. - x) a - F6 3 } = 0, 
or (x-g l )(x-g 2 )(x-g 3 )-xe* + Sevre = ............ (3). 

From each value of x found from this equation the corre 
sponding value of p is given by (2) in the form 

P H V {(9, - ) , - FeoJ {(g, - x) 2 - Fea 2 }, 

II (& - ) (0, - *0 8 + (02 - ) ^ V - (01 ~ ) a 2^i e ~ ^^S 6 

II (0i ~ ) (02 - ) 8 + ^ a 3 (w - 0) e 

* Proc. ^. S. E., 187980. 


132 QUATERNIONS. [l88. 

The simplest method of dealing with (3) seems to be to find the 
limiting value of Te, Ue being given, that the roots may be all real. 
They are obviously real when Te = 0. It is clear from the pro 
perties of -or that the extreme values of S . UetzUe (which will be 
called f ) are g l and g y 

Trace the curve 

and draw the (unique) tangent to it from the point x = J; , y = 0, 
f having any assigned value from g 3 to g^. Let this tangent make 
an angle with the axis of x. Suppose a simple shear to be 
applied to the figure so as to make this tangent turn round the 
point f, 0, and become the x axis, while the y axis is unchanged. 
The value of y will be increased by (x f ) tan 0. Comparing this 
with (3) we see that tan 6 is the desired limiting value of (Tef. 

188. We have shewn, at some length, how a linear and vector 
equation containing an unknown vector is to be solved in the most 
general case ; and this, by 150, shews how to find an unknown 
quaternion from any sufficiently general linear equation containing 
it. That such an equation may be sufficiently general it must 
have both scalar and vector parts : the first gives one, and the 
second three, scalar equations ; and these are required to determine 
completely the four scalar elements of the unknown quaternion. 

Thus Tq = a 

being but one scalar equation, gives 

q a Ur, 
where r is any quaternion whatever. 

Similarly Sq = a 

gives q = a + 0, 

where is any vector whatever. In each of these cases, only one 
scalar condition being given, the solution contains three scalar in- 
determinates. A similar remark applies to the following : 

TVq = a 

gives q = x + ad ; 

and SUq = cos A, 

gives q = x0 2A/7r , 

in each of which x is any scalar, and 6 any unit vector. 


189. Again, the reader may easily prove that 

V.aVq = /3, 
where a is a given vector, gives, by putting Sq = x, 

Vaq = @ + xa. 

Hence, assuming Saq = y, 

we have aq = y + XOL + ft, 

or q = x + yen 1 + a" 1 /?. 

Here, the given equation being equivalent to two scalar con 
ditions, the solution contains two scalar indeterminates. 

190. Next take the equation 

Vq = /3. 
Operating by S. a" 1 , we get 

Sq = Sa 1 ^, 
so that the given equation becomes 


From this, by 170, we see that 

Vq = QL- l (x+<xVar*p), 
whence q = Sa^ft + a" 1 (x + aFa 1 

and, the given equation being equivalent to three scalar conditions, 
but one undetermined scalar remains in the value of q. 

This solution might have been obtained at once, since our 
equation gives merely the vector of the quaternion aq, and leaves 
its scalar undetermined. 

Hence, taking x for the scalar, we have 
aq = Saq + Vaq 

191. Finally, of course, from 

= & 

which is equivalent to four scalar equations, we obtain a definite 
value of the unknown quaternion in the form 

2 = - . 

192. Before taking leave of linear equations, we may mention 

134 QUATERNIONS. [ I 93- 

that Hamilton has shewn how to solve any linear equation con 
taining an unknown quaternion, by a process analogous to that 
which he employed to determine an unknown vector from a linear 
and vector equation ; and to which a large part of this Chapter has 
been devoted. Besides the increased complexity, the peculiar fea 
ture disclosed by this beautiful discovery is that the symbolic 
equation for a linear quaternion function, corresponding to the cubic 
in (/> of 174, is a biquadratic, so that the inverse function is given 
in terms of the first, second, and third powers of the direct function. 
In an elementary work like the present the discussion of such a 
question would be out of place : although it is not very difficult to 
derive the more general result by an application of processes 
already explained. But it forms a curious example of the well- 
known fact that a biquadratic equation depends for its solution 
upon a cubic. The reader is therefore referred to the Elements of 
Quaternions, p. 491. 

193. As an example of the solution of the linear equation in 
quaternions, let us take the problem of finding the differential of 
the n th root of a quaternion. This comes to finding dq in terms of 
dr when 

q = r. 

[Here n may obviously be treated as an integer ; for, if it were 
fractional, both sides could be raised to the power expressed by the 
denominator of the fraction.] 

This gives 

q n ~ l dq + q n - 2 dq.q+...+ dq.q n - 1 = </>(dq) = dr (1), 

and from this equation dq is to be found ; < being now a linear 
and quaternion function. 

Multiply by q, and then into q, and subtract. We obtain 

q n dq dq. q n = qdr dr.q, 

or 2V.Vq n Vdq=2V. VqVdr (2). 

But, from the equation 

q = Sq+ Vq, 
we have at once Vr = Vq n = Q n Vq, 

where = * 

[The value of Q n is obvious from 116, but we keep the 
present form.] 


With this (2) becomes 

Q n V.VqVdq= V.VqVdr, 
whence Q n Vdq = Vdr + x Vq, 

x being an undetermined scalar. 

Adding another such scalar, so as to introduce Sdq and Sdr, we 

Q n dq = (y + xVq) + dr ..................... (3). 

Substitute in (1) and we have 

Q n dr = nf- 1 (y + xVq} + $ (dr), 
or, by (3) again, 

Q n dr = nq n - 1 (Q n dq - dr) + j> (dr) ; 
so that, finally, 

.............. (4). 

Thus dq is completely determined. 

It is interesting to form, in this case, an equation for <. This 
is easily done by eliminating dr from (4) by the help of (1). We 
thus obtain 

This might have been foreseen from the nature of $, as defined in 
(1) ; because it is clear that its effect on a scalar, or on a vector 
parallel to the axis of q (which is commutative with q), is the same 
as multiplication by n(f~ l ; while for any vector in the plane of q 
it is equivalent to the scalar factor Q n . 

It is left to the student to solve the equation (1) by putting it 
in the form 

dr = p + qpq 1 + (fpq~* + . . . + q n ~ l pq~ n+l 
or dr - qdrq~ l =p- q n pq~ n , 

where p = dqq n ~ l . 

The nature of the operator q ( ) q~ l was considered in 119 

194. The question just treated involves the solution of a 
particular case only of the following equation : 

<!>(q)=2aqa = b ........................ (1), 

where a, a , &c. are coplanar quaternions. 

Let = r + , 


where r is a quaternion coplanar with the as, and p a vector in 
their plane. Then, for any a, 

ra = ar, 

while pa = Ka . p. 

Thus the given equation takes the form 

(/> (q) = 2 (aa ) . r + 2 (a#a) . p, 
so that the functional equation becomes in its turn 
{(/> - 2 (aa )} (0-2 (aJTa )J = 0. 

If a be the unit-vector perpendicular to the plane of the as, we 

-b = 

and the required solution is obviously 

q = (2 aa ) 1 (Sb - aSab) - (2 a^Ta )" 1 a Fa Vb. 

In the case ( 193) of the differential of the n ih root of a quaternion, 
s, we have 

2(aa )=?wr 1 , 

2 (aKaf) = 2(8. s n ~ l + T*sS. s n ~ 3 +...) 

The last expression (in which, it must be noticed, the last term is 
not to be doubled when n is odd) is the Q n of the former solution, 
though the form in which it is expressed is different. It will 
be a good exercise for the student to prove directly that they are 

195. The solution of the following frequently-occurring par 
ticular form of linear quaternion equation 

aq+qb = c, 

where a, b, and c are any given quaternions, has been effected by 
Hamilton by an ingenious process, which was applied in 140 (5) 
above to a simple case. 

Multiply the whole by Ka, and (separately) into b, and we 

T 2 a . q -f Ka . qb = Ka . c, 

and a . qb + qb* = cb. 

Adding, we have 

q (Ta + 6 2 -f 2Sa . b) = Ka . c + cb, 
from which q is at once found. 


To this form any equation such as 

a qb + c qd = e 
can of course be reduced, by multiplication by c " 1 and into b ~ l . 

196. To shew some of the characteristic peculiarities in the 
solution of quaternion equations even of the first degree when they 
are not sufficiently general, let us take the very simple one 

aq qb, 
and give every step of the solution, as practice in transformations. 

Apply Hamilton s process ( 195), and we get 

qtf = aqb. 

These give q (Ta + 6 2 - 2bSa) = 0, 

so that the equation gives no real finite value for q unless 

or = a 

where j3 is some unit-vector. This gives Sa = Sb. 
By a similar process we may evidently shew that 

a. being another unit-vector. 
But, by the given equation, 

Ta = Tb, 
or S*a + T 2 Va = S*b + T 2 Vb; 

from which, and the above values of a and b, we see that we may 

Sa Sb 

Thus we may write 

a = a + a, 6 = 
where a and /3 are unit-vectors. 

If, then, we separate q into its scalar and vector parts, thus 

q = u + p, 
the given equation becomes 

(a + a)( + p) = (w + /C ))(a + /9) ............... (1). 

Multiplying out we have 

u (a-0)=pj3- a/o, 

which gives 8 (a - /3) p = 0, 

and therefore p = Vy (a - {3), 

where 7 is an undetermined vector. 


We have now 
u (a. ft) = p/3 ap 

Having thus determined u, we have 

- 7 ( + 0) + 7 ( - / 
= - 27 - 27/3. 

Here, of course, we may change the sign of 7, and write the solu 
tion of 

aq = qb 

in the form q = ay + 7/3, 

where 7 is any vector, and 

a=UVa, /3=UVb. 
To verify this solution, we see by (1) that we require only to 

shew that 

aq = q@. 

But their common value is evidently 

- 7 + ay/3. 

An apparent increase of generality of this solution may be 
obtained by writing 

q = ar + r/3 

where r is any quaternion. But this is easily seen to be equiva 
lent to adding to 7 (which is any vector] a term of the form xVa/3. 
It will be excellent practice for the student to represent the 
terms of this equation by versor-arcs, as in 54, and to deduce the 
above solution from the diagram annexed : 

The vector of the intersection of the plane of q, with that of aq 


and qb, is evidently symmetrically situated with regard to the 
great circles of a and b. Hence it is parallel to 

(a-f /3) Fa/3, i.e. to a-/3. 
Let 7 be any vector in the plane of a. 
Then #x 7 (-)> 

oc cty+ry/3, 

because 8a.y = 0, and thus ya. = ay. 

Another simple form of solution consists in writing the equa 
tion as 

a = qbq~\ 

and applying the results of 119. 

197. No general quaternion method of solving equations of 
the second or higher degrees has yet been found ; in fact, as will be 
shewn immediately, even those of the second degree involve (in 
their most general form) algebraic equations of the sixteenth degree. 
Hence, in the few remaining sections of this Chapter we shall con 
fine ourselves to one or two of the simpler forms for the treatment 
of which a definite process has been devised. But first, let us 
consider how many roots an equation of the second degree in an 
unknown quaternion must generally have. 

If we substitute for the quaternion the expression 

w + ix + jy + kz (80), 

and treat the quaternion constants in the same way, we shall have 
on development ( 80) four equations, generally of the second 
degree, to determine w, x, y, z. The number of roots will therefore 
be 2 4 or 16. And similar reasoning shews us that a quaternion 
equation of the mth degree has m 4 roots. It is easy to see, how 
ever, from some of the simple examples given above ( 188 190, 
&c.) that, unless the given equation is equivalent to four inde 
pendent scalar equations, the roots will contain one or more 
indeterminate quantities. 

198. Hamilton has effected in a simple way the solution of 
the quadratic 

q 2 = qa + b, 

or the following, which is virtually the same (as we see by taking 
the conjugate of each side), 


140 QUATERNIONS. [l99- 

He puts q = J (a + w + p), 

where w is a scalar, and p a vector. 

Substituting this value in the first written form of the equation, 
we get 

a 2 + (w + pf + 2iua + ap + pa = 2 (a 2 + wa + /oa) + 46, 
or (w + /o) 2 + ap - pa = a 2 + 46. 

If we put Fa = a, $ (a 2 + 46) = c, F(a 2 + 46) = 2y, this becomes 

which, by equating separately the scalar and vector parts, may be 
broken up into the two equations 

W 1 + /> = (!, 

V(iu+ a.) p =7. 

The latter of these can be solved for p by the process of 168 ; or 
more simply by operating at once by $ . a, which gives the value 
of 8 (w + a) p. If we substitute the resulting value of p in the 
former we obtain, as the reader may easily prove, the equation 

(w* - a 2 ) (w 4 - cw* + 7 2 ) - ( Fa 7 ) 2 = 0. 

The solution of this scalar cubic gives six values of w, for each of 
which we find a value of p, and thence a value of q. 

Hamilton shews (Lectures, p. 633) that only two of these values 
are real quaternions, the remaining four being biquaternions, and 
the other ten roots of the given equation being infinite. 

Hamilton farther remarks that the above process leads, as the 
reader may easily see, to the solution of the two simultaneous 


q + r = a, 

qr = -b; 

and he connects it also with the evaluation of certain continued 
fractions with quaternion constituents. (See the Miscellaneous 
Examples at the end of this volume.) 

199. The equation <f = aq + qb, 

though apparently of the second degree, is easily reduced to the 
first degree by multiplying by, and into, q~ l , when it becomes 

1 = q~ l a + bq~ l , 

and may be treated by the process of 195. 
The equation 


where a and /3 are given vectors, is easily seen to require for a 
real (i.e. a non biquaternion) solution that q shall be a vector. 
Hence we may write it as 

whence, at once, 

a + Vj3p = xp. 

Assume S{3p = y, 

and we have 

- (y - a ) = - ) p, 

or -( a} + l3)(y-a)=( a ?-F)p. 

The condition that p is a vector gives 

xy - Sa/3 = 0, 
so that the value of p, containing one scalar indeterminate, is 

To determine p completely we require one additional scalar 

If we have, for instance, 

Syp = e, 

x is given by the cubic equation 

- 8. *. 


But if the condition be that p is a vector-radius of the unit 
sphere (a result which will be required below) we have the 


This gives two real values of # 2 , but they have opposite signs ; 
so that there are always two, and only two, real values of x. 

200. The equation q m = aqb, 
where a and b are given quaternions, gives 

and, by 54, it is evident that the planes of q and aqb must coin 
cide. A little consideration (after the manner of the latter part 
of 196) will shew that the solution depends upon drawing two 
arcs which shall intercept given arcs upon each of two great 
circles; while one of them bisects the other, and is divided by 
it in the proportion of m : 1. The equation treated in 196 is 
the special case of this when m = l. 



1. Solve the following equations : 

(6) apPp = pap/3. 

(d) S . afip + /3Sap - a Vftp = 7. 

(e) p + ap(3 = a/3. 

Do any of these impose any restriction on the generality of a and /3? 
2. Suppose p = ix +jy + kz, 

and ~ ( t ) P = aiBip + bj&jp + ckSkp ; 

put into Cartesian coordinates the following equations : 
(a) T<t>p=l. 

(c) S . p ((/> 

(d) Tp = I 

3. If X, /A, v be aw/ three non-coplanar vectors, and 

q = Fw-z^ . <i>X -f- FfX . (fa/A -f- K Xyu< . cpz , 

shew that g is necessarily divisible by S.\pv. 
Also shew that the quotient is 

m 2 -2e, 

where Fe/o is the non-commutative part of </>. 

Hamilton, Elements, p. 442. 

4. Solve the simultaneous equations : 

= 0,1 





5. If fo = 1,13 Sap + Vrp, 
where r is a given quaternion, shew that 

m = 2 (5 . W .S . 0J3J3J + 2S (r Va^ . V/B J3J + SrZS . */3r 

- 2 (SarSjSr) + SrTr\ 

and m0-V = 2 (Fa^flf.fl&cr) + 27. *V(V/3(r.r) + V<rrSr- VrSar. 

Lectures, p. 561. 

6. If [pq] denote pq qp, 

(pqr) S.p[qr], 

[pqr] (pqr) + [rq] Sp + [pr] Sq + [qp] Sr, 

and (pqrs) S.p [qrs] ; 

shew that the following relations exist among any five quaternions 

0=p (qrst) + q (rstp) + r (stpq) + s (tpqr) + t (pqrs), 
and # (prst) = [rst] Spq - [stp] Srq + [tpr] Ssq [prs] Stq. 

Elements, p. 492. 

7. Shew that if <f>, ty be any linear and vector functions, and 
a, /3, 7 rectangular unit-vectors, the vector 

6=V (</>^a + 0/3^/3 + 7 f 7) 

is an invariant. [This will be immediately seen if we write it in 
the form 6 = V. </>V^p, 

which is independent of the directions of a, /3, 7. But it is good 
practice to dispense with V, when possible.] 

If *p = 

and typ = 

shew that this invariant may be expressed as 
-SFi^K or SF^. 
Shew also that ^^V ^^ p = Vdp. 

The scalar of the same quaternion is also an invariant, and may be 
written as 

8. Shew that if $p = aSap + /380p + 
where a, /3, 7 are any three vectors, then 

where x = V/3y, &c. 


9. Shew that any self-conjugate linear and vector function may 
in general be expressed in terms of two given ones, the expression 
involving terms of the second order. 

Shew also that we may write 

+ z = a (w + xY + b (a + x) (to + y) + c (to + y)\ 
where a, b, c, x, y, z are scalars, and ta and o> the two given func 
tions. What character of generality is necessary in tzr and o> ? How 
is the solution affected by non-self-conjugation in one or both ? 

10. Solve the equations : 

(a) q* = 5qi+IOj. 
(6) g 2 = 2g + t. " 

(c) qaq = bq + c. 

(d) aq = qr = rb. 

11. Shew that 

12. If (/> be self-conjugate, and a, /3, 7 a rectangular system, 

13. 0^ and ^< give the same values of the invariants m t 
Wj, m 2 . 

14. If ft be conjugate to </>, (j><f) is self-conjugate. 

15. Shew that ( Fa<9) 2 + ( F/3<9) 2 + ( Fy0) 2 = 26> 2 
if a, /3, 7 be rectangular unit-vectors. 

1 6. Prove that V 2 ($-g)p = - pV 2 g + 2V g. 

17. Solve the equations : 

(a) $* = &; 

(^ * + *:: 

where one, or two, unknown linear and vector functions are given 
in terms of known ones. (Tait, Proc. R. 8. E. 1870-71.) 

18. If </> be a self-conjugate linear and vector function, f and TJ 
two vectors, the two following equations are consequences one of 
the other, viz. : 


From either of them we obtain the equation 

This, taken along with one of the others, gives a singular theorem 
when translated into ordinary algebra. What property does it give 
of the surface 

= I? [Ibid.] 

19. Solve the equation 

qaq = /3q@. 

Shew that it has a vector solution, involving the trisection of an 
angle : and find the condition that it shall admit of a real 
quaternion solution. 

20. Solve 

bqaq = qbqa, 

and state the corresponding geometrical problem ; shewing that 
when a and b are equal vectors, q is equal to each. 

21. Given (/>, a self-conjugate linear and vector function, and 
a vector e ; find the cubic in ty, where 

typ = <f>p + Vep. 

22. Investigate the simplest expressions for any linear and 
vector function in terms of given ones: and point out what 
degree of generality is necessary in the latter. 

Why cannot the conjugate of a linear and vector function be 
generally expressed in powers of the function itself ? 

T. Q. I. 10 



(a) Expression, Addition, and Multiplication. 

BY what precedes we are led to an analytical theory of the 
Quaternion q = w + ix + jy + kz, where the imaginary symbols 
i, jy k are such that 

i 2 = 1, f = 1, k 2 = 1, jk kj = i, ki = - ik =j, ij ji = k. 
The Tensor Tq is = Jw* + x* + f + z 2 ] and 

the Versor Uq is = . = (w -\-ix+jy + kz), 

x/w 2 + a? + 

which, or the quaternion itself when Tq = 1, may be expressed in 
the form 

cos 8 -f sin 8 (ia +jb + kc) where a 2 + 6 2 + c 2 = 1 ; 

such a quaternion is a Unit Quaternion. The squared tensor 
w 2 + 3? -f y* + 2 2 is called the Norm. 

The scalar part Sq is = w, and the vector part Fg, or say a 

Vector, \$ = ix-\-jy + kz. The Length is = ^sc* + 2/ 2 + z 2 , and the 

quotient ._. g ___ ^=^ (^ + J2/ + ^z), or say a vector ix + jy + kz 
v x \ y \ z 

where a? + if + z 2 = 1, is a Unit Vector. 

The quaternions w + ix +jy + &z and w ix jy kz are said 
to be Conjugates, each of the other. Conjugate quaternions have 
the same norm ; and the product of the conjugate quaternions is 
the norm of either of them. The conjugate of a quaternion is 
denoted by q, or Kq. 


Quaternions q = w + ix+jy + kz, q w 4- ix 4- jy 4- kz are 
added by the formula 

q + q = w + w + i(x + x) 4- j (y + y ) + k(z + z\ 
the operation being commutative and associative. 
They are multiplied by the formula 

qq = ww xx yy zz 
+ i (iox 4- xw 4- yz zy f ) 
+j (wy + yw 4- zx z x) 
+ k (wz f 4- zw 4- xy x y\ 
where observe that the norm is 

= O 2 4- x* 4- 2/ 2 4- s 8 ) (uP + x* 4- 7/ /2 4- **) 
the product of the norms of q and q. 

The multiplication is not commutative, q q =}= qq ; but it is 
associative, qq q" = q q q" = ##V > ^ Ct ^ n combination with 
addition it is distributive, q (q f 4- q } qq 4- qq , &c. 

(6) Imaginary Quaternions. Nullitats. 

The components w, x, y, z of a quaternion are usually real, 
but they may be imaginary of the form a + b J 1, where J 1 is 
the imaginary of ordinary algebra: we cannot (as in ordinary 
algebra) represent this by the letter i, but when occasion requires 
another letter, say 0, may be adopted (the meaning, = J 1, being 
explained). An imaginary quaternion is thus a quaternion of the 
form (w 4- 0-w^ +i(x + Ooc^) +j(y + Oy^) 4- k (z 4- 6z^, or, what is the 
same thing, if q, q^ be the real quaternions w+ix+jy+kz t 
w l + ix l +jy l + kz l , it is a quaternion q-\-Oq l \ this algebraical 
imaginary 6 = J 1 is commutative with each of the symbols 
i, j, k : or, what comes to the same thing, it is not in general 
necessary to explicitly introduce 6 at all, but we work with the 
quaternion w 4- ix +jy 4- kz, in exactly the same way as if w, x, y, z 
were real values. A quaternion of the above form, q + Oq^, was 
termed by Hamilton a " biquaternion " but it seems preferable to 
speak of it simply as a quaternion, using the term biquaternion 
only for a like expression q 4- Oq l , wherein 6 is not the J 1 of 
ordinary algebra. 

It may be noticed that, for an imaginary quaternion, the squared 
tensor or norm w* 4- x* 4- y 2 + z 1 may be = ; when this is so, the 
quaternion .is said to be a " Nullitat " ; the case is one to be 
separately considered. 




(c) Quaternion as a Matrix. 

Quaternions have an intimate connection with Matrices. 
Suppose that 6, = J 1, is the J 1 of ordinary algebra, and in 
place of i, j, k consider the new imaginaries x, y, z, w which are 
such that 

x = i ( 1 6i\ or conversely 1 = x + w, 

2 = I (- j - 0fy> j = (y- z \ 

so that a, b, c, d being scalars, ax + by +cz + dw denotes the 
imaginary quaternion 

We obtain for x, y, z, w the laws of combination 
x y z w 


that is x* = x, xy = y, xz= 0, xw 

and consequently for the product of two linear forms in (x, y, z, w) 

we have 

(ax + by + cz + dw) (a x + b y -t c z + d w) 

(aa + be) x + (ab + bd) y 4- (ca + dc ) z + (cb + dd) w ; 
and this is precisely the form for the product of two matrices, viz. 
we have 

(a, c ) (& , d) 

a b 
c d 

a, b 
c , d 

= (a, b) 
(c, d) 

aa + be , ab + bd 
I ca + dc , cb + dd 

and hence the linear form ax + by + cz + dw, and the matrix 
a, b 

c, d 

may be regarded as equivalent symbols. This identifica 

tion was established by the remark and footnote " Peirce s Linear 
Associative Algebra," Amer. Math. Jour. t. 4 (1881), p. 132. 

(d) The Quaternion Equation 

In ordinary algebra, an equation of the first degree, or linear 
equation with one unknown quantity x, is merely an equation of 
the form ax = b, and it gives at once x = a~ l b. 


But the case is very different with quaternions; the general 
form of a linear equation with one unknown quaternion q is 

AtfBt + A z qB 2 -K . .= C, or say ^AqB = C, 

where G and the several coefficients A and B are given qua 

Considering the expression on the left-hand side, and assuming 
(f=w + i+jy -+- kz, it is obvious that the expression is in effect of 
the form 

S W + OLX +/3 y + 7 z 

+ i (S^v + QLjX + /^y + 7^) 


where the coefficients 8, a, /3, 7 &c. are given scalar magnitudes : 
if then this is equal to a given quaternion (7, say this is 

X + i\ +j\ + k\, 
we have for the determination of w, x, y, z the four equations 

7 3 s - X a , 

and we thence have w, x, y, z, each of them as a fraction with a 
given numerator, and with the common denominator 


S, a, 

8 a 3 ft 7, 

viz. this is the determinant formed with the coefficients 8, a, ft 7, 
&c. Of course if A = 0, then either the equations are inconsistent, 
or they reduce themselves to fewer than four independent 

The number of these coefficients is = 16, and it is thus clear 
that, whatever be the number of the terms A^qB^ A 2 qB> 2) &c. we 
only in effect introduce into the equation 16 coefficients. A single 
term such as A 1 qB J may be regarded as containing seven coefficients, 
for we may without loss of generality write it in the form 

g (1 + ia + jb + kc) q(l+ id +je + kf\ 
and thus we do not obtain the general form of linear equation 


by taking a single term A l qB l (for this contains seven coefficients 
only) nor by taking two terms A t qB v A 2 qB z (for these contain 14 
coefficients only) ; but we do, it would seem, obtain the general 
form by taking three terms (viz. these contain 21 coefficients, 
which must in effect reduce themselves to 16) : that is, a form 
A 1 qS l + A^qB t2 + A 3 qB 3 is, or seems to be, capable of representing 
the above written quaternion form with any values whatever of the 
16 coefficients S, a, (3, 7 &c. But the further theory of this reduc 
tion to 16 coefficients is not here considered. 

The most simple case of course is that of a single term, say we 
have AqB = C : here multiplying on the left by A~ l and on the 
right by B~ l , we obtain at once q = A~ l CB~ l . 

(e) The Nivellator, and its Matrix. 

In the general case, a solution, equivalent to the foregoing, but 
differing from it very much in form may be obtained by means 
of the following considerations. 

A symbol of the above form 2-4 ( ) B, operating upon a 
quaternion q so as to change it into 2-4 (q) B, is termed by 
Prof. Sylvester a "Nivellator:" it may be represented by a single 
letter, say we have < = 2.4 ( ) B ; the effect of it, as has just 
been seen, is to convert the components (w, x t y, 2), into four linear 
functions (w v x v y v z^ which may be expressed by the equation 

(w v x v y lt ^) - | 8, a, 

(w, x, y t 

2 > 2 > & 7 2 

3 a s> ft> 7 3 

or say by the multiplication of (w, x, y, z) by a matrix which may 
be called the matrix of the nivellator; and the theory of the 
solution of the linear equation in quaternions thus enters into 
relation with that of the solution of the linear equation in 

The operation denoted by c/> admits of repetition : we have for 


and similarly for more than two terms, and for higher powers. 

Considering < in connexion with its matrix M, we have 
M 2 (w, x, y, z) for the components of <t>*(q), M* (w, x, y, z) for 



those of c/> 3 (q), and so on. Hence also we have the negative 
powers <$>~ l , &c. of the operation (. The mode in which <"* can 
be calculated will presently appear: but assuming for the moment 
that it can be calculated, the given equation is (f> (q) C, that 
is we have q = (f>~ 1 ((7), the solution of the equation. 

A matrix M of any order satisfies identically an equation of 
the same order : viz. for the foregoing matrix M of the fourth 
order we have 

B-M, v, 0, y 

viz. this is 



/3 3 , 



where h is the before mentioned determinant 

S, a, ft, 7 i , say this is, h = A. 

Jf, in its operation on the components (w, x, y, z) of q, exactly 
represents (f> in its operation on q : we thus have 

viz. this means that operating successively with on the arbitrary 
quaternion Q we have identically 

f (Q) - ff (Q) +f<F (Q) -g<l>(Q) + hQ = 0; 

where observe that the coefficients e, /, g, h have their foregoing 
values, calculated by means of the minors of the determinant : but 
that their values may also be calculated quite independently of 
this determinant: viz. the equation shews that there is an identical 
linear relation connecting the values 4 (Q), </> 8 (Q), < 2 (Q), < (Q) 
and Q : and from the values (assumed to be known) of these 
quantities, we can calculate the identical equation which connects 
them. But in whatever way they are found, the coefficients 
e, f, g, h are to be regarded as known scalar functions. 
Writing in the equation c/T 1 Q in place of Q, we have 

<t> s (Q) - <><? (Q) +/* (Q) -9Q + A*" (Q) = o, 

viz. this equation gives (/T 1 (Q) as a linear function of Q, <p (Q), 2 (Q) 
and < 3 (Q) : and hence for the arbitrary quaternion Q writing the 
value 0, we have q, = ^T 1 (6 ) given as a linear function of 


(7, <(0), 2 (C) and 
given linear equation. 


(C) : we have thus the solution of the 

(/) The Vector Equation 

The theory is similar if, instead of quaternions, we have vectors. 
As to this observe in the first place that, even if A, q, B are each 
of them a vector, the product AqB will be in general, not a vector, 
but a quaternion. Hence in the equation %AqB = 0, if C and the 
several coefficients A and B be all of them vectors, the quantity q 
as determined by this equation will be in general a quaternion : 
and even if it should come out to be a vector, still in the process 
of solution it will be necessary to take account, not only of the 
vector components, but also of the scalar part ; so that there is 
here no simplification of the foregoing general theory. 

But the several coefficients A, B may be vectors so related to 
each other that the sum ^ApB, where p is an arbitrary vector, 
is always a vector 1 ; and in this case, if be also a vector, the 
equation ^ApB = G will determine p as a vector : and there is 
here a material simplification. Writing p = ix +jy + kz, then 
is in effect of the form 

viz. we have these three linear functions of (x t y, z) to be equalled 
to given scalar values \ t , X 2 , \ 3 , and here #, y, z have to be 
determined by the solution of the three linear equations thus 
obtained. And for the second form of solution, writing as before 
(f> = 2A ( )B, then </> is connected with the more simple matrix 

^=1 !, &, 7, 
2 8 > 7 2 

s> & 7 3 

and it thus (instead of a biquadratic equation) satisfies the cubic 



- M, 



= 0, 


1 Thus, if A, B are conjugate quaternions, ApB is a vector a: this is in fact 
the form which presents itself in the theory of rotation. 


We have therefore for < the cubic equation 

-.**+/< -0 = 0, 

and thus cf)~ l (Q) is given as a linear function of Q, c#>(Q), <J> 2 (Q), or, 
what is the same thing, <f>~ l (C) as a linear function of C, $(C), <t>*(C) 
and (this being so) then for the solution of the given equation 
(f) (p) = (7, we have p <f>~ 1 (C), a given linear function of C, <f> (C), 

(g) Nullitats. 

Simplifications and specialities present themselves in particular 
cases, for instance in the cases Aq + qB=C, and Aq = qB, which 
are afterwards considered. 

The product of a quaternion into its conjugate is equal to 
the squared tensor, or norm; aa = T*(a) , and thus the reciprocal 
of a quaternion is equal to the conjugate divided by the norm ; 
hence if the norm be = 0, or say if the quaternion be a nullitat, 
there is no reciprocal. In particular, 0, qua quaternion, is a 

The equation aqb = c, where a, b, c are given quaternions, 
q the quaternion sought for, is at once solvable ; we have 
q a~ l cb~ 1 , but the solution fails if a, or 6, or each of them, is a 
nullitat. And when this is so, then whatever be the value of q, 
we have aqb a nullitat, and thus the equation has no solution 
unless also c be a nullitat. 

If a and c are nullitats, but b is not a nullitat, then the equa 
tion gives aq = cb~ l , which is of the form aq = c , and similarly if 
b and c are nullitats but a is not a nullitat, then the equation 
gives qb = a~ 1 c, which is of the form qb = c: thus the forms to 
be considered are aq = c, qb = c, and aqb = c, where in the first 
equation a and c, in the second equation b and c, and in the 
third equation a, b, c, are nullitats. 

The equation aq = c, a and c nullitats, does not in general 
admit of solution, but when it does so, the solution is indeter 
minate; viz. if Q be a solution, then Q + aR (where R is an 
arbitrary quaternion) is also a solution. Similarly for the equation 
qb = c, if Q be a solution, then Q + Sb (S an arbitrary quaternion) 
is also a solution : and in like manner for the equation aqb = c, if 
Q be a solution then also Q i-aR + Sb (R, S arbitrary quaternions) 
is a solution. 


(h) Conditions of Consistency, ivheti some Coefficients 
are Nullitats. 

Consider first the equation aq = c; writing a 
(a* 4- a* + a* + a 3 = 0) and C = c 4 +ic 1 +jc 2 + ^c 3 (c 4 2 + c 1 2 + c 2 2 -f c 3 2 =0); 
also q = w 4- la; 4- jy 4- &, the equation gives 

c 4 - 4 w - a^x - a z y - a 3 z, 

c l = cijW 4- a 4 x a 3 y 4- a/, 

c 2 = a 2 w 4- a 8 # 4- a$ a^z, 

c s = a s w - a 2 # 4- a^y 4- a^z, 

equations which are only consistent with each other when two 
of the c s are determinate linear functions of the other two c s ; 
and when this is so, the equations reduce themselves to two 
independent equations. Thus from the first, second and third 
equations, multiplying by a l a 3 2 a 4 , a 4 a 3 a^, and a 2 2 a 3 2 , 
and adding, we obtain 

(a^ - 8 a 4 ) c 4 - (a 3 a 4 4- a^) c t - (a 2 2 4- a*) C 2 = ; 
similarly from the first, second and fourth equations, multiplying 
by a 3 a 4 aji^ a^a 3 + a 2 a 4 , a 2 2 a 3 2 , and adding, we have 

- (a^ + a 3 a 4 ) c 4 - (a,a 8 - a 2 a 4 ) c, - (a 2 2 + a 3 2 ) c 3 - 0, 
and when these two equations are satisfied, the original equations 
are equivalent to two independent equations ; so that we ha,ve for 
instance a solution Q = w + ix where c 4 = a^w a^jc y c t = a^w 4- a 4 #, 

, . a.c. 4- tt.c, a,c. 4- a.c, , , , . 

that is w = --n- o , x= V -- 2 4 - 1 ; and the general solution is 
af + a* a 4 * + a* 

then obtained as above. 

The equations connecting the as and the c s may be presented 

in a variety of different forms, all of them of course equivalent in 

virtue of the relations <x 4 2 4 a* + a* -f a 3 2 = 0, c 4 2 4- c* + c 2 2 + c s 2 = ; 

viz. writing 

A l = a* -f a 4 2 = - a 2 2 a s 2 , F^ a 2 a 3 + a t a 4 , F^ = a z a 3 a^, 
A 2 = a^ + a* = - a 3 - a* , F 2 = a 3 a 1 4- a 2 a 4 , F^ = a g a 1 - a 2 a 4 , 
A 3 = a* + a* = - a* - a*, F 3 = a^ + a s a 4 , F 3 = a x a 2 - a 3 a 4 , 

then the relation between any three of the c s may be expressed 
in three different forms, with coefficients out of the sets A l ,A 2 ,A 3 ; 
F v F z , F 3 ; F{, F^ y F s . Obviously the relation between the c s is 
satisfied if c = : the equation then is aq = 0, satisfied by q = aR, 
R an arbitrary quaternion. 


We have a precisely similar theory for the equation qb=c; 
any two of the c s must be determinate linear functions of the 
other two of them; and we have then only two independent 
equations for the determination of the w, an, y, z. 

In the case of the equation aqb = c (a, b, c all nullitats) the 
analysis is somewhat more complicated, but the final result is a 
simple and remarkable one ; from the condition that a, b are 
nullitats, it follows that ab, aib, ajb, akb are scalar (in general 
imaginary scalar) multiples of one and the same nullitat, say of 
ab: the condition to be satisfied by c then is that c shall be a 
scalar multiple of this same nullitat, say c = \ab; the equation 
aqb = \ab has then a solution q = X, and the general solution is 
q = X -j- a R + Sb, where R, S are arbitrary quaternions. 

(i) The Linear Equations, aq qb = 0, and aqqb = c. 

The foregoing considerations explain a point which presents 
itself in regard to the equation aq-qb = 0, (a, b given quaternions, 
q a quaternion sought for): clearly the equation is not solvable 
(otherwise than by the value q = 0) unless a condition be satisfied 
by the given quaternions a, b; but this condition is not (what 
at first sight it would appear to be) T 2 a = T 2 b. The condition 
(say ft) may be satisfied although T*a=t= T 2 b } and being satisfied, 
there exists a determinate quaternion q, which must evidently 
be a nullitat (for from the given equation aq = qb we have 
(To, - T*b) T*q = 0, that is T 2 q = 0). If in addition to the con 
dition ft we have also T z a T*b = 0, then (as will appear) we 
have an indeterminate solution q, which is not in general a 

Take the more general equation aq qb = c: this may be 
solved by a process (due to Hamilton) as follows: multiplying 
on the left hand by a and on the right hand by b, we have 
daq aqb dc, aqb qb* = cb, whence subtracting 

aaq (a + a) qb + qb 2 = ac cb, 

or since ad, a + d are scalars q {ad (a 4- d) b + 6 2 } = ac cb: viz. 
this is an equation of the form qB = C (B, C given quaternions), 
having a solution q = CB ~ 1 . 

Suppose c = 0, then also (7 = 0; and unless B is a nullitat, the 
equation qB = (representing the original equation aq = qb), has 
only the solution g = 0; viz. the condition in order that the 


equation aq = qb may have a solution other than q = 0, is B = nul 
litat, that is aa (a + a) b + 6 2 = nullitat ; viz. we must have 

+ b? + 26 4 (ib, +j\ + kb 3 ) - 6, a - &, - 6 3 8 = nullitat, 
that is 

^ + < + a, 8 4- a 8 8 - 2</. 4 & 4 + 6 4 8 - &/ - 6 8 8 - 6 3 2 

+ 2 (6 4 - a 4 ) (t 6, + j/> 2 4- &6 3 ) = nullitat. 
The condition ft thus is 

that is 

j(a 4 - 6 4 ) 8 + < + a./ 4- a 3 2 - 6 1 3 -6 8 8 -6 8 7 + 4(a 4 -6 4 ) 8 (6 1 s +6;+6 8 s )=0, 

or, as this may also be written, 

( 4 - 6 4 ) 4 4- 2 ( 4 - 6 4 ) 2 ( ai 8 + < + 3 2 + b: 4 6, 8 + 6 3 2 ) 

+ + < + < - 6 t 2 - 6.; - 6 8 8 )* = 0. 
Writing herein 

< + a, 9 + a? + < - ^L 8 , 6 4 2 + 6 a 2 + 6 2 2 + 6 3 2 = ^ 2 , 
the condition is 

(a 4 - 6 4 ) 4 + 2 ( 4 - 6 4 ) 8 (4* + 2 - a 4 8 - & 4 ") + (^L 2 - B 2 - a* + 6 4 2 ) 2 = 0, 
which is easily reduced to 

4 (a, - 6 4 ) (a 4 ^ 2 - 6 4 ^1 2 ) + (A* - BJ = 0, 

and, as already noticed, this is different from T 2 a T 2 b = 0, that is 
A*-& = 0. 

If the equation A* B* = Q is satisfied, then the condition 1 
reduces itself to a 4 6 4 = 0; we then have a = a- 4 -|-a, 6 = a 4 + /5, 
where a, y5 are vectors, and the equation is therefore aq = q/3 
where (since A 2 -B\ = a x 2 + a 2 2 + ct 3 2 - b* - b* - 6 3 2 , = 0), the tensors 
are equal, or we may without loss of generality take a, /3 to be 
given unit vectors, viz. we have a 2 = ] , /3 2 = 1 : and this being 
so, we obtain at once the solution q = X (a + /3) + //,(! - a/3) (X, IJL, 
arbitrary scalars): in fact this value gives 

aq = X (- 1 + a/3) + p (a + 0) = q@. 

Reverting to the general equation oq qb = c, the conjugate of 
ad(a-\-d)b + b 2 is ad (a + a)b + b \ and we thus obtain the 

2 {* K- 6 4 ) K^ 2 - M a ) + (^ 2 -^ 2 ) 2 } = (c - cb) [ad -( a + a)b + 6 2 }, 


but this solution fails if ad (a + a)b + 6 2 is a imllitat : supposing 
it to be so, the equation is only solvable when C satisfies the 
condition which expresses that the equation qB = C is solvable 
when B, C are nullitats. 

The equation aq qb = c, could it is clear be in like manner 
reduced to the form Aq= C. 

(j) The Quadric Equation <f 2aq+b = 0. 

We consider the quadric equation (f 2aq + 6 = 0; a and b 
given quaternions, q the quaternion sought for. The solution 
which follows is that given by Prof. Sylvester for a quadric 
equation in binary matrices. 

In general if q be any quaternion, = w 4- ix + jy + kz, then 
(q-w) 2 4- x?+ y*+z* = 0, that is q* - 2qw + n? 4- a? + y 2 + z~ = 0, or say 
(f 2q (seal, q) + norm q = : viz. this is an identical relation 
connecting a quaternion with its scalar and its norm. 

Writing as above q = w + i&+jy + kz t and t = w* + a? -f?/ 2 + 2 
for the norm, we thus have 

cf - 2wq + t = 0, 
and combining this with the given equation 

q*--2aq + b = 0, 
we find 2 (a - w) q-(b-t) = 0, that is 2q = (a - w) 1 (b - t), 

an expression for q in terms of the scalar and norm w, t, and of the 
known quaternions a and b. 

2q as thus determined satisfies the identical equation 

(2#) 2 - 2 (2q) seal, {(a - lu) 1 (&-*)} + norm {(a - w)~ l (b - 1)} = 0, 

and we have 

\-i/7 *\) seal, {(a w) (& $)! 

seal, {(a- w) l (b-t)\ =- -, 

norm (a - w) 

\-i/i ,M norm (6-^) 

norm\(a-w) l (b-t)}= ^ 

norm (a w) 

(a the conjugate of a). 

The equation thus becomes 

4g 2 norm (a w) ^q (seal, (d w)(b t)} -f norm (b t) = : 
this must agree with 

(f -2qw + t =0, 


or say the function is = 4X (q z 2qw + t) ; we thus have 

norm (a w) X, 
seal, (a w}(b t} 2Xw, 
norm (b t) - 4X, 

three equations for the determination of X, w, t ; and then, w, t 
being determined, the required value of q is *2q = (a w)~ l (b t) 
as above. 

To develope the solution let the values of a, b, c, f, g, h be 
denned as follows : viz. 

norm (ax + by 4 z) = (a, .b, c, f, g, h$#, y, z) z , 
viz. writing a = a 4 + ia l +ja 2 4 ka a , 

b = b 4 + i\ 4 j6 2 4 kb s , 
then this equation is 

(a 4 x + by 4 zf + (a v x 4 6 t y)* + (<yc + 6 2 2/) 2 + (a 3 + 6 3 2/) 2 

= (a, b, c, f, g, hja?, y, 0) 2 , 

that is, a, b, c, f, g, h denote as follows 
a = a? + a* + a* + a 3 2 , f = 6 4 , 
b = 6 4 2 + ^ 2 +6 2 2 + 6 3 2 , g=a 4 , 
c = 1, h = a 4 6 4 + a A 4- a A + a A- 

We then have 

norm (a w) = (a 4 - -w;) 2 + a^ 4- 2 8 4 a s 2 , 

seal, (a -w)(b-t) = (a 4 - w) (6 4 - 4 a l b l 4 a 2 6 8 4 a s 6 8 , 
norm (b - t) = (6 4 - *) a + 6 : 2 4 6, 2 4 & 3 2 , 

or expressing these in terms of (a, b, c, f, g, h) the foregoing three 
equations become 

a - 2gw 4 cw* = X, 

h g fw + ctw = Z\w, 
b-2ft 4c^ 2 =4M, 

where c (introduced only for greater symmetry) is = 1. 

Writing moreover A, B, C, F, G, #=bc-f 2 , ca-g 2 , ab-h 2 , 
gh - af, hf - bg, fg - ch, and K = abc - af 2 - bg 2 - ch 2 + 2fgh ; also 
in place of w, t introducing into the equations n =w- g, and 
v = t f, the equations become 

u* + B=\, 
uv -H = 2\(u + g), 
tf+A= 4X (v 4 f). 


We deduce u* = X - B, 

and we thence obtain, to determine X, the cubic equation 

(X - B) (4X 2 + 4Xf- A) - (2Xg + H) 2 = 0, 
viz. this is 

4X 3 4- 4X 2 (f - a) + X {- be + f 2 + 4 (gh - af)} 

4- c (abc - af 2 - bg 2 - ch 2 + 2fgh) = 0, 
that is, 4X 3 + 4X 2 (f - a) + X (- A + 4<F) + K = 0, 

and, X being determined by this equation, then 

and then w = u + g, t = v+f; consequently 

2q = (a-g-u)- 1 (b-f-v). 
Write for a moment a g u = , then 
( + 2u) = (a-gy-u 2 = a z -2ag + &-B-u\ =-X 

(since a = g + ia t +ja 9 + ka v a = g z + a 1 2 + a* + a* and thus the 
identical equation for a is a J 2ag + a = 0): that is O 2 +2^@+X = 0, 
or X 1 = - ( -f 2u) = (a-g + u) ; that is 1 , = (a - g - u)~\ 

= -(a g+u) , and the value of q is 2q = - (ag+u)(bfv\ 

X A- 

or say it is 

where X is determined by the cubic equation, and u is = + J\ B; 
we have thus six roots of the given quadric equation q 2 2aq + 6 = 0. 



201. HAVING, in the preceding Chapters, given a brief ex 
position of the theory and properties of quaternions, we intend to 
devote the rest of the work to examples of their practical appli 
cation, commencing, of course, with the simplest curve and surface, 
the straight line and the plane. In this and the remaining Chapters 
of the work a few of the earlier examples will be wrought out in 
their fullest detail, with a reference to the previous part of the 
book whenever a transformation occurs; but, as each Chapter 
proceeds, superfluous steps will be gradually omitted, until in 
the later examples the full value of the quaternion processes is 

202. Before proceeding to the proper business of the Chapter 
we make a digression in order to give a few instances of applica 
tions to ordinary plane geometry. These the student may multiply 
indefinitely with great ease. 

(a) Euclid, I. 5. Let a and /3 be the vector sides of an iso 
sceles triangle ; /3 a is the base, and 

The proposition will evidently be proved if we shew that 

a(a-0r l = K0(i3-ar (52). 
This gives a (a - /3) 1 = ( - a) 1 ^, 

or (-a)a = (a-/9), 

or - = _*. 

(b) Euclid, I. 32. Let ABC be the triangle, and let 


where 7 is a unit-vector perpendicular to the plane of the triangle. 
If 1= 1, the angle CAB is a right angle ( 74). Hence 

A=l7r/2 (74). Let = m7r/2, C = n7r/2. We have 

Hence UBA = y m . 7 n . 7 UAB, 

Or 1 = cy*++. 

That is I -f- m + n = 2, 

or A + B + (7 = TT. 

This is, properly speaking, Legendre s proof; and might have been 
given in a far shorter form than that above. In fact we have for 
any three vectors whatever, 

U ?-i 

/3 7 a 

which contains Euclid s proposition as a mere particular case. 

(c) Euclid, I. 35. Let /3 be the common vector-base of the 
parallelograms, a. the conterminous vector-side of any one of them. 
For any other the vector-side is a + xj3 ( 28), and the proposition 
appears as 

TV$ (a + x&) = TV/301 ( 96, 98), 

which is obviously true. 

(d) In the base of a triangle find the point from which lines, 
drawn parallel to the sides and limited by them, are equal. 

If a, j3 be the sides, any point in the base has the vector 

p = (1 x) a 
For the required point 

(\-x)Ta = 
which determines x. 

Hence the point lies on the line 

which bisects the vertical angle of the triangle. 

This is not the only solution, for we should have written 

instead of the less general form above which tacitly assumes that 
1 x and x have the same sign. We leave this to the student. 
T.Q.I. 11 

162 QUATERNIONS. [ 2 O3- 

(e) If perpendiculars be erected outwards at the middle 
points of the sides of a triangle, each being proportional to the 
corresponding side, the mean point of the triangle formed by their 
extremities coincides with that of the original triangle. Find the 
ratio of each perpendicular to half the corresponding side of the 
old triangle that the new triangle may be equilateral. 

Let 2a, 2/3, and 2 (a -f 0) be the vector-sides of the triangle, 
i a unit-vector perpendicular to its plane, e the ratio in question. 
The vectors of the corners of the new triangle are (taking the 
corner opposite to 2/3 as origin) 

pt = a + eia, 

From these 

i (Pi + P* + P.) 
which proves the first part of the proposition. 
For the second part, we must have 

Substituting, expanding, and erasing terms common to all, the 
student will easily find 

3^ = 1. 

Hence, if equilateral triangles be described on the sides of any 
triangle, their mean points form an equilateral triangle. 

203. Such applications of quaternions as those just made are 
of course legitimate, but they are not always profitable. In fact, 
when applied to plane problems, quaternions often degenerate into 
mere scalars, and become ( 33) Cartesian coordinates of some 
kind, so that nothing is gained (though nothing is lost) by their 
use. Before leaving this class of questions we take, as an 
additional example, the investigation of some properties of the 

204. We have already seen (31 (&)) that the equation 

p = a cos 6 + /3 sin 

represents an ellipse, being a scalar which may have any value. 
Hence, for the vector- tangent at the extremity of p we have 

OT = ~ = a sin 6 -f- ft cos 6, 


which is easily seen to be the value of p when is increased by 
7T/2. Thus it appears that any two values of p, for which 
differs by vr/2, are conjugate diameters. The area of the 
parallelogram circumscribed to the ellipse and touching it at 
the extremities of these diameters is, therefore, by 96, 

4TVp = 4TV (a cos + ft sin 0) (- a sin d + /3 cos 0) 

a constant, as is well known. 

205. For equal conjugate diameters we must have 
T (a cos 6 + ft sin 0) = T (- a sin 6 + ft cos 0), 
or (a 2 - ft 2 ) (cos 2 - sin 2 0) H- 4a/3 cos sin = 0, 

The square of the common length of these diameters is of course 

because we see at once from 204 that the sum of the squares of 
conjugate diameters is constant. 

206. The maximum or minimum of p is thus found ; 

d0~ TpdB 

= - ~ {- (a 2 - ft 2 ) cos sin + Saft (cos 2 - sin 2 0)}. 
For a maximum or minimum this must vanish*, hence 

and therefore the longest and shortest diameters are equally 
inclined to each of the equal conjugate diameters ( 205). Hence, 
also, they are at right angles to each other. 

207. Suppose for a moment a and ft to be the greatest and 
least semidiameters, so that 


* The student must carefully notice that here we put * / = 0, and not -^ = 0. 

da do 

A little reflection will shew him that the latter equation involves an absurdity. 


164 QUATERNIONS. [208. 

Then the equations of any two tangent-lines are 

p = a cos + ft sin -\- x ( a sin 6 + ft cos 6), 
p = a cos X + (3 sin l + X I (GL sin l + /3 cos X ). 

If these tangent-lines be at right angles to each other 

8 (- a sin + cos 0) (- a sin t + cos 0J = 

or a 2 sin 6 sin X + /3 2 cos 6 cos t = 0. 

Also, for their point of intersection we have, by comparing 
coefficients of a, ft in the above values of p, 

cos 6 x sin 6 = cos a x l sin O v 
sin 6 + a? cos = sin a + ^ cos r 
Determining a^ from these equations, we easily find 

2V = -( + /S"X 

the equation of a circle ; if we take account of the above relation 
between and r 

Also, as the equations above give x x^ the tangents are 
equal multiples of the diameters parallel to them ; so that the line 
joining the points of contact is parallel to that joining the 
extremities of these diameters. 

208. Finally, when the tangents 

p = acos9 +/3sin0 +x ( asin0 +$cos0), 
p = a cos 0j + j3 sin l + x l (a sin O l + ft cos 0^, 

meet in a given point 

p = aa + bft, 

we have a = cos x sin = cos X ^ sin lf 

6 = sin + x cos = sin X + ^ cos r 
Hence 2 = a 2 + 6 2 - 1 = a;* 

and a cos + b sin = 1 = a cos t + b sin : 

determine the values of and x for the directions and lengths of 
the two tangents. The equation of the chord of contact is 

p = y (a cos + ft sin 0) + (1 - y) (a cos 6 l + ft sin 0J. 
If this pass through the point 

we have p = y cos -f (1 y) cos 1? 

# = y sin + (1 y) sin t , 


from which, by the equations which determine 6 and V we get 
ap + bq = y + l-y = l. 

Thus if either a and b, or p and q, be given, a linear relation 
connects the others. This, by 30, gives all the ordinary properties 
of poles and polars. 

209. Although, in 28 30, we have already given some of 
the equations of the line and plane, these were adduced merely for 
their applications to anharmonic coordinates and transversals ; 
and not for investigations of a higher order. Now that we are 
prepared to determine the lengths and inclinations of lines we 
may investigate these and other similar forms anew. 

210. The equation of the indefinite line drawn through the 
origin 0, of which the vector OA, = a, forms a part, is evidently 

p = xa, 

or p || a, 

or Yap = 0, 

or Up = VOL ; 

the essential characteristic of these equations being that they are 
linear, and involve one indeterminate scalar in the value of p. 

We may put this perhaps more clearly if we take any two 
vectors, /3, 7, which, along with a, form a non-coplanar system. 
Operating with S . Fa/5 and S . Vay upon any of the preceding 
equations (except the third, and on it by S . j3 and 8. 7) we get 


Separately, these are the equations of the planes containing a, ft, 
and a, 7 ; together, of course, they denote the line of intersection. 

211. Conversely, to solve equations (1), or to find p in terms 
of known quantities, we see that they may be written 

so that p is perpendicular to Va{3 and Vay, and is therefore 
parallel to the vector of their product. That is, 


or p = XOL. 

166 QUATERNIONS. [212. 

212. By putting p /3 for p we change the origin to a point 
B where OB = ft, or BO = /3 ; so that the equation of a line 
parallel to a, and passing through the extremity of a vector /3 
drawn from the origin, is 

p - /3 = xa., 

or p = /3 + xa. 

Of course any two parallel lines may be represented as 
p = ft + xu, 

or Fa<>-/3) = 0, 


213. The equation of a line, drawn through the extremity of ft, 
and meeting a perpendicularly, is thus found. Suppose it to be 
parallel to 7, its equation is 

p = /3 + any. 

To determine 7 we know, first, that it is perpendicular to a, 
which gives 

Say = 0. 

Secondly, a, /3, and 7 are in one plane, which gives 

S.a/3y = 0. 
These two equations give 

whence we have p = /3 + xaVaft. 

This might have been obtained in many other ways ; for 
instance, we see at once that 

/3 = a 1 a/3 = a 1 Sa{3 + a 1 Fa/3. 

This shews that a 1 Fa/3 (which is evidently perpendicular to a) is 
coplanar with a and /3, and is therefore the direction of the 
required line ; so that its equation is 

the same as before if we put - ~ z for x. 

214. By means of the last investigation we see that 

-a 1 Fa/3 
is the vector perpendicular drawn from the extremity of $ to the 


p = oca.. 


Changing the origin, we see that 

-of 1 Fa (-7) 

is the vector perpendicular from the extremity of /3 upon the line 

p = 7 + a?o. 

215. The vector joining B (where OB /3) with any point in 
p = 7 + XVL 

is 7 + XQL J3. 

Its length is least when 

or So. (7 + XOL - /8) = 0, 

i.e. when it is perpendicular to a. 
The last equation gives 

or XOL = - a" 1 $a (7 - 0). 

Hence the vector perpendicular is 

or cT l Va (7 - /3) = - a 1 Fa ( - 7), 

which agrees with the result of last section. 

216. To find the shortest vector distance between two lines in 

and Pi = A + !!; 

we must put dT (p - p : ) = 0, 

or S(p- P 3(dp-d Pl ) = O t 

or S o adx - adx = 0. 

Since x and x 1 are independent, this breaks up into the two 

proving the well-known truth that the required line is perpendicular 
to each of the given lines. 

Hence it is parallel to VOLOL V and therefore we have 

p-/? 1 = ^ + a?a-^ 1 ~a? l o 1 = y.Faa 1 ............ (1). 

Operate by 8 . ao^ and we get 

168 QUATERNIONS. [217. 

This determines y, and the shortest distance required is 

[Note. In the two last expressions T before S is inserted simply 
to ensure that the length be taken positively. If 

8 . attj (/3 ySJ be negative, 
then ( 89) 8 . a,a (0 - ft) is positive. 

If we omit the T, we must use in the text that one of these two 
expressions which is positive.] 

To find the extremities of this shortest distance, we must operate 
on (1) with S.OL and S.a.^ We thus obtain two equations, which 
determine x and oc v as y is already known. 

A somewhat different mode of treating this problem will be 
discussed presently. 

217. In a given tetrahedron to find a set of rectangular coordi 
nate axes, such that each axis shall pass through a pair of opposite 

Let a, /S, 7 be three (vector) edges of the tetrahedron, one 
corner being the origin. Let p be the vector of the origin of the 
sought rectangular system, which may be called i, j, k (unknown 
vectors). The condition that i, drawn from p, intersects a is 

S.ioip = Q ............................... (1). 

That it intersects the opposite edge, whose equation is 

ts-=r + a?08-7), 
the condition is 

S.;(/3-7)(/>-/3) = 0, or & {( - 7) p - 7} = ...(2). 

There are two other equations like (1), and two like (2), which can 
be at once written down. 

Put -7 = a 1 , 7-a = ft, a-^ = 7 1 , 

a, Fya = ft, Fa/3 = 7,, 

and the six become 

The two in i give i \\ aSa. 2 p - p (Saa 2 + Sa 3 p). 



- p (S/3/3, + S{3 3 p), and k \\ ySy 2 p - p (77, + S 7s p). 
The conditions of rectangularity, viz., 

at once give three equations of the fourth order, the first of which 


= SapSajSfo - SapSaj (S/3/3 2 

The required origin of the rectangular system is thus given as 
the intersection of three surfaces of the fourth order. 

218. The equation Sap = 

imposes on p the sole condition of being perpendicular to a ; and 
therefore, being satisfied by the vector drawn from the origin to 
any point in a plane through the origin and perpendicular to a, is 
the equation of that plane. 

To find this equation by a direct process similar to that usually 
employed in coordinate geometry, we may remark that, by 29, 
we may write 

p = xp + 7/7, 

where and 7 are any two vectors perpendicular to a. In this 
form the equation contains two indeterminates, and is often useful; 
but it is more usual to eliminate them, which may be done at 
once by operating by S . a, when we obtain the equation first 

It may also be written, by eliminating one of the indeter 
minates only, as 

where the form of the equation shews that Sa/3 = 0. 
Similarly we see that 

represents a plane drawn through the extremity of and perpen 
dicular to a. This, of course, may, like the last, be put into various 
equivalent forms. 

219. The line of intersection of the two planes 
8. a (p-0) =0 

and S . a, (p - /3.) = ( 


contains all points whose value of p satisfies both conditions. But 
we may write ( 92), since a, a 1? and Vaa t are not coplanar, 

pS. aa l Vaa l = Vaafi.aaj + V. a^Vaafiap + F. F(aa 1 ) a/Sfc^p, 
or, by the given equations, 

- P T 2 Vaa^ = F. ^Vaafiap + V. V(aaJ aSa 1 {3 1 + xVaa^ . .(2), 

where x, a scalar indeterminate, is put for S . aa^p which may have 
any value. In practice, however, the two definite given scalar 
equations are generally more useful than the partially indeter 
minate vector-form which we have derived from them. 

When both planes pass through the origin we have ft = ft l = 0, 
and obtain at once 

as the equation of the line of intersection. 

220. The plane passing through the origin, and through the 
line of intersection of the two planes (1), is easily seen to have the 

or 8 (aSa^ - afiajS) p = Q. 

For this is evidently the equation of a plane passing through the 
origin. And, if p be such that 

Sap = Sa/3, 

we also have Sa^ = Safi^ 

which are equations (1). 

Hence we see that the vector 

aSa l j3 l - afia(3 

is perpendicular to the vector-line of intersection (2) of the two 
planes (1), and to every vector joining the origin with a point in 
that line. 

The student may verify these statements as an exercise. 

221. To find the vector-perpendicular from the extremity of @ 
on the plane 

Sap = 0, 

we must note that it is necessarily parallel to a, and .hence that the 
value of p for its foot is 

where xa is the vector-perpendicular in question. 
Hence Sa (ft + xa) = 0, 


which gives xa? = Sa{3, 

or oca = of 1 Sa{3. 

Similarly the vector-perpendicular from the extremity of /3 on the 


may easily be shewn to be 

222. The equation of the plane which passes through the ex 
tremities of a, /3, y may be thus found. If p be the vector of any 
point in it, p a, a /3, and ft y lie in the plane, and therefore 
( 101) 

or Sp(Va/3+ V/3y + Vya) -S. a/3y = 0. 

Hence, if B = as( Va/3 + Vj3y + Vya) 

be the vector-perpendicular from the origin on the plane containing 

the extremities of a, ft, 7, we have 

S = ( Fa/3 + Vj3y + Vya.)- 1 S . a{3y. 

From this formula, whose interpretation is easy, many curious pro 
perties of a tetrahedron may be deduced by the reader. Thus, for 
instance, if we take the tensor of each side, and remember the 
result of 100, we see that 


is twice the area of the base of the tetrahedron. This may be 
more simply proved thus. The vector area of the base is 

i F(a - /3) (7 - /S) = - i ( Fa/3 + F/3 7 + F 7 a). 

Hence the sum of the vector areas of the faces of a tetrahedron, 
and therefore of any solid whatever, is zero. This is the hydrostatic 
proposition for translational equilibrium of solids immersed in a 
fluid subject to no external forces. 

223. Taking any two lines whose equations are 
p = fi + XOL, 
p = ft + a? 1 o l , 
we see that 8 . aa x (p 8) = 

is the equation of a plane parallel to both. Which plane, of course, 
depends on the value of S. 

172 QUATERNIONS. [224. 

Now if 8 = /?, the plane contains the first line ; if 8 = ft, the 

Hence, if 7/Facq be the shortest vector distance between the 
lines, we have 

S. oa^/3- ft -yFoo^O, 

or T (yVaaJ = TS . (/3 - ft) UVaa lt 

the result of 216. 

224. Find the equation of the plane, passing through the origin, 
which makes equal angles with three given lines. Also find the angles 
in question. 

Let a., ft, y be unit- vectors in the directions of the lines, and let 
the equation of the plane be 

SSp = 0. 
Then we have evidently 

SaS = S/3S $y = x, suppose, 
where - -^ 

is the sine of each of the required angles. 
But ( 92) we have 

SS . a{By = x ( Va/3 + F/3y + Fya). 
Hence 8 . p (Va/3 + F/3y + Fya) = 

is the required equation ; and the required sine is 

5 r ( Fa/3 +F/3y+ Fya) 

225. Find the locus of the middle points of a series of straight 
lines, each parallel to a given plane and having its extremities in two 
fixed straight lines. 

Let Syp = 

be the plane, and 

the fixed lines. Also let x and ac l correspond to the extremities of 

one of the variable lines, r being the vector of its middle point. 

Then, obviously, 2or = /3 + xa + ft + x^. 

Also $7 ( _ ft + xa. - x^) = 0. 

This gives a linear relation between x and x v so that, if we sub 

stitute for x t in the preceding equation, we obtain a result of the 


OT = + xe, 


where 8 and e are known vectors. The required locus is, therefore, 
a straight line. 

226. Three planes meet in a point, and through the line of 
intersection of each pair a plane is drawn perpendicular to the 
third ; prove that these planes pass through the same line. 

Let the point be taken as origin, and let the equations of the 
planes be 

Sap = 0, Sj3p = 0, Syp = 0. 

The line of intersection of the first two is || Fa/3, and therefore the 
normal to the first of the new planes is 

Hence the equation of this plane is 


and those of the other two planes may be easily formed from this 
by cyclical permutation of a, /3, 7. 

We see at once that any two of these equations give the third 
by addition or subtraction, which is the proof of the theorem. 

227. Given any number of points A, B, C, &c., whose vectors 
(from the origin) are a l} 2 , 3 , <&c., find the plane through the origin 
for which the sum of the squares of the perpendiculars let fall upon 
it from these points is a maximum or minimum. 

Let Svp = 

be the required equation, with the condition (evidently allowable) 

2V = 1. 
The perpendiculars are ( 221) ^~ l S^oi v &c. 

Hence 2$Va 

is a maximum. This gives 

2 . SvrCLScLdvr = ; 

and the condition that vr is a unit-vector gives 

Svrdvr = 0. 

Hence, as d^ may have any of an infinite number of values, 
these equations cannot be consistent unless 

where a? is a scalar. 

174 QUATERNIONS. [228 

The values of a are known, so that if we put 

cf) is a given self-conjugate linear and vector function, and therefore 
x has three values (g v g 2 , g s , 175) which correspond to three 
mutually perpendicular values of or. For one of these there is a 
maximum, for another a minimum, for the third a maximum- 
minimum, in the most general case when g v g 2 , g 3 are all different. 

228. The following beautiful problem is due to Maccullagh. 
Of a system of three rectangular vectors, passing through the origin, 
two lie on given planes, find the locus of the third. 

Let the rectangular vectors be w, p, a. Then by the conditions 
of the problem 

Svrp = Spcr = $CTCT = 0, 

and San = 0, S/3p = 0. 

The solution depends on the elimination of p and OT among these 
five equations. [This would, in general, be impossible, as p and & 
between them involve six unknown scalars ; but, as the tensors are 
(by the very form of the equations) not involved, the five given 
equations are necessary and sufficient to eliminate the four unknown 
scalars which are really involved. Formally to complete the requisite 
number of equations we might write 

Tvr = a, Tp = b, 

but a and 6 may have any values whatever.] 
From Soivr = 0, W = 0, 

we have OT = xVacr. 

Similarly, from 8/3p = 0, So-p = 0, 

we have py V/3<r. 

Substitute in the remaining equation 

Svrp = 0, 

and we have S . VacrVffcr = 0, 

or SaaSP<r-<r*Sap = 0, 

the required equation. As will be seen in next Chapter, this is a 
cone of the second degree whose circular sections are perpendicular 
to a and ft. [The disappearance of x and y in the elimination 
instructively illustrates the note above.] 



1. What propositions of Euclid are proved by the mere form 
of the equation 

P = (1-X) OL + XP, 

which denotes the line joining any two points in space ? 

2. Shew that the chord of contact, of tangents to a parabola 
which meet at right angles, passes through a fixed point. 

3. Prove the chief properties of the circle (as in Euclid, III.) 
from the equation 

p = a cos 6 + /3 sin 6 ; 

where Ta = T0, and Sa/3 = 0. 

4. What locus is represented by the equation 

where Ta = 1 ? 

5. What is the condition that the lines 

intersect ? If this is not satisfied, what is the shortest distance 
between them ? 

6. Find the equation of the plane which contains the two 
parallel lines 

7. Find the equation of the plane which contains 

F0>-/3) = 0, 
and is perpendicular to Sjp = 0. 

8. Find the equation of a straight line passing through a given 
point, and making a given angle with a given plane. 

Hence form the general equation of a right cone. 

9. What conditions must be satisfied with regard to a number 
of given lines in space that it may be possible to draw through 
each of them a plane in such a way that these planes may intersect 
in a common line ? 

10. Find the equation of the locus of a point the sum of the 
squares of whose distances from a number of given planes is 


11. Substitute "lines" for " planes" in (10). 

12. Find the equation of the plane which bisects, at right 
angles, the shortest distance between two given lines. 

Find the locus of a point in this plane which is equidistant from 
the given lines. 

13. Find the conditions that the simultaneous equations 

Sap = a, 8/3 p = b } Syp = c, 
may represent a line, and not a point. 

14. What is represented by the equations 

where a, ft, 7 are any three vectors ? 

15. Find the equation of the plane which passes through two 
given points and makes a given angle with a given plane. 

16. Find the area of the triangle whose corners have the 
vectors a, ft, 7. 

Hence form the equation of a circular cylinder whose axis and 
radius are given. 

17. (Hamilton, Bishop Law s Premium Ex., 1858.) 

(a) Assign some of the transformations of the expression 


where a and ft are the vectors of two given points A and B. 

(b) The expression represents the vector 7, or OG, of a point C 
in the straight line AB. 

(c) Assign the position of this point G. 

18. (Ibid.) 

(a) If a, ft, 7, 8 be the vectors of four points, A, B, G, D, what 
is the condition for those points being in one plane ? 

(b) When these four vectors from one origin do not thus 
terminate upon one plane, what is the expression for the volume 
of the pyramid, of which the four points are the corners ? 

(c) Express the perpendicular S let fall from the origin on 
the plane ABG, in terms of a, ft, 7. 


19. Find the locus of a point equidistant from the three 

Sap = 0, S/3p = 0, Syp = 0. 

20. If three mutually perpendicular vectors be drawn from a 
point to a plane, the sum of the reciprocals of the squares of their 
lengths is independent of their directions. 

21. Find the general form of the equation of a plane from the 
condition (which is to be assumed as a definition) that any two 
planes intersect in a single straight line. 

22. Prove that the sum of the vector areas of the faces of any 
polyhedron is zero. 

T. Q. I. 12 



229. AFTER that of the plane the equations next in order of 
simplicity are those of the sphere, and of the cone of the second 
order, To these we devote a short Chapter as a valuable prepara 
tion for the study of surfaces of the second order in general. 

230. The equation 

Tp = Ta, 

or p 2 = a 2 , 

denotes that the length of p is the same as that of a given vector a, 
and therefore belongs to a sphere of radius To, whose centre is the 
origin. In 107 several transformations of this equation were ob 
tained, some of which we will repeat here with their interpretations. 


shews that the chords drawn from any point on the sphere to the 
extremities of a diameter (whose vectors are a and ct) are at right 
angles to each other. 

shews that the rectangle under these chords is four times the area 
of the triangle two of whose sides are a and p. 

p = (p + a)" 1 a (p + a) (see 105) 

shews that the angle at the centre in any circle is double that at 
the circumference standing on the same arc. All these are easy 
consequences of the processes already explained for the interpreta 
tion of quaternion expressions, 


231. If the centre of a sphere be at the extremity of a, the 
equation may be written 

which is the most general form. 

If Tct = T/3, 

or 2 = /3 2 , 

in which case the origin is a point on the surface of the sphere, this 

f - ZSctp = 0. 
From this, in the form 

8p (p - 2a) = 

another proof that the angle in a semicircle is a right angle is 
derived at once. 

232. The converse problem is Find the locus of the feet of 
perpendiculars let fall from a given point, p = /3, on planes passing 
through the origin. 

Let Sap = 

be one of the planes, then ( 221) the vector-perpendicular is 

and, for the locus of its foot, 

= a 1 Fa/3. 

[This is an example of a peculiar form in which quaternions some 
times give us the equation of a surface. The equation is a vector 
one, or equivalent to three scalar equations ; but it involves the 
undetermined vector a in such a way as to be equivalent to only 
two indeterminates (as the tensor of a is evidently not involved). 
To put the equation in a more immediately interpretable form, a 
must be eliminated, and the remarks just made shew this to be 

Now (p-p)* = a-*S*aP, 

and (operating by S . /3 on the value of p above) 

Adding these equations, we get 


V 2/ 2 


180 QUATERNIONS. [233. 

so that, as is evident, the locus is the sphere of which /3 is a 

233. To find the intersection of the two spheres 


square the equations, and subtract, and we have 

J - 1 p = <x-a I --/ 1) 

which is the equation of a plane, perpendicular to a a,, the vector 
joining the centres of the spheres. This is always a real plane 
whether the spheres intersect or not. It is, in fact, what is called 
their Radical Plane. 

234. Find the locus of a point the ratio of whose distances from 
two given points is constant. 

Let the given points be and A, the extremities of the vector a. 
Also let P be the required point in any of its positions, and OP = p. 

Then, at once, if n be the ratio of the lengths of the two lines, 

This gives p 2 - 2Sap + cr = ?i 2 /a 2 , 

or, by an easy transformation, 


Thus the locus is a sphere whose radius is T (= J > an d whose 
centre is at B, where OB = - -- ^ a definite point in the line OA. 

235. If in any line, OP, drawn from the origin to a given plane, 
OQ be taken such that OQ . OP is constant, find the locus of Q. 

Let Sap = a 2 

be the equation of the plane, OT a vector of the required surface. 
Then, by the conditions, 

T&Tp = constant = b 2 (suppose), 
and Uvr = Up. 

VU b 2 

From these p = = -- ^ . 

_/ w t& 

Substituting in the equation of the plane, we have 

a V 2 + tfScLv = 0, 


which shews that the locus is a sphere, the origin being situated 
on it at the point farthest from the given plane. 

236. Find the locus of points the sum of the squares of whose 
distances from a set of given points is a constant quantity. Find 
also the least value of this constant, and the corresponding locus. 

Let the vectors from the origin to the given points be ct v 2 , 
...... a n , and to the sought point p, then 

/ 2a\ 2 c 2 + 2(a 2 ) (2a) 2 
Otherwise (p --- ) = -- ^^ + - ~- , 

\ r n J n n 2 

the equation of a sphere the vector of whose centre is - - , i.e. 

whose centre is the mean of the system of given points. 

Suppose the origin to be placed at the mean point, the equation 

p* = - 2 + ^ (a2) (because Sa = 0, 31 (e)). 

The right-hand side is negative, and therefore the equation denotes 
a real surface, if 

c 2 >22V, 

as might have been expected. When these quantities are equal, 
the locus becomes a point, viz. the new origin, or the mean point 
of the system. 

237. If we differentiate the equation 

we get Spdp = 0. 

Hence ( 144), p is normal to the surface at its extremity, a well- 

known property of the sphere. 

If or be any point in the plane which touches the sphere at the 
extremity of p, w p is a line in the tangent plane, and therefore 
perpendicular to p. So that 


is the equation of the tangent plane. 

238. If this plane pass through a given point B, whose vector 
is ft we have 

182 QUATERNIONS. [239. 

This is the equation of a plane, perpendicular to /3, and cutting 
from it a portion whose length is 


If this plane pass through a fixed point whose vector is 7 we must 

so that the locus of j3 is a plane. These results contain all the 
ordinary properties of poles and polars with regard to a sphere. 

239. A line drawn parallel to 7, from the extremity of /3, has 
the equation 

p = ft + x<y. 
This meets the sphere 

in points for which x has the values given by the equation 

The values of x are imaginary, that is, there is no intersection, if 

The values are equal, or the line touches the sphere, if 


This is the equation of a cone similar and similarly situated to the 
cone of tangent-lines drawn to the sphere, but its vertex is at the 
centre. That the equation represents a cone is obvious from the 
fact that it is homogeneous in Ty, i.e. that it is independent of the 
length of the vector 7. 

[It may be remarked that from the form of the above equation 
we see that, if x and x be its roots, we have 

which is Euclid, III. 35, 36, extended to a sphere.] 

240. Find the locus of the foot of the perpendicular let fall from 
a given point of a sphere on any tangent-plane. 

Taking the centre as origin, the equation of any tangent-plane 
may be written 

The perpendicular must be parallel to p, so that, if we suppose it 


drawn from the extremity of a. (which is a point on the sphere) we 

have as one value of TX 

r = a + scp. 

From these equations, with the help of that of the sphere 

P 2 = a 2 , 
we must eliminate p arid x. 

We have by operating on the vector equation by S . vr 
cr 2 = Savr + xStxp 
= Saw + xof. 
a a 2 (W a) 

Hence p = 


-si 2 Savr 

Taking the tensors, we have 

O 2 - Saw) = a 2 O - a) 2 , 
the required equation. It may be put in the form 

S 2 ^U(^-a) = -oL\ 

and the interpretation of this (viz. that the projection of vr on 
(w a) is of constant length) gives at once a characteristic 
property of the surface formed by the rotation of the Cardioid 
about its axis of symmetry. 

If the perpendiculars be let fall from a point, ft, not on the 
sphere, it is easy to see that the equation of the locus is 

> v!7(-) = -a B , 
whose interpretation is equally easy. 

241. We have seen that a sphere, referred to any point what 
ever as origin, has the equation 

Hence, to find the rectangle under the segments of a chord drawn 
through any point, we may put 

P = ^\ 
where 7 is any unit-vector whatever. This gives 

0V - 2a$e*y + a 2 = /3 2 , 
and the product of the two values of x is 

This is positive, or the vector-chords are drawn in the same direc 
tion, if 

T0 < Ta, 
i.e. if the origin is outside the sphere. 

184 QUATERNIONS. [242. 

242. A t B are fixed points ; and, being the origin and P a 
point in space, 

AP li + BP* = OP 2 ; 

find the locus of P, and explain the result when Z A OB is a right, or 
an obtuse, angle. 

Let OA=a,Ofi = P, OP = p, then 



While Sa/3 is negative, that is, while Z AOB is acute, the locus 

is a sphere whose centre has the vector a + /:?. If $a/3 = 0, or 

Z^.0.5 = 7r/2, the locus is reduced to the point 

p = a + 0. 
If Z AOB> r jrj^ there is no point which satisfies the conditions. 

243. Describe a sphere, with its centre in a given line, so as to 
pass through a given point and touch a given plane. 

Let xa., where x is an undetermined scalar, be the vector of 
the centre, r the radius of the sphere, ft the vector of the given 
point, and 

Syp = a 

the equation of the given plane. 

The vector-perpendicular from the point xa. on the given plane 
is ( 221) 

(a xSya) y l . 

Hence, to determine x and r we have the equations 

T. (a - xSya) y 1 = T(xa-fy = r, 

so that there are, in general, two solutions. It will be a good 
exercise for the student to find from these equations the condition 
that there may be no solution, or two coincident ones. 

244. Describe a sphere whose centre is in a given line, and 
which passes through two given points. 

Let the vector of the centre be xa, as in last section, and let 
the vectors of the points be ft and 7. Then, at once, 

Here there is but one sphere, except in the particular case when 
we have 

Ty = T0, and Say = 

in which case there is an infinite number. 


The student should carefully compare the results of this 
section with those of the last, so as to discover why in general two 
solutions are indicated as possible in the one problem, and only 
one in the other. 

245. A sphere touches each of two straight lines, which do not 
meet : find the locus of its centre. 

We may take the origin at the middle point of the shortest 
distance ( 216) between the given lines, and their equations will 
then be 

p = a + xft, 

where we have, of course, 

Let a be the vector of the centre, p that of any point, of one 
of the spheres, and r its radius ; its equation is 

Since the two given lines are tangents, the following equations in 
x and # t must have pairs of equal roots, 

The equality of the roots in each gives us the conditions 

-p {(*-*?+,*}, 

Eliminating r we obtain 

which is the equation of the required locus. 

[As we have not, so far, entered on the consideration of the 
quaternion form of the equations of the various surfaces of the 
second order, we may translate this into Cartesian coordinates to 
find its meaning. If we take coordinate axes of x, y, z respectively 
parallel to /3, /3 V a, it becomes at once 

(x + myf (y + mxf pz, 

where m and p are constants ; and shews that the locus is a 
hyperbolic paraboloid. Such transformations, which are exceed 
ingly simple in all cases, will be of frequent use to the student 
who is proficient in Cartesian geometry, in the early stages of his 
study of quaternions. As he acquires a practical knowledge of 

186 QUATERNIONS. [246. 

the new calculus, the need of such assistance will gradually cease 
to be felt] 

Simple as the above solution is, quaternions enable us to give one 
vastly simpler. For the problem may be thus stated Find the 
locus of the point whose distances from two given lines are equal. 
And, with the above notation, the equality of the perpendiculars is 
expressed ( 214) by 

which is easily seen to be equivalent to the equation obtained above. 

246. Two spheres being given, shew that spheres which cut them 
at given angles cut at right angles another fixed sphere. 

If c be the distance between the centres of two spheres whose 
radii are a and 6, the cosine of the angle of intersection is evidently 

a 2 + b 2 - c 2 

Hence, if a, a v and p be the vectors of the centres, and a, a v r the 
radii, of the two fixed, and of one of the variable, spheres ; A and 
A l the angles of intersection, we have 

(p a) 2 + a 2 + r 2 = 2ar cos A, 
(p - aj 2 + a x 2 + r 2 = 2^9" cos A t . 

Eliminating the first power of r, we evidently must obtain a result 
such as 

where (by what precedes) e is the vector of the centre, and e the 
radius, of a fixed sphere 

(p - e ) 2 + e 2 = 0, 

which is cut at right angles by all the varying spheres. By effect 
ing the elimination exactly we easily find e and e in terms of 
given quantities. 

247. If two vectors divide one another into parts a and ea, 
p and p, respectively, the lines joining the free ends meet in 

(l + e)p- 2ea 

1 - e 
We may write this as 

1 + e 2ea. 

lf e = a= 


For another such pair of vectors, passing through the same 
point, we have, say, 

/ 1 <r = OT + /3 1 . 

Thus the plane quadrilateral, whose diagonals are made up 
respectively of the complete a and /3 vectors, will be projected (by 
lines from the point r) as a square on the plane of p, cr, provided 

Tp=T<7, and Sp<r = 0. 
That is, provided 

.................. (1), 


These equations obviously belong to spheres which intersect 
one another at right angles. For the centres are at 

Thus the distance between the centres is 
T e * + f* ( a -ft}- 

"2(/i -O (l ft) 

and this is obviously less than the sum, and greater than the 
difference, of the radii 

Z\^L( ai -ft), and r.ifo-ft). 
J\ ~ e i 

And because its square is equal to the sum of their squares, the 
spheres intersect at right angles. Hence 

The locus of the points, from which a plane (uncrossed) quadri 
lateral can be projected as a square, is a circle whose centre is in 
the plane of the quadrilateral, and whose plane is perpendicular to 
that plane. 

To find the points (if any) of this circle from which the quadri 
lateral is seen as a square, we must introduce the additional 

Svrp = 0, >SW = 0, 

or Sw (r + OJ = 0, flr(isr + ft)=0 ............ (3). 

Hence the points lie on each of two spheres which pass through 
the origin : i.e. the intersection of the diagonals. 

[If we eliminate r among the four equations (1), (2), (3), we 
find the condition 

S(,-/8 I )(/,X + e, 1 A) = 0- 

This we leave to the student.] 

188 QUATERNIONS. [248. 

Another mode of solving the last problem, viz. to find the points 
from which a given plane quadrilateral is seen as a square, consists 
in expressing that the four portions of the diagonals subtend equal 
angles, and that the planes containing them are at right angles to 
one another. 

The first condition gives, with the notation of the beginning of 
this section, 

S . VF U (-ar - a) = S . w U (w + ea) 

= 8.vU(v-0)=8.vU(w+fl3) ...... (4). 

The second condition is 

or w^Sfa - oter/&j = ..................... (5), 

the cone whose cyclic normals are a, ft. 

[It will be excellent practice for the student to shew that (4) 
and (5) are equivalent to (1} ; (2), (3). Thus, in particular, the 
first equality in (4) 

S.vrU (sr a) = 8 .vfU(af + ea), 
is equivalent to the first of (3), viz. 

S . w (j + aj = 0.] 

It is obvious that the solution of the first problem in this 
section gives at once the means of solving the problem of projecting 
an ellipse into a circle, so that any given (internal) point may be 
projected as the centre of the circle. And numerous other con 
sequences follow, which may be left to the reader. 

248. To inscribe in a given sphere a closed polygon, plane or 
gauche, whose sides shall be parallel respectively to each of a series 
of given vectors. 

Let Tp = l 

be the sphere, a, ft, 7, ...... , tj, 6 the vectors, n in number, and let 

p lt p z , ...... p n , be the vector-radii drawn to the angles of the polygon. 

Then p 2 p l = XJL, &c., &c. 

From this, by operating by 8.(p 1t + p l ), we get 

Also = Vap 2 -Va Pl . 

Adding, we get = a/? 2 4- Kap^ = ap 2 + p^a. 

Hence p z = - oL^pfr 

or, if we please, p 2 = a/OjGf 1 . 


[This might have been written clown at once from the result of 

Similarly p 3 = - /3T l pJ3 = ft ^p^ft, &c. 

Thus, finally, since the polygon is closed, 

P m -/,-(-) 0*1 * ...... Fa pflfi ...... ri0. 

We may suppose the tensors of a, ft ...... r), 6 to be each unity. 

Hence, if 

a = aft ...... rjO, 

l ~ l 

we have a~ l = 6~rj 

which is a known quaternion ; and thus our condition becomes 

This divides itself into two cases, according as n is an even or an 
odd number. 

If n be even, we have 

a Pl = p,a. 

Removing the common part p^Sa, we have 

7/0,7(1 = 0. 

This gives one determinate direction, + Fa, for p l ; and shews that 
there are two, and only two, solutions. 
If n be odd, we have 

a Pi = ~ PA 
which (operating, for instance, by S . /oj requires that we have 

Sa = Q, 

i.e. that there may be a solution, a must be a vector. 
Hence Sap^ = 0, 

and therefore p l may be drawn to any point in the great circle of 
the unit-sphere whose pole is on the vector a. 

249. To illustrate these results, let us take first the case of 
n = 3. We must have 

or the three given vectors must (as is obvious on other grounds) be 
parallel to one plane. Here afty, which lies in this plane, is ( 106) 
the vector-tangent at the first corner of each of the inscribed tri 
angles ; and is obviously perpendicular to the vector drawn from 
the centre to that corner. 
If n = 4t } we have 

as might have been at once seen from 106. 

190 QUATERNIONS. [250. 

250. Hamilton has given (Lectures, p. 674 and Appendix C.), 
an ingenious process by which the above investigation is rendered 
applicable to the more difficult problem in which each side of the 
inscribed polygon is to pass through a given point instead of being 
parallel to a given line. His process, which (see his Life, Vol. in., 
pp. 88, 426) he evidently considered as a specially tough piece of 
analysis, depends upon the integration of a linear equation in 
finite differences. 

The gist of Hamilton s method is (briefly) as follows (Lectures, 

Let the (unit) vectors to the corners of the polygon be, as 
above, p v p# ...... p n . Also let ct v 2 , ...... a n be the points through 

which the successive sides are to pass. The sides are respectively 
parallel to the vectors 

which correspond to a, /3, ...... 0, of 248. Hence, if we write 

we have (as in that section), since the expressions are independent 
of the tensors of the qs, 

P* = -?*& *> &*> 

These give, generally, (with the condition p? 1) 

Where r m = V,n-l + *,-!. 

[We may easily eliminate s, by the use of the separable symbol D 
or 1 + A, but this leads to a troublesome species of equation of 
second differences. Hamilton ingeniously avoids this by the use 
of biquaternions.] 

Putting i for the algebraic J I, we have 

r m + t* m = ( m + (r m _, - is m _,\ 

(where, as usual, we have a second equation by changing through 
out the sign of i). 

The complete solution of this equation is, of course, obtained 


at once in the form of a finite product. But it is sufficient to 
know some of its characteristic properties only. 
The squared tensor is 

so that, by equating real and imaginary parts, we have 

But, by the value of q l above, we have r, = a. v s l = l, so that 
TV. - T\ = (TV. - 1) (ra m _, - 1) ...... (2-a, - 1), 

Thus it appears that we may write 

?. = 6 + /9 + (-l)"( + 7)ft .................. (2), 

with the condition 

bc = S/3y ............................. (3). 

But, if we write, putting i instead of p l in q n , 

we have ~- Q = \ + /-", suppose, 

where X and //, are real vectors whose values can be calculated 
from the data. And we now have 

(1 4- X 

= /_ 1\ 


When w, is odd, this gives at once 

which, since pj does not vanish, leads to the two equations 

S\ Pl = 0, 

These planes intersect in a line, whose intersections with the unit- 
sphere give the possible extremities of the required first radius. 
When n is even, we have 

V\ Pl = fi + p l 8fj,p l = V. piVppv 
or V. Pl (\ + V^ Pl ) = 0. (199.) 

With the notation of (2) the condition (3) becomes 

- c 

192 QUATERNIONS. [250*. 

For further details, see especially Appendices B and C to the 

By an immediate application of the linear and vector function 
of Chapter V., the above solutions may be at once extended to any 
central surface of the second order. 

250*. The quaternions which Hamilton employed (as above) 
were such as change the radius to one corner of the polygon into 
that to the next by a conical rotation. It may be interesting and 
useful to the student to compare with Hamilton s solution the 
following, which employs the quaternions which directly turn one 
side of the polygon to lie along the next. The successive sides are 
expressed as ratios of one of these quaternions to the next. 

Let p^ p z , &c., p n be (unit) vectors drawn from the centre of 
the sphere to the corners of the polygon; a l5 2 ,...a n , the points 
through which the successive sides are to pass. Then (by Euclid) 
we have 

(P* ~ (Pi - i) = 1 + , 2 = A> suppose. 

&c. &c. 

These equations ensure that if the tensor of any one of the ps be 
unit, those of all the others shall also be units. Thus we have 
merely to eliminate 2 , . .., p n \ and then remark that (for the 
closure of the polygon) we must have 

Pn+l = Pi 

That this elimination is possible we see from the fact already 
mentioned, which shews that the unknowns are virtually mere 
unit-vectors ; while each separate equation contains coplanar 
vectors only. In other words, when p m and cc m are given, p m+l is 
determinate without ambiguity. 

We may now write the first of the equations thus : 

(ft - <Pi ~ i) = A + (i ~ (ft ~ = ?i suppose. 
Thus the angle of q^ is the angle of the polygon itself, and in the 
same plane. By the help of the second of the above equations 
this becomes 


& = ^ 2 (Pi - a i) + K 


By the third, this becomes 

(ft ~ ) 9 2 = A 3 ^ 5 

(P4 ~ 4> & = A & + (a ~ 4 ) ? 2 = & 
The law of formation is now obvious; and, if we write 

</o = />,-,, ft =!-*> ft = 2 - &C., 

we have 

We have also, generally, 


Pm = ff^i^aa = ^g^. m -,9,,,- 2 _ P,,,- 2 > _ 

!*- S m-a ^m-2 

From (1), and the value of q , we see that all the values of q 
are linear functions of p l of the form 



Similarly ^^ - ^, t _ 2 - m ^_ 2 j 

But the first equations in (1) give at once 

P = + ^l whence ^ 
^o=-,+ pj 

and ^ 1= a 2 - a i+( 1 

p, = + q lPl 
This suggests that 

By (4) we have 

^-l = ^- 2 +n^- 

^-i = ^,n- 2 -a m ^_ 2 
Let ?>i be odd, then we should have by (5) 

PM = -* + %, 

9m- = B-A Pt ; 

T. Q. I. 

194 QUATERNIONS. [ 2 5*- 

whence p m _ l = B- Ap l + a m (A 4- BpJ, 


These agree with (5), because m - 1 is even. And similarly we 
may prove the proposition when m is even. 
If now, in (2), we put n + 1 for m, we have 

f , 
-p, U 

C-D Pl . f , 
= -=r ^p if n be odd, 
# + />! 

G and Z) being quaternions to be calculated (as above) from the 
data. The two cases require to be developed separately. 

Take first the odd polygon : 
then piD + pflp^C-Dpv 

or Pl (d + 8) + Pl (c + 7) Pl = c + 7 - (d + 3) Pl , 

if we exhibit the scalar and vector parts of the quaternions C and 
D. Cutting out the parts which cancel one another, and dividing 
by 2, this becomes 

which, as p is finite, divides itself at once into the two equations 

- c = 0. 

These planes intersect in a line which, by its intersections (if real) 
with the sphere, gives two possible positions of the first corner of 
the polygon. 

For the even polygon we have 

or Vp$ - 7 - p l Sjp 1 = ; 

which may be written 

This equation gives, as in 199 above, 

where x is to be found from 


The two values of a? have opposite signs. Hence there are two 
real values of x, equal and with opposite signs, giving two real 
points on the sphere. Thus this case of the problem is always 

251. To find the equation of a cone of revolution, whose vertex 
is the origin. 

Suppose a, where Ta= 1, to be its axis, and e the cosine of its 
semi-vertical angle ; then, if p be the vector of any point in the 

or S ap = - ey. 

252. Change the origin to the point in the axis whose vector 
is XOL, and the equation becomes 

- x 
Let the radius of the section of the cone made by 

retain a constant value b, while x changes ; this necessitates 


so that when x is infinite, e is unity. In this case the equation 

which must therefore be the equation of a circular cylinder of 
radius 6, whose axis is the vector a. To verify this we have only 
to notice that if TO be the vector of a point of such a cylinder we 
must ( 214) have 

= b, 

which is the same equation as that above. 

253. To find, generally, the equation of a cone which has a 
circular section : 

Take the origin as vertex, and let the circular section be the 
intersection of the plane 

with the sphere (passing through the origin) 


196 QUATERNIONS. [ 2 54- 

These equations may be written thus, 


Hence, eliminating Tp by multiplying the right, and left, members 
together, we find the following equation which Up must satisfy 

SaUpS{3Up = -I, 

or f - SapSj3p = 0, 

which is therefore the required equation of the cone. 

As a. and /3 are similarly involved, the mere form of this 
equation proves the existence of the subcontrary section dis 
covered by Apollonius. 

254. The equation just obtained may be written 

or, since a and (3 are perpendicular to the cyclic planes ( 59*), 
sinp sin |/ = constant, 

where p and p are arcs drawn from any point of a spherical conic 
perpendicular to the cyclic arcs. This is a well-known property of 
such curves. 

255. If we cut the cyclic cone by any plane passing through 
the origin, as 

then Vary and V/3y are the traces on the cyclic planes, so that 

p = scUVa.y+-yUVpy (24). 

Substitute in the equation of the cone, and we get 
- a? - y* + Pxy = 0, 

where P is a known scalar. Hence the values of x and y are the 
same pair of numbers. This is a very elementary proof of the 

proposition in 59*, that PL = MQ (in the last figure of that 

256. When x and y are equal, the transversal arc becomes a 
tangent to the spherical conic, and is evidently bisected at the 
point of contact. Here we have 


This is the equation of the cone whose sides are perpendiculars 
(through the origin) to the planes which touch the cyclic cone, 
and from this property the same equation may readily be deduced. 

257. It may be well to observe that the property of the 
Stereographic projection of the sphere, viz. that the projection of 
a circle is a circle, is an immediate consequence of the above form 
of the equation of a cyclic cone. 

258. That 253 gives the most general form of the equation 
of a cone of the second degree, when the vertex is taken as origin, 
follows from the early results of next Chapter. For it is shewn 
in 263 that the equation of a cone of the second degree can 
always be put in the form 

This may be written Sptyp 0, 

where </> is the self-conjugate linear and vector function 

<f) P = 2 V. ap/3 + (A + ZScijS) p. 
By 180 this may be transformed to 

< t ) P=PP+ V- V/^> 
and the general equation of the cone becomes 

( p - SXfju) f + 2S\p Sf^p = 0, 
which is the form obtained in 253. 

259. Taking the form 

Sp(f)p = 
as the simplest, we find by differentiation 

Sdpfyp + Spd^p = 0, 
or ZSdpfo = 0. 

Hence $p is perpendicular to the tangent-plane at the extremity 
of p. The equation of this plane is therefore (w being the vector 
of any point in it) 

Sj>p (*r-p) = 0, 
or, by the equation of the cone, 

S-sr$p = 0. 

260. The equation of the cone of normals to the tangent-planes 
of a given cone can be easily formed from that of the cone itself. 
For we may write the equation of the cone in the form 

198 QUATERNIONS. [261. 

and if we put tfjp = a, a vector of the new cone, the equation 

Str4>~ l tr = 0. 

Numerous curious properties of these connected cones, and of the 
corresponding spherical conies, follow at once from these equations. 
But we must leave them to the reader. 

261. As a final example, let us find the equation of a cyclic 
cone when five of its vector-sides are given i.e. find the cone of the 
second degree whose vertex is the origin, and on whose surface lie 
the vectors a, ft, 7, S, e. 

If we write, after Hamilton, 

Q = S.V(Va/3V8e)V(Vl3vVep)V(VvSVpoL) ......... (1) 

we have the equation of a cone whose vertex is the origin for the 
equation is not altered by putting xp for p. Also it is the equation 
of a cone of the second degree, since p occurs only twice. Moreover 
the vectors a, ft, 7, 8, e are sides of the cone, because if any one 
of them be put for p the equation is satisfied. Thus if we put 
ft for p the equation becomes 

= S .V (VaftVZe] [VftaS.V^Vft^Veft- VySS .Vfta VftyVeft J. 
The first term vanishes because 

and the second because 

since the three vectors Vfta., V/3y, Fe/3, being each at right angles 
to ft, must be in one plane. 

As is remarked by Hamilton, this is a very simple proof of 
Pascal s Theorem for (1) is the condition that the intersections of 
the planes of a, ft and 8, e ; ft, 7 and e, p ; 7, S and p, a ; shall lie 
in one plane ; or, making the statement for any plane section of 
the cone : In order that the points of intersection of the three pairs 
of opposite sides, of a hexagon inscribed in a curve, may always lie 
in one straight line, the curve must be a conic section. 



1. On the vector of a point P in the plane 

a point Q is taken, such that QO . OP is constant ; find the equation 
of the locus of Q. 

2. What spheres cut the loci of P and Q in (1) so that both 
circles of intersection lie on a cone whose vertex is ? 

3. A sphere touches a fixed plane, and cuts a fixed sphere. 
If the point of contact with the plane be given, the plane of the 
intersection of the spheres contains a fixed line. 

Find the locus of the centre of the variable sphere, if the plane 
of its intersection with the fixed sphere passes through a given 

4. Find the radii of the spheres which touch, simultaneously, 
the four given planes 

[What is the volume of the tetrahedron enclosed by these planes ?] 

5. If a moveable line, passing through the origin, make with 
any number of fixed lines angles 6, 6 V # 2 , &c., such that 

a cos 6 + a v cos 6 l + ...... = constant, 

where a, a v ...... are constant scalars, the line describes a right cone. 

6. Determine the conditions that 

Spcf)p = 
may represent a right cone, </> being as in 258. 

7. What property of a cone (or of a spherical conic) is given 
directly by the following particular form of its equation, 

S . ipicp = ? 

8. What are the conditions that the surfaces represented by 

Sp(j>p = 0, and S . cpfcp = 0, 
may degenerate into pairs of planes ? 


9. If arcs of great circles, drawn from any given point of a 
sphere to a fixed great circle, be bisected, find the locus of these 
middle points ; and shew that the arcs drawn from the pole of this 
fixed great circle, to that of which the given point is pole, are also 
bisected by the same locus. 

10. Find the locus of the vertices of all right cones which 
have a common ellipse as base. 

11. Two right circular cones have their axes parallel. Find 
the orthogonal projection of their curve of intersection on the 
plane containing their axes. 

12. Two spheres being given in magnitude arid position, every 
sphere which intersects them in given angles will touch two other 
fixed spheres and cut a third at right angles. 

13. If a sphere be placed on a table, the breadth of the 
elliptic shadow formed by rays diverging from a fixed point is 
independent of the position of the sphere. 

14. Form the equation of the cylinder which has a given 
circular section, and a given axis. Find the direction of the 
normal to the subcontrary section. 

15. Given the base of a spherical triangle, and the product of 
the cosines of the sides, the locus of the vertex is a spherical conic, 
the poles of whose cyclic arcs are the extremities of the given 

16. (Hamilton, Bishop Laws Premium Ex., 1858.) 

(a) What property of a sphero-conic is most immediately 
indicated by the equation 

(b) The equation 

also represents a cone of the second order ; X is a focal line, and 
//, is perpendicular to the director-plane corresponding. 

(c) What property of a sphero-conic does the equation most 
immediately indicate ? 

17. Shew that the areas of all triangles, bounded by a tangent 
to a spherical conic and by the cyclic arcs, are equal. 


18. Shew that the locus of a point, the sum of whose arcual 
distances from two given points on a sphere is constant, is a 
spherical conic. 

19. If two tangent planes be drawn to a cyclic cone, the four 
lines in which they intersect the cyclic planes are sides of a right 

20. Find the equation of the cone whose sides are the 
intersections of pairs of mutually perpendicular tangent planes 
to a given cyclic cone. 

21. Find the condition that five given points may lie on a 

22. What is the surface denoted by the equation 

p 2 = xoc + yp 4 ^ 

where p = xa. + yf$ + zy, 

a, ft, 7 being given vectors, and x, y, z variable scalars ? 

Express the equation of the surface in terms of p, a, ft, 7 alone. 

23. Find the equation of the cone whose sides bisect the 
angles between a fixed line, and any line in a given plane, which 
meets the fixed line. 

What property of a spherical conic is most directly given 
by this result ? 



262. THE general scalar equation of the second degree in a 
vector p must evidently contain a term independent of p, terms of 
the form S. apb involving p to the first degree, and others of the 
form S.apbpc involving p to the second degree, a, b, c, &c. being 
constant quaternions. Now the term S.apb may be written as 

S P V(ba), 
or as 

S.(Sa+ Va) p (Sb + Vb) = SaSpVb + SbSpVa + 8. pVbVa, 

each of which may evidently be put in the form Syp, where 7 is 
a known vector. 

Similarly* the term S . apbpc may be reduced to a set of terms, 
each of which has one of the forms 

the second being merely a particular case of the third. Thus (the 
numerical factors 2 being introduced for convenience) we may 
write the general scalar equation of the second degree as follows: 

22 . SapSffp + Ap* + 2%> = G ............... (1). 

263. Change the origin to D where OD = S, then p becomes 
p + S, and the equation takes the form 

22 . SapSQp + Ap 9 + 22 (SapSpS + SppSaS) + ZAS&p + 2Syp 

* For S . apbpc = S . capbp = S . a pbp = (2Sa Sb - Sa b) p 2 + ZSa pSbp ; and in parti 
cular cases we may have Va Vb. 


from which the first power of p disappears, that is the surface is 
referred to its centre , if 

2 (aSfiS + /3Sa8) + AS + y = (2), 

a vector equation of the first degree, which in general gives 
a single definite value for 8, by the processes of Chapter V. [It 
would lead us beyond the limits of an elementary treatise to 
consider the special cases in which (2) represents a line, or a plane, 
any point of which is a centre of the surface. The processes to be 
employed in such special cases have been amply illustrated in the 
Chapter referred to.] 

With this value of S, and putting 

D = G- 2SyS - AV - 22 . SaS3/3S, 
the equation becomes 

22 . SapS/3p + Ap 2 = D. 

If D = 0, the surface is conical (a case treated in last Chapter) ; 
if not, it is an ellipsoid or hyperboloid. Unless expressly stated 
not to be, the surface will, when D is not zero, be considered an 
ellipsoid. By this we avoid for the time some rather delicate 

By dividing by D, and thus altering only the tensors of the 
constants, we see that the equation of central surfaces of the 
second degree, referred to the centre, is (excluding cones) 

^(S a pS^p)+gp"=-l (3). 

[It is convenient to use the negative sign in the right-hand 
member, as this ensures that the important vector <frp (which we 
must soon introduce) shall make an acute angle with p ; i.e. be 
drawn, on the whole, towards the same parts.] 

264. Differentiating, we obtain 

22 {SadpS/3p + SapS/Bdp] + ZgSpdp = 0, 
or S.dp {2 (aS/3p + %>) + gp] = 0, 

and therefore, by 144, the tangent plane is 

Sfr-p) {2 (aSp P + 0S*p) + gp} = 0, 
i.e. 8 . {2 (aS/Bp + {3ScL P ) +gp} = -l,\)y (3). 

Hence, if v=*2(aS/3p + l3S*p) + gp (4), 

the tangent plane is Sv& = 1, 

and the surface itself is Svp 1, 

204 QUATERNIONS. [265. 

And, as v~ l (being perpendicular to the tangent plane, and 
satisfying its equation) is evidently the vector-perpendicular from 
the origin on the tangent plane, v is called the vector of proximity . 

265. Hamilton uses for v, which is obviously a linear and 
vector function of p, the notation (f>p, $ expressing a functional 
operator, as in Chapter V. But, for the sake of clearness, we will 
go over part of the ground again, especially in the interests of 
students who have mastered only the more elementary parts of 
that Chapter. 

We have, then, </o = 2 (aSflp + /3Sap) + gp. 
With this definition of <, it is easy to see that 

(a) </> (p + <r) = </>/> + $&, &c., for any two or more vectors. 

(b) </> (asp) = xfyp, a particular case of (a), x being a scalar. 

(c) d$p = 4>(dp}. 

(d) 3o-<f)p = S (ScL(rS/3p + S/3aSap) + gSp<r = Spfa, 
or <f) is, in this case, self-conjugate. 

This last property is of great importance in what follows. 

266. Thus the general equation of central surfaces of the 
second degree (excluding cones) may now be written 

Sp4>p= -i .......................... (i). 

Differentiating, Sdp(f>p -f Spd<j)p = 0, 

which, by applying (c) and then (d) to the last term on the left, 

= 0, 

and therefore, as in 264, though now much more simply, the 
tangent plane at the extremity of p is 

S (-DT p) <pp = 0, 

or Stztyp = Sptyp = 1. 

If this pass through A (OA = a), we have 

Sa<l>p = - 1, 
or, by (d), Spfa = - 1, 

for all possible points of contact. 

This is therefore the equation of the plane of contact of tangent 
planes drawn from A. 


207. To find the enveloping cone whose vertex is A, notice that 

(Sp<t>p + 1) +p (Sp<f>CL + I) 2 = 0, 

where p is any scalar, is the equation of a surface of the second 
degree touching the ellipsoid along its intersection with the plane. 
If this pass through A we have 

(8a<j)OL + 1) + p (Safa + I) 2 = 0, 
and p is found. Then our equation becomes 

(Sp<l)p + I)(Sai<l>oL + l)-(Sp<l>a + l) 9 = Q (1), 

which is the cone required. To assure ourselves of this, transfer 
the origin to A, by putting p + a for p. The result is, using (a) 
and (d), 

(Spfo + 28pcf)CL + Safa + 1) (Sa<t>a + 1) - (SpQa + SaQoL + I) 2 - 0, 
or Sp^p (Safct + 1) - (Sp<t>a.y = 0, 

which is homogeneous in Tp, and is therefore the equation of a 

[In the special case when A lies on the surface, we have 
$</> + 1 = 0, 

and the value of p is infinite. But this is not a case of failure, for 
the enveloping cone degenerates into the tangent plane 

/>< + 1 = 0.] 

Suppose A infinitely distant, then we may put in (1) oca. for a, 
where x is infinitely great, and, omitting all but the higher terms, 
the equation of the cylinder formed by tangent lines parallel to a is 

(Sp<j>p + 1) SOL^OL - (Spfa) 9 = 0. 
See, on this matter, Ex. 21 at end of Chapter. 

268. To study the nature of the surface more closely, let us 
find the locus of the middle points of a system of parallel chords. 

Let them be parallel to a, then, if & be the vector of the middle 
point of one of them, -57 + xa. and r xa. are values of p which 
ought simultaneously to satisfy (1) of 266. 

That is S .(or xa)<f>(iffa;a) = I. 

Hence, by (a) and (d), as before, 

+ afScvpa = 1, 

206 QUATERNIONS. [269. 

The latter equation shews that the locus of the extremity of w, 
the middle point of a chord parallel to a, is a plane through the 
centre, whose normal is (/> ; that is, a plane parallel to the tangent 
plane at the point where OA cuts the surface. And (d) shews that 
this relation is reciprocal so that if ft be any value of r, i.e. be 
any vector in the plane (1), a will be a vector in a diametral plane 
which bisects all chords parallel to ft. The equations of these 
planes are 

= 0, 

= 0, 

so that if V. (fracfrft = 7 (suppose) is their line of intersection, we have 

= = $a<^7 } 

and (1) gives Sfttya = 0= Satyft) 

Hence there is an infinite number of sets of three vectors a, ft, 7, 
such that all chords parallel to any one are bisected by the diametral 
plane containing the other two. 

269. It is evident from 23 that any vector may be expressed 
as a linear function of any three others not in the same plane ; let 

where, by last section, 

Sa^ft = Sft$a = 0, 
$#(7 = S<y(f)ct = 0, 

And let Sa<j>a = - 1 

Sft<f>ft = -l\, 

so that a, ft, and 7 are vector conjugate semi-diameters of the 
surface we are engaged on. 

Substituting the above value of p in the equation of the surface, 
and attending to the equations in a, ft, 7 and to (a), (6), and (d), 
we have 

To transform this equation to Cartesian coordinates, we notice that 
x is the ratio which the projection of p on a bears to a itself, &c. 


If therefore we take the conjugate diameters as axes of f, 77, ?, and 
their lengths as a, b, c, the above equation becomes at once 

the ordinary equation of the ellipsoid referred to conjugate 

270. If we write ^ instead of <, these equations assume an 
interesting form. We take for granted, what we shall afterwards 
prove, that this extraction of the square root of the vector function 
is lawful, and that the new linear and vector function has the same 
properties (a), (b), (c), (d) ( 265) as the old. The equation of the 
surface now becomes 

or Styptyp = 1, 

or, finally, T*fyp = 1. 

If we compare this with the equation of the unit-sphere 


we see at once the analogy between the two surfaces. The sphere 
can be changed into the ellipsoid, or vice versa, by a linear deforma 
tion of each vector, the operator being the function ty or its inverse. 
See the Chapter on Kinematics. 

271. Equations (2) 268, by 270 become 

#a^ a /3 = = <Styra^v8, &c ...................... (1), 

so that i/ra, -\/r/3, ^7, the vectors of the unit-sphere which correspond 
to semi-conjugate diameters of the ellipsoid, form a rectangular 

We may remark here, that, as the equation of the ellipsoid 
referred to its principal axes is a case of 269, we may now suppose 
i,j, and k to have these directions, and the equation is 

which, in quaternions, is 


We here tacitly assume the existence of such axes, but in all cases, 
by the help of Hamilton s method, developed in Chapter V., we at 
once arrive at the cubic equation which gives them. 

208 QUATERNIONS. [272. 

It is evident from the last-written equation that 

fiSip jSjp kSkp 

and -Jro^z- - r + -L^r + - - , 

V a 6 c 

which latter may be easily proved by shewing that 

And this expression enables us to verify the assertion of last section 
about the properties of ty. 

As Sip x, &c., a?, y, being the Cartesian coordinates 
referred to the principal axes, we have now the means of at once 
transforming any quaternion result connected with the ellipsoid 
into the ordinary one. 

272. Before proceeding to other forms of the equation of the 
ellipsoid, we may use those already given in solving a few problems. 

Find the locus of a point when the perpendicular from the centre 
on its polar plane is of constant length. 

If -or be the vector of the point, the polar plane is 

= 1, 

and the length of the perpendicular from is ( 264). 


Hence the required locus is 

T$w = C, 
or &sr<V - - C 2 , 

a concentric ellipsoid, with its axes in the same directions as those 
of the first. By 271 its Cartesian equation is 

273. Find the locus of a point whose distance from a given point 
is always in a given ratio to its distance from a given line 

Let p = xft be the given line, and A (OA = a) the given point, 
and choose the origin so that Saft = 0. Then for any one of the 
required points 

This is the equation of a surface of the second degree, which may 
be written 


Let the centre be at 8, and make it the origin, then 

and, that the first power of p may disappear, 

a linear equation for S. To solve it, note that Sa/3 = ; operate by 
S . 0, and we get 

(1 - e 2 /3 2 + e 2 /3 a ) S/3B = S/38 = 0. 

Hence 8 - a = - e*3 2 3 

Referred to this point as origin the equation becomes 


which shews that it belongs to a surface of revolution (of the 
second degree) whose axis is parallel to ft, since its intersection 
with a plane Sftp = a, perpendicular to that axis, lies also on the 


I 1 I yi- /Q2 / 

In fact, if the point be the focus of any meridian section of an 
oblate spheroid, the line is the directrix of the same. 

274. A sphere, passing through the centre of an ellipsoid, is cut 
by a series of spheres whose centres are on the ellipsoid and which 
pass through the centre thereof; find the envelop of the planes of 

Let (p a) 2 = a 2 be the first sphere, i.e. 

One of the others is p 2 2Svrp = 0, 

where 8^^ = - 1. 

The plane of intersection is 

S (to- - a.) p = 0. 

Hence, for the envelop (see next Chapter), 
jSfww = 

, ^r, where tzr = cfo, 


or <t>is = xp, {Vx = 0}, 

i.e. -BT = x<f> l p. 

T. Q. I, 14 

210 QUATERNIONS. [ 2 75- 

Hence a?Sp(f)~ l p = 1 ) 

and aSpcfr^p = Sap) 
and, eliminating x, 

a cone of the second degree. 

275. From a point in the outer of two concentric ellipsoids a 
tangent cone is drawn to the inner, find the envelop of the plane of 

If $ti7( G7 = 1 be the outer, and Spifrp = 1 be the inner, < 
and ty being any two self-conjugate linear and vector functions, 

the plane of contact is 

Stz ^rp = 1. 

Hence, for the envelop, S^ ^p = 0) 

Str tytJ = Oj 

therefore </>^ = octyp, 

or t*7 = xtpT^p. 

This gives xS . 

and xrS . 

and therefore, eliminating x, 

S.^rp^~ l ^p= -1, 

or 8 . piWtyp = 1 , 

another concentric ellipsoid, as \|r^)~ 1 i^ is a linear and vector function 
% suppose ; so that the equation may be written 

276. Find the locus of intersection of tangent planes at the 
extremities of conjugate diameters. 

If a, /3, 7 be the vector semi-diameters, the planes are 

with the conditions 271. 

Hence - ^S . ^a^fi^y = ^ = -^a + ^/3 + ^7> b y 92 
therefore Ti|rr = V3, 

since -v/ra, ^^3, -^7 form a rectangular system of unit-vectors. 

This may also evidently be written 


shewing that the locus is similar and similarly situated to the given 
ellipsoid, but larger in the ratio *J3 : 1. 

277. Find the locus of intersection of three tangent planes 
mutually at right angles. 

If p be the point of contact, 

Svnpp = - 1 
is the equation of the tangent plane. 

The vector perpendicular from the origin is ( 264) 

where a is a unit-vector. This gives 

whence, by the equation of the ellipsoid, 

Thus the perpendicular is 
a 7^ 

The sum of the squares of these, corresponding to a rectangular 
unit-system, is 

- 2 $a^~ 2 = m 2 , 
by 185. See also 279. 

278. Find the locus of the intersection of three spheres ivhose 
diameters are semi-conjugate diameters of an ellipsoid. 
If a be one of the semi-conjugate diameters 

And the corresponding sphere is 

with similar equations in @ and 7. Hence, by 92, 

^~ 1 P S . f a ^^r 7 = _ 
and, taking tensors, Tifr~ l p = 

or, finally, 

This is Fresnel s Surface of Elasticity, Chap. XII. 





279. Before going farther we may prove some useful properties 
of the function < in the form we are at present using viz. 

fiSip jSjp kSkp\ 

<PP = -~ 1 ~~a~ H 1-2 H 2 

V a b c 2 / 

We have p = ^ p JSjp kSkp, 

and it is evident that 


Also cf> 1 p = 

and so on. 

Again, if a, ft, y be any rectangular unit-vectors 

fe) 2 , (Sja? , (Sk*T 

But as 

we have 

&c. = &c. 
(%) 2 -f (%) 2 + (Skp)* = - p\ 

OL + 

V + P + ^- 

Lv (/ O 


$ . (bci(bft<}>v = - S . ( TT- -f ... I I -5- + ...)( 

/ \ a 

Sia, SJOL, Ska. 
Sift, Sjft, Skft 
Siy, Sjy, Sky 

And so on. These elementary investigations are given here for 
the benefit of those who have not read Chapter V. The student 
may easily obtain all such results in a far more simple manner by 
means of the formulae of that Chapter. 

280. Find the locus of intersection of a rectangular system of 
three tangents to an ellipsoid. 

If w be the vector of the point of intersection, a, ft, y the 


tangents, then, since CT 4- xa, must give equal values of x when 
substituted in the equation of the surface, so that 

S (OT 4- xa) O 4- xa) = - 1, 

we have (Svr^a) 2 = Sacfra (S^rfa 4-1). 

Adding this to the two similar equations in /3 and 7, we have 


or ((POT) = I 2 4~ -T2 H 2 1 (fa GHjyGr 4~ 1)> 


2 + - 2 j ($ 3T<) S7 4~1), 


an ellipsoid concentric with the first. 

281. If a rectangular system of chords be drawn through any 
point within an ellipsoid, the sum of the reciprocals of the rectangles 
under the segments into which they are divided is constant. 

With the notation of the solution of the preceding problem, nr 
giving the intersection of the vectors, it is evident that the 
product of the values of x is one of the rectangles in question 
taken negatively. 

Hence the required sum is 


This evidently depends on S^^^- only and not on the particular 
directions of a, (3, 7 : and is therefore unaltered if or be the vector 
of any point of an ellipsoid similar, and similarly situated, to the 
given one. [The expression is interpretable even if the point be 
exterior to the ellipsoid.] 

282. Shew that if any rectangular system of three vectors be 
drawn from a point of an ellipsoid, the plane containing their other 
extremities passes through a fixed point. Find the locus of the 
latter point as the former varies. 

With the same notation as before, we have 

mzr^xsr = 1, 
and S(& + xa) </> (w 4- xa) 1 ; 

therefore # = 


Hence the required plane passes through the extremity of 

- a - , 

and those of two other vectors similarly determined. It therefore 
(see 30) passes through the point whose vector is 

* ~ 


Thus the first part of the proposition is proved. 
But we have also tzr = 2 (< ^j 0, 
whence by the equation of the ellipsoid we obtain 

the equation of a concentric ellipsoid. 

283. Find the directions of the three vectors which are parallel 
to a set of conjugate diameters in each of two central surfaces of the 
second degree. 

Transferring the centres of both to the origin, let their equations 

=-I or 

and Sp^p = - 1 or 

If a, /3, 7 be vectors in the required directions, we must have ( 268) 

Sa(t>/3 = 0, Sofy/3 = j 

$807=0, /S^7=OV .................. (2). 

$y(/>a = 0, Sry^ct = J 

From these equations (/> || V{3y || tya, &c. 

Hence the three required directions are the roots of 

V.<j>p^p = ........................... (3). 

This is evident on other grounds, for it means that if one of the 
surfaces expand or contract uniformly till it meets the oilier, it will 
touch it successively at points on the three sought vectors. 
We may put (3) in either of the following forms 




and, as $ and i/r are given functions, we find the solutions (when 
they are all real, so that the problem is possible) by the processes 
of Chapter V. 

[Note. As <f>~ l ifr and ty~ l <j> are not, in general, self-conjugate 
functions, equations (4) do not signify that a, /3, 7 are vectors 
parallel to the principal axes of the surfaces 

S . p(j>~ l ^rp = - 1, S. p*^~ l <t>p = - 1. 

In these equations it does not matter whether c^rS/r is self-conjugate 
or not ; but it does most particularly matter when, as in (4), they 
are involved in such a manner that their non-conjugate parts do 
not vanish.] 

Given two surfaces of the second degree, which have parallel 
conjugate diameters, every surface of the second degree passing 
through their intersection has conjugate diameters parallel to these. 

For any surface of the second degree through the intersection 

Spfa = - 1 and 8 (p - a) i/r (p - a) = - e, 

is fSp<j>p -S(p-a.)^(p-a) = e-f, 

where e and f are scalars, of which f is variable. 
The axes of this depend only on the term 

Hence the set of conjugate diameters which are the same in all 
are parallel to the roots of 


as we might have seen without analysis. 

The locus of the centres is given by the equation 

where / is a scalar variable. 

284. Find the equation of the ellipsoid of which three conjugate 
semi-diameters are given. 

Let the vector semi-diameters be a, /3, 7, and let 

Spfo = -l 

be the equation of the ellipsoid. Then ( 269) we have 
= - 1, Sot<f>{3 = 0, 

= ; 

216 QUATERNIONS. [ 2 ^5- 

the six scalar conditions requisite (178) for the determination of 
the self-conjugate linear and vector function </>. 

They give a || VtfrPQy, 

or sea. = c/T 1 V{3y. 

Hence x xSa$a = S . afiy, 

and similarly for the other combinations. Thus, as we have 

pS . afty = aS . @yp + ftS . yap + <yS . afip, 
we find at once 

- <f>pS 2 . a{3y = VftyS . ftyp + VyaS . yap + VOL ft 8 . a/3p ; 
and the required equation may be put in the form 
S 2 . a{3y = S 2 .a{3p + S 2 .(3yp + S* . yap. 

The immediate interpretation is that if four tetrahedra be formed 
by grouping, three and three, a set of semi-conjugate vector axes of 
an ellipsoid and any fourth vector of the surface, the sum of the 
squares of the volumes of three of tliese tetrahedra is equal to the 
square of the volume of the fourth. 

285. A line moves with three of its points in given planes, find 
the locus of any fourth point. 

Let a, b, c be the distances of the three points from the fourth, 
a, ft 7 unit-vectors perpendicular to the planes respectively. If p 
be the vector of the fourth point, referred to the point of inter 
section of the planes, and a a unit-vector parallel to the line, we 
have at once 


Sap H j Sfip 

The condition Tcr = 1 gives the equation of an ellipsoid referred to 
its centre. 

We may write the equation in the form 

pSafty = aVpySao- + bVyaSffo- + cVa/3Sya 

= (pa . Sa(3y, suppose, 
and from this we find at once for the volume of the ellipsoid 

j Q 
. afty 


altogether independent of the relative inclinations of the three 
planes. This curious extension of a theorem of Monge is due to 

286. We see from 270 that (as in 31 (m)) we can write the 
equation of an ellipsoid in the elegant form 

where (/> is a self-conjugate linear and vector function, and we 
impose the condition 

Te = l. 

Hence, when the same ellipsoid is displaced by translation and 
rotation, by 119 we may write its equation as 

p = 8 + q<t>eq~\ 

with the condition that e is still a unit-vector. 

Where it touches a plane perpendicular to i, we must have, 

and = See. 

Hence e = U(f) (q~ l iq] 

at the point of contact ; and, if the plane touched be that of jk, 

= SiS-S. q~ l iq<l> U<l> (q~ l iq\ 
or = SiS + T(j> (q~ l iq). 

Thus, if we write 

q-*iq = a, q~ l jq = ft q~ l kq = 7, 
we have 

which gives the possible positions of the centre of a given ellipsoid 
when it is made to touch the fixed coordinate planes. 

We see by 279 that T8 is constant. And it forms an 
interesting, though very simple, problem, to find the region of 
this spherical surface to which the position of the centre of the 
ellipsoid is confined. This, of course, involves giving to a, ft 7 
all possible values as a rectangular unit system. 

287. For an investigation of the regions, on each of the 
coordinate planes, within which the point of contact is confined, 
see a quaternion paper by Plarr (Trans. R. 8. E. 1887). The 
difficulty of this question lies almost entirely in the eliminations, 
which are of a very formidable character. Subjoined is a mere 
sketch of one mode of solution, based on the preceding section. 


The value of p, at the point of contact with 

Sip = 0, 

or + S 

or, finally, 

Hence, in ordinary polar coordinates, the point of contact with 
the plane ofjk is 

To find the boundary of the region within which the point of 
contact must lie, we must make r a maximum or minimum, 9 
being constant, and a, /3, 7 being connected by the relations 

= Ty =1 

Differentiating (1) and (2), with the conditions 

d0 = 0, 
(because 6 is taken as constant) and 

dr = 

as the criterion of the maximum, we have eight equations which 
are linear and homogeneous in da, dj3, dy. Eliminating the two 
latter among the seven equations which contain them, we have 

= Sd* [(PSafr + &) tf/3 7l + (yScL 7l + 72 ) Sy0J ...... (3) 

where ft = <j> ( U$OL - U(j)/3), 

7. - 

But we have also, as yet unemployed, 

Sada = ........................... (4). 

Conditions (3) and (4) give the two scalar equations 
= (S/3 2J 8 - 800) Sj,/3 + -Sy^-S&T 


Theoretically, the ten equations (1), (2), and (5), suffice to 
eliminate the nine scalars involved in a, /3, 7 : and thus to leave 
a single equation in r, 0, which is that of the boundary of the 
region in question. 

The student will find it a useful exercise to work out fully the 
steps required for the deduction of (5). 

288. When the equation of a surface of the second order can 
be put in the form 

8p^p = -l ........................ (1), 

where (0 -g)(<l>- g,) ($ - g 9 ) = 0, 

we know that g, g v g z are the squares of the principal semi- 
diameters. Hence, if we put </> + h for <f> we have a second surface, 
the differences of the squares of whose principal semi axes are the 
same as for the first. That is, 

Sp(<j> + h)- l p = -l ..................... (2), 

is a surface confocal with (1). From this simple modification of 
the equation all the properties of a series of confocal surfaces may 
easily be deduced. We give a couple of examples. 

Any two confocal surfaces of the second degree, which meet, 
intersect at right angles. 

For the normal to (2) is evidently parallel to 

and that to another of the series, if it passes through the common 
point whose vector is p, is parallel to 

But a.Q + kr 

and this evidently vanishes if h and \ are. different, as they must 
be unless the surfaces are identical. 

To find the locus of the points of contact of a series of confocal 
surfaces with a series of parallel planes. 

Here the direction of the normal at the point, p, of contact is 


and is parallel to the common normal of the planes, say a, 

220 QUATERNIONS. [289. 

Thus the locus has the equation of a plane curve 
p = XOL + y$cL, 

and the relation between x and y is, by the general equation of 
the confocals, 

1 + y 2 Sct(j)CL + xyo? = 0. 

Hence the locus is a hyperbola. 

289. To find the conditions of similarity of two central surfaces 
of the second degree. 

Referring them to their centres, let their equations be 


Now the obvious conditions are that the axes of the one are 
proportional to those of the other. Hence, if 

9* - 2 <7 2 +m l g -m =0} , 

f-m\f + m d-m = v] 

be the equations for determining the squares of the reciprocals of 
the semi-axes, we must have 

m m 2 m , , 

=P, -=AC, =/* .................. (3), 

m 2 m 1 m 

where /A is an undetermined scalar. Thus it appears that there 
are but two scalar conditions necessary. Eliminating ft we have 

( . 
m * 

which are equivalent to the ordinary conditions. 

290. Find the greatest and least semi-diameters of a central 
plane section of an ellipsoid. 

Here Sp(f>p = 1 j ... , 

Sap= OJ" 

together represent the elliptic section; and our additional condition 
is that Tp is a maximum or minimum. 

Differentiating the equations of the ellipse, we have 

8<j>pdp = 0, 
Sadp = 0, 
and the maximum condition gives 

dTp = 0, 
or Spdp = 0. 


Eliminating the indeterminate vector dp we have 

........................... (2). 

This shews that the maximum or minimum vector, the normal at 
its extremity, and the perpendicular to the plane of section, lie in 
one plane. It also shews that there are but two vector-directions 
which satisfy the conditions, and that they are perpendicular to 
each other, for (2) is unchanged if ap be substituted for p. 

We have now to solve the three equations (1) and (2), to find 
the vectors of the two (four) points in which the ellipse (1) inter 
sects the cone (2). We obtain at once 

fa = xV . ((/T 1 a) Vap. 
Operating by S . p we have 

1 = Xp*Sa.(f)~ l OL. 

Sp6~ l a 
Hence _ = _ 

from which 8 . a(l +/>~ 1 a - ..................... (4); 

a quadratic equation in p 2 , from which the lengths of the maximum 
and minimum vectors are to be determined. By 184 it may be 

wip Sa^ a + p*S.a (m >2 - 0) a + a 2 - ......... (5). 

[If we had operated by $ . (f>~ 1 a. or by S . <f)~ l p, instead of by 
S . p, we should have obtained an equation apparently different 
from this, but easily reducible to it. To prove their identity is a 
good exercise for the student.] 

Substituting the values of p 2 given by (5) in (3) we obtain the 
vectors of the required diameters. [The student may easily prove 
directly that 

(l+p^- a and (I + pft)- 1 a 

are necessarily perpendicular to each other, if both be perpen 
dicular to a, and if p* and p 2 2 be different. See 288.] 

291. By (5) of last section we see that 

2 2 _ 

Pi T2 ~~ 

Hence the area of the ellipse (1) is 


222 QUATERNIONS. [292. 

Also the locus of central normals to all diametral sections of an 
ellipsoid, whose areas are equal, is the cone 

SoLf- 1 a = COL*. 
When the roots of (5) are equal, i.e. when 

(ra 2 a 2 - a0a) 2 - 4ma a a<Jf J a (6), 

the section is a circle. It is riot difficult to prove that this 
equation is satisfied by only two values of Ua, but another 
quaternion form of the equation gives the solution of this and 
similar problems by inspection. (See 292 below.) 

292. By 180 we may write the equation 

Spfp = -I 

in the new form S . \pf^p + pp 2 = 1, 

where p is a known scalar, and X and //, are definitely known (with 
the exception of their tensors, whose product alone is given) in 
terms of the constants involved in </>. [The reader is referred 
again also to 128, 129.] This may be written 

ZSXpSpp + (p - S\/JL) p" = - 1 (1). 

From this form it is obvious that the surface is cut by any plane 
perpendicular to X or /^ in a circle. For, if we put 

S\p = a, 

we have ^aSfip + (p S\fi) p 2 = 1, 

the equation of a sphere which passes through the plane curve of 

Hence X and fju of 180 are the values of a in equation (6) of 
the preceding section. 

293. Any two circular sections of a central surface of the 
second degree, whose planes are not parallel, lie on a sphere. 

For the equation 

(S\p - a} (Sfjip - b) = 0, 

where a and b are any scalar constants whatever, is that of a 
system of two non-parallel planes, cutting the surface in circles. 
Eliminating the product SXpSpp between this and equation (1) of 
last section, there remains the equation of a sphere. 

294. To find the generating lines of a central surface of the 
second degree. 

Let the equation be 


then, if a be the vector of any point on the surface, and ta a vector 
parallel to a generating line, we must have 

p = a + xty 
for all values of the scalar x. 

Hence S(a + XTZ} $ (a + mar) = 1 

gives the two equations 



The first is the equation of a plane through the origin parallel 
to the tangent plane at the extremity of a, the second is the 
equation of the asymptotic cone. The generating lines are there 
fore parallel to the intersections of these two surfaces, as is well 

From these equations we have 

2/^-DT Fate- 
where y is a scalar to be determined. Operating on this by S./3 
and S . 7, where ft and 7 are any two vectors not coplanar with a, 
we have 

S<sr (yQft+Vaft) = 0, Sw (y<fa - F 7 a) = (1). 

Hence S . <a (y<j>ft + Fa/3) (y<f>v - Fya) = 0, 

or my*S . afty SafaS . a/3y = 0. 

Thus we have the two values 

a / 1 

\f m 

belonging to the two generating lines. That they may be real 
it is clear that m must be negative : i.e. the surface must be the 
one-sheeted hyperboloid. 

295. But by equations (1) we have 

z*r = V. (y<j>ft + Fa/3) (y$y - F 7 a) 

which, according to the sign of y, gives one or other generating 

Here F/3 7 may be any vector whatever, provided it is not 
perpendicular to a (a condition assumed in last section), and we 
may write for it 0. 

Substituting the value of y before found, we have 


224 QUATERNIONS. [ 2 95- 

= V. c^aFac^- 1 0J--. F6tya, 

Y^ Wls 

or, as we may evidently write it, 

= <t>- l (V.a.V<l>oL0) J - ? * ."P0*a ............ (2). 

Put r = 

and we have 

ztx = 6~ l VOLT + A / --- . T, 

- v 

with the condition STC^OL = 0. 

[Any one of these sets of values forms the complete solution of the 
problem ; but more than one have been given, on account of their 
singular nature and the many properties of surfaces of the second 
degree which immediately follow from them. It will be excellent 
practice for the student to shew that 

is an invariant. This may most easily be done by proving that 
F. ^-0^0, = identically.] 

Perhaps, however, it is simpler to write a. for V/3y, and we thus 

ZTZ d>~ 1 V. a Vad)OL + * / -- . Facia. 
V m 

[The reader need hardly be reminded that we are dealing with the 
general equation of the central surfaces of the second degree the 
centre being origin.] 


1. Find the locus of points on the surface 

Spfo = - 1 
where the generating lines are at right angles to one another. 

2. Find the equation of the surface described by a straight 
line which revolves about an axis, which it does not meet, but 
with which it is rigidly connected. 


3. Find the conditions that 

may be a surface of revolution, with axis parallel to a given vector. 

4. Find the equations of the right cylinders which circum 
scribe a given ellipsoid. 

5. Find the equation of the locus of the extremities of per 
pendiculars to central plane sections of an ellipsoid, erected at the 
centre, their lengths being the principal semi-axes of the sections. 
[Fresnel s Wave-Surface. See Chap. XII] 

6. The cone touching central plane sections of an ellipsoid, 
which are of equal area, is asymptotic to a confocal hyperboloid. 

7. Find the envelop of all non-central plane . sections of an 
ellipsoid when their area is constant. 

8. Find the locus of the intersection of three planes, perpen 
dicular to each other, and touching, respectively, each of three 
confocal surfaces of the second degree. 

9. Find the locus of the foot of the perpendicular from the 
centre of an ellipsoid upon the plane passing through the extremi 
ties of a set of conjugate diameters. 

10. Find the points in an ellipsoid where the inclination 
of the normal to the radius-vector is greatest. 

11. If four similar and similarly situated surfaces of the 
second degree intersect, the planes of intersection of each pair pass 
through a common point. 

12. If a parallelepiped be inscribed in a central surface of the 
second degree its edges are parallel to a system of conjugate 

13. Shew that there is an infinite number of sets of axes for 
which the Cartesian equation of an ellipsoid becomes 

14. Find the equation of the surface of the second degree 
which circumscribes a given tetrahedron so that the tangent plane 
at each angular point is parallel to the opposite face ; and shew 
that its centre is the mean point of the tetrahedron. 

T.Q.I, 15 


15. Two similar and similarly situated surfaces of the second 
degree intersect in a plane curve, whose plane is conjugate to the 
vector joining their centres. 

16. Find the locus of all points on 

where the normals meet the normal at a given point. 

Also the locus of points on the surface, the normals at which 
meet a given line in space. 

17. Normals drawn at points situated on a generating line 
are parallel to a fixed plane. 

18. Find the envelop of the planes of contact of tangent 
planes drawn to an ellipsoid from points of a concentric sphere. 
Find the locus of the point from which the tangent planes are 
drawn if the envelop of the planes of contact is a sphere. 

19. The sum of the reciprocals of the squares of the perpen 
diculars from the centre upon three conjugate tangent planes is 

20. Cones are drawn, touching an ellipsoid, from any two 
points of a similar, similarly situated, and concentric ellipsoid. 
Shew that they intersect in two plane curves. 

Find the locus of the vertices of- the cones when these plane 
sections are at right angles to one another. 

21. Any two tangent cylinders to an ellipsoid intersect in two 
plane ellipses, and no other tangent cylinder can be drawn through 
either of these. 

Find the locus of these ellipses : 

(a) When the axes of the two cylinders are conjugate to each 
other, and to a given diameter. 

(6) When they are conjugate to each other, and to diameters 
lying in one plane. 

(c) When they are conjugate to each other, and to any 
diameter whatever. 

22. If a, j3, 7 be unit vectors parallel to conjugate semi- 
diameters of an ellipsoid, what is the vector 

/ ^ ~ / ~~ . - / ^ 


and what the locus of its extremity ? 


23. Find the locus of the points of contact of tangent planes 
which are equidistant from the centre of a surface of the second 

24. From a fixed point A, on the surface of a given sphere, 
draw any chord AD ; let D be the second point of intersection of 
the sphere with the secant BD drawn from any point B ; and 
take a radius vector AE, equal in length to BD , and in direction 
either coincident with, or opposite to, the chord AD: the locus 
of E is an ellipsoid, whose centre is A, and which passes through 
B. (Hamilton, Elements, p. 227.) 

25. Shew that the equation 

l z (e* -l)(e + ) = (Sap)* - ZeSapSa p + (Sap)* + (1 - e*) p\ 
where e is a variable (scalar) parameter, and a, a unit- vectors, 
represents a, system of confocal surfaces. (Ibid. p. 644.) 

26. Shew that the locus of the diameters of 

which are parallel to the chords bisected by the tangent planes to 
the cone 

Sp^p = 0, 

is the cone S . <~ 1 <> = 0. 

27. Find the equation of a cone, whose vertex is one summit 
of a given tetrahedron, and which passes through the circle 
circumscribing the opposite side. 

28. Shew that the locus of points on the. surface 

the normals at which meet that drawn at the point p = r, is on 
the cone 

= 0. 

29. Find the equation of the locus of a point the square 
of whose distance from a given line is proportional to its distance 
from a given plane. 

30. Shew that the locus of the pole of the plane 
with respect to the surface 

is a sphere, if a be subject to the condition 

2 a = C. 



31. Shew that the equation of the surface generated by lines 
drawn through the origin parallel to the normals to 

Sp(j) l p = - 1 
along its lines of intersection with 

is ^ 2 - hSw + h- l vr = 0. 

32. Common tangent planes are drawn to 

2/Skp%> + (jp-fiV)p* = -l, and Tp = h, 

find the value of h that the lines of contact with the former surface 
may be plane curves. What are they, in this case, on the sphere ? 
Discuss the case of 

p* - S V = 0. 

33. If tangent cones be drawn to 

from every point of 

the envelop of their planes of contact is 

34. Tangent cones are drawn from every point of 

S(p-a)<t>(p-cL) = -n*, 
to the similar and similarly situated surface 

fy<fr> = -l, 

shew that their planes of contact envelop the surface 

35. Find the envelop of planes which touch the parabolas 

p = a? + fit, p = au* + 7M, 
where a, /3, 7 form a rectangular system, and t and u are scalars. 

36. Find the equation of the surface on which lie the lines of 
contact of tangent cones drawn from a fixed point to a series of 
similar, similarly situated, and concentric ellipsoids. 

37. Discuss the surfaces whose equations are 

SapSj3p = Syp, 

38. Shew that the locus of the vertices of the right cones 
which touch an ellipsoid is a hyperbola. 


39. If a,, a 2 , a 3 be vector conjugate diameters of 

where < 3 m$? 4- m^ m = 0, 

shew that 

m 1 2 m 

and 2 (</>a) 2 = - ra 2 . 

40. Find the locus of the lines of contact of tangent planes 
from a given point to a series of spheres, whose centres are in one 
line and which pass through a given point in that line. 

41. Find the locus of a circle of variable radius, whose plane 
is always parallel to a given plane, and which passes through each 
of three given lines in space. 



296. WE have already seen ( 31 (I)) that the equations 

p = $t = 2 . af(t\ 

and p = (f>(t,u) = 2,. af(t, u), 

where a represents one of a set of given vectors, and f a scalar 
function of scalars t and u, represent respectively a curve and a 
surface. We commence the present too brief Chapter with a few 
of the immediate deductions from these forms of expression. We 
shall then give a number of examples, with little attempt at 
systematic development or even arrangement. 

297. What may be denoted by t and u in these equations is, 
of course, quite immaterial : but in the case of curves, considered 
geometrically, t is most conveniently taken as the length, s, of the 
curve, measured from some fixed point. In the Kinematical 
investigations of the next Chapter t may, with great convenience, 
be employed to denote time. 

298. Thus we may write the equation of any curve in 
space as 

p = (f)S, 

where is a vector function of the length, s, of the curve. Of 
course it is a linear function when, and only when, the equation 
(as in 31 (I)) represents a straight line. 

If be a periodic function, such that 

the curve is a reentrant one, generally a Knot in space. 


[In 306 it is shewn that when s is the arc certain forms of c/> 
are not admissible.] 

299. We have also seen ( 38, 39) that 

dp d , ,, 

-f- = -j- 6s = 6 s 
ds ds^ 

is a vector of unit length in the direction of the tangent at the 
extremity of p. 

At the proximate point, denoted by s 4- &s, this unit tangent 

vector becomes 

</> s + fi sSs + &c. 

But, because T<j> s 1, 

we have S . fisfi s = 0. 

Hence fi s, which is a vector in the osculating plane of the curve, 

is also perpendicular to the tangent. 

Also, if 80 be the angle between the successive tangents <j) s 
and fis + <f>"s&s + ...... , we have 

so that the tensor of <j>"s is the reciprocal of the radius of absolute 
curvature at the point s. 

300. Thus, if dP = <t>s be the vector of any point P of the 
curve, and if C be the centre of curvature at P, we have 

PC - - 

and thus 00 = 65777- 

(/> S 

is the equation of the locus of the centre of curvature. 

Hence also V. <f> s<j>"s or </> s<j>"s 

is a vector perpendicular to the osculating plane ; and therefore 

is the tortuosity of the given curve, or the rate of rotation of 
its osculating plane per unit of length. 

301. As an example of the use of these expressions let us find 
the curve whose curvature and tortuosity are both constant. 
We have curvature = Tfi s = Tp" = c. 


Hence fisfi s = p p" = ca, 

where a is a unit vector perpendicular to the osculating plane. 
This gives 

if c l represent the tortuosity. 
Integrating we get 

p> // = c/ + /3 .......................... (1), 

where /3 is a constant vector. Squaring both sides of this equation, 
we get 

c 2 = c, 2 - /3 2 - ZcfiPp 

(for by operating with 8 . p upon (1) we get f c, = $//), 


Multiply (1) by /o , remembering that 

2V- 1, 

and we obtain p" = c t + p fi, 

or, by integration, p = CjS p/3 + a- ..................... (2), 

where a is a constant quaternion. Eliminating p , we have 

of which the vector part is 

p" -p& = - erf - Fa/3. 
The complete integral of this equation is evidently 

p = ? cos. 8 T0 + 7]sm.sT/3-~(c lS /3+ Fa/3) ...... (3), 

f and ?; being any two constant vectors. We have also by (2), 

SPp = CjS + Sa, 

which requires that Sftj; = 0, Spy = 0. 
The farther test, that Tp f = 1, gives us 

2 . 

C ~T~ C. 

This requires, of course, 

= 0, Tt=Tr, = , 

so that (3) becomes the general equation of a helix traced on a 
right cylinder. (Compare 31 (m).) 


302. The vector perpendicular from the origin on the tangent 
to the curve 

p = $8 

is, of course, Vpp, or p Vpp 

(since p is a unit vector). 

To find a common property of curves whose tangents are all 
equidistant from the origin. 


which may be written p 2 S*pp = c 2 ..................... (i). 

This equation shews that, as is otherwise evident, every curve 
on a sphere whose centre is the origin satisfies the condition. For 


p z = c 2 gives Spp = 0, 

and these satisfy (1). 

If Spp does not vanish, the integral of (1) is 

c = s ........................... (2), 

an arbitrary constant not being necessary, as we may measure s 
from any point of the curve. The equation of an involute which 
commences at this assumed point is 

or = p sp. 
This gives 2V = Tp* + s* + ZsSpp 

= Tp* + s 2 - 2 5 jTp* - c 2 , by (1), 
= c 2 , by (2). 

This includes all curves whose involutes lie on a sphere about the 

303. Find the locus of the foot of the perpendicular drawn to a 
tangent to a right helix from a point in the axis. 
The equation of the helix is 

o o 

p = a cos - + j3 sin - + ys, 
a a 

where the vectors a, ft, 7 are at right angles to each other, and 

Ta=T@ = b, while 
(The latter condition is from Tp = 1.) 

234 QUATERNIONS. [34- 

The equation of the required locus is, by last section, 

f s a*-b 2 s\ O f . s d*-b* s\ b* 
= a ( cos - H -- -s sm - + p sm ---- 3 scos - + y- 2 s. 
\ a a aj \ a a 3 aj a 

This curve lies on the hyperboloid whose equation is 

as the reader may easily prove for himself. 

304. To find the least distance between consecutive tangents to a 
tortuous curve. 

Let one tangent be -or = p + xp , 
then a consecutive one, at a distance s along the curve, is 

The magnitude of the least distance between these lines is, by 
216, 223, 

TVp p"$s 

if we neglect terms of higher orders. 

It may be written, since p p" is a vector, and Tp = 1, 

But ( 140 (2)) -j^fr t = V^f-fT & = ^ p S . p p p". 

Hence ^Ls.Up"Vp p" 

is the small angle, S<, between the two successive positions of the 
osculating plane. [See also 300.] 

Thus the shortest distance between two consecutive tangents is 
expressed by the formula 


where r, = m , is the radius of absolute curvature of the tortuous 



305. Let us recur for a moment to the equation of the parabola 

Here p = ( 

whence, if we assume $a/3 = 0, 

from which the length of the arc of the curve can be derived in 
terms of t by integration. 

Again, p" = (a + ft 


<M = d L J_ ,dt8 P(+AQ 
ds 2 ds T~(a~+~ftt) ds T(a + fttf 

Hence p" = 

and therefore, for the vector of the centre of curvature we have 
( 300), 

OT = a t + ^ - (a 8 + fftj (- /3<x 

= / Y + " 
which is the quaternion equation of the evolute. 

306. One of the simplest forms of the equation of a tortuous 
curve is 

where a, ft, 7 are any three non-coplanar vectors, and the numerical 
factors are introduced for convenience. This curve lies on a para 
bolic cylinder whose generating lines are parallel to 7; and also on 
cylinders whose bases are a cubical and a semi-cubical parabola, 
their generating lines being parallel to ft and a respectively. We 
have by the equation of the curve 

Pa + ftt +-=- ) -y-, 
2 J ds 

from which, by Tp = 1, the length of the curve can be found in 
terms of t ; and 

236 QUATERNIONS. [307. 

from which p" can be expressed in terms of s. The investigation 
of various properties of this curve is very easy, and will be of great 
use to the student. 

[Note. It is to be observed that in this equation t cannot stand 
for s, the length of the curve. It is a good exercise for the student 
to shew that such an equation as 

p = as + 0s 2 + 75 3 , 
or even the simpler form 

p = as + /3s 2 , 
involves an absurdity.] 

307. The equation p = </>*e, 

where (/> is a given self-conjugate linear and vector function, t a 
scalar variable, and e an arbitrary vector constant, belongs to a 
curious class of curves. 

We have at once = ( fi ^# ^ > 

where log </> is another self-conjugate linear and vector function, 
which we may denote by ^. These functions are obviously com 
mutative, as they have the same principal set of rectangular vectors, 
hence we may write 


which of course gives -7^- = ^ 2 p, &c., 


since ^ does not involve t. 

As a verification, we should have 

where e is the base of Napier s Logarithms. 

This is obviously true if ^ = e stx , 
or </> = e*, 

or log < = x, 

which is our assumption. See 337, below, 

[The above process is, at first sight, rather startling, but the 


student may easily verify it by writing, in accordance with the 
results of Chapter V, 

<e = - g.aSae - g^S/Be - gjySyc, 
whence <f> f = g\ a.Scx.6 g\/3S/3e g\ySye. 

He will find at once 

X = - log g l aSae - log gftSpe 
and the results just given follow immediately.] 

308. That the equation 

represents a surface is obvious from the fact that it becomes the 
equation of a definite curve whenever either t or u has a particular 
value assigned to it. Hence the equation at once furnishes us with 
two systems of curves, lying wholly on the surface, and such that 
one of each system can, in general, be drawn through any assigned 
point on the surface. Tangents drawn to these curves a,t their 
point of intersection must, of course, lie in the tangent plane, whose 
equation we have thus the means of forming. [Of course, there 
may occasionally be cases of indeterminateness, as when the curves 
happen to touch one another. But the general consideration of 
singular points on surfaces is beyond the scope of this work.] 

309. By the equation we have 

where the brackets are inserted to indicate, partial differential 
coefficients. If we write this as 

the normal to the tangent plane is evidently 


and the equation of that plane 

310. Thus, as a simple example, let 
p = ta + u(3 + tuy. 

This surface is evidently to be constructed by drawing through each 
point to., of the line a, a line parallel to /3 + ty ; or through u/3, a 
line parallel to a + uy. 

238 QUATERNIONS. [3 11 - 

We may easily eliminate t and u, and obtain 
8 . @ypS .jap = S. a/3yS . a/3p ; 

and the methods of last chapter enable us to recognise a hyperbolic 

Again, suppose a straight line to move along a fixed straight 
line, remaining always perpendicular to it, while rotating about it 
through an angle proportional to the space it has advanced ; the 
equation of the ruled surface described will evidently be 

p = at + u ((3 cos t + 7 sin t) .................. (1 ) , 

where a, /3, 7 are rectangular vectors, and 

frj r\ _ fjj 

This surface evidently intersects the right cylinder 

p = a (j3 cos t + 7 sin t) + VOL, 
in a helix ( 31 (m), 301) whose equation is 

p = at + a(/3 cos t + 7 sin t). 

These equations illustrate very well the remarks made in 31 (I), 
308, as to the curves or surfaces represented by a vector equation 
according as it contains one or two scalar variables. 
From (1) we have 

dp = [a u (j3 sin t 7 cos t)] dt + (/3 cos t + 7 sin t) du, 
so that the normal at the extremity of p is 

Ta (7 cost-/3 sin t)-u T/3*Ua. 

Hence, as we proceed along a generating line of the surface, for 
which t is constant, we see that the direction of the normal changes. 
This, of course, proves that the surface is not developable. 

311. Hence the criterion for a developable surface is that if it 
be expressed by an equation of the form 

where </> and tyt are vector functions, we must have the direction 
of the normal 

independent of u. 

This requires either Vtytty t 0, 

which would reduce the surface to a cylinder, all the generating 
lines being parallel to each other ; or 


This is the criterion we seek, and it shews that we may write, for a 
developable surface in general, the equation 

p = <l>t+ ufit (1). 

Evidently p = <f)t 

is a curve (generally tortuous) and <fit is a tangent vector. Hence 
a developable surface is the locus of all tangent lines to a tortuous 

Of course the tangent plane to the surface is the osculating 
plane at the corresponding point of the curve; and this is indicated 
by the fact that the normal to (1) is parallel to 

Vtftf t. (See 300.) 

To find the form of the section of the surface made by a normal 
plane through a point in the curve. 

The equation of the surface in the neighbourhood of the 
extremity of p is approximately 

s 2 

07 = p + Sp + - p" + &C. + X (p + Sp" + &C.). 

The part of *& p which is parallel to p is 

therefore OT - p = Ap + ( | + xs ) p" - ( S - + ~ ) pVp p " + ... . 

\z / \o z / 

And, when ^4 = 0, i.e. in the normal section, we have approximately 

x = s, 

s z s 3 
so that OT p = - p" + ^ p Vp p". 

Hence the curve has an equation of the form 

a = S 2 a + s 3 /3, 
a semicubical parabola. 

312. A Geodetic line is a curve drawn on a surface so that its 
osculating plane at any point contains the normal to the surface. 
Hence, if v be the normal at the extremity of p, p and p the first 
and second differentials of the vector of the geodetic, 

S . vp p" = 0, 
which may be easily transformed into 

V.vdU P =0. 


313. In the sphere Tp = avte have 


hence S . pp p" = 0, 

which shews of course that p is confined to a plane passing through 
the origin, the centre of the sphere. 

For a formal proof, we may proceed as follows 
The above equation is equivalent to the three 

from which we see at once that 6 is a constant vector, and 
therefore the first expression, which includes the others, is the 
complete integral. 

Or we may proceed thus 

Q = - P S. pp p" + p"S . p*p = F. Vpp Vpp" = V. Vpp d Vpp, 
whence by 140 (2) we have at once 

UVpp = const. = suppose, 
which gives the same results as before. 

314. In any cone, when the vertex is taken as origin, we 
have, of course, 

since p lies in the tangent plane. But we have also 

Hence, by the general equation of 312, eliminating v we get 
= 8. pp Vp p" = SpdUp by 140 (2). 

Integrating C = 8pUp - j SdpUp = SpUp + j Tdp. 

The interpretation of this is, that the length of any arc of the 
geodetic is equal to the projection of the side of the cone (drawn 
to its extremity) upon the tangent to the geodetic. In other 
words, when the cone is developed on a plane the geodetic becomes a 
straight line. A similar result may easily be obtained for the 
geodetic lines on any developable surface whatever. 

315. To find the shortest line connecting two points on a given 

Here I Tdp is to be a minimum, subject to the condition that 
dp lies in the given surface. [We employ 8, though (in the 


notation we employ) it would naturally denote a vector, as the 
symbol of variation.] 

Now SJTdp=f STdp = - j -^-- p = ~JS. UdpdSp 

where the term in brackets vanishes at the limits, as the extreme 
points are fixed, and therefore Sp = at each. 
Hence our only conditions are 


S.Spd Udp = 0, and SvSp = 0, giving 

V.vdUdp = 0, as in 31 2. 

If the extremities of the curve are not given, but are to lie on 
given curves, we must refer to the integrated portion of the 
expression for the variation of the length of the arc. And its 


shews that the shortest line cuts each of the given curves at right 

316. The osculating plane of the curve 

is 8 . $t$"t O - p) = ...................... (1), 

and is, of course, the tangent plane to the surface 

p = ft + u<j> t ........................... (2). 

Let us attempt the converse of the process we have, so far, 
pursued, and endeavour to find (2) as the envelop of the variable 
plane (1). 

Differentiating (1) with respect to t only, we have 

8. < </> " (r-p) = 0. 
By this equation, combined with (1), we have 

or 5T = p + uty = cj) + u$ , 

which is equation (2). 

317. This leads us to the consideration of envelops generally, 
and the process just employed may easily be extended to the 
problem of finding the envelop of a series of surfaces whose 
equation contains one scalar parameter. 

T. Q. I. 16 

242 QUATERNIONS. [318. 

When the given equation is a scalar one, the process of finding 
the envelop is precisely the same as that employed in ordinary 
Cartesian geometry, though the work is often shorter and simpler. 

If the equation be given in the form 

p = Tr, U, v, 

where ^r is a vector function, t and u the scalar variables for any 
one surface, v the scalar parameter, we have for a proximate surface 

u l9 v = p r t -v u u v v. 

Hence at all points on the intersection of two successive surfaces 
of the series we have 

y t St + i/r M Su + Tjr . Sv = 0, 

which is equivalent to the following scalar equation connecting the 
quantities t, u and v ; 

s. * ,* * .=<>. 

This equation, along with 

p=ty(t t u, v), 

enables us to eliminate t, u, v, and the resulting scalar equation is 
that of the required envelop. 

318. As an example, let us find the envelop of the osculating 
plane of a tortuous curve. Here the equation of the plane is 

( 316) 

S.(n-p) $t$ t = 0, 

or OT = <f)t + x(f) t + y<j>"t = ty (cc, y, t), 

if p = fa 

be the equation of the curve. 

Our condition is, by last section, 

fl- f.fft .-O, 

or S . fit fi t [fit + xfi t + yfi"t] = 0, 

or yS.fitfi tfi"t = 0. 

Now the second factor cannot vanish, unless the given curve 
be plane, so that we must have 

2/ = 0, 

and the envelop is r = fit + xfit 

the developable surface, of which the given curve is the edge of 
regression, as in S 316. 


319. When the equation contains two scalar parameters, its 
differential coefficients with respect to them must vanish, and we 
have thus three equations from which to eliminate two numerical 

A very common form in which these two scalar parameters 
appear in quaternions is that of an arbitrary unit-vector. In this 
case the problem may be thus stated : 

Find the envelop of the surface whose scalar equation is 

where a is subject to the one condition 


Differentiating with respect to a alone, we have 
Svda = 0, Sada = 0, 

where v is a known vector function of p and a. Since da may have 
any of an infinite number of values, these equations shew that 

VOLV = 0. 

This is equivalent to two scalar conditions only, and these, in 
addition to the two given scalar equations, enable us to eliminate a. 
With the brief explanation we have given, and the examples 
which follow, the student will easily see how to treat any other set 
of data he may meet with in a question of envelops. 

320. Find the envelop of a plane whose distance from the 
origin is constant. 

Here Sap = - c, 

with the condition Ta = 1. 

Hence, by last section, Vpa = 0, 
and therefore p = ca, 

or Tp = c, 

the sphere of radius c, as was to be expected. 

If we seek the envelop of those only of the planes which are 
parallel to a given vector /3, we have the additional relation 

SOL& = 0. 

In this case the three differentiated equations are 
Spda=Q, Sada = 0, S{3da = 0, 

and they give S. afip = 0. 


244 QUATERNIONS. [3 2 I . 

Hence a=U./3V/3p, 

and the envelop is TV/3p = cT0, 

the circular cylinder of radius c and axis coinciding with /3. 

By putting Saj3 = e, where e is a constant different from zero, 
we pick out all the planes of the series which have a definite 
inclination to /3, and of course get as their envelop a right cone. 

321. The equation S"ap + 2S.a/3p = b 

represents a parabolic cylinder, whose generating lines are parallel 
to the vector aVa(3. For the equation is of the second degree, and 
is riot altered by increasing p by the vector ocaVaff , also the 
surface cuts planes perpendicular to a in one line, and planes 
perpendicular to Va/3 in two parallel lines. Its form and position 
of course depend upon the values of a, /3, and b. It is required to 
find its envelop if /3 and b be constant, and a be subject to the one 
scalar condition 

r = i. 

The process of 319 gives, by inspection, 

pSap -f Vftp = xa. 
Operating by 8 . a, we get 

which gives S . a{3p = x + b. 

But, by operating successively by S . F/3p and by 8 . p, we have 

and (p 2 x) Sap = 0. 

Omitting, for the present, the factor Sap, these three equations 
give, by elimination of x and a, 

which is the equation of the envelop required. 

This is evidently a surface of revolution of the fourth degree 
whose axis is ft ; but, to get a clearer idea of its nature, put 

c p 1 = *r, 
and the equation becomes 

which is obviously a surface of revolution of the second degree, 
referred to its centre. Hence the required envelop is the reciprocal 
of such a surface, in the sense that the rectangle under the lengths 


of condirectional radii of the two is constant : i.e. it is the Electric 

We have a curious particular case if the constants are so 
related that 

b + /3 2 = 0, 

for then the envelop breaks up into the two equal spheres, 
touching each other at the origin, 

P 2 =S/3 P , 

while the corresponding surface of the second order becomes the 
two parallel planes 

Sfa = c 2 . 

322. The particular solution above met with, viz. 

Sap = 0, 

limits the original problem, which now becomes one of finding the 
envelop of a line instead of a surface. In fact this equation, taken 
in conjunction with that of the parabolic cylinder, belongs to that 
generating line of the cylinder which is the locus of the vertices 
of the principal parabolic sections. 
Our equations become 

2S.a/3p = b, 
Sap = 0, 
Ta = l- 
whence V/3p = xa ; 

giving x = - S . a(3p = - ^ , 

and thence TV/3p = ^ ; 

so that the envelop is a circular cylinder whose axis is {3. [It is 
to be remarked that the equations above require that 

Sap = 0, 

so that the problem now solved is merely that of the envelop of a 
parabolic cylinder which rotates about its focal line. This discussion 
has been entered into merely for the sake of explaining a peculiarity 
in a former result, because of course the present results can be 
obtained immediately by an exceedingly simple process.] 

323. The equation SapS . a/3p = a 2 , 
with the condition Toe 1, 

246 QUATERNIONS. [324. 

represents a series of hyperbolic cylinders. It is required to find 
their envelop. 

As before, we have 

pS . aftp + VjSpSap = xa, 
which by operating by $ . a, 8 . p, and S .VjBp, gives 

W = -x y 
p*S . afip = xSap, 
(Vpp)*Sap = xS.oLpp. 
Eliminating a and x we have, as the equation of the envelop, 

p Wpf-tof. 

Comparing this with the equations 

f = ~ 2a 2 , 
and (V/3p) 2 = -2a\ 

which represent a sphere and one of its circumscribing cylinders, 
we see that, if condirectional radii of the three surfaces be drawn 
from the origin, that of the new surface is a geometric mean 
between those of the two others. 

324. Find the envelop of all spheres which touch one given line 
and have their centres in another. 

Let p = {3 + yry 

be the line touched by all the spheres, and let XOL be the vector of 
the centre of any one of them, the equation is (by 213, or 214) 

" --. 7 -. 

or, putting for simplicity, but without loss of generality, 

Ty = 1, Sa/3 = 0, Spy = 0, 
so that @ is the least vector distance between the given lines, 

(p - xa? = (/3- xaf + x 2 S 2 ay, 
and, finally, p 2 - /3 2 - ZxSap = x 2 S 2 aj. 

Hence, by 317, - 2Sap = 2xS 2 ay. 

[This gives no definite envelop, except the point p = /3, if 

Say = 0, 

i.e. if the line of centres is perpendicular to the line touched by 
all the spheres.] 

Eliminating x, we have for the equation of the envelop 
S*ap + S 9 *v (p 2 - *) - 0, 


which denotes a surface of revolution of the second degree, whose 
axis is a. 

Since, from the form of the equation, Tp may have any 
magnitude not less than T/3, and since the section by the plane 

Sap = 
is a real circle, on the sphere 

P -P = 0, 

the surface is a hyperboloid of one sheet. 

[It will be instructive to the student to find the signs of the 
values of g v g v g 3 as in 177, and thence to prove the above 

325. As a final example of this kind let us find the envelop 
of the hyperbolic cylinder 

SapS/3p-c = 0, 
where the vectors a and (B are subject to the conditions 

Tct = T{3 = l, 
Say = 0, S/3S = 0, 
y and 8 being given vectors. 

[It will be easily seen that two of the six scalars involved 
in a, still remain as variable parameters.] 

We have Sada = 0, Syda = 0, 

so that da = x Vay. 

Similarly d/3 = yV/38. 

But, by the equation of the cylinders, 

SapSpd/3 + SpdaSj3p = 0, 

or ySapS . /38p + asS . aypS/3p = 0. 

Now by the nature of the given equation, neither Sap nor Sftp can 
vanish, so that the independence of da and dj3 requires 

Hence a=U.yVyp, = U.S VSp, 

and the envelop is T.VypVSp - cTy& = 0, 

a surface of the fourth degree, which may be constructed by laying 
off mean proportionals between the lengths of condirectional radii 
of two equal right cylinders whose axes meet in the origin. 

326. We may now easily see the truth of the following 
general statement. 

248 QUATERNIONS. [3 2 6. 

Suppose the given equation of the series of surfaces, \vhose 
envelop is required, to contain m vector, and n scalar, parameters ; 
and that these are subject to p vector, and q scalar, conditions. 

In all there are 3m + n scalar parameters, subject to 3p + q 
scalar conditions. 

That there may be an envelop we must therefore in general 

l y or = 2. 

In the former case the enveloping surface is given as the locus of 
a series of curves, in the latter of a series of points. 

Differentiation of the equations gives us 3p + q + 1 equations, 
linear arid homogeneous in the 3m + n differentials of the scalar 
parameters, so that by the elimination of these we have one final 
scalar equation in the first case, two in the second ; and thus in 
each case we have just equations enough to eliminate all the 
arbitrary parameters. 

Sometimes a very simple consideration renders laborious cal 
culation unnecessary. Thus a rectangular system turns about the 
centre of an ellipsoid. Find the envelop of the plane which passes 
through the three points of intersection. 

If a, /3, 7 be the rectangular unit-system, the points of 
intersection with 

Spfo = - 1 

are at the extremities of 

- SOLQOL J -&&<!># -Syfa 

And if these lie in the plane 

*/.-! . ................. (1), 

we must have e = Sa\/ ScL<f>a. ........................ (2). 

It would be troublesome to work out the envelop of (1), with 
(2) and the conditions of a rectangular unit-system as the data, 
but we may proceed as follows. 

The length of the perpendicular from the centre on the 
plane (1) is 

- (Sa^a + S/3(f>l3 + 

a constant, by 185. Hence the envelop is a sphere of which this 
is the radius ; and it has the same property for all the ellipsoids 


which, having their axes in the same lines as the first, intersect it 
at the point on 

i+j + k. 

We may obtain the result in another way. By 281 the sum 
of the reciprocals of the squares of three rectangular central vectors 
of an ellipsoid is constant ; while it is easily shewn (see Ex. 20 to 
Chap. VII.) that the same sum with regard to a plane is the 
reciprocal of the square of its distance from the origin. 

327. To find the locus of the foot of the perpendicular drawn 
from the origin to a tangent plane to any surface. 

If Svdp = 

be the differentiated equation of the surface, the equation of the 
tangent plane is 

S(<SF-p)v = 0. 

We may introduce the condition 

Svp = - 1, 

which in general alters the tensor of v, so that v~ l becomes the 
required vector perpendicular, as it satisfies the equation 

Svrv = - 1. 

It remains that we eliminate p between the equation of the 
given surface, and the vector equation 

37 = V~ l . 

The result is the scalar equation (in -BT) required. 
For example, if the given surface be the ellipsoid 

we have trr* 1 = v = 

so that the required equation is 


which is Fresnel s Surface of Elasticity. ( 278.) 

It is well to remark that this equation is derived from that of 
the reciprocal ellipsoid 

Sp^p = -l 

by putting iar~ l for p. 

328. To find the reciprocal of a given surface with respect to 
the unit sphere whose centre is the origin. 

With the condition Spv 1, 

250 QUATERNIONS. [S 2 9- 

of last section, we see that v is the vector of the pole of the 
tangent plane 

S(vr-p)v = Q. 

Hence we must put -BT = v, 

and eliminate p by the help of the equation of the given surface. 
Take the ellipsoid of last section, and we have 

VT = <f)p, 

so that the reciprocal surface is represented by 

8vp*v = -l. 

It is obvious that the former ellipsoid can be produced from 
this by a second application of the process. 
And the property is general, for 

gives, by differentiation, and attention to the condition 

Svdp = 
the new relation Spdv = 0, 

so that p and v are corresponding vectors of the two surfaces : 
either being that of the pole of a tangent plane drawn at the 
extremity of the other. 

329. If the given surface be a cone with its vertex at the 
origin, we have a peculiar case. For here every tangent plane 
passes through the origin, and therefore the required locus is 
wholly at an infinite distance. The difficulty consists in Spv 
becoming in this case a numerical multiple of the quantity which 
is equated to zero in the equation of the cone, so that of course 
we cannot put as above 

330. The properties of the normal vector v enable us to write 
the partial differential equations of families of surfaces in a very 
simple form. 

Thus the distinguishing property of Cylinders is that all their 
generating lines are parallel. Hence all positions of v must be 
parallel to a given plane or 

Sav = 0, 
which is the quaternion form of the well-known equation 

. dF dF dF 
I -= f- ra j -t- n , 0. 
dx ay dz 


To integrate it, remember that we have always 

Svdp = 0, 

and that as v is perpendicular to a it may be expressed in terms 
of any two vectors, fi and 7, each perpendicular to a. 

Hence v = x/3 + yy, 

and xS{3dp + ySydp = 0. 

This shews that S0p and Syp are together constant or together 
variable, so that 


where /is any scalar function whatever. 

331. In Surfaces of Revolution the normal intersects the axis. 
Hence, taking the origin in the axis a, we have 

S.apv = 0, 
or v = xa + yp. 

Hence xSadp + ySpdp = 0, 

whence the integral Tp =f(Sap). 

The more common form, which is easily derived from that just 
written, is 

In Cones we have Svp = 0, 

and therefore 

Svdp = S.v(TpdUp+ UpdTp) = TpSvd Up. 

Hence SvdUp = 0, 

so that v must be a function of Up, and therefore the integral is 

which simply expresses the fact that the equation does not involve 
the tensor of p, i.e. that in Cartesian coordinates it is homogeneous. 

332. If equal lengths be laid off on the normals drawn to any 
surface, the new surface formed by their extremities is normal to the 
same lines. 

For we have -BT = ^ + a Uv, 

and Svd-v = Svdp + aSvd Uv = 0, 

which proves the proposition. 

Take, for example, the surface 

S P<I>P = - 1 5 

252 QUATERNIONS. [333- 

the above equation becomes 

= f >+ > 

so that a = I -=-r- +1 OT 


and the equation of the new surface is to be found by eliminating 
mT (written x) between the equations 


333. It appears from last section that if one orthogonal 
surface can be drawn cutting a given system of straight lines, an 
indefinitely great number may be drawn : and that the portions of 
these lines intercepted between any two selected surfaces of the 
series are all equal. 

Let p = a -f XT, 

where a and r are vector functions of p, and x is any scalar, be 
the general equation of a system of lines : we have 

Srdp = = S(p-o-)dp 

as the differentiated equation of the series of orthogonal surfaces, 
if it exist. Hence the following problem. 

334. It is required to find the criterion of integrability of the 

Svdp = .............................. (1) 

as the complete differential of the equation of a series of surfaces. 

Hamilton has given (Elements, p. 702) an extremely elegant 
solution of this problem, by means of the properties of linear and 
vector functions. We adopt a different and somewhat less rapid 
process, on account of some results it offers which will be useful to 
us later ; and also because it will shew the student the connection 
of our methods with those of ordinary differential equations. 

If we assume Fp = C 

to be the integral, we have by 144, 


Comparing with the given equation, (1), we see that the latter 
represents a series of surfaces if v, or a scalar multiple of it, can be 
expressed as VF. 


If v = VF, 

_ _ fd*F d*F d*F\ 
we have Vz; = V 2 F = - -^~ 2 4- -, + -j-y , 

V dx dy dz* / 

and the last-written quantities are necessarily scalars, so that the 
only requisite condition of the integrability of (1) is 

FVi/ = 0....: ......................... (2). 

If v do not satisfy this criterion, it may when multiplied by a 
scalar. Hence the farther condition 

7V (wv) = 0, 
which may be written 

Vrfw-iuWv = b ...................... (3). 

This requires that 

If then (2) be not satisfied, we must try (4). If (4) be satisfied w 
will be found from (3) ; and in either case (1) is at once integrable. 

[If we put dv = (frdp, 

where is a linear and vector function, not necessarily self- 
conjugate, we have 

by 185. Thus, if (/> be self-conjugate, e = 0, and the criterion (2) 
is satisfied. If <f> be not self-conjugate we have by (4) for the 

Sev = 0. 

These results accord with Hamilton s, lately referred to, but the 
mode of obtaining them is quite different from his.] 

335. As a simple example let us first take lines diverging 
from a point. Here v || p, and we see that if v = p 

Vy = -3, 

so that (2) is satisfied. And the equation is 

Spdp = 0, 

whose integral Tp 2 = C 

gives a series of concentric spheres. 

Lines perpendicular to, and intersecting, a fixed line. 
If a be the fixed line, (3 any of the others, we have 

0, S/3dp = 0. 

254 QUATERNIONS. [335. 

Here v | a. Vap, 

and therefore equal to it, because (2) is satisfied. 

Hence S . dp a. Vap = 0, 

or S. VapVadp = 0, 

whose integral is the equation of a series of right cylinders 

To find the orthogonal trajectories of a series of circles whose 
centres are in, and their planes perpendicular to, a given line. 

Let a be a unit- vector in the direction of the line, then one of 
the circles has the equations 

Tp = C] 
Sap = C \ 

where C and C are any constant scalars whatever. 
Hence, for the required surfaces 

v 1| d lP 1| Vap, 

where d^ is an element of one of the circles, v the normal to the 
orthogonal surface. Now let dp be an element of a tangent to 
the orthogonal surface, and we have 

Svdp = S . apdp = 0. 

This shews that dp is in the same plane as a and p, i.e. that the 
orthogonal surfaces are planes passing through the common ax 
[To integrate the equation 

S.apdp = Q 

evidently requires, by 334, the introduction of a factor. For 
= - aSVp + SaV . p ( 90, (1)) 

so that the first criterion is not satisfied. But 

8. VapV.VVap=2S.aVcLp = 0, 

so that the second criterion holds. It gives, by (3) of 334, 
- V . VapVw + 2wa = 0, 

or pSaVw aSpVw + 2wa = 0. 

That is ^ , . 

= 2w 

These equations are satisfied by 




But a simpler mode of integration is easily seen. Our equation 
may be written 

. , 

P Up /3 

which is immediately integrable, (3 being an arbitrary but constant 

As we have not introduced into this work the logarithms of 
versors, nor the corresponding angles of quaternions, we must refer 
to Hamilton s Elements for a further development of this point.] 

336. As another example, let us find series of surfaces wJ tick, 
together, divide space into cubes. 

If p be the vector of one series which has the required property, 
a that of a second, it is clear that (u being a scalar) 

da = uq~ l dpq ........................... (1), 

where u and q are functions of p. For, to values of dp belonging 
to edges of one cube correspond values of dcr belonging to edges of 
another. Operate by S . a, where a is any constant vector, then 

Sada = iiS . qaq l dp. 

As the left-hand member is a complete differential, we have by 

FV (uqaq~ l ) = 0. 

This is easily put in the form {Chap. IV., Ex. (4), and 140 (8)} 


V. qaq- 1 = - 2qaq~ l S . Vqq~ l + 2S (qaq~ l V) q.q 1 .. .(2). 

Multiply by qaq~ l , and add together three equations of the resulting 
form, in which the values of a. form a rectangular unit system. 


+ 2 -- = + QS .Vq q 1 - 2V q q 1 . 

This shews that 

Take account of this result in (2), and put dp for qaq~ l , which 
may be any vector. Thus 

From this we see at once that 

q = a Up, 

256 QUATERNIONS. [33^- 

where a is any constant versor. Then (3) gives 

Vu__2Up du_ ZSpdp 
V = "Yp 1T = ^y~ 

so that u = ffj- z . 

1 P 
Thus, from (1), we have 

a = a - a~ l + 8. 

This gives the Electric Image transformation, with any subsequent 
rotation, followed (or, as is easily seen, preceded) by a translation. 
Hence the only series of surfaces which satisfy the question, are 
mutually perpendicular planes ; and their images, which are series 
of spheres, passing through a common point and having their 
centres on three rectangular lines passing through that point. 
[For another mode of solution see Proc. R. 8. E., Dec. 1877.] 
In some respects analogous to this is the celebrated physical 
problem of finding series of Orthogonal Isothermal Surfaces. We 
give a slight sketch of it here. 

If three such series of surfaces be denoted by their temperatures 

thus : 

F^T,, F,= T t , F S = T S , 

the conditions of orthogonality are fully expressed by putting for 
the respective values of the flux of heat in each series 

where a, /3, 7 form a constant rectangular unit vector system. 
But the isothermal conditions are simply 

v*F t = v*F 9 = v*F 3 = o. 

Hence we have three simultaneous equations, of which the 

first is 

= ........................... (a). 

[In the previous problem a might be any vector whatever, the 
values of u were equal, and the vector part only of the left-hand 
member was equated to zero. These conditions led to an unique 
form of solution. Nothing of the kind is to be expected here.] 

From equation (a) we have at once 

where v l has been put for logu v and a for qaq~\ 


From the group of six equations, of which (6) gives two, we 

with three of the type 

S.* S(a V)qq* = 0. 

We also obtain without difficulty 

2Vv + 4tfqq- l = 0, 

which may be put in the form 

V.(u l uji s )*q = 0. 

But when we attempt to find the value of dq we are led to 
expressions such as 

27. a S (a dp) Vw, + Zdqq- 1 = 0, 

which, in consequence of the three different values of v, are 
comparatively unmanageable. (See Ex. 24 at end of Chapter.) 
They become, however, comparatively simple when one of the 
three families is assumed. The student will find it useful to work 
out the problem when F l represents a series of parallel planes, so 
that the others are cylinders; or when ^represents planes passing 
through a line, the others being surfaces of revolution ; &c. 

337. To find the orthogonal trajectories of a given series of 

If the equation Fp = C 

give Svdp = 0, 

the equation of the orthogonal curves is 

This is equivalent to two scalar differential equations ( 210), 
which, when the problem is possible, belong to surfaces on each of 
which the required lines lie. The finding of the requisite criterion 
we leave to the student. [He has only to operate on the last- 
written equation by 8. a, where a is any constant vector ; and, 
bearing this italicized word in mind, proceed as in 334.] 

Let the surfaces be concentric spheres. 

Here p* = C, 

and therefore Vpdp = 0. 

Hence Tp*d Up = -Up Vpdp = 0, 

and the integral is Up = constant, 

denoting straight lines through the origin. 

T. Q. I. 17 

258 QUATERNIONS. [338. 

Let the surfaces be spheres touching each other at a common 
point. The equation is ( 235) 

whence V. pctpdp = 0. 

The integrals may be written 

the first (/3 being any vector) is a plane through the common 
diameter ; the second represents a series of rings or tores ( 340) 
formed by the revolution, about a, of circles touching that line 
at the point common to the spheres. 

Let the surfaces be similar, similarly situated, and concentric, 
surfaces of the second degree. 

Here &PXP ~ @, 

therefore V^pdp 0. 

But, by | 307, the integral of this equation is 

where and % are related to each other, as in 307 ; and e is any 
constant vector. 

338. To integrate the linear partial differential equation of a 
family of surfaces. 

The equation (see 330) 

P du + du + R du = Q 
dx dy dz 

may be put in the very simple form 

8(aV)u = Q ........................... (1), 

if we write a = iP +jQ + kR, 

. d . d , d 

and V =i -=- + i -J- +K j- . 

dx J dy dz 

[From this we see that the meaning of the differential 
equation is that at every point of the surface 

u = const. 

the corresponding vector, <r, is a tangent line. Thus we have a 
suggestion of the ordinary method of solving such equations.] 

(1) gives, at once, Vu 


where m is a scalar and 6 a vector (in whose tensor m might have 
been included, but is kept separate for a special purpose). Hence 

= mS . Ocrdp 

if we put dr = m Vadp 

so that m is an integrating factor of V . crdp. If a value of m can 
be found, it is obvious, from the form of the above equation, that 
6 must be a function of r alone ; and the integral is therefore 

u = F (r) = const. 

where F is an arbitrary scalar function. 

Thus the differential equation of Cylinders is 

S(aV)u = 0, 
where a is a constant vector. Here m= 1, and 

u = F( Vap) const. 
That of Cones referred to the vertex is 

Here the expression to be made integrable is 


But Hamilton long ago shewed that ( 140 (2)) 
d Up _ ydp _ Vpdp 


which indicates the value of m, and gives 
u = F( Up) = const. 

It is obvious that the above is only one of a great number of 
different processes which may be applied to integrate the differen 
tial equation. It is quite easy, for instance, to pass from it to the 
assumption of a vector integrating factor instead of the scalar m, 
and to derive the usual criterion of integrability. There is no 
difficulty in modifying the process to suit the case when the right- 
hand member is a multiple of u. In fact it seems to throw a very 
clear light upon the whole subject of the integration of partial 
differential equations. If, instead of 8 (o"V), we employ other 
operators as 8 (crV) 8 (TV), 8 . crVrV, &c. (where V may or may not 
operate on u alone), we can pass to linear partial differential 


260 QUATERNIONS. [339- 

equations of the second and higher orders. Similar theorems can 
be obtained from vector operators, as F(crV)*. 

339. Find the general equation of surf aces described by a line 
which always meets, at right angles, a fixed line. 

If a be the fixed line, /3 and 7 forming with it a rectangular 
unit system, then 

where y may have all values, but x and z are mutually dependent, 
is one form of the equation. 

Another, expressing the arbitrary relation between x and z, is 

But we may also write 

as it obviously expresses the same conditions. 

The simplest case is when F (x) = hx. The surface is one which 
cuts, in a right helix, every cylinder which has a. for its axis. 

340. The centre of a sphere moves in a given circle, find the 
equation of the ring described. 

Let a be the unit-vector axis of the circle, its centre the origin, 
r its radius, a that of the sphere. 

Then (p-fi)* = -a z 

is the equation of the sphere in any position, where 

These give ( 326) S . a/3p = 0, and /3 must now be eliminated. 
The immediate result is that 

giving (p 2 -r* + a 2 ) a = 4r 2 T 2 Vap, 

which is the required equation. It may easily be changed to 

(p*-a* + f)* = -4ia*p*-4?r*S*ap ............... (1), 

and in this form it enables us to give a very simple proof of the 
singular property of the ring (or tore) discovered by Villarceau. 


(a \ 
a i 2 2 j = 0, 

Tait, Proc. E. S. E., 1869-70. 


which together are represented by 

evidently pass through the origin and touch (and cut) the ring. 

The latter equation may be written 

r*S 2 ap - a* (tfap + S 2 p U0) = 0, 
or r*S oLp + a*(p* + S .oipUp) = ................ (2). 

The plane intersections of (1) and (2) lie obviously on the new 

which consists of two spheres of radius r, as we see by writing its 
separate factors in the form 

341. It may be instructive to work out this problem from a 
different point of view, especially as it affords excellent practice in 

A circle revolves about an axis passing within it, the perpen 
dicular from the centre on the axis lying in the plane of the circle: 
shew that, for a certain position of the axis, the same solid may be 
traced out by a circle revolving about an external axis in its own 

Let a Jb* + c 2 be the radius of the circle, i the vector axis of 
rotation, COL (where To. = 1) the vector perpendicular from the 
centre on the axis i, and let the vector 

bi + da 

be perpendicular to the plane of the circle. 
The equations of the circle are 

Also - /o 2 = S 2 ip + tfctp 4 S 2 . ictp, 

= S*ip + S*ctp + -,S*ip 

by the second of the equations of the circle. But, by the first, 

(p 2 + by = 4c 2 S 2 a / D = - 4 ( C y + a*S*ip), 
which is easily transformed into 


262 QUATERNIONS. [342. 

If we put this in the forms 

and (p_ a ) + c = 0, 

where (3 is a unit-vector perpendicular to i and in the plane of i 
and p, we see at once that the surface will be traced out by, a 
circle of radius c, revolving about i, an axis in its own plane, 
distant a from its centre. 

[This problem is not well adapted to shew the gain in brevity 
and distinctness which generally attends the use of quaternions ; 
as, from its very nature, it hints at the adoption of rectangular 
axes and scalar equations for its treatment, so that the solution 
we have given is but little different from an ordinary Cartesian 

342. A surface is generated by a straight line which intersects 
two fixed straight lines : find the general equation. 

If the given lines intersect, there is no surface but the plane 
containing them. 

Let then their equations be, 

p = a + x$, p = a l + xfi r 
Hence every point of the surface satisfies the condition, 30, 

P = y(* + x&) + (l-y) (a, + *A) ............... (1). 

Obviously y may have any value whatever : so that to specify a 
particular surface we must have a relation between x and # r By 
the help of this, x l may be eliminated from (1), which then takes 
the usual form of the equation of a surface 

p = <l>(x> y\ 

Or we may operate on (1) by F. (a + x/3 a x xfi^) t so that we 
get a vector equation equivalent to two scalar equations ( 98, 123), 
and not containing y. From this a? and x l may easily be found in 
terms of p, and the general equation of the possible surfaces may 
be written 

f(x, x^ = 0, 

where f is an arbitrary scalar function, and the values of x and x^ 
are expressed in terms of p. 

This process is obviously applicable if we have, instead of two 
straight lines, any two given curves through which the line must 
pass; and even when the tracing line is itself a given curve, 
situated in a given manner. But an example or two will make the 
whole process clear. 


343. Suppose the moveable line to be restricted by the condition 
that it is always parallel to a fixed plane. 

Then, in addition to (1), we have the condition 

y being a vector perpendicular to the fixed plane. 

We lose no generality by assuming a and a v which are any 
vectors drawn from the origin to the fixed lines, to be each per 
pendicular to 7 ; for, if for instance we could not assume Sya. = 0, 
it would follow that Syj3 = 0, and the required surface would either 
be impossible, or would be a plane, cases which we need not con 
sider. Hence 

= 0. 

Eliminating x v by the help of this equation, from (1) of last section, 
we have 

= y (a + a/3) + (1 - y) , + a/3, 

Operating by any three non-coplanar vectors and with the charac 
teristic S, we obtain three equations from which to eliminate x and 
y. Operating by 8 . y we find 

Syp = xSfty. 
Eliminating x by means of this, we have finally 

which appears to be of the third degree. It is really, however, only 
of the second degree : since, in consequence of our assumptions, 
we have 

and therefore Syp is a spurious factor of the left-hand side. 

344. Let the fixed lines be perpendicular to each other, and let 
the moveable line pass through the circumference of a circle, whose 
centre is in the common perpendicular, and whose plane bisects that 
line at right angles. 

Here the equations of the fixed lines may be written 

where a, /3, y form a rectangular system, and we may assume the 
two latter to be unit-vectors. 
The circle has the equations 

p 2 = - a 2 , Sap = 0. 

264 QUATERNIONS. [345- 

Equation (1) of 342 becomes 

Hence SoC 1 p = y (ly) = 0, or y = -J. 

Also f = - a 2 = (2y - 1) 2 a 2 - *y - < (1 - y)\ 

or 4a 2 

so that if we now suppose the tensors of /3 and 7 to be each 2a, we 
may put x = cos 0, x^ = sin 0, from which 

For this specially simple case the solution is not better than the 
ordinary Cartesian one ; but the student will easily see that we 
may by very slight changes adapt the above to data far less sym 
metrical than those from which we started. Suppose, for instance, 
/3 and 7 not to be at right angles to one another ; and suppose the 
plane of the circle not to be parallel to their plane, &c., &c. But 
farther, operate on every line in space by the linear and vector 
function <f>, and we distort the circle into an ellipse, the straight 
lines remaining straight. If we choose a form of < whose principal 
axes are parallel to a, /3. 7, the data will remain symmetrical, but 
not unless. This subject will be considered again in the next 

345. To find the curvature of a normal section of a central 
surface of the second degree. 

In this, and the few similar investigations which follow, it will 
be simpler to employ infinitesimals than differentials ; though for 
a thorough treatment of the subject the latter method, as it may 
be seen in Hamilton s Elements, is preferable. 

We have, of course, Sp(f>p = 1, 

and, if p + Sp be also a vector of the surface, we have rigorously, 
whatever be the tensor of Sp, 

Hence ZSSpfo + SSp&p = ..................... (1). 

Now (j>p is normal to the tangent plane at the extremity of p, 
so that if t denote the distance of the point p + Sp from that plane 


and (1) may therefore be written 

2tT<l>p - T 2 SpS . U8p<t> USp = 0. 
But the curvature of the section is evidently 

or, by the last equation, 

In the limit, 8p is a vector in the tangent plane ; let is be the 
vector semidiameter of the surface which is parallel to it, and the 
equation of the surface gives 

so that the curvature of the normal section, at the point p, in the 
direction of -or, is 


directly as the perpendicular from the centre on the tangent plane, 
and inversely as the square of the semidiameter parallel to the 
tangent line, a well-known theorem. 

346. By the help of the known properties of the central 
section parallel to the tangent plane, this theorem gives us all 
the ordinary properties of the directions of maximum and mini 
mum curvature, their being at right angles to each other, the 
curvature in any normal section in terms of the chief curvatures 
and the inclination to their planes, &c., &c., without farther 
analysis. And when, in a future section, we shew how to find 
an osculating surface of the second degree at any point of a given 
surface, the same properties will be at once established for surfaces 
in general. Meanwhile we may prove another curious property 
of the surfaces of the second degree, which similar reasoning 
extends to all surfaces. 

The equation of the normal at the point p + Sp on the surface 
treated in last section is 

This intersects the normal at p if ( 216, 223) 

8 . $p(f)p(f>$p = 0, 
that is, by the result of 290, if Bp be parallel to the maximum or 

266 QUATERNIONS. [346. 

minimum diameter of the central section parallel to the tangent 

Let 0-j and cr 2 be those diameters, then we may write in general 

where p and q are scalars, infinitely small. 

If we draw through a point P in the normal at p a line parallel 
to cTj, we may write its equation 

tff = p + CKpp + ycr^ 
The proximate normal (1) passes this line at a distance (see 216) 

or, neglecting terms of the second order, 

TV^ff) ^ pS ^" 1 ^ " 1 + aq8 

The first term in the bracket vanishes because o^ is a principal 
vector of the section parallel to the tangent plane, and thus the 
expression becomes 

Hence, if we take a = To-*, the distance of the normal from the 
new line is of the second order only. This makes the distance of 
P from the point of contact T<f)p Tcr*, i.e. the principal radius of 
curvature along the tangent line parallel to cr 2 . That is, the group 
of normals drawn near a point of a central surface of the second 
degree pass ultimately through two lines each parallel to the tangent 
to one principal section, and drawn through the centre of curvature 
of the other. The student may form a notion of the nature of this 
proposition by considering a small square plate, with normals 
drawn at every point, to be slightly bent, but by different amounts, 
in planes perpendicular to its edges. The first bending will make 
all the normals pass through the axis of the cylinder of which the 
plate now forms part; the second bending will not sensibly disturb 
this arrangement, except by lengthening or shortening the line in 
which the normals meet, but it will make them meet also in the 
axis of the new cylinder, at right angles to the first. A small 
pencil of light, with its focal lines, presents this appearance, due 
to the fact that a series of rays originally normal to a surface 
remain normals to a surface after any number of reflections and 
(ordinary) refractions. (See 332.) 


347. To extend these theorems to surfaces in general, it is 
only necessary, as Hamilton has shewn, to prove that if we write 

dv = (f)dp, 

(f) is a self-conjugate function ; and then the properties of <, as ex 
plained in preceding Chapters, are applicable to the question. 

As the reader will easily see, this is merely another form of the 
investigation contained in 334. But it is again cited here to 
shew what a number of very simple demonstrations may be given 
of almost all quaternion theorems. 

The vector v is defined by an equation of the form 

dfp = Svdp, 

where / is a scalar function. Operating on this by another inde 
pendent symbol of differentiation, 8, we have 

Bdfp = SSvdp + SvSdp. 
In the same way we have 

dSfp = SdvSp + SvdSp. 

But, as d and 8 are independent, the left-hand members of these 
equations, as well as the second terms on the right (if these exist 
at all), are equal, so that we have 

SdvSp = SSvdp. 

This becomes, putting dv = (frdp, 

and therefore $v = <f>Sp, 

S&p(f)dp = Sdp<f>Sp, 
which proves the proposition. 

348. If we write the differential of the equation of a surface 
in the form 

dfp = 2Svdp, 

then it is easy to see that 

f(p + dp) =fp + ZSvdp + Sdvdp + &c., 

the remaining terms containing as factors the third and higher 
powers of Tdp. To the second order, then, we may write, except 
for certain singular points, 

and, as before, ( 345), the curvature of the normal section whose 
tangent line is dp is 

_1_ ydv 

TV dp 

268 QUATERNIONS. [349- 

349. The step taken" in last section, although a very simple 
one, virtually implies that the first three terms of the expansion of 
f(p + dp) are to be formed in accordance with Taylor s Theorem, 
whose applicability to the expansion of scalar functions of quater 
nions has not been proved in this work (see 142) ; we therefore 
give another investigation of the curvature of a normal section, 
employing for that purpose the formulae of 299. 

We have, treating dp as an element of a curve, 

Svdp = 0, 

or, making s the independent variable, 

Svp = 0. 
From this, by a second differentiation, 

qdv , ~ lr A 

ds p + p 
The curvature is, therefore, since v \\ p" and Tp = 1, 

m it 1 ci dv ,o 1 a dv , f, 

Tp" = - -^- S -j- p 2 = jfr S -y- , as before. 
TV dp r TV dp 

350. Since we have shewn that 

dv = <pdp 

where $ is a self-conjugate linear and vector function, whose con 
stants depend only upon the nature of the surface, and the position 
of the point of contact of the tangent plane ; so long as we do not 
alter these we must consider < as possessing the properties explained 
in Chapter V. 

Hence, as the expression for Tp" does not involve the tensor of 
dp, we may put for dp any unit-vector r, subject of course to the 

Svr = Q .............................. (1). 

And the curvature of the normal section whose tangent is r is 

If we consider the central section of the surface of the second degree 

made by the plane Svvr = 0, 

we see at once that the curvature of the given surface along the 
normal section touched by r is inversely as the square of the 
parallel radius in the auxiliary surface. This, of course, includes 
Euler s and other well-known Theorems. 


351. To find the directions of maximum and minimum 

curvature, we have 

ST(J)T = max. or min. 

with the conditions, Svr = 0, 

Tr= 1. 
By differentiation, as in 290, we obtain the farther equation 

>Sf.^T</)T = ........................... (1). 

If r be one of the two required directions, r =rUv is the other, 
for the last-written equation may be put in the form 

or S . vr <T = 0. 

Hence the sections of greatest and least curvature are perpendicular 
to one another. 

We easily obtain, as in 290, the following equation 
S.v(<t> + Srct>T)- 1 v = 0, 

which gives two values of $T<T, and these divided by TV are 
the required curvatures. 

352. Before leaving this very brief introduction to a subject, 
an exhaustive treatment of which will be found in Hamilton s 
ElementSy we may make a remark on equation (1) of last section 

S.VT(f)T = ) 

or, as it may be written, by returning to the notation of 350, 

S . vdpdv = 0. 

This is the general equation of lines of curvature. For, if we 
define a line of curvature on any surface as a line such that 
normals drawn at contiguous points in it intersect, then, Bp being 
an element of such a line, the normals 

tx = p + xv and w = p + Bp + y (v + 8v) 
must intersect. This gives, by 216, the condition 

as above. 



1. Find the length of any arc of a curve drawn on a sphere 
so as to make a constant angle with a fixed diameter. 

2. Shew that, if the normal plane of a curve always contains 
a fixed line, the curve is a circle. 

3. Find the radius of spherical curvature of the curve 

Also find the equation of the locus of the centre of spherical 

4. (Hamilton, Bishop Law s Premium Examination, 1854.) 

(a) If p be the variable vector of a curve in space, and if the 
differential die be treated as = 0, then the equation 

dT(p-tc) = Q 

obliges ic to be the vector of some point in the normal plane to 
the curve. 

(6) In like manner the system of two equations, where d/c 
and d 2 tc are each =0, 

dT(p - K) = 0, d 2 T(p - K) = 0, 

represents the axis of the element, or the right line drawn through 
the centre of the osculating circle, perpendicular to the osculating 

(c) The system of the three equations, in which K is treated 
as constant, 

dT(p- K ) = 0, d*T(p-K) = Q, d 5 T(p- K ) = 0, 
determines the vector K of the centre of the osculating sphere. 

(d) For the three last equations we may substitute the 
following : 

S.(p- K )dp = 0, 

(e) Hence, generally, whatever the independent and scalar 
variable may be, on which the variable vector p of the curve 


depends, the vector K of the centre of the osculating sphere admits 
of being thus expressed : 

3V. dptfpS . dpd*p - dp*V. dpd 5 p 
~ p ~ S.dpd*pd 3 p 

(/) In general 

d(p-*V.dpUp) = d( Tp~ z V.pdp) 

= Tp~ 5 (3V. pdpS.pdp- P *V.pd*p) ; 

and, therefore, the recent expression for K admits of being thus 


(g) If the length of the element of the curve be constant, 
dTdp = 0, this last expression for the vector of the centre of the 
osculating sphere to a curve of double curvature becomes, more 


d . d 2 pdp* 



or K = p + 


(h) Verify that this expression gives K = 0, for a curve 
described on a sphere which has its centre at the origin of vectors ; 
or shew that whenever dTp = 0, d?Tp = 0, d?Tp = 0, as well as 
dTdp = 0, then 

5. Find the curve from every point of which three given 
spheres appear of equal magnitude. 

6. Shew that the locus of a point, the difference of whose 
distances from each two of three given points is constant, is a 
plane curve. 

7. Find the equation of the curve which cuts at a given angle 
all the sides of a cone of the second degree. 

Find the length of any arc of this curve in terms of the 
distances of its extremities from the vertex. 

8. Why is the centre of spherical curvature, of a curve 
described on a sphere, not necessarily the centre of the sphere ? 


9. Find the equation of the developable surface whose gene 
rating lines are the intersections of successive normal planes to a 
given tortuous curve. 

10. Find the length of an arc of a tortuous curve whose 
normal planes are equidistant from the origin. 

11. The reciprocals of the perpendiculars from the origin on 
the tangent planes to a developable surface are vectors of a 
tortuous curve ; from whose osculating planes the cusp-edge of 
the original surface may be reproduced by the same process. 

12. The equation p=Va t {3, 

where a is a unit-vector not perpendicular to /3, represents an 
ellipse. If we put 7 = Fa/3, shew that the equations of the locus 
of the centre of curvature are 

13. Find the radius of absolute curvature of a spherical 

14. If a cone be cut in a circle by a plane perpendicular to a 
side, the axis of the right cone which osculates it, along that side, 
passes through the centre of the section. 

15. Shew how to find the vector of an umbilicus. Apply 
your method to the surfaces whose equations are 

and SapSfipSyp = 1. 

16. Find the locus of the umbilici of the surfaces represented 
by the equation 

s P (<f>+hr P =-i, 

where h is an arbitrary parameter. 

17. Shew how to find the equation of a tangent plane which 
touches a surface along a line, straight or curved. Find such 
planes for the following surfaces 

and (p* - a 2 + 6 2 ) 2 + 4 (a> 2 + V&ap) = 0. 


18. Find the condition that the equation 

(,> + )</> = - 1, 

where <p is a self-conjugate linear and vector function, may 
represent a cone. 

19. Shew from the general equation that cones and cylinders 
are the only developable surfaces of the second degree. 

20. Find the equation of the envelop of planes drawn at each 
point of an ellipsoid perpendicular to the radius vector from the 

21. Find the equation of the envelop of spheres whose centres 
lie on a given sphere, and which pass through a given point. 

22. Find the locus of the foot of the perpendicular from the 
centre to the tangent plane of a hyperboloid of one, or of two, 

23. Hamilton, Bishop Law s Premium Examination, 1852. 

(a) If p be the vector of a curve in space, the length of the 
element of that curve is Tdp ; and the variation of the length of a 
finite arc of the curve is 

SfTdp = -JSUdpSdp = - ASUdpSp + JSdUdpBp. 

(b) Hence, if the curve be a shortest line on a given surface, 
for which the normal vector is v t so that SvSp = 0, this shortest or 
geodetic curve must satisfy the differential equation, 

Also, for the extremities of the arc, we have the limiting 

Interpret these results. 

(c) For a spheric surface, Vvp = 0, V.pdUdp = , the inte 
grated equation of the geodetics is Vp Udp = r, giving Svrp = Q 
(great circle). 

For an arbitrary cylindric surface, 

the integral shews that the geodetic is generally a helix, making 
a constant angle with the generating lines of the cylinder. 

(d) For an arbitrary conic surface, 

T, Q. I, 18 


integrate this differential equation, so as to deduce from it, 
TVpUdp = const. 

Interpret this result; shew that the perpendicular from the 
vertex of the cone on the tangent to a given geodetic line is 
constant ; this gives the rectilinear development. 

When the cone is of the second degree, the same property is a 
particular case of a theorem respecting confocal surfaces. 

(e) For a surface of revolution, 

S.apv = 0, S . apdUdp = ; 
integration gives, 

const. = 8 . ap Udp = TVapS U(Vap. dp) 

the perpendicular distance of a point on a geodetic line from the 
axis of revolution varies inversely as the cosine of the angle under 
which the geodetic crosses a parallel (or circle) on the surface. 

(/) The differential equation, S . apd Udp = 0, is satisfied not 
only by the geodetics, but also by the circles, on a surface of 
revolution ; give the explanation of this fact of calculation, and 
shew that it arises from the coincidence between the normal plane 
to the circle and the plane of the meridian of the surface. 

(g) For any arbitrary surface, the equation of the geodetic 
may be thus transformed, S.vdpd*p = Q] deduce this form, and 
shew that it expresses the normal property of the osculating plane. 

(h) If the element of the geodetic be constant, dTdp = 0, 
then the general equation formerly assigned may be reduced to 
V. vd*p = 0. 

Under the same condition, d?p = v~ l Sdvdp. 

(i) If the equation of a central surface of the second order be 
put under the form fp = 1, where the function / is scalar, and 
homogeneous of the second dimension, then the differential of that 
function is of the form dfp = 2$ . vdp, where the normal vector, 
v = cf)p ) is a distributive function of p (homogeneous of the first 
dimension), dv = d<f>p = <f>dp. 

This normal vector v may be called the vector of proximity 
(namely, of the element of the surface to the centre) ; because its 
reciprocal, jT l , represents in length and in direction the perpen 
dicular let fall from the centre on the tangent plane to the surface. 

(k) If we make Sacfrp =/(&, p) this function f is commu 
tative with respect to the two vectors on which it depends, 


f(p, o-) =f(cr, p) ; it is also connected with the former function /, 
of a single vector p, by the relation, /(/o, p) =fp : so that 

fp = Spfp. 

fdp = Sdpdv ; dfdp = 2S . dvd*p ; for a geodetic, with constant 

this equation is immediately integrable, and gives const. 
= Tv\/ (fUdp) = reciprocal of Joachimstal s product, PD. 

(I) If we give the name of "Didonia" to the curve (discussed 
by Delaunay) which, on a given surface and with a given perimeter, 
contains the greatest area, then for such a Didonian curve we have 
by quaternions the formula, 

fS. Uvdp8p + c8fTdp = 0, 
where c is an arbitrary constant. 

Derive hence the differential equation of the second order, 
equivalent (through the constant c) to one of the third order, 

C - l dp=V. UvdUdp. 

Geodetics are, therefore, that limiting case of Didonias for 
which the constant c is infinite. 

On a plane, the Didonia is a circle, of which the equation, 
obtained by integration from the general form, is 

p = TX + c Uvdp, 
w being vector of centre, and c being radius of circle. 

(m) Operating by 8. Udp, the general differential equation of 
the Didonia takes easily the following forms : 

c- l Tdp = 

AJ ~~ 

f-f j 

(n) The vector o>, of the centre of the osculating circle to a 
curve in space, of which the element Tdp is constant, has for 



Hence for the general Didonia, 

Tr 7 


(o) Hence, the radius of curvature of any one Didonia varies, 
in general, proportionally to the cosine of the inclination of the 
osculating plane of the curve to the tangent plane of the surface. 

And hence, by Meusnier s theorem, the difference of the 
squares of the curvatures of curve and surface is constant ; the 
curvature of the surface meaning here the reciprocal of the radius 
of the sphere which osculates in the direction of the element of 
the Didonia. 

(p) In general, for any curve on any surface, if f denote the 
vector of the intersection of the axis of the element (or the axis of 
the circle osculating to the curve) with the tangent plane to the 
surface, then 


Hence, for the general Didonia, with the same signification of the 

= p c Uvdp ; 

and the constant c expresses the length of the interval p , 
intercepted on the tangent plane, between the point of the curve 
and the axis of the osculating circle. 

(q) If, then, a sphere be described, which shall have its 
centre on the tangent plane, and shall contain the osculating 
circle, the radius of this sphere shall always be equal to c. 

(r) The recent expression for f, combined with the first form 
of the general differential equation of the Didonia, gives 

(s) Hence, or from the geometrical signification of the con 
stant c, the known property may be proved, that if a developable 
surface be circumscribed about the arbitrary surface, so as to touch 
it along a Didonia, and if this developable be then unfolded into a 
plane, the curve will at the same time be flattened (generally) 
into a circular arc, with radius = c. 


24. Find the condition that the equation 

may give three real values of /for any given value of p. If /be a 
function of a scalar parameter f, shew how to find the form of this 
function in order that we may have 

_Vf-* + ** + *f_o 

* ~ da? c% 2 ^ d* 8 

Prove that the following is the relation between /and f, 


in the notation of 159. (Tait, Trans. R. S. E. 1873.) 

25. Shew, after Hamilton, that the proof of Dupin s theorem, 
that " each member of one of three series of orthogonal surfaces 
cuts each member of each of the other series along its lines of 
curvature," may be expressed in quaternion notation as follows : 

If Svdp = Q, Sv dp = Q, = Q 

be integrable, and if 

Svv = 0, then Vv dp = 0, makes S . vvdv = 0. 

Or, as follows, 

If 8vVv = Q, &/Vi/ = 0, &/ Vj;" = 0, and 
then 8.v"(8v / V)v = O t 

where V-t-+j +.*. 

dx J dy dz 

26. Shew that the equation 

represents the line of intersection of a cylinder and cone, of the 
second order, which have ft as a common generating line. 

27. Two spheres are described, with centres at A, B, where 
OA = a, OB = @, and radii a, b. Any line, OPQ, drawn from the 
origin, cuts them in P, Q respectively. Shew that the equation 
of the locus of intersection of AP, BQ has the form 

Shew that this involves 8 . a{3p = 0, 

and therefore that the left side is a scalar multiple of V . aj3, so 
that the locus is a plane curve, 


Also shew that in the particular case 

the locus is the surface formed by the revolution of a Cartesian 
oval about its axis. 

28. Integrate the equations 

Shew that each represents a series of circles in space. What 
is the common property of the circles of each series? [See 140, 
(10), (11).] 

29. Express the general equation of a knot of any kind, on 
an endless cord, in the form 

P = (*) 
pointing out precisely the nature of the function <. 

What are the conditions to which (f> must be subject, when 
possible distortions of the knot are to be represented ? What are 
the conditions that the string may be capable of being brought 
into a mere ring form ? 

30. Find the envelop of the planes of equilateral triangles 
whose vertices are situated in three given lines in space. What 
does it become when two, or all three, of these lines intersect ? 

31. Form the equation of the surface described by a circle, 
when two given points in its axis are constrained to move on given 
straight lines. Also when the constraining lines are two concentric 
circles in one plane. 



353. IN the present Chapter it is not proposed to give a 
connected account of even the elements of so extensive a subject 
as that indicated by the Title. All that is contemplated is to 
treat a few branches of the subject in such a way as to shew the 
student how to apply the processes of Quaternions. 

And, with a view to the next Chapter, the portions selected 
for treatment will be those of most direct interest in their physical 

A. Kinematics of a Point. 

354. When a point s vector, p, is a function of the time t, we 
have seen ( 36) that its vector-velocity is expressed by -fa or, in 
Newton s notation, by p. 

That is, if p = (f>t 

be the equation of an orbit, containing (as the reader may see) not 
merely the form of the orbit, but the law of its description also, 

gives at once the form of the Hodograph and the law of its 

This shews immediately that the vector-acceleration of a point s 



280 QUATERNIONS. [ 3 5 5 

is the vector-velocity in the hodograph. Thus the fundamental pro 
perties of the hodograph are proved almost intuitively. 

355. Changing the independent variable, we have 

dp ds , 
p= cfedi = ^ 

if we employ the dash, as before, to denote -=- . 

This merely shews, in another form, that p expresses the 
velocity in magnitude and direction. But a second differentiation 

p = vp + v*p". 

This shews that the vector-acceleration can be resolved into two 
components, the first, vp, being in the direction of motion and 

equal in magnitude to the acceleration of the speed, v or -7- ; 

the second, v 2 // , being in the direction of the radius of absolute 
curvature, and having for its amount the square of the speed 
multiplied by the curvature. 

[It is scarcely conceivable that this important fundamental 
proposition can be proved more elegantly than by the process 
just given.] 

356. If the motion be in a plane curve, we may write the 
equation as follows, so as to introduce the usual polar coordinates, 
r and 0, 

p=rJ > "ft, 

where a is a unit-vector perpendicular to, jB a unit-vector in, the 
plane of the curve. 

Here, of course, r and 6 may be considered as connected by one 
scalar equation ; or better, each may be looked on as a function of 
t. By differentiation we get 

which shews at once that r is the velocity along, r6 that perpen- 
pendicular to, the radius vector. Again, 

p = (r- r<9 2 ) ^ $ + (2r<9 + r&) wP v ft 

which gives, by inspection, the components of acceleration along, 
and perpendicular to, the radius vector. 

357. For uniform acceleration in a constant direction, we have 
at once 

358.] KINEMATICS. 281 

Whence p = at -f {3, 

where ft is the vector-velocity at epoch. This shews that the 
hodograph is a straight line described uniformly. 

Also p = h fit, 

no constant being added if the origin be assumed to be the position 
of the moving point at epoch. 

Since the resolved parts of p, parallel to ft and a, vary respect 
ively as the first and second powers of t, the curve is evidently a 
parabola ( 31 (/)). 

But we may easily deduce from the equation the following 

T(p + fioL~ l ft) = -SUa(p + ^ oT 1 } , 

the equation of a paraboloid of revolution, whose axis is a. Also 

8 . ctftp = 0, 

and therefore the distance of any point in the path from the 
point ^ftoC l ft is equal to its distance from the line whose 
equation is 

Thus we recognise the focus and directrix property. [The 
student should remark here how the distances of the point of 
projection (which may, of course, be any point of the path) from 
the focus and from the directrix are represented in magnitude and 
direction by the two similar but different expressions 

- Jcf and - J0V 1 , 

or -i(#O and -tftfa 1 ). 

This is an excellent example of the non-commutative character of 
quaternion multiplication.] 

358. That the moving point may reach a point 7 (where 7 is, 
of course, coplanar with a and /3) we must have, for some real 
value of t, 

Now suppose T/3, the speed of projection, to be given = v, and, 
for shortness, write ts for Up. 

Then y = ^^ + vtvf ........................ (a). 

282 QUATERNIONS. [359. 

Since 2V = 1, 

m 2^4 

we have - - (V - Say) f + T 7 2 = 0. 


The values of f are real if 

is positive. Now, as TaTy is never less than Say, this condition 
evidently requires that v* Say also shall be positive. Hence, 
when they are real, both values of f are positive. Thus we have 
four values of t which satisfy the conditions, and it is easy to see 
that since, disregarding the signs, they are equal two and two, 
each pair refer to the same path, but described in opposite direc 
tions between the origin and the extremity of 7. There are 
therefore, if any, in general two parabolas which satisfy the 
conditions. The directions of projection are (of course) given by 
the corresponding values of -ST. These, in turn, are obtained at 
once from (a) in the form 

1 t 

where t has one or other of the values previously found. 

359. The envelop of all the trajectories possible, with a given 
speed, evidently corresponds to 

for then y is the vector of intersection of two indefinitely close 
paths in the same vertical plane. 

Now v 2 - Say = TaTy 

is evidently the equation of a paraboloid of revolution of which 
the origin is the focus, the axis parallel to a, and the directrix 

plane at a distance - . 

All the ordinary problems connected with parabolic motion are 
easily solved by means of the above formulae. Some, however, are 
even more easily treated by assuming a horizontal unit-vector in 
the plane of motion, and expressing /3 in terms of it and of a. 
But this must be left to the student. 

360. For acceleration directed to or from a fixed point, we 
have, taking that point as origin, and putting P for the magnitude 
of the central acceleration, 

P = PU P . 

36 1.] KINEMATICS. 283 

From this, at once, 

Integrating Vpp = 7 = 3. constant vector. 

The interpretation of this simple formula is first, p and p are 
in a plane perpendicular to 7, hence the path is in a plane (of 
course passing through the origin) ; second, the doubled area of 
the triangle, two of whose sides are p and p (that is, the moment 
of the velocity) is constant. 

[It is scarcely possible to imagine that a more simple proof 
than this can be given of the fundamental facts, that a central 
orbit is a plane curve, and that equal areas are described by the 
radius vector in equal times.] 

361. When the law of acceleration to or from the origin is 
that of the inverse square of the distance, we have 


~T P * 
where m is negative if the acceleration be directed to the origin. 

Hence p = -^rr . 

The following beautiful method of integration is due to Hamil 
ton. (See 140, (2).) 

dUp U P . Vpp Up.y 
Generally, -^ - ^ -^ , 

therefore py = m ~- , 

and p7 = e m Up, 

where e is a constant vector, perpendicular to 7, because 

Syp = 0. 

Hence, in this case, we have for the hodograph, 
p = 67" 1 m Up . 7" 1 . 

Of the two parts of this expression, which are both vectors, the 
first is constant, and the second is constant in length. Hence the 
locus of the extremity of p is a circle in a plane perpendicular to 7 
(i.e. parallel to the plane of the orbit), whose radius is 1 , 
and whose centre is at the extremity of the vector ey~ l . 

[This equation contains the whole theory of the Circular 
Hodograph. Its consequences are developed at length in Hamil 
ton s Elements.] 

284 QUATERNIONS. [362. 

362. We may write the equations of this circle in the form 

(a sphere), and Sjp = 

(a plane through the origin, and through the centre of the sphere). 
The equation of the orbit is found by operating by V.p upon 
that of the hodograph. We thus obtain 

or ry 

or mTp = Se (<fe* - p) ; 

in which last form we at once recognise the focus and directrix 
property. This is in fact the equation of a conicoid of revolution 
about its principal axis (e), and the origin is one of the foci. The 
orbit is found by combining it with the equation of its plane, 

Syp = 0. 

We see at once that 7V 1 is the vector distance of the directrix 
from the focus ; and similarly that the excentricity is T. em" 1 , and 


the major axis 2T 2 . 
m 2 + e 2 

363. To take a simpler case : let the acceleration vary as the 
distance from the origin. 

Then p i = + m z p, 

the upper or lower sign being used according as the acceleration is 
from or to the centre. 

This is 

Hence p = ae 

or p = a. cos mt + ft sin mt, 

where a and ft are arbitrary, but constant, vectors ; and e is the 

base of Napier s logarithms. 

The first is the equation of a hyperbola ( 31, k) of which a 
and ft are the directions of the asymptotes ; the second, that of an 
ellipse of which a and ft are semi-conjugate diameters. 

Since p = m{ae mt - fte~ mt ], 

or =m { asmmt + @cosmt}, 

the hodograph is again a hyperbola or ellipse. But in the first 
case it is, if we neglect the change of dimensions indicated by the 

364.] KINEMATICS. 285 

scalar factor m, conjugate to the orbit ; in the case of the ellipse 
it is similar and similarly situated. 

364 Again, let the acceleration be as the inverse third power of 

the distance, we have 

.. mUp 

Of course, we have, as usual, 

Vpp = 7. 
Also, operating by S . p, 

... mSpp 

A 2 r m 
p - --*, 

of which the integral is 

the equation of energy. 

A a - m 

Again, Spp = - t . 

Hence Spp + /3 2 = C, 

or Spp = Ct, 

no constant being added if we reckon the time from the passage 
through the apse, where Spp = 0. 

We have, therefore, by a second integration, 

p * = ce + c .......................... (i). 

[To determine C , remark that 

pp = Ct + 7, 

or p*p*=C*f-y*. 

But p 2 /3 2 = Cp 2 m (by the equation of energy), 


To complete the solution, we have, by 140 (2), 
T7 p dUp, TT ,_! d , Up 

= -- 

where /9 is a unit- vector in the plane of the orbit. 

But r=-2,. 

p p 


*-*-", s^. 71 f^ .< 

v t 

286 QUATERNIONS. [365. 

The elimination of t between this equation and (1) gives Tp in 
terms of Up, or the required equation of the path. 

We may remark that if 6 be the ordinary polar angle in the 

Hence we have = -Ty J ^ ^,1 

and 7* = -(Cf + C ) j 

from which the ordinary equations of Cotes spirals can be at once 

found. [See Tait and Steele s Dynamics of a Particle, Appendix 


365. To find the conditions that a given curve may be the hodo- 
graph corresponding to a central orbit. 

If -or be its vector, given as a function of the time, fadt is that 
of the orbit ; hence the requisite conditions are given by 

V<srfadt = y ........................... (1), 

where 7 is a constant vector. 

We may transform this into other shapes more resembling the 
Cartesian ones. 

Thus F*r/W<ft = ........................... (2), 

and Vtirfadt + F^OT - 0. 

From (2) fadt = x-fr, 

and therefore by (1) x Vvriz- = 7, 

or the curve is plane. And 

ajF-BJ-ar + FBT-BT = ; 

or eliminating x, 7 Vizix = ( Vvrdr)*. 

Now if v be the velocity in the hodograph, R its radius of curva 
ture, p the perpendicular on the tangent ; this equation gives at 


M = R p", 

which agrees with known results. 

366. The equation of an epitrochoid or hypotrochoid, referred 
to the centre of the fixed circle, is evidently 

where a is a unit vector in the plane of the curve and i another 

367.] KINEMATICS. 287 

perpendicular to it. Here o> and co l are the angular velocities in 
the two circles, and t is the time elapsed since the tracing point 
and the centres of the two circles were in one straight line. 
Hence, for the length of an arc of such a curve, 

s=fTpdt = fdt V{o>V + 2o>o> 1 a& cos (to -<o^t + to?!?}, 

= fdt Aua + >,&) 4^06 ^ "^ t\ , 

j ) 

which is, of course, an elliptic function. 

But when the curve becomes an epicycloid or a hypocycloid, 
coa T to) = 0, and 

which can be expressed in finite terms, as was first shewn by 

The hodograph is another curve of the same class, whose equa 
tion is 

and the acceleration is denoted in magnitude and direction by the 

Of course the equations of the common Cycloid and Trochoid 
may be easily deduced from these forms by making a indefinitely 
great and a indefinitely small, but the product aw finite ; and 
transferring the origin to the point 

= aa. 

B. Kinematics of a Rigid System. 

367. Let i be the normal-vector to any plane. 

Let TS and p be the vectors of any two points in a rigid plate 
in contact with the plane. 

After any small displacement of the rigid plate in its plane, let 
fc and dp be the increments of r and p. 

Then Sidvr 0, Sidp = ; and, since T (& p) is constant, 

And we may evidently assume, consistently with these equations, 

dp = a)i (p - r), 
dm = a)i (sr T) ; 

288 QUATERNIONS. [368. 

where of course r is the vector of some point in the plane, to a 
rotation co about which the displacement is therefore equivalent. 
Eliminating r, we have 

. d (OT p) 

(01 = - ^ , 

57 p 

which gives ew, and thence r is at once found. 
For any other point a in the plane figure 

Sida- = 0, 

S (p a) (dp da) = 0. Hence dp da = w l i (p a). 
S (a iv) (dm da) = 0. Hence da cZ-cr = &> 2 % (a -or). 
From which, at once, a) l = &> 2 = &>, and 

da = col (a r), 

or this point also is displaced by a rotation co about an axis through 
the extremity of r and parallel to i. 

368. In the case of a rigid body moving about a fixed point 
let OT, p, a denote the vectors of any three points of the body ; the 
fixed point being origin. 

Then CT 2 , p*, a z are constant, and so are Svrp, Spa, and Savr. 

After any small displacement we have, for za- and p, 

........................... (1). 

S-ndp + Spd = OJ 
Now these three equations are satisfied by 

dt& Vavr, dp = Fa/o, 
where a is any vector whatever. But if d^ and dp are given, then 

Vdvrdp = V. VO.TX Vap = ctS . ap-&. 
Operate by S .Vvrp, and remember (1), 

Vd&dp Vdpdtn- 

Hence a = c , , = -aT~ .................. (") 


Now consider a, Sada = 

Spda = Sadp J- , 
Svrda = Sad 

do- = Ya.a satisfies them all, by (2), and we have thus the proposi 
tion that any small displacement of a rigid body about a fixed 
point is equivalent to a rotation. 

3 70.] KINEMATICS. 289 

369. To represent the rotation of a rigid body about a given 
axis, through a given finite angle. [This is a work of supererogation, 
if we consider the results of 119. But it may be interesting to 
obtain these results in another manner.] 

Let a be a unit-vector in the direction of the axis, p the vector 
of any point in the body with reference to a fixed point in the axis, 
and 6 the angle of rotation. 

Then p = of 1 Sap + a~ l Yap, 

= aSap a Vap. 

The rotation leaves, of course, the first part unaffected, but the 
second evidently becomes 

or a Vap cos 6 + Vap sin 0. 

Hence p becomes 

p l = aSap aVap cos 9 + Vap sin 6, 

= (cos 6/2 + a sin 0/2) p (cos 0/2 - a sin 0/2), 
-a* pa *. 

370. Hence to compound two rotations about axes which meet, 
we may evidently write, as the effect of an additional rotation </> 
about the unit vector (3, 

P^^ pfi-* 1 . 
Hence p^tf 1 * of 1 pa.-" 1 * $-* *. 

If the /3-rotation had been first, arid then the a-rotation, we should 
have had 

p , = a >lf ft* p^ l " a -" , 

and the non-commutative property of quaternion multiplication 
shews that we have not, in general, 

P* = P* 

If a, (3, 7 be radii of the unit sphere to the corners of a spherical 
triangle whose angles are 0/2, <j)/2, tfr/2, we know that 

yfr/- */ a /ir = _ L (Hamilton, Lectures, p. 2G7.) 

Hence /3* /7r a e/7T = - 7 "* 7 , 

and we may write p z = y * f * py^, 

or, successive rotations about radii to two corners of a spherical 

triangle, and through angles double of those of the triangle, are 

T. Q. I. 19 

290 QUATERNIONS. [371. 

equivalent to a single rotation about the radius to the third corner, 
and through an angle double of the exterior angle of the triangle. 

Thus any number of successive finite rotations of a system, of 
which one point is fixed, may be compounded into a single rotation 
about a definite axis. 

371. When the rotations are indefinitely small, the effect of 
one is, by 369, 

and for the two, neglecting products of small quantities, 

a and b representing the angles of rotation about the unit-vectors 
a. and /3 respectively. 

But this is equivalent to 

P, = P + T(aoL + b/3) VU(aoL + b/3) p, 

representing a rotation through an angle T (aa + b/3), about the 
unit-vector U(aa + b/3). Now the latter is the direction, and the 
former the length, of the diagonal of the parallelogram whose sides 
are aa and b/3. 

We may write these results more simply, by putting a for aa, 
P for b/3, where a and /3 are now no longer unit-vectors, but repre 
sent by their versors the axes, and by their tensors the angles 
(small), of rotation. 

Thus = 

372. Given the instantaneous axis in terms of the time, it is 
required to find the single rotation which will bring the body from 
any initial position to its position at a given time. 

If a be the initial vector of any point of the body, w the value 
of the same at time t, and q the required quaternion, we have 

by 119 

-53- = qaq l ............................... (1). 

Differentiating with respect to t, this gives 
tb- = qaq~ l qaq~ l qq~ l , 

But tzr = Fe-cr = F. eqctq~\ 




Hence, as qaq l may be any vector whatever in the displaced 
body, we must have 

e = 2Vqq* ............................... (2). 

This result may be stated in even a simpler form than (2), for 
we have always, whatever quaternion q may be, 

and, therefore, if we suppose the tensor of q, which, as it is not 
involved in q ( ) q~ l , may have any value whatever, to be a 
constant (unity, for instance), we may write (2) in the form 

7 = 2j ................................. (3). 

An immediate consequence, which will be of use to us later, is 
q.q~ 1 eq = 2q ............................... (4). 

373. To express q in terms of the usual angles -fy, 6, ^>. 

Here the vectors i, j, k in the original position of the body 
correspond to OA, OB, OC, re 
spectively, at time t. The trans 
position is defined to be effected 
by first, a rotation ^r about lc\ 
second, a rotation about the new 
position of the line originally 
coinciding with j ; third, a rotation 
<f> about the final position of the 
line at first coinciding with Jc. 

This selection of angles, in 
terms of which the quaternion is 
to be expressed, is essentially 

unsymmetrical, and therefore the results cannot be expected to be 

The rotation ty about k has the operator 

This converts^ into T/, where 

,7 = A-* " 1 jk~* lir =j cos ^ - i sin ^. 

The body next rotates about rj through an angle 0. This has 
the operator 


292 QUATERNIONS. [374. 

It converts k into 

OC = f = v e/7T kv~ e/1T = (cos 0/2 + 77 sin (9/2) A; (cos 0/2 - 77 sin (9/2) 
= k cos 4- sin (i cos -^ + j sin i|r). 

The body now turns through the angle (/> about the operator 

t,0/7T/ X C.-0/7T 

Hence, omitting a few reductions, which we leave as excellent 
practice for the reader, we find 

q = f */V /ff A;* /ir 

= (cos 0/2 + ? sin 0/2) (cos 0/2 + r? sin 0/2) (cos i|r/2 + A; sin ^r/2) 
= cos (0 + -^)/2 . cos 0/2 + i sin (0 - ^)/2 . sin 0/2 + 

j cos (0 - ^)/2 . sin 0/2 + k sin (0 + ^)/2 . cos 0/2, 
which is, of course, essentially unsyrnmetrical. 

374. To find the usual equations connecting ty, 0, with the 
angular velocities about three rectangular axes fixed in the body. 

HaviDg the value of q in last section in terms of the three 
angles, it may be useful to employ it, in conjunction with equation 
(3) of 372, partly as a verification of that equation. Of course, 
this is an exceedingly roundabout process, and does not in the . 
least resemble the simple one which is immediately suggested by 

We have 2q = eq = {co l OA + o> 2 O + o> 3 OC}q, 
whence 2q~*q = q~ l {co l OA + o> 2 OB + o> 3 OC} q, 

or 2q = q (ico 1 +jo 2 + kco s ). 

This breaks up into the four (equivalent to three independent) 

2 ~ [cos (0 + ^)/2 . cos 0/2] 

= -to i sin (</> - ^)/2 . sin 0/2 - w 2 cos (0 - ^)/2 . sin 0/2 

- &) 3 sin (0 + i/r)/2 . cos 0/2, 

2 [sin (0-^/2. an 0/2] 

= &), cos ((/> 4- ^)/2 . cos 0/2 - w, sin (</> + ^r)/2 . cos 0/2 

+ w 3 cos (c/) - f )/2 . sin 0/2, 

375-] KINEMATICS. 293 

. cos 0/2 + o> 2 cos (< + ^)/2 . cos 0/2 

- o> 3 sin (< - i/r)/2 . sin 0/2, 

= - o> : cos ((/> - i/r)/2 . sin 0/2 + 2 sin (<f> - i|r)/2 . sin 0/2 

-f o> 8 cos (< + ^)/2 . cos 0/2. 

From the second and third, eliminate (j> ^jr, and we get by 

cos 0/2 . = (o) t sin </> -f ct> 2 cos <) cos 0/2, 

or = co 1 sin < -fo> 2 cos $ ...................... (1). 

Similarly, by eliminating between the same two equations, 
sin 0/2 . ((j> fy = &> s sin 0/2 + co 1 cos <^> cos 0/2 o> 2 sin (/> cos 0/2. 

And from the first and last of the group of four 
cos 0/2 . ((j> + -vfr) = co 3 cos 0/2 w l cos (/> sin 0/2 + o> 2 sin </> sin 0/2. 
These last two equations give 

< + ^jr cos = w 3 ..................... (2). 

cos + ^ = ( o) i cos cf) + o> 2 sin <) sin + &) 3 cos 0. 
From the last two we have 

T/T sin = co 1 cos </> -I- o> 2 sin ^> ............ (3). 

(1), (2), (3) are the forms in which the equations are usually given. 

375. To deduce expressions for the direction-cosines of a set of 
rectangular axes, in any position, in terms of rational functions of 
three quantities only. 

Let a, /3, 7 be unit-vectors in the directions of these axes. Let q 
be, as in 372, the requisite quaternion operator for turning the 
coordinate axes into the position of this rectangular system. Then 

q = w + xi + yj + zk, 
where, as in 372, we may write 

1 = w z + x 2 + ?/ 2 + z\ 

Th en we have q~ l = w xi yj zk, 

and therefore 

a - qiq 1 - (wi - x - yk + zj} (w - xi - yj - zk) 

= (w 2 4- a? 2 - / - z 2 -) i+2(wz + xy)j 4 2 (xz - wy) k, 

294 QUATERNIONS. [3/6. 

where the coefficients of i, j, k are the direction-cosines of a as 
required. A. similar process gives by inspection those of /3 and 7. 

As given by Cayley*, after Rodrigues, they have a slightly 
different and somewhat less simple form to which, however, they 
are easily reduced by putting 

w = as/\ = yip = z/v = 1/K*. 

The geometrical interpretation of either set is obvious from the 
nature of quaternions. For (taking Cayley s notation) if 6 be the 
angle of rotation : cosy, cos g, cos h, the direction-cosines of the 
axis, we have 

q = w + ari + yj + zk = cos 6/2 + sin 0/2 . (i cos/+ j cos g + k cos h), 
so that w = cos 0/2, 

x = sin 0/2 . cos f } 
y = sin 0/2 . cos g, 
z = sin 0/2 . cos h. 

From these we pass at once to Rodrigues subsidiary formulae, 
* = l/w 2 = sec 2 0/2, 
X = xjw = tan 0/2 . cos /*, 
&c. = &c. 

C. Kinematics of a Deformable System. 

376. By the definition of Homogeneous Strain, it is evident 
that if we take any three (non-co planar) unit-vectors a, ft, 7 in 
an unstrained mass, they become after the strain other vectors, 
not necessarily unit-vectors, a lf ft v 7 r 

Hence any other given vector, which of course may be thus 

expressed, p = XOL + yft + zy, 

becomes p l = osa 1 + yft t + zy v 

and is therefore known if a v ft v y 1 be given. 
More precisely 

pS.a/By = aS.jSyp + /SS.yap 4 yS.a/3p 

pfi . CL(3y = fpS . OL(3y - o^S . /%> + Pfi . yap + 7^ . aft p. 
Thus the properties of <J>, as in Chapter V., enable us to study 
with great simplicity homogeneous strains in a solid or liquid. 

* Camb. and Dub> Math. Journal Vol. i, (1846). 

37&] KINEMATICS. 295 

For instance, to find a vector whose direction is unchanged by 
the strain, is to solve the equation 

Vpcf)p = 0, or ^>p=gp (1), 

where g is a scalar unknown. 

[This vector equation is equivalent to three scalar equations, and 
contains only three unknown quantities ; viz. two for the direction 
of p (the tensor does not enter, or, rather, is a factor of each side), 
and the scalar g.] 

We have seen that every such equation leads to a cubic in g 
which may be written 

g* - m 2 g 2 + m^g - ra = 0, 

where m 2 , m v m are scalars depending in a known manner on the 
constant vectors involved in <. This must have one real root, and 
may have three. 

377. For simplicity let us assume that a, /3, 7 form a rectangular 
system, then we may operate on (1) by S . a, S . ft, and $.7; and 
thus at once obtain the equation for g, in the form 

i =0 

W, + <7, 

To reduce this we have, for the term independent of g, 

8aa lt SOL ft v Say 1 , = -$. aftyS . a 1 j3 l y 1 (Ex. 9, Chap. III.), 

which, if the mass be rigid, becomes 1. 
The coefficient of the first power of g is, 

Thus we can at once form the equation; which becomes, 
for the special case of a rigid system, 

-1-0 (Saa, + Sflfr + SyyJ + g 2 (Saa, + S/3/3, + 8y 7l ) + g* = 0, 
or -!( 

378. If we take Tp = C we consider a portion of the mass 
initially spherical. This becomes of course 


an ellipsoid, in the strained state of the body. 

296 QUATERNIONS. [379- 

Or if we consider a portion which is spherical after the strain, i.e. 

2>, = C-. 

its initial form was T(j)p = C, 

another ellipsoid. The relation between these ellipsoids is obvious 
from their equations. (See 327.) 

In either case the axes of the ellipsoid correspond to a rect 
angular set of three diameters of the sphere ( 271). But we must 
carefully separate the cases in which these corresponding lines in 
the two surfaces are, and are not, coincident. For, in the former 
case there is pure strain, in the latter the strain is accompanied 
by rotation. Here we have at once the distinction pointed out by 
Stokes* and Helmholtz-f- between the cases of fluid motion in 
which there is, or is not, a velocity-potential. In ordinary fluid 
motion the distortion is of the nature of a pure strain, i.e. is differ 
entially non-rotational ; while in vortex motion it is essentially 
accompanied by rotation. But the resultant of two pure strains is 
generally a strain accompanied by rotation. The question before 
us beautifully illustrates the properties of the linear and vector 

379. To find the criterion of a pure strain. Take a, /3, 7 now 
as unit-vectors parallel to the axes of the strain-ellipsoid, they 
become after the strain, ace, 6/3, cy, provided the strain be pure. 

Hence p 1 = <f)p = aaSap b{3S/3p cjSyp. 
And we have, for the criterion of a pure strain, the property of 
the function ^>, that it is self-conjugate, i.e. 

Spcfxr = So-(j)p. 

380. Two pure strains, in succession, generally give a strain 
accompanied by rotation. For if </>, ty represent the strains, since 
they are pure we have 

But for the compound strain we have 

Pi = XP 
and we have not generally 

* Cambridge Phil Trans. 1845. 

t Crclle, vol. Iv. 1857. See also PhiL Mag. (Supplement) June 1867. 

381.] KINEMATICS. 297 


by (1), and -*//( is not generally the same as <f>*fy. (See Ex. 7 to 
Chapter V.) 

To find the lines which are most altered in length by the strain. 

Here T$p is a maximum or minimum, while Tp is constant ; 
so that 

Sdp$$p = 0, Spdp = 0. 

Hence fy fyp ~ X P> 

and the required lines are the principal vectors of <j> (f), which 

( 381) is obviously self-conjugate; i.e. denotes a pure strain. 

381. The simplicity of this view of the question leads us to 
suppose that we may easily separate the pure strain from the 
rotation in any case, and exhibit the corresponding functions. 

When the linear and vector function expressing a strain is 
self-conjugate the strain is pure. When not self-conjugate, it may 
be broken up into pure and rotational parts in various ways (ana 
logous to the separation of a quaternion into the sum of a scalar 
and a vector part, or into the product of a tensor and a versor 
part), of which two are particularly noticeable. Denoting by a 
bar a self-conjugate function, we have thus either 

</> = ^ f or < = q~, 

where e is a vector, and q a quaternion (which may obviously be 
regarded as a mere versor). [The student must remark that, 
although the same letters have been employed (from habit) in 
writing the two last formulae, one is not a transformation of the 
other. In the first a pure strain is succeeded by a rotation, in the 
second the rotation is followed by the pure strain.] 

That this is possible is seen from the fact that (j> involves nine 
independent constants, while ty and r each involve six, and e and q 
each three. If <j> be the function conjugate to <, we have 

f-?-F.e< ), 

so that 2t|r = < + < , 

and 2 V. e ( ) = < -</> , 

which completely determine the first decomposition. This is, of 
course, perfectly well known in quaternions, but it does not seem 
to have been noticed as a theorem in the kinematics of strains that 

293 QUATERNIONS. [381. 

there is always one, and but one, mode of resolving a strain into 
the geometrical composition of the separate effects of (1) a pure 
strain, and (2) a rotation accompanied by uniform dilatation 
perpendicular to its axis, the dilatation being measured by 
(sec. 1) where 6 is the angle of rotation. 

In the second form (whose solution does not appear to have 
been attempted), we have 

where the pure strain precedes the rotation, and from this 

* =3?.*( )q, 

or in the conjugate strain the rotation (reversed) is followed by the 
pure strain. From these 

and w is to be found by the solution of a biquadratic equation*. 
It is evident, indeed, from the identical equation 

that the operator </> (/> is self-conjugate. 
In the same way 

or q 1 (<< q = t? (q l pq) = <j> <f> (q l pq), 

* Suppose the cubic in "w to be 

Now ro 2 is equal to 0, a known function, which we may call u. Thus 

W 2 W, 

and therefore ~m and w are commutative in multiplication. 
Eliminating 5 between these equations we have, first, 

(w - g z ) u + ^W -g = Q=i&((a + g l )- g z w - g, 
and finally w 3 + (2g^ - g^} tf + (g^ - 2gg z ) u - # 2 = 0. 

This must agree with the (known) cubic in w, 

co 3 - ??i 2 w 2 + mj<ta - m = 0, suppose ; 
so that, by comparison of coefficients we have 

and thus g is known, and r/ 2 = - 


The values of the quantities g being found, OT is given in terms of w by the equation 

383.] KINEMATICS. 299 

which shew the relations between <fxf) f , fy fy, aid q. 
To determine q we have 

<l>p.q= q^p 
whatever be p, so that 

S.Vq(<l>-v)p = 0, 

or S.p((f> - =) Vq = 0, 

which gives (</> r) F<? = 0. 

The former equation gives evidently 

Vq\\ V.(4>-v)a(4>--S)0 

whatever be a and ft ; and the rest of the solution follows at once. 
A similar process gives us the solution when the rotation precedes 
the pure strain. [Proc. R. S. E. 18701.] 

382. In general, if 

the angle between any two lines, say p and a, becomes in the 
altered state of the body 

cosT 1 (-8. UfoUjxr). 
The plane $(J> = becomes (with the notation of 157) 

S^ ^p = 8&-*p = 0. 
[For if X, IJL be any two vectors in it, 

But they become <f>\, fa, and the line perpendicular to both is 

Hence the angle between the planes Sp = 0, and Srjp = 0, which 
is cos"^ S . UUr)), becomes 

cos 1 (- 8 . UQ-^UjT 1 *!). 
The locus of lines equally elongated is, of course, 

or Tfa = eTp, 

a cone of the second degree. 

383. In the case of a Simple Shear, we have, obviously, 

Pi = <t>P = P + 
where a is a unit vector, and 


The vectors which are unaltered in length are given by 


or 2S/3pSoip + /3*S 2 ap = 0, 

which breaks up into S . ap = 0, 


The intersection of this plane with the plane of a, {3 is perpen 
dicular to 2/3 + /3 2 a. Let it be a + x/3, then 

.e. 2#-l=0. 

Hence the intersection required is 


For the axes of the strain, one is of course a/3, and the others 
are found by making T(f)Up a maximum and minimum. 

Let p = a + x/3, 

then p l = (j)p = a + x/B /3, 

T Pl 

and T=P = max. or mm., 


gives a; 2 - ic + g* = 0, 

from which the values of x (say oc^ and a? 8 ) are found. 

Also, as a verification, we must show that the lines of the 
body which become most altered in length are perpendicular to 
one another both before and after the shear. 

Thus S.(a + xfi) (OL + xfi) = - 1 + /3 V? 2 , 

should be = 0. It is so, since, by the equation in x t 


X \ X * ^2 


fif(a + (^-l)/9} fa+(^-l)/3}= - 1 +/3 2 [x^- (^ +a; 9 ) + 1}, 
ought also to be zero. And, in fact, 

by the equation for x ; so that this also is verified. 

384. We regret that our limits do not allow us to enter farther 
upon this very beautiful application. [The reader is referred to 

385.] KINEMATICS. 301 

Chapter X. of Kelland and Tail s Introduction to Quaternions ; in 
which the treatment of linear and vector equations is based upon 
the theory of homogeneous strain ; which, in its turn, is much more 
fully developed than in the present work.] 

But it may be interesting to consider briefly the effects of any 
continuous displacements (of the particles of a body) by the help of 
the operator V. 

We have seen ( 148) that the effect of the operator SdpV, 
upon any scalar function of the vector of a point, is to produce 
total differentiation due to the passage from p to p + dp. 

Hence if a be the displacement of p, that of p + $p is 

Thus the strain of the group of particles near p is such that 
T = T-(TV)<r. 

[Here we virtually assume that a is a continuous function of p.] 
But if this correspond to a linear dilatation e, combined with 
a rotation whose vector-axis is e, both being infinitesimal, 

< T = T (1 + e ) + Ver. 
Thus, for all values of T, each with its proper e, 

This gives at once (for instance by putting in succession for T 
any three rectangular unit vectors) 

from which we conclude as follows : 

If a (a continuous function of p) represent the vector displace 
ment of a point situated at the extremity of the vector p (drawn 
from the origin) 

SVo- represents the consequent cubical compression of the 
group of points in the vicinity of that considered, and 

FVcr represents twice the vector axis of rotation of the same 
group of points. 

385. As an illustration, suppose we fix our attention upon a 
group of points which originally filled a small sphere about the 
extremity of p as centre, whose equation referred to that point is 

Tco = c ................................. (1). 

After displacement p becomes p + cr, and, by last section, p + w 
becomes p + co -f a ($o>V) a. Hence the vector of the new surface 

302 QUATERNIONS. [386. 

which encloses the group of points (drawn from the extremity of 
p + a-) is 

ft), = ft)-(ASfctV) <r (2). 

Hence & is a homogeneous linear and vector function of co l ; or 

ft) = (f>CO l} 

and therefore, by (1), T^co^ = c, 

the equation of the new surface, which is evidently a central surface 
of the second degree, and therefore, of course, an ellipsoid. 

We may solve (2) with great ease by approximation, if we as 
sume that TVar is very small, and therefore that in the small term 
we may put co l for CD ; i.e. omit squares of small quantities ; thus 
&) = co 1 + ($ft> t V) a. 

386. If the vector displacement of each point of a medium is in 
the direction of, and proportional to, the attraction exerted at that 
point by any system of material masses, the displacement is effected 
without rotation. 

For if Fp = C be the potential surface, we have Sadp a complete 
differential ; and, by 334, 

FVo- = 0. 

Conversely, if there be no rotation, the displacements are in the 
direction of, and proportional to } the normal vectors to a series of 

For = - V. dp FVcj = - (SdpV) a- + VSadp, 

where, in the last term, V acts on <r alone. 

Now, of the two terms on the right, the first is ( 149, (4)) 
the complete differential da, and therefore the remaining term 
must be a complete differential. This, of course, means that 

is a complete differential. 

Thus, in a distorted system, there is no compression if 

SVa = 0, 

and no rotation if FVcr = ; 

and evidently merely transference if <r = a= a constant vector, 
which is one case of 


In the important case of a = VFp 
there is (as proved already) no rotation, since 

387.] KINEMATICS. 303 

is evidently a scalar. In this case, then, there are only translation 
and compression, and the latter is at each point proportional to the 
density of a distribution of matter, which would give the potential 
Fp. For if r be such density, we have at once 

D. Axes of Inertia. 

387. The Moment of Inertia of a body about a unit vector a 
as axis is evidently 

where p is the vector of the element ra of the mass, and the origin 
of p is in the axis. [The letter h has, for an obvious reason, been 
put here in place of the k which is usually employed for the radius 
of gyration.] 

Hence if we put /5 = e^ajh, where e is constant, we have, as 
locus of the extremity of /3, 

Me = - 2m(Vj3p) 2 = - MS/3<t>/3 (suppose), 

the well-known ellipsoid. The linear and vector function, <, 
depends only upon the distribution of matter about the (temporary) 

If r be the vector of the centre of inertia, cr the vector of in 
with respect to it, we have 

p = <& + o-, 

where (31 (e)) 2m<r = 0; 

and therefore Mh* = - 2m {( Fern) 2 + ( Few) 2 J 

Here fa is the unique value of < which corresponds to the 
distribution of matter relative to the centre of inertia. The 
equation last written gives the well-known relation between the 
moment of inertia about any line, and that about a parallel line 
through the centre of inertia. 

Hence, to find the principal axes of inertia at any point (the 
origin, whose vector from the centre of inertia is w), note that h 
is to be made max., min., or max.-min., with the condition 

a = -l. 

Thus we have SOL (-crFa-or + fact) = 0, 

Sa a = 0; 

* Proc, R, S, E., 18623. 


therefore (/^a + Tz-Vavr pa.= 7i 2 a (by operating by S . a). 

Hence (^ - tf - w*) a = - vrSam .................. (1), 

determines the values of a, k 2 being found from the cubic 

Si* (fa - W - w*)- 1 <GF = - 1 .................. (2). 

Now the normal to 

S<r(<f> l -h*- B*)- l <r = -l .................. (3), 

at the point a is parallel to 

(*,-A -rr. 

But (3) passes through OT, by (2), and there the normal is 

fa-h* -<**)*, 

which, by (1), is parallel to one of the required values of a. (3) is, 
of course ( 288), one of the surfaces confocal with the ellipsoid 

S.o- ( j)- 1 (T = -l. 

Thus we prove Binet s theorem that the principal axes at any 
point are normals to the three surfaces of the second degree, confocal 
with the central ellipsoid, which pass through that point. 


1. Form, from kinematical principles, the equation of the 
cycloid ; and employ it to prove the well-known elementary 
properties of the arc, tangent, radius of curvature, and evolute, 
of the curve. 

2. Interpret, kinematically, the equation 

where /3 is a given vector, and a a given scalar. 

Shew that it represents a plane curve ; and give it in an 
integrated form independent of t. 

3. If we write -or = (3t p, 

the equation in (2) becomes 

/3 is = a Uvr. 
Interpret this kinematically ; and find an integral. 

What is the nature of the step we have taken in transforming 
from the equation of (2) to that of the present question ? 


4. The motion of a point in a plane being given, refer it to 

(a) Fixed rectangular vectors in the plane. 

(b) Rectangular vectors in the plane, revolving uniformly 
about a fixed point. 

(c) Vectors, in the plane, revolving with different, but uni 
form, angular velocities. 

(d) The vector radius of a fixed circle, drawn to the point of 
contact of a tangent from the moving point. 

In each case translate the result into Cartesian coordinates. 

5. Any point of a line of given length, whose extremities move 
in fixed lines in a given plane, describes an ellipse. 

Shew how to find the centre, and axes, of this ellipse ; and 
the angular velocity, about the centre of the ellipse, of the tracing 
point when the describing line rotates uniformly. 

Transform this construction so as to shew that the ellipse is a 

When the fixed lines are not in one plane, what is the locus ? 

6. A point, A, moves uniformly round one circular section of 
a cone ; find the angular velocity of the point, a, in which the 
generating line passing through A meets a subcontrary section, 
about the centre of that section. 

7. Solve, generally, the problem of finding the path by which 
a point will pass in the least time from one given point to another, 
the speed at the point of space whose vector is p being expressed 
by the given scalar function 


Take also the following particular cases : 

( a ) fp = a while Sap > 1, 
fp=b while Sap < 1. 

(b) fp = TSap. 

(c) fp = -p 2 . (Tait, Trans. R. S. E. t 1865.) 

8. If, in the preceding question,//? be such a function of Tp 
that any one swiftest path is a circle, every other such path is a 
circle, and all paths diverging from one point converge accurately 
in another. (Maxwell, Camb. and Dub. Math. Journal, IX. p. 9.) 

T. Q. I. 20 


9. Interpret, as results of the composition of successive conical 
rotations, the apparent truisms 



a K i 8 7 /3 

and -- z ...... - 5 - = 1. 

K L 6 7 p a 

(Hamilton, Lectures, p. 334.) 

10. Interpret, in the same way, the quaternion operators 

-Qffl "> 

11. Find the axis and angle of rotation by which one given 
rectangular set of unit-vectors a, /3, 7 is changed into another 
given set 15 yS 1? 7 r 

12. Shew that, if (f>p = p + Fe/>, 

the linear and vector operator $ denotes rotation about the vector 
e, together with uniform expansion in all directions perpendicular 
to it. 

Prove this also by forming the operator which produces the 
expansion without the rotation, and that producing the rotation 
without the expansion ; and finding their joint effect. 

13. Express by quaternions the motion of a side of one right 
cone rolling uniformly upon another which is fixed, the vertices of 
the two being coincident. 

14. Given the simultaneous angular velocities of a body about 
the principal axes through its centre of inertia, find the position 
of these axes in space at any assigned instant. 

15. Find the linear and vector function, and also the quater 
nion operator, by which we may pass, in any simple crystal of the 
cubical system, from the normal to one given face to that to 
another. How can we use them to distinguish a series of faces 
belonging to the same zone ? 

16. Classify the simple forms of the cubical system by the 
properties of the linear and vector function, or of the quaternion 
operator, mentioned in (15) above. 


17. Find the vector normal of a face which truncates symme 
trically the edge formed by the intersection of two given faces. 

18. Find the normals of a pair of faces symmetrically truncat 
ing the given edge. 

19. Find the normal of a face which is equally inclined to 
three given faces. 

20. Shew that the rhombic dodecahedron may be derived from 
the cube, or from the octahedron, by truncation of the edges. 

21. Find the form whose faces replace, symmetrically, the 
edges of the rhombic dodecahedron. 

22. Shew how the two kinds of hemihedral forms are indi 
cated by the quaternion expressions. 

23. Shew that the cube may be produced by truncating the 
edges of the regular tetrahedron. If an octahedron be cut from a 
cube, and cubes from its tetrahedra, all by truncation of edges, the 
two latter cubes coincide. 

24. Point out the modifications in the auxiliary vector func 
tion required in passing to the pyramidal and prismatic systems 

25. In the rhombohedral system the auxiliary quaternion 
operator assumes a singularly simple form. Give this form, and 
point out the results indicated by it. 

26. Shew that if the hodograph be a circle, and the accelera 
tion be directed to a fixed point ; the orbit must be a conic section, 
which is limited to being a circle if the acceleration follow any 
other law than that of gravity. 

27. In the hodograph corresponding to acceleration f(D ) 
directed towards a fixed centre, the curvature is inversely as 

28. If two circular hodographs, having a common chord, which 
passes through, or tends towards, a common centre of force, be cut 
by any two common orthogonal s, the sum of the two times of 
hodographically describing the two intercepted arcs (small or large) 
will be the same for the two hodographs. (Hamilton, Elements, 
p. 725 ) 



29. Employ the last theorem to prove, after Lambert, that the 
time of describing any arc of an elliptic orbit may be expressed in 
terms of the chord of the arc and the extreme radii vectores. 

30. If q ( ) q~ l be the operator which turns one set of rect 
angular unit-vectors a, /3, 7 into another set a 4 , @ lt y lt shew that 
there are three equations of the form 

31. If a ray, a, fall on a fine, polished, wire 7, shew that on 
reflection it forms the surface 

a right cone. 

32. Find the path of a point, and the manner of its descrip 
tion, when 

33. In the first problem of 336 shew that 

Vq~*q = Vv, or V . uq~ l 0. 
Also that (Vv) 2 = - 2V 2 <y, or 4tTVw* = 0. 
Again, shew that there are three equations of the form 

dv _ ._ 2 ^ dv 

_ Vv = iV*v + V -j-. 
doc dx 

From these last deduce, by a semi-Cartesian process, the result 

as in the text. 

34. Give the exact solution of 

to, = ft> - Sort . a: ( 385.) 
[Note that we may assume, cr being given, 

do- = <j)dp, 

where the constituents of (f> are known functions of p. Thus we 
have what is wanted for the problem above : viz. 

with various other important results, such as 

V<7 = 2^, &c.] 



388. WE propose to conclude the work by giving a few in 
stances of the ready applicability of quaternions to questions of 
mathematical physics, upon which, even more than on the Geo 
metrical or Kinematical applications, the real usefulness of the 
Calculus must mainly depend except, of course, in the eyes of 
that section of mathematicians for whom Transversals and Anhar- 
monic Pencils, &c. have a to us incomprehensible charm. Of course 
we cannot attempt to give examples in all branches of physics, nor 
even to carry very far our investigations in any one branch : this 
Chapter is not intended to teach Physics, but merely to shew by 
a few examples how expressly and naturally quaternions seem to 
be fitted for attacking the problems it presents. 

We commence with a few general theorems in Dynamics the 
formation of the equations of equilibrium and motion of a rigid 
system, some properties of the central axis, and the motion of a 
solid about its centre of inertia. The student may profitably 
compare, with the processes in the text, those adopted by Hamilton 
in his Elements (Book III., Chap. III., Section 8). 

A. Statics of a Rigid System. 

389. When any forces act on a rigid body, the force y8 at the 
point whose vector is a, &c., then, if the body be slightly displaced, 
so that a becomes a -f Sa, the whole work done against the forces is 


310 QUATERNIONS. [39- 

This must vanish if the forces are such as to maintain equilibrium. 
Hence the condition of equilibrium of a rigid body is 

For a displacement of translation Sa is any constant vector, hence 

2/3 = .............................. (1). 

For a rotation-displacement, we have by 371, e being the axis, 
and Te being indefinitely small, 

SOL = Fea, 

and 28.ftVect = 2S.eVaft = S.6$(VcLfi) = 0, 

whatever be e, hence 2 . = ........................... (2). 

These equations, (1) and (2), are equivalent to the ordinary six 
equations of equilibrium. 

390. In general, for any set of forces, let 

2/3 = ft, 
2. Fa/3 = a,, 

it is required to find the points for which the couple a t has its axis 
coincident with the resultant force /3 r Let 7 be the vector of such 
a point. 

Then for it the axis of the couple is 

and by condition xft l = a l 

Operate by 8ft l ; therefore 

and Vyft, = a, - ft~ l Sa^ = -ft, Vafi-\ 

or y = Va 1 ft- l + uft v 

a straight line (the Central Axis) parallel to the resultant force. 

[If the resultant force and couple be replaced by an equivalent 
in the form of two forces, ft at a, and ft at of, we have 

The volume of the tetrahedron whose opposite edges are @, 
(acting as above stated) is as 8 . ftV(a a) ft. But 

so that the volume is as 8^ (& + ft) = Sa^ a constant whatever 
pair of equivalent forces be taken.] 


391. To find the points about which the couple is least. 
Here T (a x - Fy^) = minimum. 

Therefore 8 . (a, - Fy/^) VPrf = 0, 

where 7 is any vector whatever. It is useless to try 7 = ft v but 
we may put it in succession equal to OL I and to Vaft^ Thus 

and ( FaA) 2 - {3*S . 7 FA = 0. 

Hence 7 = x VOL ft \ -f yfi v 

and by operating with S . Va.J3 v we get 

or 7 = 1 / 

the same locus as in last section. 

392. The couple vanishes if 

This necessitates $,, = 0, 

or the force must be in the plane of the couple. If this be the 


7 = 1 /3," + */3 I > 
still the central axis. 

To assign the values of forces f, f t , to act at e, e p and be 
equivalent to the given system. 

Hence Fef + Fe, (/9, - f) = a,, 

and f = (6 - e,)- 1 (a, - F^/3,) + x (e - 6 t ). 

Similarly for f r The indefinite terms may be omitted, as they 
must evidently be equal and opposite. In fact the}^ are any equal 
and opposite forces whatever acting in the line joining the given 

393. If a system of parallel forces act on a rigid body, say 

x/3 at a, &c. 
they have the single resultant /3S (a), at a, such that 

312 QUATERNIONS. [393- 

Hence, whatever be the common direction of the forces, the 
resultant passes through 

_ 2 pa) 

If 5 (#) = 0, the resultant is simply the couple 

By the help of these expressions for systems of parallel forces 
we can easily proceed to the case of forces generally. 

Thus if any system of forces, /3, act at points, a, of a rigid body ; 
and if i, j, k be a system of rectangular unit vectors such that 

the resultant force is 


bk acting at , or 

as we may write it. Take this as origin, then <& = 0. 
The resultant couple, in the same way, is 

or V(i(f)i+j(f)j). 

Now </>i, $7, <A? are invariants, in the sense that they retain 
the same values however the forces and (with them) the system 
i, j, k be made to rotate : provided they preserve their mutual 
inclinations, and the forces their points of application. For the 
as are constant, and quantities of the form S/3i, S/3j t or S/3k are 
not altered by the rotation. 

We may select the positions of i and j so that (f>i and <fyj 
shall be perpendicular to one another. For this requires only 

8 . ty <f>j = 0, or S.ipthj = 0. 

But ( 381) </> (/> is a self-conjugate function; and, by our change of 
origin, k is parallel to one of its chief vectors. The desired result 
is secured if we take i, j as the two others. 

With these preliminaries we may easily prove Minding s 
Theorem : 

If a system of forces, applied at given points of a rigid body, 
have their directions changed in any way consistent with the 
preservation of their mutual inclinations, they have in an infinite 
number of positions a single force as resultant. The lines of 
action of all such single forces intersect each of two curves fixed 
in space. 


The condition for the resultant s being a single force in the 
line whose vector is p is 

which may be written as 

bp = xk J4>i + i<f)j. 

That the two last terms, together, form a vector is seen by 
operating on the former equation by S.k , for we thus have 

We may write these equations for convenience as 

bp = xk jo. + i/3 ....................... ,(1), 

Sja-Sij3 = () ....................... (2). 

[The student must carefully observe that a and /3 are now 
used in a sense totally different from that in which they first 
appeared, but for which they are no longer required. If this 
should puzzle him, he may change a into 7, and (3 into 8, in the last 
two equations and throughout the remainder of this section.] 

Our object now must be to express i and j in terms of the single 
variable k, which is afterwards to be eliminated for the final 

From (2) we find at once 

whence we easily arrive at either of the following 

or _ y = a * + + (Ska)* + (Sk/3) z + 28 . kcfi] 

Substituting for i and^ in (1) their values (3), we have 

= (vr-z)k-a/3 ........................... (5), 

where w t which is now used for a linear and vector function, is 
defined by the equation 

srp = aSap ftSffp. 

Obviously r (a/3) = 0, 

so that (w - z)~ l (a/3) = - - a/3. 


Thus -yb(^-2)- l p = k+- ..................... (6). 

314 QUATERNIONS. [394- 

Multiply together the respective members of (5) and (6), and take 
the scalar, and we have 

or, by (4), = f + z + a* + /3 2 + 


= /+ 


which, for z a 2 , or z = - /3 2 , gives as the required curves the 
focal conies of the system 

394. The preceding investigation was based on the properties 
of a system of parallel forces ; and thus has a somewhat composite, 
semi-Cartesian, character. 

That which follows is much more purely quaternionic. It is 
taken from the Trans. R. S. E. 1880. 

When any number of forces act on a rigid system ; (3 l at the 
point GL V /3 2 at 2 , &c., their resultant consists of the single force 

/8 = 2/3 
acting at the origin, and the couple 

* = -2F/3a ........................... (1). 

If these can be reduced to a single force, the equation of the 
line in which that force acts is evidently 

F/8p = 2F/3a .............................. (2). 

Now suppose the system of forces to turn about, preserving 
their magnitudes, their points of application, and their mutual 
inclinations, and let us find the fixed curves in space, each of 
which is intersected by the line (2) in every one of the infinite 
number of its positions. 

Operating on (2) by F. /3, it becomes 

with the notation of Chap. V. Now, however the forces may 
turn, 0/3 = 2a/3/3 

is an absolute constant ; for each scalar factor as S/3J3 is unaltered 
by rotation. Let us therefore change the origin, i.e. the value of 
each a, so as to make 

= ........................ (3). 


This shews that / is one of the three principal vectors of $, and 
we see in consequence that $ may be expressed in the form 

where 7 and S are unit vectors, forming with JB a rectangular 
system. They may obviously be so chosen that 7 and 8 shall be 
at right angles to one another, but these (though constants) are 
not necessarily unit vectors. 

Equation (2) is now 

&F/3 P =F77 +FSS ..................... (2 ), 

where b is the tensor, and ft the versor, of ft. 

The condition that the force shall lie in the plane of the couple 
is, of course, included in this, and is found by operating by S .0. 

Thus 8 (&/ - 78 ) = ........................... (4). 

395. We have here all the data of the problem, and solutions 
can only differ from one another in the mode of attacking 
and (4). 

Writing (4) in the form 

7 (8 

we have at once ty = F/3S + /3 Vfiy, } ,.,. 

whence tS = - V@y + ft V/3& j" 

where t is an undetermined scalar. 

By means of these we may put (2 ) in the form 

where * f = -ySy ( )-S SS ( ). 

Let the tensors of 7 and 8" be e v e 2 respectively, and let be a 
unit vector perpendicular to them, then we may write 

btp^xp-e.efl + isp ........................ (5). 

Operating by (-or + a;)" 1 , and noting that 

we have bt (cr + = /3 - -- ................. (5 ). 


Taking the scalar of the product of (5) and (5 ) we have 
+ x)~ l p = - (sc/3 - e&p 

316 QUATERNIONS. [39$. 

But by (4 ) we have 

so that, finally, 


Equation (7), in which f is given by (6) in terms of ft, is true 
for every point of every single resultant. But we get an immense 
simplification by assuming for x either of the particular values 
e* or e*. For then the right-hand side of (7) is reduced 
to unity, and the equation represents one or other of the focal 
conies of the system of confocal surfaces 

a point of each of which must therefore lie on the line (5). 

396. A singular form, in which Minding s Theorem can be 
expressed, appears at once from equation (2 ). For that equation 
is obviously the condition that the linear and vector function 

-bpS0( ) + y8y( ) + m( ) 
shall denote a pure strain. 

Hence the following problem : Given a set of rectangular unit 
vectors, which may take any initial position : let two of them, after 
a homogeneous strain, become given vectors at right angles to one 
another, find what the third must become that the strain may be 
pure. The locus of the extremity of the third is, for every initial 
position, one of the single resultants of Minding s system ; and 
therefore passes through each of the fixed conies. 

Thus we see another very remarkable analogy between strains 
and couples, which is in fact suggested at once by the general 
expression for the impure part of a linear and vector function. 

397. The scalar t, which was introduced in equations (4 ), is 
shewn by (6) to be a function of ft alone. In this connection it 
is interesting to study the surface of the fourth degree 

- (e* + e*) r 2 - 2e 1 e 2 2Wr = 1, 

where r = - ft. 


But this may be left as an exercise. 

Another form of t (by 4 ) is $77 + SSB . 


Meanwhile (6) shews that for any assumed value of ft there 
are but two corresponding Minding lines. If, on the other hand, 
p be given there are in general four values of /3. 

398. For variety, and with the view of further exploring this 
very interesting question, we may take a different mode of 
attacking equations (4) and (2 ), which contain the whole matter. 
In what follows b will be merged in p, so that the scale of the 
result will be altered. 

Operating by V. ft we transform (2 ) into 

p + pS0p=-(<ySyi3 + 8SVl3) (2"). 

Squaring both sides we have 

p* + S*/3p = S/3/3 (8). 

Since /3 is a unit vector, this may be taken as the equation of a 
cyclic cone; and every central axis through the point p lies upon it. 
For we have not yet taken account of (4), which is the condition 
that there shall be no couple. 

To introduce (4), operate on (2") by S.y and by S.8 . We 
thus have, by a double employment of (4), 

Next, multiplying (8) by Sftvrfi, and adding to it the squares of 
(9), we have 

p*Sfa/3 - 2S0pSfap - Spvp = - Sfa P (10). 

This is a second cyclic cone, intersecting (8) in the four directions 
/3. Of course it is obvious that (8) and (10) are unaltered by the 
substitution of p + y/3 for p. 

If we look on ft as given, while p is to be found, (8) is the 
equation of a right cylinder, and (10) that of a central surface of 
the second degree. 

399. A curious transformation of these equations may be made 
by assuming p 1 to be any other point on one of the Minding lines 
represented by (8) and (10). Introducing the factor -/3 2 ( = 1) in 
the terms where @ does not appear, and then putting throughout 

II Pi-P (11) 

(8) becomes - p 2 Pl 2 + S*pp 1 = S ( Pl - p) <& ( Pl - p) (8 ). 

As this is symmetrical in p, p v we should obtain only the same 

318 QUATERNIONS. [400. 

result by putting p l for p in (8), and substituting again for /3 as 

From (10) we obtain the corresponding symmetrical result 
(p 2 - SppJ Sp l ^p l + (p, 2 - SppJ Spvp = - Sppfi (p, - p) VF (p l - p) 

-S( pi -p)^( Pl -p) ...... (10 ). 

These equations become very much simplified if we assume p and p l 
to lie respectively in any two conjugate planes ; specially in the 
planes of the focal conies, so that SS p = 0, and Sy f p l = 0. 

For if the planes be conjugate we have 

i = 0, 

and if, besides, they be those of the focal conies, 
Bph SffpBffto 
8pvr 2 p = e*Spwp, &c., 
and the equations are 

P 2 p 2 

and p*Sp l &p l + p*Spwp = 8p 1 vr 3 p l Spiv 2 p ......... (1 0"). 

From these we have at once the equations of the two Minding 
curves in a variety of different ways. Thus, for instance, let 

p t =p* 

and eliminate p between the equations. We get the focal conic in 
the plane of ft , 7 . In this way we see that Minding lines pass 
through each point of each of the two curves ; and by a similar 
process that every line joining two points, one on the one curve, 
the other on the other, is a Minding line. 

400. Another process is more instructive. Note that, by the 
equations of condition above, we have 

Then our equations become 

+ Shmpt = 0> 

and (p 2 + e 2 ) Sp^p, + (p* + e 2 ) Spvp = 0. 

If we eliminate p 2 or p 2 from these equations, the resultant 
obviously becomes divisible by Spivp or Sp^p v and we at once 
obtain the equation of one of the focal conies. 


401. In passing it may be well to notice that equation (10) 
may be written in the simpler form 

8 . p/3pvr/3 + Sp-vp = Sj3<n*l3. 
Also it is easy to see that if we put 

we have (8) in the form S/36 = 0, 
and by the help of this (10) becomes 

This gives another elegant mode of attacking the problem. 

402. Another valuable transformation of (2") is obtained by 
considering the linear and vector function, x suppose, by which fi, 
7, S are derived from the system ft , Uy, US . For then we have 

P = *>XP + X**XP ..................... (2 ")- 

This represents any central axis, and the corresponding form of the 
Minding condition is 

S.Jxvr-*V = S.&x SF ~*y .................. (4"). 

Most of the preceding formulae may be looked upon as results of 
the elimination of the function x from these equations. This forms 
probably the most important feature of such investigations, so far 
at least as the quaternion calculus is concerned. 

403. It is evident from (2 ") that the vector-perpendicular 
from the origin on the central axis parallel to xfi is expressed by 

But there is an infinite number of values of % for which Ur is a 
given versor. Hence the problem ; to find the maximum and 
minimum values of Tr, when Ur is given i.e. to find the surface 
bounding the region which is filled with the feet of perpendiculars on 
central axes. 

We have 2V = - S . 

= TrS.x/3 Ur. 
Hence = S . 

But as Tj3 is constant = 8. 

320 QUATERNIONS. [404. 

These three equations give at sight 

where u, u are unknown scalars. Operate by S . %/3 and we have 

- T 2 r-u=0, 
so that ST (sr + r 2 ) 1 r = 0. 

This differs from the equation of Fresnel s wave-surface only in 
having w + r 2 instead of TZ + r~ 2 , and denotes therefore the reciprocal 
of that surface. In the statical problem, however, we have izft = 0, 
and thus the corresponding wave-surface has zero for one of its 
parameters. [See 435.] 

[If this restriction be not imposed, the locus of the point 

where (/> is now any given linear and vector function whatever, 
will be found, by a process precisely similar to that just given, to be 

8.(r- f ) (f < + T")- (T - f /3 ) = 0, 
where < is the conjugate of </>.] 

B. Kinetics of a Rigid System. 
404. For the motion of a rigid system, we have of course 

by the general equation of Lagrange. 

Suppose the displacements &a to correspond to a mere transla 
tion, then So. is any constant vector, hence 

2 (ma - /3) = 0, 
or, if a x be the vector of the centre of inertia, and therefore 

we have at once a^m 2/3 = 0, 

and the centre of inertia moves as if the whole mass were 
concentrated in it, and acted upon by all the applied forces. 

405. Again, let the displacements Sa correspond to a rotation 
about an axis e, passing through the origin, then 

it being assumed that Te is indefinitely small. 


Hence 28 . e Vet (ma - ) = 0, 

for all values of e, and therefore 

2 . Fa (ma - /9) = 0, 
which contains the three remaining ordinary equations of motion. 

Transfer the origin to the centre of inertia, i.e. put a = ^ + CT, 
then our equation becomes 

2F(a, + tsr) (ma, + WOT - ft) = ; 
or, since 2mar = 0, 

2 Far (m# - ) + Fa, (a\2m - 2y3) = 0. 
But ( 404) 3,2m - 2)9 = 0, 

hence our equation is simply 

Now 2 Fsr/3 is the couple, about the centre of inertia, produced 
by the applied forces ; call it f, then 

? ........................... (1). 

406. Integrating once, 

2roFM!r = 7 + /f<ft ..................... (2). 

Again, as the motion considered is relative to the centre of 
inertia, it must be of the nature of rotation about some axis, in 
general variable. Let e denote at once the direction of, and the 
angular velocity about, this axis. Then, evidently, 

CT = Fe-sr. 

Hence, the last equation may be written 
2mcr Fear = 7 + fdt. 
Operating by S . e, we get 

2m(Vei*y = Sev + Sft;dt .................. (3). 

But, by operating directly by 2fSedt upon the equation (1), we 

2m(FS7) 2 = -/^ 2 + 2/>Sfe^ ............... (4). 

Equations (2) and (4) contain the usual four integrals of the first 
order, h being here an arbitrary constant, whose value depends 
upon the initial kinetic energy of the system. By 387 we see 
how the principal moments of inertia are involved in the left 

T. Q. I. 21 

322 QUATERNIONS. [407. 

407. When no forces act on the body, we have f=0, and 

Sm-crFe sr = <y ........................ (5), 

2mr 8 = 2m(F6w) 8 = -^ .................. (6), 

and, from (5) and (6), Se 7 = -/i 2 ........................... (7). 

One interpretation of (6) is, that the kinetic energy of rotation 
remains unchanged : another is, that the vector e terminates in an 
ellipsoid whose centre is the origin, and which therefore assigns 
the angular velocity when the direction of the axis is given ; 
(7) shews that the extremity of the instantaneous axis is always 
in a plane fixed in space. 

Also, by (5), (7) is the equation of the tangent plane to (6) at 
the extremity of the vector e. Hence the ellipsoid (6) rolls on the 
plane (7). 

From (5) and (6), we have at once, as an equation which e 
must satisfy, 

7 2 2 . m (Fe^r) 2 = - h* (2 . mw Fesr) 2 . 

This belongs to a cone of the second degree fixed in the body. 
Thus all the results of Poinsot regarding the motion of a rigid body 
under the action of no forces, the centre of inertia being fixed, are 
deduced almost intuitively: and the only difficulties to be met 
with in more complex properties of such motion are those of 
integration, which are inherent to the subject, and appear what 
ever analytical method is employed. (Hamilton, Proc. R. /. A. 

If we write (5) as 

+" 1 6 = 7 ........................... (5), 

the special notation <}> indicating that this linear and vector 
function is related to the principal axes of the body, and not to 
lines fixed in space, and consider the ellipsoid (of which e is a 

Se+^e^-h* ........................ (6), 

we may write the equation of a confocal ellipsoid (also fixed in the 
body) as 

Any tangent plane to this is 

S.<r(4, + P r P = -h > ..................... (9). 


If this plane be perpendicular to 7, we may write 
so that, by (5) 


The plane (9) intercepts on 7 a quantity h*/xTy, which is constant 
by (8) and (10). 

The vector velocity of the point p is 

Vep px Vey = p Vyp 

(by two applications of (10)). Hence the point of contact, p, 
revolves about 7 with angular velocity pTy. That is, if the 
plane (9) be rough, and can turn about 7 as an axis, the ellipsoid (8) 
instead of sliding upon it, will make it rotate with uniform angular 
velocity. This is a very simple mode of obtaining one of Sylvester s 
remarkable results. (Phil. Trans. 1866.) 

408. For a more formal treatment of the problem of the 
rotation of a rigid body, we may proceed as follows : 

Let ct be the initial position of r, q the quaternion by which 
the body can be at one step transferred from its initial position to 
its position at time t. Then 

w = qctq 1 
and Hamilton s equation (5) of last section becomes 

S . mqaq" 1 V . eqaq~ l = 7, 
or S . mq {aS . aq~ l eq q~ l eqc?} q~ l = 7. 

The vector 7 is now written for 7 + f%dt of 406, as f is required 
for a new purpose. Thus 7 represents the resultant moment of 
momentum, and will be constant only if there is no applied couple. 

Let (f>p = S m (aSap 2 /o) .................. (1), 

where $ (compare 387) is a self-conjugate linear and vector 
function, whose constituent vectors are fixed in the body in its 
initial position. Then the previous equation may be written 


For simplicity let us write 

<T ( 1 = 
q~ l yq = 

Then Hamilton s dynamical equation becomes simply 

0i? = ? (3). 


324 QUATERNIONS. [409. 

409. It is easy to see what the new vectors rj and f represent. 
For we may write (2) in the form 

from which it is obvious that 77 is that vector in the initial position 
of the body which, at time t, becomes the instantaneous axis in the 
moving body. When no forces act, 7 is constant, and f is the 
initial position of the vector which, at time t, is perpendicular to 
the invariable plane. 

410. The complete statement of the problem is contained in 
equations (2), (3) above, and (4) of 372*. Writing them again, 
we have 

7? = <Z? .............................. (2), 

*-? = ? .............................. (3)- 

We have only to eliminate f and 77, and we get 

2<7=</<r(<r 7<?) ........................ (-5), 

in which q is now the only unknown; 7, if variable, being supposed 
given in terms of q and t. 

It is hardly conceivable that any simpler, or more easily 
interpretable, expression for the motion of a rigid body can be 
presented until symbols are devised far more comprehensive in 
their meaning than any we yet have. 

411. Before entering into considerations as to the integration 
of this equation, we may investigate some other consequences of 
the group of equations in 410. Thus, for instance, differentiating 
(2), we have 

and, eliminating q by means of (4), 

* To these it is unnecessary to add 

Tq = constant, 

as this constancy of Tq is proved by the form of (4). For, had Tq been variable, 
there must have been a quaternion in the place of the vector 17. In fact, 



whence, eliminating y by the help of (2), 

which gives, in the case when no forces act, the forms 

= W? .............................. (0). 

and (as f =(/>??) 

(/>7) = -F.77c/>?7 ........................ (7). 

To each of these the term q~ lf yq must be added on the right, if 
forces act, 

412. It is now desirable to examine the formation of the func 
tion c/>. By its definition 408, (1), we have 

<frp = 2 . m (aSap - a. 2 p), 

= S. maVap. 
Hence - Sp<j>p = 2.m (TVap}\ 

so that Sp<j)p is the moment of inertia of the body about the 
vector p, multiplied by the square of the tensor of p. Compare 
387. Thus the equation 

evidently belongs to an ellipsoid, of which the radii-vectores are 
inversely as the square roots of the moments of inertia about them; 
so that, if i, j, k be taken as unit-vectors in the directions of its 
axes respectively, we have 


A, B, G, being the principal moments of inertia. Consequently 
<j>p = -{AiSip + BjSjp + CkSkp] ..................... (9). 

Thus the equation (7) for 77 breaks up, if we put 

it] = ia) l -\-j(o 2 + &o) 3 , 
into the three following scalar equations 

which are the same as those of Euler. Only, it is to be understood 
that the equations just written are not primarily to be considered 
as equations of rotation. They rather express, with reference to 

326 QUATERNIONS. [4-13- 

fixed axes in the initial position of the body, the motion of the 
extremity, a) lt o> 2 , &> 3 , of the vector corresponding to the instan 
taneous axis in the moving body. If, however, we consider c^, o^, 
o> 3 as standing for their values in terms of w t x, y, z ( 416 below), 
or any other coordinates employed to refer the body to fixed axes, 
they are the equations of motion. 

Similar remarks apply to the equation which determines , for 
if we put 

(6) may be reduced to three scalar equations of the form 


in I 23 

413. Euler s equations in their usual form are easily deduced 
from what precedes. For, let 

whatever be p ; that is, let <}> represent with reference to the moving 
principal axes what < represents with reference to the principal 
axes in the initial position of the body, and we have 


= -V. 

which is the required expression. 

But perhaps the simplest mode of obtaining this equation is to 
start with Hamilton s unintegrated equation, which for the case 
of no forces is simply 

S . m Vvriir = 0. 

But from TS 

we deduce OT 

= -sre 2 eSetz + 
so that S . m ( Vevr Set? etrr 2 + TV Set?) = 0. 

If we look at equation (1), and remember that <j> differs from </> 
simply in having trr substituted for a, we see that this may bo 

<j>e = 0, 


the equation before obtained. The first mode of arriving at it has 
been given because it leads to an interesting set of transformations, 
for which reason we append other two. 
By (2) 

7 = qq \ 

therefore = qq~\ qgq 1 + qgq 1 - qq~ l qq~ l , 

or ^ 

But, by the beginning of this section, and by (5) of 407, this 
is again the equation lately proved. 

Perhaps, however, the following is neater. It occurs in Hamil 
ton s Elements. 

By (5) of 407 <|>e = 7. 
Hence <{>e = <j>e = 2 . m (& 

= V. e^L . 

414. However they are obtained, such equations as those of 
412 were shewn long ago by Euler to be integrable as follows, 

Putting Zfco^^co^dt = s, 

we have Aco* = AQ* + (B-C) s, 

with other two equations of the same form. Hence 

9/7*= ^f 

. C-A \*/^ A-B 

so that t is known in terms of s by an elliptic integral. Thus, 
finally, rj or f may be expressed in terms of t ; and in some of the 
succeeding investigations for q we shall suppose this to have been 
done. It is with this integration, or an equivalent one, that most 
writers on the farther development of the subject have commenced 
their investigations. 

415. By 406, 7 is evidently the vector moment of momen 
tum of the rigid body ; and the kinetic energy is 

- 2 . mtsr* = - 
But #67 = S . q- l eqq~ l y 

328 QUATERNIONS. [4 1 6. 

so that when no forces act 

But, by (2), we have also 

T=T % or 

so that we have, for the equations of the cones described in the 
initial position of the body by rj and f, that is, for the cones de 
scribed in the moving body by the instantaneous axis and by the 
perpendicular to the invariable plane, 

This is on the supposition that 7 and h are constants. If forces act, 
these quantities are functions of t, and the equations of the cones 
then described in the body must be found by eliminating t between 
the respective equations. The final results to which such a process 
will lead must, of course, depend entirely upon the way in which t 
is involved in these equations, and therefore no general statement 
on the subject can be made. 

416. Recurring to our equations for the determination of q, and 
taking first the case of no forces, we see that, if we assume rj to 
have been found (as in 414) by means of elliptic integrals, we 
have to solve the equation 

* To get an idea of the nature of this equation, let us integrate it on the suppo 
sition that ?j is a constant vector. By differentiation and substitution, we get 

Hence q = Q l cos - t + Q. 2 sin ~- t. 

Substituting in the given equation we have 

T-n (-Q lB in ^t + Q. 2 cos^t^ = (Q lC os T J t +^ t) r,. 

Hence ^ . Q 2 = Q^, 


which are virtually the same equation, and thus 

)* - 

And the interpretation of q ( ) q~ l will obviously then be a rotation about tj 
through the angle tT-rj, together with any other arbitrary rotation whatever. Thus 
any position whatever may be taken as the initial one of the body, and Q 1 ( ) Q^ 1 
brings it to its required position at time i = 0. 


that is, we have to integrate a system of four other differential 
equations harder than the first, 

Putting, as in 412, 

where o> lt co l2) w 3 are supposed to be known functions of t, and 
q = w + ix +jy + kz, 

1 ,. dw dx dii dz 
this system is 9 W = ~X = ~Y = ~7 y 

where W = w^o w^y &>/, 

X ft)jW + <DJ ft)./, 
Y = 

Z = 
or, as suggested by Cayley to bring out the skew symmetry, 

X = . ft) 3 7/ ft)/ + W V W, 

Y = ft> 3 . + ft)/ + (0> 2 W, 

Z = ft) 2 wjj . 4- o) 3 w, 
W = a)^ a) 2 y - &)/ 
Here, of course, one integral is 

w 2 + x z + y* + z 2 = constant. 

It may suffice thus to have alluded to a possible mode of 
solution, which, except for very simple values of rj, involves very 
great difficulties. The quaternion solution, when TJ is of constant 
length and revolves uniformly in a right cone, will be given 

417. If, on the other hand, we eliminate TJ, we have to 

q^(q- l yq) = 2q, 

so that one integration theoretically suffices. But, in consequence 
of the present imperfect development of the quaternion calculus, 
the only known method of effecting this is to reduce the quaternion 
equation to a set of four ordinary differential equations of the first 
order. It may be interesting to form these equations. 

Put q = w + ix +jy + kz, 

7 ia +jb +kc, 

330 QUATERNIONS. [4 1 ?- 

then, by ordinary quaternion multiplication, we easily reduce the 
given equation to the following set : 

dt _ dw _ dx _ dy _ dz 
2~~ W~X = ~~ Y ~T 

W = -x&-y3$-z<& or X = . 

X= W + y(-z3$ Y=-X<& 

Y= iM + z&-x<$, Z= 


<& = 1 [ a (w* - x *-f- /) + 2x(ax + by + cz) + 2w (bz - cy)\ 

3$ = n[b(w 2 -x 2 -y z - z 2 } + 2y(ax + by+ cz) + 2w (ex - at)], 

= I [ c (w 2 - a? -f- z 2 ) + 2z(ax + by+ cz) + 2w (ay-bx~)]. 

W, X, Y, Z are thus homogeneous functions of w, x, y, z of the third 

Perhaps the simplest way of obtaining these equations is to 
translate the group of 410 into w, x, y, z at once, instead of 
using the equation from which f and r\ are eliminated. 

We thus see that 

One obvious integral of these equations ought to be 

w z + x 2 4- y 2 + z* = constant, 
which has been assumed all along. In fact, we see at once that 

identically, which leads to the above integral. 

These equations appear to be worthy of attention, partly 
because of the homogeneity of the denominators W, X, Y, Z, but 
particularly as they afford (what does not appear to have been 
sought) the means of solving this celebrated problem at one step, 
that is, without the previous integration of Euler s equations 
( 2). 

A set of equations identical with these, but not in a homo 
geneous form (being expressed, in fact, in terms of K, \, p, v of 
375, instead of w, x, y, z), is given by Cayley (Camb. and Dub. 


Math. Journal, vol. i. 1846), and completely integrated (in the 
sense of being reduced to quadratures) by assuming Euler s 
equations to have been previously integrated. (Compare 416.) 

Cayley s method may be even more easily applied to the above 
equations than to his own ; and I therefore leave this part of the 
development to the reader, who will at once see (as in 416) that 
&, 23, ( correspond to <D V o> 2 , o) 3 of the 77 type, 412. 

418. It may be well to notice, in connection with the formulae 
for direction cosines in 375 above, that we may write 

= \[a (w*+ x 2 -f - * 2 ) + 26 (xy + wz) + 2c (xz - wy}\ 

3$ = [2a (xy -wz) + b (w* -x* + f- z 2 ) + 2c (yz + wx)\ 

= i [2a (xz + wy) + 26 (yz -wx) + c (w* - x* -y* + z*)]. 

These expressions may be considerably simplified by the usual 
assumption, that one of the fixed unit-vectors (i suppose) is 
perpendicular to the invariable plane, which amounts to assigning 
definitely the initial position of one line in the body ; and which 
gives the relations 

6-0, c = 0. 

419. When forces act, 7 is variable, and the quantities a, 6, c 
will in general involve all the variables w, x, y y z, t, so that the 
equations of last section become much more complicated. The 
type, however, remains the same if 7 involves t only ; if it involve 
q we must differentiate the equation, put in the form 

7 = 

and we thus easily obtain the differential equation of the second 

^ = 4F. & (q- q) 3- + 2q<t> ( V. q^j) g- ; 

if we recollect that, because q~ l q is a vector, we have 

Though the above formula is remarkably simple, it must, in the 
present state of the development of quaternions, be looked on as 
intractable, except in certain very particular cases. 

332 QUATERNIONS. [420. 

420. Another mode of attacking the problem, at first sight 
entirely different from that in 408, but in reality identical with 
it, is to seek the linear and vector function which expresses the 
Homogeneous Strain which the body must undergo to pass from its 
initial position to its position at time t. 

Let w = x&, 

a being (as in 408) the initial position of a vector of the body, 
OT its position at time t. In this case % is a linear and vector 
function. (See 376.) 

Then, obviously, we have, ^ l being the vector of some other 
point, which had initially the value a t , 

$57^7 1 = S . 
(a particular case of which is 

and FOTOT I = V. 

These are necessary properties of the strain-function x> depending 

on the fact that in the present application the system is rigid. 

421. The kinernatical equation 

tzr = Veiff 

becomes %a = V. e^a 

(the function % being formed from % by the differentiation of its 
constituents with respect to t). 

Hamilton s kinetic equation 

2 . miff Vein = 7, 
becomes 2 . vti^p. V. e^a = 7. 

This may be written 

2 . m (xaS . e^a - ea") = 7, 

or 2 . m (aS . a% e % -1 e . a 2 ) = %~ J 7> 

where % is the conjugate of %. 

But, because S . x a X a i = ^ aa i> 

we have $, = S . % %!> 

whatever be a and a 1? so that 

x =x" 

Hence 2 . m (aS . a%~ J e X~ IG a ") = X J 7> 
or, by 408 ^" ^X V 


422. Thus we have, as the analogues of the equations in 
408, 409, 

and the former result %a = F. 

becomes %a = F. X^IX - = %F??a. 

This is our equation to determine %, 77 being supposed known. 
To find rj we may remark that 

** = fc 
and ?=X~V 

But %%~ IQC = a > 

so that %x~ a + XX~ 1 a = - 

Hence t = ~ X~ l %X~ l V 

or (>?) = 

These are the equations we obtained before. Having found 
from the last we have to find from the condition 

423. We might, however, have eliminated 77 so as to obtain 
an equation containing ^ alone, and corresponding to that of 
410. For this purpose we have 

so that, finally, % -1 % a = ^ < ~ 1 X~ 1 7 a 

or ^a=^.%"W 1 % 

which may easily be formed from the preceding equation by 

putting -% IOL f r a > an d attending to the value of ^ given in last 

424. We have given this process, though really a disguised 
form of that in 408, 410, and though the final equations to 
which it leads are not quite so easily attacked in the way of 
integration as those there arrived at, mainly to shew how free a 
use we can make of symbolic functional operators in quaternions 

334 QUATERNIONS. [4 2 5- 

without risk of error. It would be very interesting, however, 
to have the problem worked out afresh from this point of view by 
the help of the old analytical methods : as several new forms of 
long-known equations, and some useful transformations, would 
certainly be obtained. 

425. As a verification, let us now try to pass from the final 
equation, in ^ alone, of 423 to that of 410 in q alone. 

We have, obviously, 

OT = qaq~ l = ^a, 

which gives the relation between q and ^. 
[It shews, for instance, that, as 

while S./3xoL = S. jSqaq 1 = S . aq~ l /3q, 

we have % /3 = q^ffq, 

and therefore that XX $ = # ( ( f l @<l) tf" 1 = & 
or x = X~ l > as above.] 

Differentiating, we have 

qaq 1 - qaq~ l qq~ l = tfa. 
Hence % -1 % a = <f l <i a "~ a f l q 

Also c/r 1 x~ l y = F 1 (f l 

so that the equation of 423 becomes 

or, as a may have any value whatever, 

which, if we put Tq = constant 

as was originally assumed, may be written 

2q = q^ 1 (q- l <yq), 
as in 410. 

C. Special Kinetic Problems. 

426. To form the equation for Precession and Nutation. Let 
cr be the vector, from the centre of inertia of the earth, to a particle 
m of its mass: and let p be the vector of the disturbing body, whose 
mass is M. The vector-couple produced is evidently 




no farther terms being necessary, since ^- is always small in the 

actual cases presented in nature. But, because a is measured from 
the centre of inertia, 

S . ma = 0. 

Also, as in 408, </>/> = S . m (aSap <r 2 p). 

Thus the vector-couple required is 

L r 

Referred to coordinates moving with the body, $ becomes <j> as in 
413, and 413 gives 

Simplifying the value of <J> by assuming that the earth has two 
principal axes of equal moment of inertia, we have 

Be - (A - B) aSae = vector-constant + 3M (A - B) dt. 

This gives Sae = const. = H, 

whence e = Ha + ad, 

so that, finally, 

BVc&-AOa = ^(A-B) VotpSap. 

The most striking peculiarity of this equation is that the form 
of the solution is entirely changed, not modified as in ordinary cases 
of disturbed motion, according to the nature of the value of p. 

Thus, when the right-hand side vanishes, we have an equation 
which, in the case of the earth, would represent the rolling of a 
cone fixed in the earth on one fixed in space, the angles of both 
being exceedingly small. 

336 QUATERNIONS. [4 2 7- 

If p be finite, but constant, we have a case nearly the same as 
that of a top, the axis on the whole revolving conically about p. 

But if we assume the expression 

pr(j cos mt + k sin mi), 

(which represents a circular orbit described with uniform speed,) 
a revolves on the whole conically about the vector i, perpendicular 
to the plane in which p lies. ( 408426, Trans. R 8. E., 
18G8 9.) 

427. To form the equation of motion of a simple pendulum, 
taking account of the earth s rotation. Let a be the vector (from 
the earth s centre) of the point of suspension, X its inclination to 
the plane of the equator, a the earth s radius drawn to that point; 
and let the unit- vectors i,j,kbe fixed in space, so that i is parallel 
to the earth s axis of rotation ; then, if co be the angular velocity 
of that rotation 

a = a [i sin X 4- (j cos a)t + k sin at) cos X] ......... (1). 

This gives a. = an (j sin cot + k cos cot) cos X 

= coVia .......................................... (2). 

Similarly a = CD Via = &> 2 (a ai sin X) .................. (3). 

428. Let p be the vector of the bob m referred to the point of 
suspension, R the tension of the string, then if a t be the direction 

of pure gravity 

m(a + p) = mg UOL^ R Up .................. (4), 

which may be written 

-Vw ............... (5). 

To this must be added, since r (the length of the string) is 


Tp = r .............................. (6), 

and the equations of motion are complete. 

429. These two equations (5) and (6) contain every possible case 
of the motion, from the most infinitesimal oscillations to the most 
rapid rotation about the point of suspension, so that it is necessary 
to adapt different processes for their solution in different cases. 
We take here only the ordinary Foucault case, to the degree of 
approximation usually given. 


430. Here we neglect terms involving o> 2 . Thus we write 

8 = 0, 

and we write a for a t , as the difference depends upon the ellipticity 
of the earth. Also, attending to this, we have 


where (by (6)) &m = .............................. (8), 

and terms of the order TX* are neglected. 
With (7), (5) becomes 

so that, if we write - = n z .............................. (9), 

we have Fa(w + ?iV) = ..................... (10). 

Now, the two vectors ai a sin X and Via. 

have, as is easily seen, equal tensors ; the first is parallel to the line 
drawn horizontally northwards from the point of suspension, the 
second horizontally eastwards. 

Let, therefore, w = x (ai a sin X) + y Via ............ (11), 

which (x and y being very small) is consistent with (6). 

From this we have (employing (2) and (3), and omitting &> 2 ) 
CT = x (ai a sin X) + y Via. xw sin X Via. yco (a ai sin X), 
Hs x (ai OL sin X) + y Via. 2xa) sin X Via %yw (a ai sin X). 
With this (10) becomes 

Va \x (ai a sin X) + y Via 2xo) sin X Via 2?/w (a ai sin X) 

+ n*x (ai a sin X) + tfyVia] = 0, 

or, if we note that V.aVia = a (ai a sin X), 

(as Zi/Q) sin X n z x] aVia + (y 2xco sin X +n z y)a(ai asinX) = 0. 

This gives at once x + n 2 x + 2&>?/ sin X = 0] 

y + tfy 2co% sin X = OJ 

which are the equations usually obtained; and of which the solution 
is as follows : 

If we transform to a set of axes revolving in the horizontal plane 
at the point of suspension, the direction of motion being from the 
T. Q. I. 22 

338 QUATERNIONS. [43 1. 

positive (northward) axis of as to the positive (eastward) axis of y, 
with angular velocity fl, so that 

x = f cos fit r) sin fit] ._ _. 

y = (f sin 1U + ?? cos fltfj 

and omit the terms in H 2 and in wO (a process justified by the 
results, see equation (15)), we have 

( + ri*t; ) cos fit - (77 + ft 2 ??) sin fit - 2y (1 - w sin X) = 

sin &+(*? + ^) cos HZ + 2#(H - G) sin X) = 

So that, if we put H = o> sin X (15), 

we have simply |f + n z j; = 

1 + 11*11 = OJ 

the usual equations of elliptic motion about a centre of force in the 
centre of the ellipse. (Proc. R. 8. E., 1869.) 

D. Geometrical and Physical Optics. 

431. To construct a reflecting surface from which rays, emitted 
from a point, shall after reflection diverge uniformly, but horizontally. 

Using the ordinary property of a reflecting surface, we easily 
obtain the equation 

(j3 + oLVap\% 
P ~ P = 

By Hamilton s grand Theory of Systems of Rays, we at once 
write down the second form 

Tp-T(/3 + a Vap) = constant. 

The connection between these is easily shewn thus. Let TX and 
r be any two vectors whose tensors are equal, then 

l (1 + SW 1 ) (Chapter III. Ex. 2), 
whence, to a scalar factor pres, we have 

(^\2 T + VT 
T) T 

Hence, putting vr= U(fi + aVcLp) and T= ?7/?, we have from the 
first equation above 

8. dp [Up +U(/3 + aVap)] = 0. 


But d(/3 + a Vap) = a Vadp = -dp- aSadp, 

and S. 

so that we have finally 

which is the differential of the second equation above. A curious 
particular case is a parabolic cylinder, as may be easily seen 
geometrically. The general surface has a parabolic section in the 
plane of a, /3 ; and a hyperbolic section in the plane of ft, aft. 

It is easy to see that this is but a single case of a large class of 
integrable scalar functions, whose general type is 

$ . dp ( 

V P 

the equation of the reflecting surface ; while 

is the equation of the surface of the reflected wave : the integral 
of the former being, by the help of the latter, at once obtained in 
the form 

TpT(tr-p) = constant*. 

432. We next take Fresnel s Theory of Double Refraction, but 
merely for the purpose of shewing how quaternions simplify the 
processes required, and in no way to discuss the plausibility of the 
physical assumptions. 

Let far be the vector displacement of a portion of the ether, 
with the condition 

^ 2 =-i a), 

the force of restitution, on Fresnel s assumption, is 
t (cfiSi + VjSjiff + c Mtar) = t<jysr t 

using the notation of Chapter V. Here the function </> is the 
negative of that of Chapter IX. (the force of restitution and the 
displacement being on the whole towards opposite parts), and it is 
clearly self-conjugate, a 2 , b* t c 2 are optical constants depending on 
the crystalline medium, and on the wave-length of the light, and 
may be considered as given. 

Fresnel s second assumption is that the ether is incompressible, 
or that vibrations normal to a wave front are inadmissible. If, 

* Proc. E. S. E., 1870-71. 


340 QUATERNIONS. [433- 

then, a be the unit normal to a plane wave in the crystal, we have 
of course 

2 = -l .............................. (2), 

and >SW = ............................... (3); 

but, and in addition, we have 

or $ . a-cr^-cr = ........................... (4). 

This equation (4) is the embodiment of Fresnel s second assump 
tion, but it may evidently be read as meaning, the normal to the 
front, the direction of vibration, and that of the force of restitution 
are in one plane. 

433. Equations (3) and (4), if satisfied by OT, are also satisfied 
by CTQC, so that the plane (3) intersects the cone (4) in two lines 
at right angles to each other. That is, for any given wave front 
there are two directions of vibration, and they are perpendicular to 
each other. 

434. The square of the normal speed of propagation of a 
plane wave is proportional to the ratio of the resolved part of the 
force of restitution in the direction of vibration, to the amount of 
displacement, hence 

v 2 = Stff^vf. 

Hence Fresnel s Wave-surface is the envelop of the plane 

Sap = Svr<l>vr ........................ (5), 

with the conditions or 2 = 1 .............................. (1), 

2 = -l .............................. (2), 

Sav = Q ................................. (3),<w = ....................... . ......... (4). 

Formidable as this problem appears, it is easy enough. From (3) 
and (4) we get at once, 

Hence, operating by S . r, 

X = 

Therefore (< + v*) & = 

and ^f.a(0 + z; 5 )- 1 a = ..................... (C). 


In passing, we may remark that this equation gives the normal 
speeds of the two rays whose fronts are perpendicular to a. In 
Cartesian coordinates it is the well-known equation 

70 9 9 

p +_?!!_ +^!_ = o. 

By this elimination of -cr, our equations are reduced to 

= ........................ (6), 

v = -Sp ........................... (5), 

2 = -l .............................. (2). 

They give at once, by 326, 

(< + V^OL + vpSa (0 + v*)- 2 a = ha. 
Operating by S . a we have 

Substituting for h, and remarking that 

Sa(<l> + v 2 ra = - 
because (/> is self-conjugate, we have 

This gives at once, by rearrangement, 

v (t + vT 1 * = (<#- pTp- 

Hence >-> = 

Operating by S .p on this equation we have 

SpQ-ffp l ..................... (7), 

which is the required equation. 

[It will be a good exercise for the student to translate the last 
ten formulae into Cartesian coordinates. He will thus reproduce 
almost exactly the steps by which Archibald Smith* first arrived 
at a simple and symmetrical mode of effecting the elimination. 
Yet, as we shall presently see, the above process is far from being 
the shortest and easiest to which quaternions conduct us.] 

435. The Cartesian form of the equation (7) is not the usual 
one. It is, of course, 

* Cambridge Phil. Trans., 1835. 


But write (7) in the form 

and we have the usual expression 

oV by cV_ 

The last-written quaternion equation can also be put into either 
of the new forms 


436. By applying the results of 183, 184 we may introduce 
a multitude of new forms. We must confine ourselves to the most 
simple ; but the student may easily investigate others by a process 
precisely similar to that which follows. 

Writing the equation of the wave as 

where we have g = p~*, 

we see that it may be changed to 

s p (^ 1 + hr l p = o, 

if mSp<j>p = glip* = h. 

Thus the new form is 

Sp((F l -mSp<l>prp = ..................... (1). 

Here m = -,--,, Sp^p = aV + Vtf + cV, 

Ct C 

and the equation of the wave in Cartesian coordinates is, putting 

a? y* z* 

^ + - * + ** - 

c - 

437. By means of equation (1) of last section we may easily 
prove Pliicker s Theorem : 

The Wave- Surf ace is its own reciprocal with respect to the 
ellipsoid whose equation is 


The equation of the plane of contact of tangents to this surface 
from the point whose vector is p is 

The reciprocal of this plane, with respect to the unit-sphere 
about the origin, has therefore a vector cr where 

Hence p = 


and when this is substituted in the equation of the wave we have 
for the reciprocal (with respect to the unit-sphere) of the reciprocal 
of the wave with respect to the above ellipsoid, 

1 \ -1 

O(J(p (T 1 (T == 0. 

in / 

This differs from the equation (1) of last section solely in having 
0" 1 instead of (/>, and (consistently with this) l/m instead of m. 
Hence it represents the index-surface. The required reciprocal 
of the wave with reference to the ellipsoid is therefore the wave 

438. Hamilton has given a remarkably simple investigation 
of the form of the equation of the wave-surface, in his Elements, 
p. 736, which the reader may consult with advantage. The 
following is essentially the same, but several steps of the process, 
which a skilled analyst would not require to write down, are 
retained for the benefit of the learner. 

Let %> = -! (1) 

be the equation of any tangent plane to the wave, i.e. of any wave- 
front. Then //, is the vector of wave-slowness, and the normal 
velocity of propagation is therefore 1/Tfj,. Hence, if TZ be the vector 
direction of displacement, /J,~*TZ- is the effective component of the 
force of restitution. Hence, (/>OT denoting the whole force of 
restitution, we have 

and, as CT is in the plane of the wave-front, 

= 0, 
" 2 )~ l p = (2). 

344 QUATERNIONS. [439. 

This is, in reality, equation (6) of 434. It appears here, 
however, as the equation of the Index- Surf ace, the polar reciprocal 
of the wave with respect to a unit-sphere about the origin. Of 
course the optical part of the problem is now solved, all that 
remains being the geometrical process of 328. 

439. Equation (2) of last section may be at once transformed, 
by the process of 435, into 

Let us employ an auxiliary vector 

T=G* 2 -.->, 

whence ^ = (/** - (/T^r ........................ (1). 

The equation now becomes 

SHT = ! .............................. (2), 

or,by(l), ^-Sr^r = l ........................ (3). 

Differentiating (3), subtract its half from the result obtained by 
operating with S . r on the differential of (1). The remainder is 

But we have also ( 328) 

Spdfi = 0, 

and therefore, since dp has an infinite number of values, 

Xp = yL6T 2 T, 

where x is a scalar. 

This equation, with (2), shews that 

Srp = () .............................. (4). 

Hence, operating on it by 8 . p, we have by (1) of last section 

V = -T* 

and therefore p 1 = //, + r" 1 . 

This gives p~ 2 = ^ - T " 2 . 

Substituting from these equations in (1) above, it becomes 

T- -^^+T- 2 -<->, 

or T-($-*-p*)-V. 

Finally, we have for the required equation, by (4), 


or, by a transformation already employed, 




440. It may assist the student in the practice of quaternion 
analysis, which is our main object, if we give a few of these 
investigations by a somewhat varied process. 

Thus, in 432, let us write as in 180, 

cfiSi-v + tfjSjw + tfkSk-v = \ Sp!^ + p S^v? p r ix. 
We have, by the same processes as in 432, 

8 . uroWSp v + S . isroLfi SXfw = 0. 

This may be written, 50 far as the generating lines we require are 

S.vaV. \ V/JL = = 8. 
since ora is a vector. 
Or we may write V. OTX -Bra = = 8. 

Equations (1) denote two cones of the second order which pass 
through the intersections of (3) and (4) of 432. Hence their 
intersections are the directions of vibration. 

441. By (1) we have 


Hence rXV, a, fjf are coplanar ; and, as w is perpendicular to a, 
it is equally inclined to FX a and F//a. 

For, if L, M, A be the projections of X , ///, a on the unit 
sphere, BG the great circle whose 
pole is A, we are to find for the ^ 

projections of the values of OT on 
the sphere points P and P , such 
that if LP be produced till 

Q may lie on the great circle AM. 
Hence, evidently, 


which proves the proposition, since 

the projections of FX a and F//a on the sphere are points b and 

c in BC, distant by quadrants from G and B respectively. 


442. Or thus, Si* a. = 0, 

S . vrV . aX vT/jf = 0, 
therefore XTZ = V. aF. 

= - F. XV// - 

Hence (x> - a?) w = (V + aaX ) / + (/ 
Operate by $ . X , and we have 

(x + SXaSfi a) SxV = [X V - SVa 

Hence by symmetry, 

J _ 

and as S-&OL = 0, 

/ a UVpfa). 

443. The optical interpretation of the common result of the 
last two sections is that the planes of polarization of the two rays 
whose wave-fronts are parallel, bisect the angles contained by planes 
passing through the normal to the wave-front and the vectors (optic 
axes) X , fju. 

444. As in 434, the normal speed is given by 

[This transformation, effected by means of the value of r in 
442, is left to the reader.] 

Hence, if v iy v 2 be the velocities of the two waves whose 
normal is a, 

oc sin X a sin /u/a. 

That is, the difference of the squares of the speeds of the two 
waves varies as the product of the sines of the angles between the 
normal to the wave-front and the optic axes (X , //). 


445. We have, obviously, 

(T 2 - S 2 ) . V\ OL Vp a =TV. V\ OL VH OL = S 2 . \ 
Hence v* =p + (T S) . VXaVpa. 

The equation of the index surface, for which 

is therefore 1 = - p p* + (T S) . VX p Vp p. 

This will, of course, become the equation of the reciprocal of the 
index-surface, i.e. the wave-surface, if we put for the function </> 
its reciprocal : i.e. if in the values of X , //, , p we put I/a, 1/6, 1/c 
for a, 6, c respectively. We have then, and indeed it might have 
been deduced even more simply as a transformation of 434 (7), 

as another form of the equation of Fresnel s wave. 

If we employ the i, K transformation of 128, this may be 
written, as the student may easily prove, in the form 

(V - ij = S*(i- K )p + (TVi P 

446. We may now, in furtherance of our object, which is to 
give varied examples of quaternions, not complete treatment of any 
one subject, proceed to deduce some of the properties of the wave- 
surface from the different forms of its equation which we have 

447. Fresnel s construction of the wave by points. 

From 290 (4) we see at once that the lengths of the principal 
semidiameters of the central section of the ellipsoid 

Spj- p = 1, 

by the plane Sap = 0, 

are determined by the equation 

3.a(<l>- l -p-*)- 1 a = 0. 

If these lengths be laid off along a, the central perpendicular to the 
cutting plane, their extremities lie on a surface for which a = Up, 
and Tp has values determined by the equation. 

Hence the equation of the locus is 

as in 55 434, 439. 

348 QUATERNIONS. [44$. 

Of course the index-surface is derived from the reciprocal ellip 

SpQp = 1 
by the same construction. 

448. Again, in the equation 

l=-pp*+(TS).V\ P V f j,p, 
suppose V\p = 0, or Vjj,p = 0, 

we obviously have 

,U\ tfyu, 

P = -7- r P = , , 
\/p *Jp 

and there are therefore four singular points. 

To find the nature of the surface near these points put 


P = -/- + , 

where TVT is very small, and reject terms above the first order in 
TOT. The equation of the wave becomes, in the neighbourhood of 
the singular point, 

2pS\v + S.&V. \V\fji = T. V\iz V\p, 
which belongs to a cone of the second order. 

449. From the similarity of its equation to that of the wave, it 
is obvious that the index-surface also has four conical cusps. As 
an infinite number of tangent planes can be drawn at such a point, 
the reciprocal surface must be capable of being touched by a plane 
at an infinite number of points ; so that the wave-surface has four 
tangent planes which touch it along ridges. 

To find their form, let us employ the last form of equation of 
the wave in 445. If we put 

TK P = TV Kp (1), 

we have the equation of a cone of the second degree. It meets the 
wave at its intersections with the planes 

fl((.-*)/> = ("- ) (2). 

Now the wave-surface is touched by these planes, because we 
cannot have the quantity on the first side of this equation greater 
in absolute magnitude than that on the second, so long as p 
satisfies the equation of the wave. 


That the curves of contact are circles appears at once from (1) 
and (2), for they give in combination 

)p ........................ (3), 

the equations of two spheres on which the curves in question are 

The diameter of this circular ridge is 

= (a* - V) (b* - <f). 

[Simple as these processes are, the student will find on trial 
that the equation 

V = 0, 

gives the results quite as simply. For we have only to examine 
the cases in which p~ z has the value of one of the roots of the 
symbolical cubic in </>~\ In the present case Tp = b is the only 
one which requires to be studied.] 

450. By 438, we see that the auxiliary vector of the succeed 
ing section, viz. 

T =(/* -- )> = (- -p-)V, 

is parallel to the direction of the force of restitution, ^>t7. Hence, 
as Hamilton has shewn, the equation of the wave, in the form 

(4) of 439, indicates that the direction of the force of restitution is 
perpendicular to the ray. 

Again, as for any one versor of a vector of the wave there are 
two values of the tensor, which are found from the equation 

we see by 447 that the lines of vibration for a given plane front 
are parallel to the axes of any section of the ellipsoid 

made by a plane parallel to the front ; or to the tangents to the lines 
of curvature at a point where the tangent plane is parallel to the 

451. Again, a curve which is drawn on the wave-surface so as 
to touch at each point the corresponding line of vibration has 

350 QUATERNIONS. [452. 

Hence Sfypdp = 0, or Sp(f>p = C, 

so that such curves are the intersections of the wave with a series 
of ellipsoids concentric with it. 

452. For curves cutting at right angles the lines of vibration we 

Hence Spdp = 0, or Tp = C, 

so that the curves in question lie on concentric spheres. 

They are also spherical conies, because where 

Tp = C 
the equation of the wave becomes 

s. P (4>- + <ryv = o, 

the equation of a cyclic cone, whose vertex is at the common centre 
of the sphere and the wave-surface, and which cuts them in their 
curve of intersection. ( 432 452, Quarterly Math. Journal, 1859.) 
The student may profitably compare, with the preceding investi 
gations, the (generally) very different processes which Hamilton 
(in his Elements) applies to this problem. 

E. Electrodynamics. 

453. As another example we take the case of the action of 
electric currents on one another or on magnets; and the mutual 
action of permanent magnets. 

A comparison between the processes we employ and those of 
Ampere (Theorie des Phenomenes Electrodynamiques) will at once 
shew how much is gained in simplicity and directness by the use 
of quaternions. 

The same gain in simplicity will be noticed in the investiga 
tions of the mutual effects of permanent magnets, where the 
resultant forces and couples are at once introduced in their most 
natural and direct forms. 

454. Ampere s experimental laws may be stated as follows : 

I. Equal and opposite currents in the same conductor produce 
equal and opposite effects on other conductors. 


II. The effect of a conductor bent or twisted in any manner is 
equivalent to that of a straight one, provided that the two are 
traversed by equal currents, and the former nearly coincides with 
the latter. 

III. No closed circuit can set in motion an element of a 
circular conductor about an axis through the centre of the circle 
and perpendicular to its plane. 

IV. In similar systems traversed by equal currents the forces 
are equal. 

To these we add the assumption that the action between two 
elements of currents is in the straight line joining them. [In a 
later section ( 473) other assumptions will be made in place of 
this.] We also take for granted that the effect of any element of 
a current on another is directly as the product of the strengths of 
the currents, and of the lengths of the elements. 

455. Let there be two closed currents whose strengths are 
a and a l ; let a, a x be elements of these, a being the vector 
joining their middle points. Then the effect of a on a l must, 
when resolved along a lf be a complete differential with respect 
to a (i.e. with respect to the three independent variables involved 
in a), since the total resolved effect of the closed circuit of which 
a is an element is zero by III. 

Also by I, II, the effect is a function of Ta, Sa.a, Saa.^ and 
SCL CL^ since these are sufficient to resolve a and a l into elements 
parallel and perpendicular to each other and to a. Hence the 
mutual effect is 

a^Uaf^a, Saa , SOL\), 

and the resolved effect parallel to a l is 


Also, that action and reaction may be equal in absolute magnitude, 
/ must be symmetrical in 8a.a and $aa r Again, a (as differential 
of a) can enter only to the first power, and must appear in each 
term of f. 

Hence /= A8a\ + BScm Sou^. 

But, by IV, this must be independent of the dimensions of the 
system. Hence A is of 2 and B of 4 dimensions in To.. 

- [ASaa.Safa, + BSaa &aaJ 

352 QUATERNIONS. [45 6. 

is a complete differential, with respect to a, if da. = of. Let 


where G is a constant depending on the units employed, therefore 

G B 

and the resolved effect 

Cad -. S O.O.. 

= To? = * TaTc? 

^ + 

The factor in brackets is evidently proportional in the ordinary 
notation to 

sin 6 sin & cos co | cos 6 cos 6 . * 

456. Thus the whole force is 

Caa^a j $ 2 aa t _ Caa^a. , 8*0.0. 
~ l ~ 3 > 

as we should expect, d^a. being = a r [This may easily be trans 
formed into 

which is the quaternion expression for Ampere s well-known form.] 

457. The whole effect on a t of the closed circuit, of which a is 
an element, is therefore 


rv >a J ZV) 

between proper limits. As the integrated part is the same at both 
limits, the effect is 

and depends on the form of the closed circuit. 


458. This vector 0, which is of great importance in the whole 
theory of the effects of closed or indefinitely extended circuits, cor 
responds to the line which is called by Ampere " directrice de 
fraction electrodynamique." It has a definite value at each point 
of space, independent of the existence of any other current. 

Consider the circuit a polygon whose sides are indefinitely 
small ; join its angular points with any assumed point, erect at 
the latter, perpendicular to the plane of each elementary triangle 
so formed, a vector whose length is a>/r, where &> is the vertical 
angle of the triangle and r the length of one of the containing 
sides ; the sum of such vectors is the " directrice " at the assumed 

[We may anticipate here so far as to give another expression 
for this important vector in terms of processes to be explained 

We have, by the formula (for a closed curve) of 497 below, 

(where ds is an element of any surface bounded by the circuit, Uv 
its unit normal) 

But the last integral is obviously the whole spherical angle, II 
suppose, subtended by the circuit at the origin, and (unless Tp = 0) 
we have ( 145) 

Hence, generally, 


Thus II may be considered as representing a potential, for which 
ft is the corresponding force. 

This is a " many- valued " function, altering by 4-Tr whenever we 
pass through a surface closing the circuit. For if o- be the vector 
of a closed curve, the work done against /3 during the circuit is 

fSjBdo- = - J SdoVQ, = fdtt. 

The last term is zero if the curve is not linked with the 
circuit, but increases by + 4-n- for each linkage with the circuit.] 
T. Q. I. 23 

354 QUATERNIONS. [459- 

459. The mere/orm of the result of 457 shews at once that 
if the element a x be turned about its middle point, the direction of the 
resultant action is confined to the plane whose normal is {3. 

Suppose that the element a t is forced to remain perpendicular 
to some given vector 8, we have 

Saf = 0, 

and the whole action in its plane of motion is proportional to 


But V . SVafi = - afilSS. 

Hence the action is evidently constant for all possible positions 
of otj ; or 

The effect of any system of closed currents on an element of a 
conductor which is restricted to a given plane is (in that plane) 
independent of the direction of the element. 

460. Let the closed current be plane and very small. Let e 
(where Te = 1) be its normal, and let 7 be the vector of any point 
within it (as the centre of inertia of its area) ; the middle point of 
4 being the origin of vectors. 

Let a. = 7 + p ; therefore a = p, 

to a sufficient approximation. 

Now (between limits) jVpp = 
where A is the area of the closed circuit. 
Also generally 

fVyp Syp = J (%>F 7 p + y 7. 

= (between limits) A 7 Vye. 
Hence for this case 


461. If, instead of one small plane closed current, there be a 
series of such, of equal area, disposed regularly in a tubular form, 
let x be the distance between two consecutive currents measured 
along the axis of the tube ; then, putting y = xe, we have for the 
whole effect of such a set of currents on a 

CAaa f/y 

2x 1 A?V + Ty 5 

CAaa. Va..y ,, ,. . 

^ -^ V (between proper limits). 

If the axis of the tubular arrangement be a closed curve this will 
evidently vanish. Hence a closed solenoid exerts no influence on an 
element of a conductor. The same is evidently true if the solenoid 
be indefinite in both directions. 

It the axis extend to infinity in one direction, and y n be the 
vector of the other extremity, the effect is 

and is therefore perpendicular to the element and to the line joining 
it with the extremity of the solenoid. It is evidently inversely as 
Ty* and directly as the sine of the angle contained between the 
direction of the element and that of the line joining the latter with 
the extremity of the solenoid. It is also inversely as x, and there 
fore directly as the number of currents in a unit of the axis of the 

462. To find the effect of the whole circuit whose element is 
otj on the extremity of the solenoid, we must change the sign of 
the above and put a l = y , therefore the effect is 


l f V % % 

] Ty* 

2x Ty* 

an integral of the species considered in 458, whose value is easily 
assigned in particular cases. 

463. Suppose the conductor to be straight) and indefinitely 
extended in both directions. 

Let hO be the vector perpendicular to it from the extremity of 
the solenoid, and let the conductor be || 77, where T6 = Trj = 1. 

Therefore y = h& + yrj (where y is a scalar), 





and the integral in 462 is 


The whole effect is therefore 

_CAaa J 


and is thus perpendicular to the plane passing through the conductor 
and the extremity of the solenoid, and varies inversely as the distance 
of the latter from the conductor. 

This is exactly the observed effect of an indefinite straight 
current on a magnetic pole, or particle of free magnetism. 

464. Suppose the conductor to be circular, and the pole nearly 
in its axis. [This is not a proper subject for Quaternions.] 

Let EPD be the conductor, AB its axis, and G the pole ; BG 
perpendicular to AB, and small in comparison with AE = h the 
radius of the circle. 

Let A B be aj, BC=bk, AP = h(jx + ley) 
where \ = \ . \EAP=\ . \ B. 

y] l sm j l sm j 

Then CP = y = a^i -\-bk-h (jx + ky). 

And the effect on G x l-fJ-j- t 

where the integral extends to the whole circuit. 


465. Suppose in particular C to be one pole of a small magnet 
or solenoid CO whose length is 21, and whose middle point is at G 
and distant a from the centre of the conductor. 

Let /.CGB = A. Then evidently 

a x = a 4- 1 cos A, 
b = I sin A. 

Also the effect on C becomes, if a/ 2 -f b 2 + h 2 = A 2 , 

1 f/r 


J 3 I ^ I V * . //" < -" 1 <~/ I w,.i/it/ i t-"-I 

since for the whole circuit 

If we suppose the centre of the magnet fixed, the vector axis 
of the couple produced by the action of the current on C is 

IV. (i cos A 4- & sin A) I -^^ 
sin A . L 36 2 . 15 W 3a,6 cos A 

A 3 A* 2 

If A } &c. be now developed in powers of , this at once becomes 

vh*l sin A . f _ 6aZ cos A 15aT cos 2 A 

/ I 9 . T > 

(a 2 -f ^) "I a 2 + W (a 2 + A 2 ) 2 a 2 + /i 2 

_3^sin 2 A 15 /iT sin 2 A (a + Zcos A)^cos A > 5a^cosA\) 
~ aM- A 2 h (a 2 + A 2 ) 2 ~ a* + h* ( ~?"+l ~ / j 

Putting ^ for Z and changing the sign of the whole to get that 
for pole C , we have for the vector axis of the complete couple 

A . 

which is almost exactly proportional to sin A, if 2a be nearly 
equal to h and I be small. (See Ex. 15 at end of Chapter.) 

On this depends a modification of the tangent galvanometer. 
(Bravais, Ann. de Ghimie, xxxviii. 309.) 

358 QUATERNIONS. [466. 

466. As before, the effect of an indefinite solenoid on a t is 

CAaa 1 VOL*/ 

Now suppose a t to be an element of a small plane circuit, 8 the 
vector of the centre of inertia of its area, the pole of the solenoid 
being origin. 

Let 7 = 8 -f p, then a x = p. 

The whole effect is therefore 

CAaa 1 


where A l and e t are, for the new circuit, what A and e were for 
the former. 

Let the new circuit also belong to an indefinite solenoid, and 
let S be the vector joining the poles of the two solenoids. Then 
the mutual effect is 



_CAA l aa l S UB 


which is exactly the mutual effect of two magnetic poles. Two 
finite solenoids, therefore, act on each other exactly as two magnets, 
and the pole of an indefinite solenoid acts as a particle of free 

467. The mutual attraction of two indefinitely small plane 
closed circuits, whose normals are e and e lt may evidently be 
deduced by twice differentiating the expression U8/T& for the 
mutual action of the poles of two indefinite solenoids, making dS 
in one differentiation [| e and in the other || e lt 

But it may also be calculated directly by a process which will 
give us in addition the couple impressed on one of the circuits by 
the other, supposing for simplicity the first to be circular. [In 
the sketch we are supposed to be looking at the faces turned 
towards one another.] 


Let A and B be the centres of inertia of the areas of A and B, 

e and e, vectors normal to their planes, cr any vector radius of B 


Then whole effect on cr , by 457, 460, 

-A V A 
* V7e - 

But between proper limits, 

for generally fV(r rjS0a =-%( VrjaSOo- +V.<rjV. 6/Vaa). 

Hence, after a reduction or two, we find that the whole force 
exerted by A on the centre of inertia of the area of B 


This, as already observed, may be at once found by twice 


differentiating 7*7 - In the same way the vector moment, due to 
A, about the centre of inertia of B, 

C - 777 

These expressions for the whole force of one small magnet on 
the centre of inertia of another, and the couple about the latter, 
seem to be the simplest that can be given. It is easy to deduce 

360 QUATERNIONS. [468. 

from them the ordinary forms. For instance, the whole resultant 
couple on the second magnet 

may easily be shewn to coincide with that given by Ellis (Gamb. 
Math. Journal, iv. 95), though it seems to lose in simplicity and 
capability of interpretation by such modifications. 

468. The above formulae shew that the whole force exerted 
by one small magnet M, on the centre of inertia of another m 
consists of four terms which are, in order, 

1st. In the line joining the magnets, and proportional to the 
cosine of their mutual inclination. 

2nd. In the same line, and proportional to five times the product 
of the cosines of their respective inclinations to this line. 

3rd and 4th. Parallel to ^ and proportional to the cosine of 
the inclination of \ ! to the joining line. 

All these forces are, in addition, inversely as the fourth power 
of the distance between the magnets. 

For the couples about the centre of inertia of m we have 

1st. A couple whose axis is perpendicular to each magnet, and 
which is as the sine of their mutual inclination. 

2nd A couple whose axis is perpendicular to m and to the line 
joining the magnets, and whose moment is as three times the product 
of the sine of the inclination of m, and the cosine of the inclination 
of M, to the joining line. 

In addition these couples vary inversely as the third power of 
the distance between the magnets. 

[These results afford a good example of what has been called 
the internal nature of the methods of quaternions, reducing, as 
they do at once, the forces and couples to others independent of 
any lines of reference, other than those necessarily belonging to 
the system under consideration. To shew their ready applicability, 
let us take a Theorem due to Gauss.] 


469. If two small magnets be at right angles to each other, the 
moment of rotation of the first is approximately twice as great when 
the axis of the second passes through the centre of the first, as when 
the axis of the first passes through the centre of the second. 

In the first case e II $ _!_ e t ; 

C 2C" 

therefore moment = T (ee l - 3e^) = ^ ^ee r 

In the second e x || @ J_ e ; 


therefore moment = ^- Tee Hence the theorem. 

470. Again, we may easily reproduce the results of 467, if 
for the two small circuits we suppose two small magnets perpen 
dicular to their planes to be substituted. /3 is then the vector 
joining the middle points of these magnets, and by changing the 
tensors we may take 2e and 2e x as the vector lengths of the 

Hence evidently the mutual effect 


which is easily reducible to 

as before, if smaller terms be omitted. 

If we operate with F. e x on the two first terms of the unreduced 
expression, and take the difference between this result and the 
same with the sign of e : changed, we have the whole vector axis of 
the couple on the magnet 2e v which is therefore, as before, seen to 
be proportional to 

471. Let F (7) be the potential of any system upon a unit 
particle at the extremity of 7. Then we have 

Svdy = 0, 

where v is a vector normal giving the force in direction and 
magnitude ( 148). 

362 QUATERNIONS. [47 2. 

Now by 460 we have for the vector force exerted by a small 
plane closed circuit on a particle of free magnetism the expression 

A / SySye\ 


merging in A the factors depending on the strength of the current 
and the strength of magnetism of the particle. 

Hence the potential is 

area of circuit projected perpendicular to 7 

^ 7p 2 


x spherical opening subtended by circuit. 

The constant is omitted in the integration, as the potential must 
evidently vanish for infinite values of Ty. 

By means of Ampere s idea of breaking up a finite circuit into 
an indefinite number of indefinitely small ones, it is evident that 
the above result may be at once ex 
tended to the case of such a finite closed 

472. Quaternions give a simple 
method of deducing the well-known 
property of the Magnetic Curves. 

Let A, A be two equal magnetic 

poles, whose vector distance, 2a, is bi- A~~ o ~A f 
sected in 0, QQ an indefinitely small 
magnet whose length is 2/o , where p = OP. Then evidently, 
taking moments, 

a)j , V(p-*)p 

where the upper or lower sign is to be taken according as the poles 
are like or unlike. 

[This equation may also be obtained at once by differentiating 
the equation of the equipotential surface, 

+ T, X = COnSt " 

T (p + a) T (p - a 
and taking p parallel to its normal ( 148).] 


Operate by 8 . Vap, 

Sap (p + a) 2 So. (p + a) So (p + a) .., 

~.n/ -T 3 - = same with - a}, 

-L (p + a) 

or <S.aFf- J CT(/o + a)= {same with -a}, 

i.e. SadU(p + a) = + $acZ U(p - a), 

a {#"(/> + a) + tT(p - a)} = const., 
or cos /. OAP cos OA P = const, 

the property referred to. 

If the poles be unequal, one of the terms to the left must be 
multiplied by the ratio of their strengths. 

( 453472, Quarterly Math. Journal, 1860.) 

F. General Expressions for the Action between Linear 

473. The following general investigation of different possible 
expressions for the mutual action between elements of linear 
conductors is taken from Proc. R. S. E. 1873 4. 

Ampere s data for closed currents are briefly given in 454 
above, and are here referred to as I, II, III, IV, respectively. 

(a) First, let us investigate the expression for t\\e force exerted 
by one element on another. 

Let a be the vector joining the elements a p a , of two circuits ; 
then, by I, II, the action of a. v on a is linear in each of a t , a , 
and may, therefore, be expressed as 


where </> is a linear and vector function, into each of whose con 
stituents otj enters linearly. 

The resolved part of this along a is 
8. Z7a <K 

and, by III, this must be a complete differential as regards the 
circuit of which a, is an element. Hence, 

</>a = -(8. a,V) ^a + Fa x P 

where \|r and ^ are linear and vector functions whose constituents 
involve a only. That this is the case follows from the fact that 
(f>a is homogeneous and linear in each of a l5 a . It farther 

364 QUATERNIONS. [473- 

follows, from IV, that the part of </>a which does not disappear 
after integration round each of the closed circuits is of no dimen 
sions in To., To. , Ta^ Hence % is of 2 dimensions in To., and 

i - ~2V " 5V ~W~ 
where jo, q, r are numbers. 
Hence we have 

ofaSaoL, qV*^ rV.a Vaa, 

Change the sign of a in this, and interchange of and a,, and we 
get the action of a! on a r This, with a and a t again interchanged, 
and the sign of the whole changed, should reproduce the original 
expression since the effect depends on the relative, not the abso 
lute, positions of a, a. v a . This gives at once, 

p = 0, q = 0, 

r V. a Fact 

with the condition that the first term changes its sign with a, and 

thus that 

i|ra = aSaa F (To) + a! 

which, by change of F, may be written 

where /and F are any scalar functions whatever. 

<j>a = -S (cqV) [aS (a V)f(Ta) + a F (To)] + ~^- 1 
which is the general expression required. 

(6) The simplest possible form for the action of one current- 
element on another is, therefore, 


Here it is to be observed that Ampere s directrice for the circuit 

a. is 



the integral extending round the circuit ; so that, finally, 


(c) We may obtain from the general expression above the 
absolutely symmetrical form, 

if we assume 

/(2k) = const, 

Here the action of a on a t is parallel and equal to that of a t on a. 
The forces, in fact, form a couple, for a is to be taken negatively 
for the second and their common direction is the vector drawn 
to the corner a of a spherical triangle abc, whose sides ab, be, ca in 
order are bisected by the extremities of the vectors Ua, Ua, Ua^ 
Compare Hamilton s Lectures on Quaternions, 223 227. 

(d) To obtain Ampere s form for the effect of one element on 
another write, in the general formula above, 

and we have 

!. - a.* [_s*r[^r.*v mi 


V. a Faa, 

* / 20 3 o o "\ 
g I Ct oCtjtt Q ottft oOCOCj 1 , 

I \ -^ / 

To 5 


which are the usual forms. 

(e) The remainder of the expression, containing the arbitrary 
terms, is of course still of the form 

- S (a,V, [aS (a V)/(Ta) + a F (2V.)]. 

In the ordinary notation this expresses a force whose com 
ponents are proportional to 

(Note that, in this expression, r is the distance between the 

(2) Parallel to a ^ . 

366 QUATERNIONS. [473. 

(3) Parallel to - ^ . 


If we assume /= F= Q, we obtain the result given by Clerk- 
Maxwell (Electricity and Magnetism, 525), which differs from the 
above only because he assumes that the force exerted by one 
element on another, when the first is parallel and the second per 
pendicular to the line joining them, is equal to that exerted when 
the first is perpendicular and the second parallel to that line. 

(/) What precedes is, of course, only a particular case of the 
following interesting problem : 

Required the most general expression for the mutual action of 
two rectilinear elements, each of which has dipolar symmetry in the 
direction of its length, and which may be resolved and compounded 
according to the usual kinematical law. 

The data involved in this statement are equivalent to I and 
II of Ampere s data above quoted. Hence, keeping the same 
notation as in (a) above, the force exerted by ctj on a must be 
expressible as 


where (f> is a linear and vector function, whose constituents are 
linear and homogeneous in a t ; and, besides, involve only a. 

By interchanging a t and a , and changing the sign of a, we get 
the force exerted by a on a r If in this we again interchange a t 
and a, and change the sign of the whole, we must obviously repro 
duce </> . Hence we must have $ changing its sign with a, or 

(/> = PaSa/ + QaSaaficta. + Rafiaa! + RaSctct v 
where P, Q, R, R are functions of Ta only. 

(g) The vector couple exerted by a t on a must obviously be 
expressible in the form 

V , a waj, 

where or is a new linear and vector function depending on a alone. 
Hence its most general form is 

where P and Q are functions of To. only. The form of these func 
tions, whether in the expression for the force or for the couple, 
depends on the special data for each particular case. Symmetry 
shews that there is no term such as 


(ti) As an example, let a l and a be elements of solenoids or of 
uniformly and linearly magnetised wires, it is obvious that, as a 
closed solenoid or ring-magnet exerts no external action, 

Thus we have introduced a different datum in place of Ampere s 
No. III. But in the case of solenoids the Third Law of Newton 
holds hence 

where % is a linear and vector function, and can therefore be of no 
other form than 

Now two solenoids, each extended to infinity in one direction, act 
on one another like two magnetic poles, so that (this being our 
equivalent for Ampere s datum No. TV.) 

Hence the vector force exerted by one small magnet on another is 


(i) For the couple exerted by one element of a solenoid, or of 
a uniformly and longitudinally magnetised wire, on another, we 
have of course the expression 

V. * ** 

where & is some linear and vector function. 
Here, in the first place, it is obvious that 

for the couple vanishes for a closed circuit of which a t is an 
element, and the integral of wo^ must be a linear and vector 
function of a alone. It is easy to see that in this case 

F(Ta) oc (Taf. 

(j) If, again, a t be an element of a solenoid, and a an element 
of current, the force is 

< = _ Sa^V . Tfra t 

^a = Pa + QaSacL + RVoaf. 

368 QUATERNIONS. [473. 

But no portion of a solenoid can produce a force on an element of 
current in the direction of the element, so that 

whence P = 0, Q = 0, 

and we have < = Sa t V (RVaa. ). 

This must be of 1 linear dimensions when we integrate for the 
effect of one pole of a solenoid, so that 

7? P 

** = 7/r~3 
If the current be straight and infinite each way, its equation being 

where Ty = l and Sfty = 0, 

we have, for the whole force exerted on it by the pole of a solenoid, 
the expression 

/+ 00 dx 
Pftv I ^ = ~ 2 PP 7i 

which agrees with known facts. 

(&) Similarly, for the couple produced by an element of a 
solenoid on an element of a current we have 


and it is easily seen that 


(I) In the case first treated, the couple exerted by one current- 
element on another is, by (g), 

V. CL K^, 

where, of course, ^^ are the vector forces applied at either end 
of a . Hence the work done when a changes its direction is 

with the condition S . a So! = 0. 

So far, therefore, as change of direction of a alone is concerned, 
the mutual potential energy of the two elements is of the form 

8 . 


This gives, by the expression for OT in (g), the following value 

Hence, integrating round the circuit of which cq is an element, we 
have ( 495 below) 

f(PSa\ + QSaa SaaJ =ffdsfi. Up,V (Pa! + QctSaa), 

where < = _ + Q. 

Integrating this round the other circuit we have for the mutual 
potential energy of the two, so far as it depends on the expression 
above, the value 

ffdsfi. UpJVota. 3> 

= -ffd Sl S. UpJJds V. 

UP (24> + TOL&) + Sa Uv Sa UP. 

But, by Ampere s result, that two closed circuits act on one another 
as two magnetic shells, it should be 

ffds, ffds S . Uv.VS . Uv V ~ 

= ffds, ffds S . Uv, UP + 3^a UP SOL UP, 
Comparing, we have 

gving <J) = - * = &> 

which are consistent with one another, and which lead to 

Hence, if we put 

1 n 

weget p 

T. Q. I. 24 

370 QUATERNIONS. [474. 

and the mutual potential of two elements is of the form 

, Sao, , . Sao. Seta. 

which is the expression employed by Helmholtz in his paper Ueber 
die Bewegungsgleichungen der Electricitdt, Crelle, 1870, p. 76. 

G. Application of V to certain Physical Analogies. 

474. The chief elementary results into which V enters, in 
connection with displacements, are given in 384 above. The 
following are direct applications. 

Thus, if o- be the vector-displacement of that point of a homo 
geneous elastic solid whose vector is p, we have, p being the 
consequent pressure produced, 

whence S&pV 2 a = SpVp = $p, a complete differential (2). 

Also, generally, p = cSVa, 

and if the solid be incompressible 

SVo- = Q (3). 

Thomson has shewn (Camb. and Dub. Math. Journal, ii. p. 62), 
that the forces produced by given distributions of matter, electricity, 
magnetism, or galvanic currents, can be represented at every point 
by displacements of such a solid producible by external forces. It 
may be useful to give his analysis, with some additions, in a 
quaternion form, to shew the insight gained by the simplicity of 
the present method. 

475. Thus, if Sa-Sp = 5 , we may write each equal to 

This gives 

the vector-force exerted by one particle of matter or free electricity 
on another. This value of cr evidently satisfies (2) and (3). 

Again, if S . SpVa = 8 ^ , either is equal to 


Here a particular case is 

~ Tf 

which is the vector-force exerted by an element a of a current upon 
a particle of magnetism at p. ( 461.) 

476. Also, by 146 (3), 

and we see by 460, 461 that this is the vector- force exerted by 
a small plane current at the origin (its plane being perpendicular 
to a) upon a magnetic particle, or pole of a solenoid, at p. This 
expression, being a pure vector, denotes an elementary rotation 
caused by the distortion of the solid, and it is evident that the 
above value of cr satisfies the equations (2), (3), and the distortion 
is therefore producible by external forces. Thus the effect of an 
element of a current on a magnetic particle is expressed directly 
by the displacement, while that of a small closed current or 
magnet is represented by the vector-axis of the rotation caused by 
the displacement. 

477. Again, let SS P V^ = S^- 

It is evident that a satisfies (2), and that the right-hand side of 
the above equation may be written 

Hence a particular case is 

and this satisfies (3) also. 

Hence the corresponding displacement is producible by external 
forces, and V<r is the rotation axis of the element at p, and is seen 
as before to represent the vector-force exerted on a particle of 
magnetism at p by an element a of a current at the origin. 

478. It is interesting to observe that a particular value of cr 
in this case is 

as may easily be proved by substitution. 


372 QUATERNIONS. [479. 

- ~ 

j~ , 

Again, if S8p<r = - 

we have evidently cr = V _- . 

Now, as ^~ is the potential of a small magnet a, at the origin, 

on a particle of free magnetism at p, cr is the resultant magnetic 
force, and represents also a possible distortion of the elastic solid 
by external forces, since Vcr = W = 0, and thus (2) and (3) are 
both satisfied. (Proc. R S, E. 1862.) 

H. Elementary Properties of V. 

479. In the next succeeding sections we commence with a 
form of definition of the operator V somewhat different from that 
of Hamilton ( 145), as we shall thus entirely avoid the use of 
Cartesian coordinates. For this purpose we write 

where a is any unit-vector, the meaning of the right-hand operator 
(neglecting its sign) being the rate of change of the function to 
which it is applied per unit of length in the direction of the 
vector a. If a be not a unit-vector we may treat it as a vector- 
velocity, and then the right-hand operator means the rate of 
change per unit of time due to the change of position. 

Let a, /3, 7 be any rectangular system of unit-vectors, then by 
a fundamental quaternion transformation 

V - - aSaV - 3SSV - ^ V = ad 

- - aa - // - 77 = a a + ft + y y , 

which is identical with Hamilton s form so often given above. 
(Lectures, 620.) 

480. This mode of viewing the subject enables us to see at 
once that V is an Invariant, and that the effect of applying it to 
any scalar function of the position of a point is to give its vector of 
most rapid increase. Hence, when it is applied to a potential u, 
we have the corresponding vector-force. From a velocity-potential 
we obtain the velocity of the fluid element at p ; and from the 
temperature of a conducting solid we obtain the temperature- 


gradient in the direction of the flux of heat. Finally, whatever 
series of surfaces is represented by 

u = G, 

the vector Vu is the normal at the point p, and its length is 
inversely as the normal distance at that point between two con 
secutive surfaces of the series. 

Hence it is evident that 

S . dpVa = du, 
or, as it may be written, 

the left-hand member therefore expresses total differentiation in 
virtue of any arbitrary, but small, displacement dp. 

These results have been already given above, but they were 
not obtained in such a direct manner. 

Many very curious and useful transformations may easily be 
derived (see Ex. 34, Chap. XI.) from the assumption 

da = (frdp, or </> = SaV . cr, 

where the constituents of < are known functions of p. For 
instance, if we write 

_ . d . d j d 

v - 3f + *i + *3r 

where a- = i% + jrj + k, 

we find at once V = V ff , or V^ = ^ ^V ; 

a formula which contains the whole basis of the theory of the 
change of independent variables from x, y, z to f , 77, f, or vice versa. 
The reader may easily develop this application. Its primary 
interest is, of course, purely mathematical : but it has most 
important uses in applied mathematics. Our limits, however, 
do not permit us to reach the regions of its special physical 

481. To interpret the operator V. aV, let us apply it to a 
potential function u. Then we easily see that u may be taken 
under the vector sign, and the expression 

denotes the vector-couple due to the force at p about a point whose 
relative vector is a. 

374 QUATERNIONS. [482. 

Again, if cr be any vector function of p, we have by ordinary 
quaternion operations 

F (aV) . o- = 8 . a FVo- + aV<r - Va<7. 

The meaning of the third term (in which it is of course understood 
that V operates on a alone) is obvious from what precedes. The 
other terms were explained in 384. 

J. Applications of V to Line-, Surface-, and Volume- Integrals. 

482. In what follows we have constantly to deal with integrals 
extended over a closed surface, compared with others taken through 
the space enclosed by such a surface ; or with integrals over a 
limited surface, compared with others taken round its bounding 
curve. The notation employed is as follows. If Q per unit of 
length, of surface, or of volume, at the point p, Q being any 
quaternion, be the quantity to be summed, these sums will be 
denoted by 

ffQds and fflQds, 

when comparing integrals over a closed surface with others through 
the enclosed space ; and by 

ffQds and fQTdp, 

when comparing integrals over an unclosed surface with others 
round its boundary. No ambiguity is likely to arise from the 

double use of 


for its meaning in any case will be obvious from the integral with 
which it is compared. What follows is mainly from Trans. R. S. E. 
186970. See also Proc. R. 8. E. 18623. 

483. We have shewn in 384 that, if a- be the vector displace 
ment of a point originally situated at 

then S.Va 


expresses the increase of density of aggregation of the points of 
the system caused by the displacement. 

484. Suppose, now, space to be uniformly filled with points, 
and a closed surface 5 to be drawn, through which the points can 
freely move when displaced. 


Then it is clear that the increase of number of points within 
the space S, caused by a displacement, may be obtained by either 
of two processes by taking account of the increase of density at 
all points within 2, or by estimating the excess of those which 
pass inwards through the surface over those which pass outwards. 
These are the principles usually employed (for a mere element of 
volume) in forming the so-called Equation of Continuity. 

Let v be the normal to 2 at the point p, drawn outwards, then 
we have at once (by equating the two different expressions of the 
same quantity above explained) the equation 

which is our fundamental equation so long as we deal with triple 
integrals. [It will be shewn later ( 500) that the corresponding 
relation between the single and the double integral can be deduced 
directly from this.] 

As a first and very simple example of its use, let p be written 
for o: It becomes 

i.e. the volume of any closed space is the sum of the elements of 
area of its surface, each multiplied by one-third of the perpendicular 
from the origin on its plane. 

485. Next, suppose a to represent the vector force exerted 
upon a unit particle at p (of ordinary matter, electricity, or 
magnetism) by any distribution of attracting matter, electricity, 
or magnetism partly outside, partly inside 2. Then, if P be the 
potential at p, 

and if r be the density of the attracting matter, &c., at p, 

by Poisson s extension of Laplace s equation. 

Substituting in the fundamental equation, we have 
47r ///rcfe = 4-rf = ffS . VP Uvds, 

where M denotes the whole quantity of matter, &c., inside X. 
This is a well-known theorem. 

486. Let P and P x be any scalar functions of p, we can of 
course find the distribution of matter, &c., requisite to make either 

376 QUATERNIONS. [48 7- 

of them the potential at p ; for, if the necessary densities be r and 
r l respectively, we have as before 

Now V (PVPJ = VP VP l 

Hence, if in the formula of 484 we put 

we obtain 

JJJS . VP VP^ 9 = - ///P V 2 P^9 + ffPS . V P l Uvds, 
= _ fffP^Pds + JJPfi . VP Uvds, 

which are the common forms of Green s Theorem. Sir W. Thomson s 
extension of it follows at once from the same proof. 

487. If P l be a many-valued function, but V P 1 single-valued, 
and if 2 be a multiply-connected* space, the above expressions 
require a modification which was first shewn to be necessary by 
Helmholtz, and first supplied by Thomson. For simplicity, suppose 
2) to be doubly-connected (as a ring or endless rod, whether knotted 
or not). Then if it be cut through by a surface s, it will become 
simply-connected, but the surface -integrals have to be increased 
by terms depending upon the portions just added to the whole 
surface. In the first form of Green s Theorem, just given, the 
only term altered is the last : and it is obvious that if p l be the 
increase of P after a complete circuit of the ring, the portion to be 
added to the right-hand side of the equation is 


taken over the cutting surface only. A similar modification is 
easily seen to be produced by each additional complexity in the 
space 2. 

488. The immediate consequences of Green s theorem are 
well known, so that I take only a few examples. 

Let P and P 1 be the potentials of one and the same distribution 
of matter, and let none of it be within S. Then we have 

2 ds = J/PS . VP Uvds, 

* Called by Helmholtz, after Biemann, mehrfach susammenhangend. In trans 
lating Helmholtz s paper (Phil. Mag. 1867) I used the above as an English 
equivalent. Sir W. Thomson in his great paper on Vortex Motion (Trans. R. S> E. 
1868) uses the expression "multiply-continuous." 


so that if VP is zero all over the surface of 2, it is zero all through 
the interior, i.e., the potential is constant inside 2. If P be the 
velocity-potential in the irrotational motion of an incompressible 
fluid, this equation shews that there can be no such motion of the 
fluid unless there is a normal motion at some part of the bounding 
surface, so long at least as 2 is simply-connected. 
Again, if 2 is an equipotential surface, 

///(VP/ ds = PJfti . V PUvds = P ///V 2 Pd? 

by the fundamental theorem. But there is by hypothesis no 
matter inside 2, so this shews that the potential is constant 
throughout the interior. Thus there can be no equipotential 
surface, not including some of the attracting matter, within which 
the potential can change. Thus it cannot have a maximum or 
minimum value at points unoccupied by matter. 

489. Again, in an isotropic body whose thermal conductivity 
does not vary with temperature, the equation of heat-conduction is 


dt c 

where (for the moment) k and c represent as usual the conductivity, 
and the water equivalent of unit volume. 

The surface condition (assuming Newton s Law of Cooling) is 

Assuming, after Fourier, that a particular integral is 

= e~ mt u, 

we have V 2 w-w = 0, 


Let u m be a particular integral of the first of these linear 
differential equations. Substitute it for u in the second ; and we 
obtain (with the aid of the equation of the bounding surface) a 
scalar equation giving the admissible values of m. 

Suppose the distribution of temperature when = to be 
given ; it may be expressed linearly in terms of the various values 
of u, thus 

w = 

For if u iy u^ be any two of these particular integrals, we have 
by Green s Theorem, and the differential equations, 


Hence, unless m A = m 2 , we have 


Thus we have A m fffu* m d<: = ffjwu m ds. 
A m being thus found, we have generally 
v = I,A m e-""u m . 

490. If, in the fundamental theorem, we suppose 

o- = Vr, 
which imposes the condition that 

i.e., that the cr displacement is effected without condensation, it 

ffS . Vr Uvds = fffSV 2 rck = 0. 

Suppose any closed curve to be traced on the surface X, dividing 
it into two parts. This equation shews that the surface-integral is 
the same for both parts, the difference of sign being due to the 
fact that the normal is drawn in opposite directions on the two 
parts. Hence we see that, with the above limitation of the value 
of cr, the double integral is the same for all surfaces bounded by a 
given closed curve. It must therefore be expressible by a single 
integral taken round the curve. The value of this integral will 
presently be determined ( 495). 

491. The theorem of 485 may be written 

!ffV 2 Pck =JfSUvVPd8=ffS(UvV) Pds. 
From this we conclude at once that if 

(which may, of course, represent any vector whatever) we have 
or, if 

This gives us the means of representing, by a surface-integral, a 
vector-integral taken through a definite space. We have already 
seen how to do the same for a scalar-integral so that we can now 
express in this way, subject, however, to an ambiguity presently 
to be mentioned, the general integral 


where q is any quaternion whatever. It is evident that it is only 
in certain classes of cases that we can expect a perfectly definite 
expression of such a volume-integral in terms of a surface-integral. 

492. In the above formula for a vector-integral there may 
present itself an ambiguity introduced by the inverse operation 

to which we must devote a few words. The assumption 

VV = T 

is tantamount to saying that, if the constituents of a- are the 
potentials of certain distributions of matter, &c., those of r are the 
corresponding densities each multiplied by 4?r. 

If, therefore, r be given throughout the space enclosed by S, 
cr is given by this equation so far only as it depends upon the 
distribution within 2, and must be completed by an arbitrary vector 
depending on three potentials of mutually independent distributions 
exterior to . 

But, if a- be given, r is perfectly definite ; and as 
Vo- = VV, 

the value of V 1 is also completely defined. These remarks must 
be carefully attended to in using the theorem above : since they 
involve as particular cases of their application many curious 
theorems in Fluid Motion, &c. 

493. As a very special case, the equation 

of course gives Vcr = u, a scalar. 

Now, if v be the potential of a distribution whose density is u, we 

We know that when u is assigned this equation gives one, and but 
one, definite value for v. We have in fact, by the definition of a 

where the integration (confined to u^ and p t ) extends to all space 
in which u differs from zero. 

380 QUATERNIONS. [494. 

Thus there is no ambiguity in 

and therefore a = Vv 


is also determinate. 

494. This shews the nature of the arbitrary term which must 
be introduced into the solution of the equation 

FV<7 = T. 

To solve this equation is ( 384) to find the displacement of any 
one of a group of points when the consequent rotation is given. 

Here SVr = S. VFV<7 = VV = 0; 

so that, omitting the arbitrary term ( 493), we have 

W = V T , 

and each constituent of cr is, as above, determinate. Compare 

Thomson* has put the solution in a form which may be written 

if we understand by /( ) dp integrating the term in dx as if y 
and z were constants, &c. Bearing this in mind, we have as 

J27i \V-ri + fV^ dp 

495. We now come to relations between the results of integra 
tion extended over a non-closed surface and round its boundary. 

Let cr be any vector function of the position of a point. The 
line-integral whose value we seek as a fundamental theorem is 


where T is the vector of any point in a small closed curve, drawn 
from a point within it, and in its plane. 

* Electrostatics and Magnetism, 521, or Phil. Trans., 1852. 


Let (7 be the value of a at the origin of r, then 

o- = a- -S(rV)a- 0) 

so that fSadr =JS.{(r -S (rV) a Q ] dr. 

But fdr = 0, 

because the curve is closed ; and (ante, 467) we have generally 

fSrVS(7 dr = JV (T&TOT - <T O fVrdr). 
Here the integrated part vanishes for a closed circuit, and 

where ds is the area of the small closed curve, and Uv is a unit- 
vector perpendicular to its plane. Hence 

fSo- dT = S.V<T Q Uv.ds. 

Now, any finite portion of a surface may be broken up into small 
elements such as we have just treated, and the sign only of the 
integral along each portion of a bounding curve is changed when 
we go round it in the opposite direction. Hence, just as Ampere 
did with electric currents, substituting for a finite closed circuit 
a network of an infinite number of infinitely small ones, in each 
contiguous pair of which the common boundary is described by 
equal currents in opposite directions, we have for a finite unclosed 

There is no difficulty in extending this result to cases in which 
the bounding curve consists of detached ovals, or possesses multiple 

This theorem seems to have been first given by Stokes (Smith s 
Prize Exam. 1854), in the form 

rrj (/ (<h dfi\ (da dy\ fd/3 da 

= Jjds ll(-r--r) + m(- r --j L ) + n(-^ --:- 

( \dy dz) \dz dxj \dx dy 

It solves the problem suggested by the result of 490 above. It 
will be shewn, however, in a later section that the equation above, 
though apparently quite different from that of 484, is merely a 
particular case of it. 

[If we recur to the case of an infinitely small area, it is clear 

is a maximum when 

F. Z 

382 QUATERNIONS. [49 6. 

Hence FVcr is, at every point, perpendicular to a small area 
for which 

is a maximum.] 

496. If <7 represent the vector force acting on a particle of 
matter at p, S . adp represents the work done by it while the 
particle is displaced along dp, so that the single integral 


of last section, taken with a negative sign, represents the work 
done during a complete cycle. When this integral vanishes it is 
evident that, if the path be divided into any two parts, the work 
spent during the particle s motion through one part is equal to 
that gained in the other. Hence the system of forces must be 
conservative, i.e., must do the same amount of work for all paths 
having the same extremities. 

But the equivalent double integral must also vanish. Hence a 
conservative system is such that 

whatever be the form of the finite portion of surface of which ds 
is an element. Hence, as Ver has a fixed value at each point of 
space, while Uv may be altered at will, we must have 

or Vo- = scalar. 

If we call X, Y, Z the component forces parallel to rectangular 
axes, this extremely simple equation is equivalent to the well- 
known conditions 

dX dY dY dZ dZ dX _ 

~7 -- ~7 - ^J ~7 --- 7 ^> ~J -- 7 - - - 

dy dx dz dy ax dz 

Returning to the quaternion form, as far less complex, we see 


Vcr = scalar = 47rr, suppose, 

implies that cr = VP, 

where P is a scalar such that 

that is, P is the potential of a distribution of matter, magnetism, 
or statical electricity, of volume-density r. 


Hence, for a non-closed path, under conservative forces, 

depending solely on the values of P at the extremities of the path. 

497. A vector theorem, which is of great use, and which cor 
responds to the scalar theorem of 491, may easily be obtained. 
Thus, with the notation already employed, 

Now F ( F. V F. rdr) o- = - S(rV) V. <r dr - S (drV) VT<T O , 
and d {S ( T V) FO- O T} 8 (rV) F. a Q dr + S (drV) F<T O T. 

Subtracting, and omitting the term which is the same at both 
limits, we have 

Extended as above to any closed curve, this takes at once the form 
JF. adp = -ffdsV. (FZTi/V) <r. 

Of course, in many cases of the attempted representation of a 
quaternion surface-integral by another taken round its bounding 
curve, we are met by ambiguities as in the case of the space- 
integral, 492 : but their origin, both analytically and physically, 
is in general obvious. 

498. The following short investigation gives, in a complete 
form, the kernel of the whole of this part of the subject. But 
495 7 have still some interest of their own. 

If P be any scalar function of p, we have (by the process of 
495, above) 

fPdr=J{P -S(TV)P 9 }dT 

= -fS.rVP Q .dr. 

But F. V F. rdr = drS .rV-rS. drV, 

and d (rSrV) = drS .rV + rS. drV. 

These give 

fPdr = - } {rSrV - F. JF(rdr) V} P = dsV. UvVP Q . 
Hence, for a closed curve of any form, we have 

from which the theorems of 495, 497 may easily be deduced. 

384 QUATERNIONS. [499- 

Multiply into any constant vector, and we have, by adding 
three such results 

fdpa- = ffdsV(UvV) <r. [See 497.] 

Hence at once (by adding together the corresponding members 
of the two last equations, and putting 

where q is any quaternion whatever. 

[For the reason why we have no corresponding formula, with S 
instead of V in the right-hand member, see remark in [ ] in 

499. Commencing afresh with the fundamental integral 

put o- = u& 

and we have f/fS{3Vuds = ffuS/3 Uvds ; 

from which at once fffVuds = ffuUvds ........................ (1), 

or fJfVTd*=ffU V .Td8 ..................... (2), 

which gives 

JJ/FV Yards = Vf}UvV<rrds = ff(rSUv(r - crSUvr) ds. 

Equation (2) gives a remarkable expression for the surface of a 
space in terms of a volume-integral. For take 

T =Uv= UVP, 

where P = const. 

is the scalar equation of the closed bounding surface. Then 

_ jfds = fJUv Uvds = ///V U VPds. 
(Note that this implies 

which in itself is remarkable.) 

Thus the surface of an ellipsoid 


the integration being carried on throughout the enclosed space. 
(Compare 485.) 


Again, in (2), putting u^r for T, and taking the scalar, we have 

ffJSrV Ul + w,fifV T ) d* = f 
whence fff[S(rV) a + aSVr\ ds = ffaSrUvds .............. (3). 

The sum of (1) and (2) gives, for any quaternion, 

The final formulae in this, and in the preceding, section give 
expressions in terms of surface integrals for the volume, and the 
line, integrals of a quaternion. The latter is perfectly general, but 
(for a reason pointed out in 492) the former is definite only when 
the quaternion has the form Vq. 

500. The fundamental form of the Volume and Surface 
Integral is (as in 499 (1)) 

fffVuds = ffUvuds. 

Apply it to a space consisting of a very thin transverse slice of 
a cylinder. Let t be the thickness of the slice, A the area of one 
end, and a a unit-vector perpendicular to the plane of the end. 
The above equation gives at once 

where dl is the length of an element of the bounding curve of the 
section, and the only values of Uv left are parallel to the plane of 
the section and normal to the bounding curve. If we now put p 
as the vector of a point in that curve, it is plain that 

V.aUv=Udp, dl = Tdp, 
and the expression becomes 

V(oiV)u.A =fudp. 

By juxtaposition of an infinite number of these infinitely small 
directed elements, a (now to be called Uv) being the normal vector 
of the area A (now to be called ds), we have at once 

ffV (UvV) uds = fudp, 

which is the fundamental form of the Surface and Line Integral, 
as given in 498. Hence, as stated in 484, these relations are 
not independent. 

In fact, as the first of these expressions can be derived at once 
from the ordinary equation of "continuity," so the second is merely 
the particular case corresponding to displacements confined to a 
T. Q. I. 25 


given surface. It is left to the student to obtain it, simply and 
directly, (in the form of 495) from this consideration. 

[Note. A remark of some importance must be made here. It 
may be asked : Why not adopt for the proof of the fundamental 
theorems of the present subject the obvious Newtonian process (as 
applied, for instance, in Thomson and Tait s Natural Philosophy, 
194, or in Clerk-Maxwell s Electricity, 591) ? 

The reply is that, while one great object of the present work is 
(as far as possible) to banish artifice, and to shew the "perfect 
naturalness of Quaternions/ the chief merit of the beautiful 
process alluded to is that it forms one of the most intensely 
artificial applications of an essentially artificial system. Cartesian 
and Semi-Cartesian methods may be compared to a primitive 
telegraphic code, in which the different signals are assigned to 
the various letters at hap-hazard; Quaternions to the natural 
system, in which the simplest signals are reserved for the most 
frequently recurring letters. In the former system some one word, 
or even sentence, may occasionally be more simply expressed than 
in the latter : though there can be no doubt as to which system 
is to be preferred. But, even were it not so, the methods we have 
adopted in the present case give a truly marvellous insight into 
the real meaning and " inner nature " of the formulae obtained.] 

501. As another example of the important results derived 
from the simple formulae of 499, take the following, viz.: 

ffV. V (a Uv) rds = ffaSr Uvds - JfUvSards, 

where by (3) and (1) of that section we see that the right-hand 
member may be written 

= /// {# (TV) a- + o-SVr - VSar] ds 

= -///F.F(Vo-)T<fc (4). 

In this expression the student must remark that V operates on r 
as well as on a. Had it operated on r alone, we should have 
inverted the order of V and <r, and changed the sign of the whole ; 
or we might have had recourse to the notation of the end of 133. 

This, and similar formulae, are easily applied to find the 
potential and the vector-force due to various distributions of 
magnetism. To shew how they are to be introduced, we briefly 
sketch the mode of expressing the potential of a distribution. 


K. Application of the V Integrals to Magnetic &c. Problems. 

502. Let cr be the vector expressing the direction and intensity 
of magnetisation, per unit of volume, at the element ds. Then if 
the magnet be placed in a field of magnetic force whose potential 
is u, we have for its potential energy 

This shews at once that the magnetism may be resolved into a 
volume-density SVcr, and a surface-density - SaUv. Hence, for 
a solenoid al distribution, 

SVa = 0. 

What Thomson has called a lamellar distribution (Phil. Trans. 
1852), obviously requires that 

be integrable without a factor ; i.e., that 

FVo- = 0. 

A complex lamellar distribution requires that the same expression 
be integrable by the aid of a factor. If this be u, we have at once 

FV (ua-) = 0, 

or S . <rVcr = 0. 

But we easily see that (4) of 501 may be written 
- J/F ( F<7 UP) rds = - J//F. rVVo-ds - JJ/F aVrds + JffSaV . rds. 

Now, if T = V (- 

where r is the distance between any external point and the element 
cfc, the last term on the right is the vector-force exerted by the 
magnet on a unit-pole placed at the point. The second term on 
the right vanishes by Laplace s equation, and the first vanishes as 
above if the distribution of magnetism be lamellar, thus giving 
Thomson s result in the form of a surface integral. 

503. As another instance, let a be the vector of magnetic 
induction, /3 the vector potential, at any point. Then we know, 
physically, that 

fSpdp=ffSUv* t d8. 


388 QUATERNIONS. [504. 

But. by the theorem of S 496, we have 

t/ <J 

fSI3dp=JfS. UvV/3ds. 

Since the boundary and the enclosed surface may be any what 
ever, we must have 

and, as a consequence, 

Va = 0. 

Hence V/5 = a + u, 

= V~ l a + V" 1 ^ + Vq, 
where q is a quaternion satisfying the equations 

To interpret the other terms, let 
V~ l u = o-, 

so fchat Wo- = ; 

and o- = Vv, 

where v is a scalar such that 

V*v = u. 

Thus v is the potential of a distribution w/4-Tr, and can there 
fore be found without ambiguity when u is given. And of course 

V" 1 ^ = Vv 
is also found without ambiguity. 

Again, as SVa. = 0, 

we have Va = 7, suppose. 

Hence, for any assumed value of 7, we have 

so that this term of the above value of /3 is also found without 

The auxiliary quaternion q depends upon potentials of arbitrary 
distributions wholly outside the space to which the investigation 
may be limited. [Compare this section with 494.] 

504. An application may be made of similar transformations 
to Ampere s Directrice de I action electrodynamique, which, 458 
above, is the vector-integral 

Tp 3 


where dp is an element of a closed circuit, and the integration 
extends round the circuit. This may be written 

so that its value as a surface integral is 
ffs (UvV) V i ds - 

Of this the last term vanishes, unless the origin is in, or infinitely 
near to, the surface over which the double integration extends. 
The value of the first term is seen (by what precedes) to be the 
vector-force due to uniform normal magnetisation of the same 
surface. Thus we see the reason why the Directrice can be 
expressed in terms of the spherical opening of the circuit, as in 
459, 471. 

505. The following result is obviously but one of an extensive 
class of useful transformations. Since 

we obtain at once from 484 

a curious expression for the gravitation potential of a homogeneous 
body in terms of a surface-integral. 

[The right-hand member may be written as 

and an examination into its nature shews us why we ought not to 
expect to have a general expression for 


in terms of a line-integral. It will be excellent practice for the 
student to make this examination himself. Of course, a more 
general method presents itself in finding the volume-integral 
which is equivalent to the last written surface-integral extended 
to the surface of a closed space.] 

From this, by differentiation with respect to a, after putting 
p + a for p, or by expanding in ascending powers of To. (both of 
which tacitly assume that the origin is external to the space 

390 QUATERNIONS. [506. 

integrated through, i.e., that Tp nowhere vanishes within the 
limits), we have 

_ f r v. u p v. u.Up 

-~ ~~ 2 

and this, again, involves 

"^ = II ^ SUvUpds. 

506. The interpretation of these, and of more complex formulae 
of a similar kind, leads to many curious theorems in attraction and 
in potentials. Thus, from (1) of 499, we have 

///**-///^*-n-* < 

which gives the attraction of a mass of density t in terms of the 
potentials of volume distributions and surface distributions. Putting 

this becomes 

[[[Vo-ds _ rrr Up . o-cfc _[[Uv. < 

JJJ Tp ])] Tp*"-}) Tp 

By putting a = p, and taking the scalar, we recover a formula 
given above ; and by taking the vector we have 

This may be easily verified from the formula 

by remembering that VTp= Up. 

Again if, in the fundamental integral, we put 

we have ffj^d<-2 fjj%= jf tSUvUpds. 

[It is curious how closely, in fact to a numerical factor of one 
term pres, this equation resembles what we should get by operating 
on (1) by S.p, and supposing that we could put p under the signs 
of integration.] 


L. Application of V to the Stress-Function. 

507. As another application, let us consider briefly the Stress- 
function in an elastic solid. 

At any point of a strained body let X be the vector stress per 
unit of area perpendicular to i, ft and v the same for planes per 
pendicular to j and k respectively. 

Then, by considering an indefinitely small tetrahedron, we 
have for the stress per unit of area perpendicular to a unit- vector 
co the expression 

\Sia) + pSjw + vSkw = (pco, 

so that the stress across any plane is represented by a linear and 
vector function of the unit normal to the plane. 

But if we consider the equilibrium, as regards rotation, of an 
infinitely small parallelepiped whose edges are parallel to i, j, k 
respectively, we have (supposing there are no molecular couples) 


or, supposing V to apply to p alone, 

This shews ( 185) that in the present case <p is self-conjugate, and 
thus involves not nine distinct constants but only six. 

508. Consider next the equilibrium, as regards translation, of 
any portion of the solid filling a simply-connected closed space. 
Let u be the potential of the external forces. Then the condition 
is obviously 

where v is the normal vector of the element of surface ds. Here 
the double integral extends over the whole boundary of the closed 
space, and the triple integral throughout the whole interior. 

To reduce this to a form to which the method of 485 is 
directly applicable, operate by S . a where a is any constant vector 
whatever, and we have 

JfS . </>a Uvds + {ffdsSaVu = 

392 QUATERNIONS. [5o8. 

by taking advantage of the self-conjugateness of $. This may be 
written (by transforming the surface-integral into a volume- 

fff<k (S . V</> + S . aVu) = 0, 

and, as the limits of integration may be any whatever, 

S.V<f>oL + SaVu = ...................... (1). 

This is the required equation, the indeterminateness of a rendering- 
it equivalent to three scalar conditions. 

There are various modes of expressing this without the a. 
Thus, if A be used for V when the constituents of <p are considered, 
we may write 

Vu = 

It is easy to see that the right-hand member may be put in 
either of the equivalent forms 


In integrating this expression through a given space, we must 
remark that V and p are merely temporary symbols of construction, 
and therefore are not to be looked on as variables in the integral. 

Instead of transforming the surface-integral, we might have 
begun by transforming the volume-integral. Thus the first equa 
tion of this section gives 

//((/> + u) Uvds = 0. 
From this we have at once 

fJS. Uv (< + u) ads = 0. 
Thus, by the result of 490, whatever be a we have 

.V(( + w)a = 0, 
which is the condition obtained by the former process. 

As a verification, it may perhaps be well to shew that from this 
equation we can get the condition of equilibrium, as regards 
rotation, of a simply-connected portion of the body, which can be 
written by inspection as 

JJF. p<l> ( Uv) ds + /// VpVuds = 0. 
This is easily done as follows : (1) gives 

if, and only if, <j satisfy the condition 


Now this condition is satisfied by 

a = Yap 
where a is any constant vector. For 

8. (V) Vap = -S.aV(l> (V)p = S. aW^ = 0, 
in consequence of the self-conjugateness of (/>. Hence 

fffds (S . 7< Vap + 8 . apVu) = 0, 
or JfdsS . ap(f> Uv + fffdsS . apVu = 0. 

Multiplying by a, and adding the results obtained by making a in 
succession each of three rectangular vectors, we obtain the required 

509. To find the stress-function in terms of the displacement at 
each point of an isotropic solid, when the resulting strain is small, 
we may conveniently apply the approximate method of 384. As 
the displacement is supposed to be continuous, the strain in the 
immediate neighbourhood of any point may be treated as homo 
geneous. Thus, round each point, there is one series of rectangular 
parallelepipeds, each of which remains rectangular after the strain. 
Let a, /3, 7 ; a 15 {3 V <y l ; be unit vectors parallel to their edges 
before, and after, the strain respectively ; and let e v e 2 , e 3 be the 
elongations of unit edges parallel to these lines. We shall not 
have occasion to determine these quantities, as they will be 
eliminated after having served to form the requisite equations. 

Since the solid is isotropic and homogeneous, the stress is 
perpendicular to each face in the strained parallelepipeds ; and 
its amount (per unit area) can be expressed as 

P l = 2we 1 + (c - f n) 2e, &c (1), 

where n and c are, respectively, the rigidity and the resistance to 

Next, as in 384, let a be the displacement at p. The strain- 
function is 

i^tzr = OT $33-V. a 

so that at once 

S e = -SV<r; (2), 

and, if q ( ) q 1 be the operator which turns a into a,, &c v we 

*r (3). 

394 QUATERNIONS. [509. 

Thus, if (f> be the stress-function, we have (as in 507) 

<< = - 2 . P^&VB (4). 

But >/r&> = 2 . (1 + e a ) a 

so that -\J/ft> = 2 . (1 + e x 

and qty wq 1 = 2.(1 +ej afia.^ (5). 

By the help of (1), (2), and (5), (4) becomes 

00) = Zriqifr toq" 1 2nco (c f ??) coSVa ; 

and, to the degree of approximation employed, (3) shews that this 
may be written 

(j)0) = -n (SwV. a + VScoa) - (c - f n) wSVa (G), 

which is the required expression, the function c/> being obviously 

As an example of its use, suppose the strain to be a uniform 
dilatation. Here 

o- = ep, 

and (/>&) = Zneay + 3 (c f n) ea) = Scew ; 

denoting traction See, uniform in all directions. If e be negative, 
there is uniform condensation, and the stress is simply hydrostatic 

Again, let a eaSap, 

which denotes uniform extension in one direction, unaccompanied 
by transversal displacement. We have (a being a unit vector) 

(f)a) = 2neaS(oa. + (c f n) ew. 
Thus along a there is traction 

(c + jn)e, 
but in all directions perpendicular to a there is also traction 

Finally, take the displacement 

a = ea.S/3p. 

It gives ^o) = - ne (aSco/3 + fiSua) -(c- f ?i) ewSaff. 
This displacement gives a simple shear if the unit vectors a and ft 
are at right angles to one another, and then 

(f)a) = ne (a.Sco/3 + fiStoa), 

which agrees with the well-known results. In particular, it shews 
that the stress is wholly tangential on planes perpendicular either 


to a. or to {3 ; and wholly normal on planes equally inclined to them 
and perpendicular to their plane. The symmetry shews that the 
stress will not be affected by interchange of the unit vectors, 
a and /3, in the expression for the displacement. 

510. The work done by the stress on any simply connected 
portion of the solid is obviously 

because $ (Uv) is the vector force overcome per unit of area on 
the element ds. [The displacement at any moment may be written 
XCT ; and, as the stress is always proportional to the strain, the 
factor xdx has to be integrated from to 1.] This is easily trans 
formed to 

511. We may easily obtain the general expression for the 
work corresponding to a strain in any elastic solid. The physical 
principles on which we proceed are those explained in Appendix 
G to Thomson and Tait s Natural Philosophy. The mode in 
which they are introduced, however, is entirely different; and a 
comparison will shew the superiority of the Quaternion notation, 
alike in compactness and in intelligibility and suggestiveness. 

If the strain, due to the displacement a, viz. 

^T = T SrV . a 

be a mere rotation, in which case of course no work is stored up 
by the stress, we have at once 

S . ^TW^TT= S(i)T 

for all values of co and r. We may write this as 
S . a) (i|rS/r - 1) r = Sw^r = 0, 

where % is ( 380) a self-conjugate linear and vector function, whose 
complete value is 

X r = - SrV . a - VSro- + V^rVSo-ov 
The last term of this may, in many cases, be neglected. 

When the strain is very small, the work (per unit volume) 
must thus obviously be a homogeneous function, of the second 
degree, of the various independent values of the expression 

396 QUATERNIONS. [5 I 2. 

On account of the self-conjugateness of ^ there are but six 
such values : viz. 

Their homogeneous products of the second degree are therefore 
twenty-one in number, and this is the number of elastic coefficients 
which must appear in the general expression for the work. In 
the most general form of the problem these coefficients are to be 
regarded as given functions of p. 

At and near any one point of the body, however, we may take 
i, j, k as the chief vectors of ^ at that point, and then the work 
for a small element is expressible in terms of the six homogeneous 
products, of the second degree, of the three quantities 

Si W> S JXJ> Sk X k - 

This statement will of course extend to a portion of the body of 
any size if (whether isotropic or not) it be homogeneous and 
homogeneously strained. From this follow at once all the 
elementary properties of homogeneous stress. 

M. The Hydrokinetic Equations. 

512. As another application, let us form the hydrokinetic 
equations, on the hypothesis that a perfect fluid is not a molecular 
assemblage but a continuous medium. 

Let a be the vector-velocity of a very small part of the fluid at 
p ; e the density there, taken to be a function of the pressure, p, 
alone ; i.e. supposing that the fluid is homogeneous when the 
pressure is the same throughout ; P the potential energy of unit 
mass at the point p. 

The equation of " continuity" is to be found by expressing the 
fact that the increase of mass in a small fixed space is equal 
to the excess of the fluid which has entered over that which 
has escaped. If we take the volume of this space as unit, the 
condition is 

.................... (1). 

We may put this, if we please, in the form 

where 3 expresses total differentiation, or, in other words, that we 
follow a definite portion of the fluid in its motion. 


The expression might at once have been written in the form 
(2) from the comparison of the results of two different methods of 
representing the rate of increase of density of a small portion of 
the fluid as it moves along. Both forms reduce to 

when there is no change of density ( 384). 

Similarly, for the rate of increase of the whole momentum 
within the fixed unit space, we have 

*&1 = - e vp- Vp +SJSUvo- . eads ; 

where the meanings of the first two terms are obvious, and the 
third is the excess of momentum of the fluid which enters, over 
that of the fluid which leaves, the unit space. 

The value of the double integral is, by 49.9 (3), 

(eer) + eSo-V . o- = o- ^ + eSa-V . o-, by (1). 
Thus we have, for the equation of motion, 

or, finally 

fa i * i ^ / ~~ T ^ () 

This, in its turn, might have been even more easily obtained by 
dealing with a small definite portion of the fluid. 

It is necessary to observe that in what precedes we have 
tacitly assumed that a is continuous throughout the part of the 
fluid to which the investigation applies : i.e. that there is neither 
rupture nor finite sliding. 

513. There are many ways of dealing with the equation (3) 
of fluid motion. We select a few of those which, while of historic 
interest, best illustrate quaternion methods. 

We may write (3) as 

Now we have always 

F. o-FVo- - SoV . a - V^cr = SaV . a - JV . a 2 . 



Hence, if the motion be irrotational, so that ( 384) 

Wo- = 0, 
the equation becomes 

But, if w be the velocity-potential, 

0- = 

and we obtain (by substituting this in the first term, and operating 
on the whole by S . dp) the common form 

7 dw , 7 2 ir\ 
d -T. + d . v 2 = - dQ, 

where l^ 2 (= |o- 2 ) is the kinetic energy of unit mass of the fluid. 

If the fluid be incompressible, we have Laplace s equation 
for w, viz. 

V% = SVa = 0. 

When there is no velocity-potential, we may adopt Helmholtz 
method. But first note the following quaternion transformation 
(Proc. R. 8. E. 186970) 

[The expression on the right has many remarkable forms, the 
finding of which we leave, as an exercise, to the student. For our 
present purpose it is sufficient to know that its vector part is 

This premised, operate on (3) by V. V, and we have 

Hence at once, if the fluid be compressible, 

= -S. VV . <7 + VVa . SVa- = V. V V. 

But if the fluid be incompressible 

Either form shews that when the vector-rotation vanishes, its 
rate of change also vanishes. In other words, those elements of 
the fluid, which were originally devoid of rotation, remain so during 
the motion. 


Thomson s mode of dealing with (3) is to introduce the 


which he calls the " flow " along the arc from the point a to the 
point p ; these being points which move with the fluid. 

Operating on (3) with S . dp, we have as above 

so that, integrating along any definite line in the fluid from a. to 
p, we have 

which gives the rate at which the flow along that line increases, as 
it swims along with the fluid. 

If we integrate round a closed curve, the value of df/dt vanishes, 
because Q is essentially a single-valued function. In this case the 
quantity / is called the " circulation," and the result is stated in 
the form that the circulation round any definite path in the fluid 
retains a constant value. 

Since the circulation is expressed by the complete integral 


it can also be expressed by the corresponding double integral 

-//. UvVads, 

so that it is only when there is at least one vortex-filament passing 
through the closed circuit that the circulation can have a finite 

N. Use of V in connection with Taylor s Theorem. 

514. Since the algebraic operator 

when applied to any function of x, simply changes x into x + h, it 
is obvious that if a be a vector not acted on by 

. d . d d 
V = ^ + ; ,-+&,-, 
dx ay dz 

we have e ~ S(rV f (p) =f (p + <r), 

400 QUATERNIONS. [515. 

whatever function f may be. From this it is easy to deduce 
Taylor s theorem in one important quaternion form. 

If A bear to the constituents of or the same relation as V bears 
to those of p, and if / and F be any two functions which satisfy 
the commutative law in multiplication, this theorem takes the 
curious form 

of which a particular case is (in Cartesian symbols) 

The modifications which the general expression undergoes, when 
/ and F are not commutative, are easily seen. 

If one of these be an inverse function, such as, for instance, 
may occur in the solution of a linear differential equation, these 
theorems of course do not give the arbitrary part of the integral, 
but they often materially aid in the determination of the rest. 

One of the chief uses of operators such as $aV, and various 
scalar functions of them, is to derive from 1/Tp the various orders 
of Spherical Harmonics. This, however, is a very simple matter. 

515. But there are among them results which appear startling 
from the excessively free use made of the separation of symbols. 
Of these one is quite sufficient to shew their general nature. 

Let P be any scalar function of p. It is required to find the 
difference between the value of P at p, and its mean value 
throughout a very small sphere, of radius r and volume v t which has 
the extremity of p as centre. This, of course, can be answered at 
once from the formula of 485. But the somewhat prolix method 
we are going to adopt is given for its own sake as a singular piece 
of analysis, not for the sake of the problem. 

From what is said above, it is easy to see that we have the 
following expression for the required result : 

;///<-* -*>** 

where a is the vector joining the centre of the sphere with the 
element of volume cfc, and the integration (which relates to <r 
and ds alone) extends through the whole volume of the sphere. 


Expanding the exponential, we may write this expression in the 

1 JJJ (SaVf Pd,-&c., 


higher terms being omitted on account of the smallness of r, the 
limit of To-. 

Now, symmetry shews at once that 
JIM? = 0. 

Also, whatever constant vector be denoted by a, 
/// (Scwr) 1 & = - a 2 /// (OrUay tk. 

Since the integration extends throughout a sphere, it is obvious 
that the integral on the right is half of what we may call the 
moment of inertia of the volume about a diameter. Hence 

If we now write V for a, as the integration does not refer to V, 
we have by the foregoing results (neglecting higher powers of r) 

<-**-!>**-- jS^. 

which is the expression given by Clerk-Maxwell *. Although, for 
simplicity, P has here been supposed a scalar, it is obvious that in 
the result above it may at once be written as a quaternion. 

516. As another illustration, let us apply this process to the 
finding of the potential of a surface-distribution. If p be the 
vector of the element ds, where the surface density is fp, the 
potential at a is 


F being the potential function, which may have any form whatever. 
By .the preceding, 514, this may be transformed into 

or, far more conveniently for the integration, into 

* London Math. Soc. Proc., vol. iii, no. 34, 1871. 
T. Q. I. 26 

402 QUATERNIONS. [5 1 7- 

where A depends on the constituents of a in the same manner as 
V depends on those of p. 

A still farther simplification may be introduced by using a 
vector CT O , which is finally to be made zero, along with its corre 
sponding operator A , for the above expression then becomes 

where p appears in a comparatively manageable form. It is obvious 
that, so far, our formulae might be made applicable to any dis 
tribution. We now restrict them to a superficial one. 

517. Integration of this last/orm can always be easily effected 
in the case of a surface of revolution, the origin being a point in 
the axis. For the expression, so far as the integration is concerned, 
can in that case be exhibited as a single integral 

where F may be any scalar function, and x depends on the cosine of 
the inclination of p to the axis. And 

da) a 

As the interpretation of the general results is a little trouble 
some, let us take the case of a spherical shell, the origin being the 
centre and the density unity, which, while simple, sufficiently 
illustrates the proposed mode of treating the subject. 

We easily see that in the above simple case, a. being any 
constant vector whatever, and a being the radius of the sphere, 

C+a 977V7 

ffds e s *P = 2-Tra e xTa dx = ^ (e aTa - e- aTa ). 
J -a -L & 

Now, it appears that we are at liberty to treat A as a has just 
been treated. It is necessary, therefore, to find the effects of such 
operators as TA, e arA , &c., which seem to be novel, upon a scalar 
function of To- ; or , as we may for the present call it. 


Now <TA) 2 F = - tfF = F" + ~ , 

whence it is easy to guess at a particular form of TA. To make 
sure that it is the only one, assume 


where f and fare scalar functions of ^ to be found. This gives 

= ?F" + (ft + f f + ff ) F + (ff + f 2 ) F. 
Comparing, we have 

From the first, f = + 1, 

whence the second gives f = + ==. ; 

the signs of f and f being alike. The third is satisfied identically. 

That is, finally, ^ = ^ + 1. 

Also, an easy induction shews that 

Hence wo have at once 

n / d 

l 1 


+&c - 

by the help of which we easily arrive at the well-known results. 
This we leave to the student*. 

O. Applications of V in connection with Calculus of Variations. 

518. We conclude with a few elementary examples of the use 
of V in connection with the Calculus of Variations. These depend, 
for the most part, on the simple relation 

Let us first consider the expression 

where Tdp is an element of a finite arc along which the inte 
gration extends, and the quaternion Q is a function of p, generally 
a scalar. 

* Proc.R. S. E. t 1871-2. 


404 QUATERNIONS. [5 1 9. 

To shew the nature of the enquiry, note that if Q be the 
speed of a unit particle, A will be what is called the Action. If 
Q be the potential energy per unit length of a chain, A is the 
total potential energy. Such quantities are known to assume 
minimum, or at least "stationary", values in various physical 

We have for the variation of the above quantity 
BA = f (SQTdp + QSTdp) 

= - [QSUdpSp] + f {BQTdp + S.S P d (Q Udp)}, 

where the portion in square brackets refers to the limits only, and 
gives the terminal conditions. The remaining portion may easily 
be put in the form 


If the curve is to be determined by the condition that the varia 
tion of A shall vanish, we must have, as Bp may have any direction, 

or, with the notation of Chap. X, 

This simple equation shews that (when Q is a scalar) 

(1) The osculating plane of the sought curve contains the 
vector VQ. 

(2) The curvature at any point is inversely as Q, and directly 
as the component of VQ parallel to the radius of absolute curvature. 

519. As a first application, suppose A to represent the Action 
of a unit particle moving freely under a system of forces which 
have a potential ; so that 

Q = Tp, 

and p* = 2(P-H), 

where P is the potential, H the energy-constant. 

These give TpVTp = QVQ = - VP, 

and Qp = p, 

so that the equation above becomes simply 


which is obviously the ordinary equation of motion of a free 

520. If we look to the superior limit only, the first expression 
for SA becomes in the present case 

If we suppose a variation of the constant H, we get the following 
term from the unintegrated part 


Hence we have at once Hamilton s equations of Varying Action in 

the forms 



and -j^. = t. 


The first of these gives, by the help of the condition above, 

the well-known partial differential equation of the first order and 
second degree. 

521. To shew that, if A be any solution whatever of this 
equation, the vector VA represents the velocity in a free path 
capable of being described under the action of the given system of 
forces, we have 

= -S(VA.V)VA. 
But ~.VA = -S(p 


A comparison shews at once that the equality 

is consistent with each of these vector equations. 

522. Again, if 3 refer to the constants only, 
d(VA)* = S.VAdVA = -dH 
by the differential equation. 

But we have also ^v = t, 


which gives %-(dA)=*-S (/3V) dA = dff. 


406 QUATERNIONS. [523. 

These two expressions for dH again agree in giving 

and thus shew that the differential coefficients of A with regard to 
the two constants of integration must, themselves, be constants. 
We thus have the equations of two surfaces whose intersection 
determines the path. 

523. Let us suppose next that A represents the Time of 
passage, so that the brachistochrone is required. Here we have 

the other condition being as in 519, and we have 

which may be reduced to the symmetrical form 

It is very instructive to compare this equation with that of the 
free path as above, 519 ; noting how the force VP is, as it 
were, reflected on the tangent of the path ( 105). This is the 
well-known characteristic of such brachistochrones. 

The application of Hamilton s method may be easily made, as 
in the preceding example. (Tait, Trans. R. S. E., 1805.) 

524. As a particular case, let us suppose gravity to be the 
only force, then 

a constant vector, so that 

d ._! ,_ 2 

jt p p a=0 - 

The form of this equation suggests the assumption 

where p and q are scalar constants, and 

Substituting, we get 

-pq sec 2 qt + ( - /3 2 - pV tan 2 qt) = 0, 
which gives pq = T*$ = p*Fa. 


Now let p{3~ 1 a. = 7 ; 

this must be a unit-vector perpendicular to a and /3, so that 


p~ l = . (cos at 7 sin otf), 

cos g v 

whence p = cos qt (cos qt + y sin gtf) /3" 1 

(which may be verified at once by multiplication). 

Finally, taking the origin so that the constant of integration 
may vanish, we have 

2/o/S = t + ^- (sin 2qt - 7 cos 2q), 

which is obviously the equation of a cycloid referred to its vertex. 
The tangent at the vertex is parallel to/3, and the axis of symmetry 
to a. The equation, it should be noted, gives the law of descrip 
tion of the path. 

525. In the case of a chain hanging under the action of given 
forces we have, as the quantity whose line-integral is to be a 

Q = Pr, 

where P is the potential, r the mass per unit-length. 
Here we have also, of course, 

the length of the chain being given. 

It is easy to see that this leads, by the method above, to the 

where u is a scalar multiplier. 

526. As a simple case, suppose the chain to be uniform. 
Then we may put ru for u, and divide by r. Suppose farther 
that gravity is the only force, then 

P = Sap, VP = - a, 
and -y- {(Sap + ?/) p\ + a = 0. 

408 QUATERNIONS. [527. 

Differentiating, and operating by Sp, we find 

or, since p is a unit-vector, 


which shews that u is constant, and may therefore be allowed for 
by change of origin. 

The curve lies obviously in a plane parallel to a, and its equation is 
(Sap) 2 + a 2 s 2 = const., 

which is a well-known form of the equation of the common 

When the quantity Q of 518 is a vector or a quaternion, we 
have simply an equation (like that there given) for each of the 

527. Suppose P and the constituents of cr to be functions 
which vanish at the bounding surface of a simply-connected space 
2, or such at least that either P or the constituents vanish there, 
the others (or other) not becoming infinite. 

Then, by 484, 

fJfdsS . V(Pa) = // ds PScr Uv = 0, 
if the integrals be taken through and over 2. 
Thus fffdsS.aVP = -fffd<;PSV<7. 

By the help of this expression we may easily prove a very 
remarkable proposition of Thomson (Cam. and Dub. Math. Journal, 
Jan. 1848, or Reprint of Papers on Electrostatics, 206). 

To shew that there is one, and but one, solution of the equation 
8V (e*Vu) = 477T 

where r vanishes at an infinite distance, and e is any real scalar 
whatever, continuous or discontinuous. 

Let v be the potential of a distribution of density r, so that 

and consider the integral 


That Q may be a minimum as depending on the value of u (which 
is obviously possible since it cannot be negative, and since it may 
have any positive value, however large, if only greater than this 
minimum) we must have 

= JSQ = - fffdiS . (e 2 Vu - Vfl) VBu 
By the lemma given above this may be written 

Thus any value of u which satisfies the given equation is such as to 
make Q a minimum. 

But there is only one value of u which makes Q a minimum ; 
for, let Q l be the value of Q when 

u l = u + w 
is substituted for this value of u, and we have 

Q, = - jj/* (* v (u+w)-^vv y 

= Q - 2 ///cfc S (e 2 Vu - Vw) Vw - fffds e*(Vw) 2 . 

The middle term of this expression may, by the proposition at the 
beginning of this section, be written 

and therefore vanishes. The last term is essentially positive. Thus 
if MJ anywhere differ from u (except, of course, by a constant 
quantity) it cannot make Q a minimum: and therefore u is a 
unique solution. 


1. The expression 

Fa/3 FyS + Fa 7 FS/3 + FaS F/3 7 
denotes a vector. What vector ? 

2. If two surfaces intersect along a common line of curvature, 
they meet at a constant angle. 


3. By the help of the quaternion formulae of rotation, translate 
into a new form the solution (given in 248) of the problem of 
inscribing in a sphere a closed polygon the directions of whose sides 
are given. 

4. Find the point, the sum of whose distances from any 
number of given points is a minimum. 

If p be the sought point, a l5 2 , &c. the given points: shew 

Give a dynamical illustration of this solution. 

(Proc. R S. E. 1866-7.) 
5. Shew that 


6. Express, in terms of the masses, and geocentric vectors of 
the sun and moon, the sun s vector disturbing force on the moon, 
and expand it to terms of the second order; pointing out the 
magnitudes and directions of the separate components. 

(Hamilton, Lectures, p. 615.) 

7. lfq = r*, shew that 

2 dq = 2dr* = %(dr + Kqdrq 1 ) Sq~ l = %(dr + q^drKq) Sq~ l 

= (drq + Kqdr) q~ l (q + Kq)~ l = (drq + Kqdr) (r + Tr)~ l 

dr + U^drU- 1 

q~\qdr+Trdrq~*) _drUq+ Uq l dr _ 
~ WT Tq(l+ Ur) 1 + Ur 

- d.r + V. Vdr ~qq- 1 = dr - V 

q q q 

= drq~ l + V(Vq~\Vdr} (l + 

and give geometrical interpretations of these varied expressions 
for the same quantity. (Ibid. p. 628.) 

8. Shew that the equation of motion of a homogeneous solid 
of revolution about a point in its axis, which is not its centre 
of gravity, is 

where II is a constant. (Trans. R. S. E., 1869.) 


9. Find the point P, such that, if A^ A 2 , &c. be any fixed 
points in space, and m 1? m 2 , &c. positive numerical quantities, 
S.mAP shall be a minimum. 

Shew that a closed (gauche) polygon can be constructed whose 
sides are parallel to PA^ PA^ &c. while their lengths are as m v m 2 , 
&c., respectively. 

If T2, . mAP is to be a minimum, what is the result ? 

10. Form the quaternion condition that the lines joining the 
middle points of the sides of a closed polygon (plane or gauche) 
may form a similar polygon. 

When this condition is satisfied, find the quaternion operator 
which must be applied to the second polygon to make it similarly 
situated with the first. 

11. Solve the equations in linear and vector functions; -or 
being given, while $ and ^ are to be found : 

%+ $ = 

12. Put the equation of a Minding line ( 394) directly into 
the normal form for a line passing through each of two fixed 
curves : 

p = x $t + (1 - x) ^u, 

where < and -v/r are definite vector functions of the arbitrary 
scalars t, and u, respectively. 

13. Shew that 

are different expressions for the same scalar, and give a number of 
other forms. ( 513.) How are the values of these quantities 
respectively affected, if the V without suffix acts on a^ as well 
as on <7? 

14. If (as in 511) 


for all values of CD and of r, shew by actual transformation, not by 
the obvious geometrical reasoning, that we have also 


15. Shew that, to the third order in Tfi, the couple, due to 
any closed circuit, on a small magnet, 2/3, whose centre is at the 
origin, is proportional to 

.p f / sr/8 1 US ! /3 

r v ~ 2 v + T 

Simplify, by means of this formula, the quasi-Cartesian investiga 
tion of 464. 

16. Calculate the value of 

J Tp 3 

for a circular current, the origin being taken at any point. 

17. Find an approximate formula for the potential, near the 
centre of the field, when two equal, circular, currents have a 
common axis, and are at a distance equal to their common 
radius. (This is v. Helmholtz Tangent Galvanometer.) 

18. Deduce the various forms of Spherical Harmonics, and 
their relations, from the results of the application of scalar 
functions of $aV, and similar operators, to l/Tp. 

19. Integrate the differential equations : 

/ \ dq , 

(a) -?- + aq = 6, 


where a and b are given quaternions, and ( and ^ given linear 
and vector functions. (Tait, Proc. E. S. E., 1870-1.) 

20. Derive (4) of 92 directly from (3) of 91. 

21. Find the successive values of the continued fraction 


where i and j have their quaternion significations, and x has the 
values 1, 2, 3, &c. (Hamilton, Lectures, p. 645.) 


22. If we have 


where c is a given quaternion, find the successive values. 

For what values of c does u become constant ? (Ibid. p. 652.) 

23. Prove that the moment of hydrostatic pressures on the 
faces of any polyhedron is zero, (a) when the fluid pressure is the 
same throughout, (6) when it is due to any set of forces which 
have a potential. 

24. What vector is given, in terms of two known vectors, by 
the relation 

p-^Ma- +r 1 )? 

Shew that the origin lies on the circle which passes through the 
extremities of these three vectors. 

25. Tait, Trans, and Proc. R S. E., 1870-3. 
With the notation of 484, 495, prove 

(a) fffS (aV) rd9 - ffrSot Uvds. 

(b) If 8 (pV) r = - nr, 

(n + 3) JJ/rd? = - ffrSp Uvds. 

(c) With the additional restriction W = 0, 

ffS . UP {2np + (n + 3) /> 2 V j . rds = 0. 

(d) Express the value of the last integral over a non-closed 
surface by a line-integral. 

(e) -JTdp=ffdsS.UvV<r, 
if a- = Udp all round the curve. 

(/) For any portion of surface whose bounding edge lies 
wholly on a sphere with the origin as centre 

whatever be the vector a. 

(g) JVdpV. a = ffds { UvV 2 - S ( UvV)V] <r, 
whatever be cr. 

26. Tait, Trans. E. S. E., 1873. 

Interpret the equation 

da = uqdpq~ l , 


and shew that it leads to the following results 

VV = qVuq-\ 

V. uq~ l = 0, 

V 2 . u* = 0. 

Hence shew that the only sets of surfaces which, together, cut 
space into cubes are planes and their electric images. 

27. What problem has its conditions stated in the following 
six equations, from which f, ij, are to be determined as scalar 
functions of x } y, z, or of 

p = ix +jy + kz ? 

__ . d . d 7 d 

where V= l j-+Jj-+k-r 

dx J dy dz 

Shew that they give the farther equations 

Shew that (with a change of origin) the general solution of these 
equations may be put in the form 

where (/> is a self-conjugate linear and vector function, and f, 77, f 
are to be found respectively from the three values of / at any 
point by relations similar to those in Ex. 24 to Chapter X. 

28. Shew that, if p be a planet s radius vector, the potential 
P of masses external to the solar system introduces into the 
equation of motion a term of the form S (pV)VP. 

Shew that this is a self-conjugate linear and vector function 
of p, and that it involves only^ye independent constants. 

Supposing the undisturbed motion to be circular, find the chief 
effects which this disturbance can produce. 

29. In 430 above, we have the equations 

where &> 2 is neglected. Shew that with the assumptions 

= qrrr~ l q~ 1 

we have 

/3 = 0, T/3 = l, S{3r = 0, 
provided w/Sfw - ^ = 0. Hence deduce the behaviour of the 


Foucault pendulum without the x, y, and f, rj transformations 
in the text. 

Apply analogous methods to the problems proposed at the end 
of 426 of the text. 

30. Hamilton, Bishop Law s Premium Examination, 1862. 

(a) If OABP be four points of space, whereof the three first 
are given, and not collinear ; if also 

OA = a, OB = /3, OP = p ; 
and if, in the equation 

a a 

the characteristic of operation F be replaced by S, the locus of P 
is a plane. What plane ? 

(6) In the same general equation, if F be replaced by F, 
the locus is an indefinite right line. What line ? 

(c) If F be changed to K, the locus of P is a point. What 
point ? 

(d) If F be made = V, the locus is an indefinite half-line, 
or ray. What ray ? 

(e) If F be replaced by T, the locus is a sphere. What 
sphere ? 

(/) If F be changed to TV, the locus is a cylinder of 
revolution. What cylinder ? 

(g) If F be made TVU, the locus is a cone of revolution. 
What cone ? 

(h) If SU be substituted for F, the locus is one sheet of 
such a cone. Of what cone ? and which sheet ? 

(i) If F be changed to F7, the locus is a pair of rays. 
Which pair ? 

31. Hamilton, Bishop Law s Premium Examination, 1863. 
(a) The equation 

expresses that /> and p are the vectors of two points p and p , 
which are conjugate with respect to the sphere 

p 2 + a 2 = 0; 
or of which one is on the polar plane of the other. 

(b) Prove by quaternions that if the right line PP 7 , connect 
ing two such points, intersect the sphere, it is cut harmonically 


(c) If P be a given external point, the cone of tangents 
drawn from it is represented by the equation, 

and the orthogonal cone, concentric with the sphere, by 

(d) Prove and interpret the equation, 

T(np-a) = T(p-na), if Tp = Ta. 

(e) Transform and interpret the equation of the ellipsoid, 

(/) The equation 

(V - ij = (t 2 + * 2 ) Spp + 

expresses that p and p are values of conjugate points, with respect 
to the same ellipsoid. 

(g) The equation of the ellipsoid may also be thus written, 
Svp = 1, if O 2 - * 2 )V = (i - K)*P + ZiSicp + 2fc8ip. 

(h) The last equation gives also, 

(i) With the same signification of v, the differential equa 
tions of the ellipsoid and its reciprocal become 

Svdp = 0, Spdv = 0. 
(j) Eliminate p between the four scalar equations, 

Sap = a, Sfip = b, Syp = c, Sep = e. 

32. Hamilton, Bishop Law s Premium Examination, 1864. 

(a) Let AJ$v Afl^ . . . A n B n be any given system of posited 
right lines, the 2n points being all given; and let their vector 

AB = Afr + A& + ...+ A n B n , 
be a line which does not vanish. Then a point H, and a scalar h, 
can be determined, which shall satisfy the quaternion equation, 

namely by assuming any origin 0, and writing, 
nrr v Q^^ l + .. 


(b) For any assumed point C, let 

Q =CA l .A l B t + ... + CA..AJ.; 
then this quaternion sum may be transformed as follows, 
Qc=Qn + 

and therefore its tensor is 

in which AB and CH denote lengths. 

(c) The least value of this tensor TQ C is obtained by 
placing the point C at H; if then a quaternion be said to be a 
minimum when its tensor is such, we may write 

min. Q c =Q H =h.AB; 
so that this minimum of Q c is a vector. 

(d) The equation 

TQ C =c = any scalar constant > TQ H 

expresses that the locus of the variable point C is a spheric 
surface, with its centre at the fixed point H, and with a radius r, 
or CH, such that 

r.AB=(T<? - T<$$ = (6 - h* . AB*)* 

so that H, as being thus the common centre of a series of 
concentric spheres, determined by the given system of right- lines, 
may be said to be the Central Point, or simply the Centre, of that 

(e) The equation 

TVQc = c 1 = any scalar constant > TQ H 
represents a right cylinder, of which the radius 

divided by AB, and of which the axis of revolution is the line, 

wherefore this last right line, as being the common axis of a series 
of such right cylinders, may be called the Central Axis of the 

(/) The equation 

SQc ~ c 2 an y scalar constant 

represents a plane ; and all such planes are parallel to the Central 
Plane, of which the equation is 

SQ C =Q. 

T. Q. I. 27 


(g) Prove that the central axis intersects the central plane 
perpendicularly, in the central point of the system. 

(h) When the n given vectors A l B l ,...A n B n are parallel, 
and are therefore proportional to n scalars, b v ... b n , the scalar h 
and the vector Q H vanish ; and the centre H is then determined 
by the equation 

or by the expression, 

where is again an arbitrary origin. 

33. Hamilton, Bishop Laws Premium Examination, 1860. 

(a) The normal at the end of the variable vector p, to the 
surface of revolution of the sixth dimension, which is represented 
by the equation 

or by the system of the two equations, 

and the tangent to the meridian at that point, are respectively 
parallel to the two vectors, 

v = 2 (p CL) tp, 
and r - 2 (1 - 2$) (p-a) + tp ; 

so that they intersect the axis a, in points of which the vectors 
are, respectively, 

AOL i u \ JL --- 

-, and ^- 

(6) If dp be in the same meridian plane as p, then 

, and 

(c) Under the same condition, 

(d) The vector of the centre of curvature of the meridian, 
at the end of the vector p, is, therefore, 

3 v 6a-4-$5 


(e) The expressions in (a) give 

v 2 = a 2 * 2 (1 - t)\ r 2 = aY 5 (1 - O 2 (4 - 1) ; 

x 2 9 _ , 2 9a 2 ,, 2 

hence (cr p) = 7 a r, and ap = . dt ; 

4 4> t 

the radius of curvature of the meridian is, therefore, 

R = T(a-p)= 3 2 tTa; 
and the length of an element of arc of that curve is 

* ^M. 

(/) The same expressions give 

thus the auxiliary scalar is confined between the limits and 4, 
and we may write t = 2 vers 0, where is a real angle, which varies 
continuously from to 2?r ; the recent expression for the element 
of arc becomes, therefore, 


and gives by integration 

if the arc s be measured from the point, say F, for which p a, 
and which is common to all the meridians; and the total periphery 
of any one such curve is = 127rTa. 

(g) The value of cr gives 

4 (a 2 - a 2 ) = 3a 2 * (4 - t\ 16 (Fao-) 2 = - aV (4 - $)" ; 

if, then, we set aside the axis of revolution a, which is crossed by 
all the normals to the surface (a), the surface of centres of 
curvature which is touched by all those normals is represented by 
the equation, 

4(c7 2 -a 2 ) 3 + 27a 2 (Fa<7) 2 = ........ .......... (b). 

(h) The point F is common to the two surfaces (a) and 
(b), and is a singular point on each of them, being a triple point 
on (a), and a double point on (b) ; there is also at it an infinitely 
sharp cusp on (b), which tends to coincide with the axis a, but 
a determined tangent plane to (a), which is perpendicular to that 
axis, and to that cusp ; and the point, say F t of which the vector 
= a, is another and an exactly similar cusp on (b), but does not 
belong to (a) . 


(i) Besides the three universally coincident intersections of 
the surface (a), with any transversal, drawn through its triple 
point F, in any given direction ft, there are always three other 
real intersections, of which indeed one coincides with F if the 
transversal be perpendicular to the axis, and for which the 
following is a general formula : 

(j) The point, say F, of which the vector is p = 2a, is a 
double point of (a), near which that surface has a cusp, which 
coincides nearly with its tangent cone at that point; and the 

/ IT 

semi-angle of this cone is = ^ . 
Auxiliary Equations : 

f Svp = - eft (1 -t)(l- 20, 
\28v (P-CL) = a 2 ? (1 - 1). 

f SpT = oi*t*(l-t)(4i-t), 
\28(p-OL)T = aV (1 - t) (4 - 1). 

34. A homogeneous function of p, of the wth degree, is 
changed into the same function of the constant vector, a, by the 

35. If r be the cube root of the quaternion q, shew that 

dr =p + ( V . r 2 + r Vr) Vq 1 (rp - pr), 
where p ^r~*dq. (Hamilton, Lectures, p. 629.) 

36. Shew from 509 that the term to be added to VQ of 
512 (3), in consequence of viscosity, is proportional to 

37. Shew, from structural considerations, that 

must be a linear and vector function of Fa/3. Also prove, directly, 
that its value is 


.38. The spherical opening subtended at a by the sphere 

- ds -^ r0 

according as To. a < r. 

Hence shew, without further integration, that (with the same 

f f ds 4-Trr 2 
1 1 TjT, x = 477T, or , 

whence, of course, 

lldsU(p- a )_ 


&lso that 

ds 4-Trr 4-Trr 2 

39. Find also the values of 

Uvds Uvds 

the integration extending over the surface of the sphere Tp = r. 

40. Shew that the potential at /3, due to mass m at a, is to 
the potential at /3" 1 , due to mass m at of 1 , as 1 : Tfi, provided 
m : m :: 1 : Ta. 

Hence, by the results of 38 above, shew that, with a < r, 
ds 4?rr 4?rr 2 

(p)a) 3 ~ (r> - a?) T (/3 - a) 

according as T/3 < r, the limits of integration being as before. 

Obtain these results directly from Green s Theorem ( 486) 
without employing the Electric Image transformation. 

41. Shew that the quaternion 

[jvS ( UvV) rds - jjja-V *rd<; 

(in the notation of 482) is changed into its own conjugate by 
interchange of a and r. Express its value (by 133, 486) as a 
single volume-integral. 


42. Prove that 

jjj(o V V + K Va . V r) d? = jf<r UvVrds, 

and thence that 

43. Find the distribution of matter on a given closed surface 
which will produce, in its interior, the same potential as does 
a given distribution of matter outside it. 

Hence shew that there is one, and only one, distribution of 
matter over a surface, which will produce, at each point of it, any 
arbitrarily assigned potential. 


By the same Author. 

A Treatise on Natural Philosophy. By Sir W. THOMSON, LL.D., 

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and P. G. TAIT, M. A. , Professor of Natural Philosophy in the University of 
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Elements of Natural Philosophy. By Professors Sir W. THOMSON 

and P. G. TAIT. Demy 8vo. 9s. 

iLontum: C. J. CLAY AND SONS, 


Lectures on some Recent Advances in Physical Science. 

With Illustrations. Revised and enlarged, with the Lecture on Force delivered 
before the British Association. By P. G. TAIT, M.A. Third Edition. Crown 
8vo. 9s. 

Heat. With numerous Illustrations. By P. G. TAIT, M.A. Crown 
8vo. 6s. 

A Treatise on Dynamics of a Particle. With numerous Examples. 

By P. G. TAIT, M.A., and the late W. J. STEELE, B.A., Fellow of St Peter s 
College, Cambridge. Sixth Edition, carefully Revised. Crown 8vo. 12s. 

The Unseen Universe ; or, Physical Speculations on a Future 

State. By Professors BALFOUR STEWART and P. G. TAIT. Fifteenth Edition. 
Crown 8vo. 6s. 


Light. By P. G. TAIT, M.A., Sec. R.S.E. Second Edition. Revised 
and Enlarged. Crown 8vo. 6s. 

Properties of Matter. By P. G. TAIT, M.A., Sec. R.S.E. Crown 

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