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% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
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% Trinity College, 1999.
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\centerline{\Largebf COPY OF A LETTER}
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\centerline{\Largebf FROM SIR WILLIAM R. HAMILTON}
\vskip12pt
\centerline{\Largebf TO JOHN T. GRAVES, ESQ.}
\vskip24pt
\centerline{\largerm (Philosophical Magazine, 3rd series,
25 (1844), pp.\ 489--95.)}
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\vfill
\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 1999}
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\noindent
{\largeit On Quaternions; or on a new System of Imaginaries
in Algebra. By {\largesc Sir William Rowan Hamilton}
LL.D, P.R.I.A., F.R.A.S., Hon. M.\ R.\ Soc.\ Ed.\ and Dub.,
Hon.\ or Corr.\ M.\ of the Royal or Imperial Academies of
St.\ Petersburgh, Berlin, Turin and Paris, Member of the American
Academy of Arts and Sciences, and of other Scientific Societies
at Home and Abroad, Andrews' Prof.\ of Astronomy in the
University of Dublin, and Royal Astronomer of Ireland.}
\vskip12pt
\noindent
[{\it The London, Edinburgh and Dublin Philosophical Magazine
and Journal of Science}
vol.~xxv (1844), pp.~489--95]
\vskip12pt
\centerline{\it To the Editors of the Philosophical Magazine and Journal}
\noindent
{\sc Gentlemen},
I have been induced to think that the account contained in the
following letter, of the considerations which led me to conceive
that theory of quaternions, a part of which you have done me the
honour to publish in two recent Numbers (for July and October) of
your Magazine, might not be without interest to some of your
readers. Should you think proper to insert it, a public
acknowledgement (very pleasing to my own feelings) will have been
rendered, on the one hand to the Rev.\ Mr.~Warren, whose work on
the Geometrical Representation of the Square Roots of Negative
Quantities (printed at Cambridge in 1828) long since attracted my
attention and influenced my thoughts; and on the other hand to
the gentleman (John T.\ Graves, Esq.) to whom the letter was
addressed, and with whom I had been engaged, at intervals, for
many years in a correspondence, very instructive and suggestive
to me, on subjects connected therewith. Nor am I without hope
that Mr.~Graves may thus be led to communicate through you to
mathematicians some of the extensions which he has made of
results of mine, with some of those other speculations which are
still more fully his own. On some future occasion I may perhaps
be allowed to mention any other quarters from which I may be
conscious of having derived more recent assistance, in my
investigations on the same mathematical subject, many of which
are hitherto unpublished.
\vskip12pt
\line{\hfill I have the honour to be, Gentlemen,}
\line{\hfill Your obedient Servant,}
\line{\hfill {\sc William Rowan Hamilton}}
\vskip12pt
\line{Observatory of Trinity College, Dublin,\hfill}
\line{November~20, 1844\hfill}
\vfill\eject
\noindent
{\largeit (Copy of a) Letter from Sir William R.\ Hamilton to
John T.\ Graves, Esq. on Quaternions}
\line{\hfill Observatory, October~17, 1843}
{\sc My dear Graves},---A very curious train of mathematical
speculation occurred to me yesterday, which I cannot but hope will
prove of interest to you. You know that I have long wished, and
I believe that you have felt the same desire, to possess a Theory
of Triplets, analogous to my published Theory of Couplets, and
also to Mr.~Warren's geometrical representation of imaginary
quantities. Now I think that I discovered\footnote{${}^1$}{The reader is
requested to pardon this expression for the reason mentioned in
another note.}
yesterday a {\it theory of quaternions} which includes such a
theory of {\it triplets}.\footnote{${}^2$}{Prof.\ De Morgan has most
obligingly given the writer a sketch of a Memoir on Triple
Algebra, which he has lately presented to the Cambridge
Philosophical Society, and which he is pleased to say was not
begun till after the publication of the first part of the paper on
Quaternions in this Magazine.}
My train of thoughts was of this kind. Since $\sqrt{-1}$ is in a
certain well-known sense, a line perpendicular to the line~1, it
seemed natural that there should be some other imaginary to
express a line perpendicular to the former; and because the
rotation from this to this also being doubled conducts to $-1$,
it ought also to be a square root of negative unity, though not
to be confounded with the former. Calling the old root, as the
Germans often do, $i$, and the new one $j$, I inquired what laws
ought to be assumed for multiplying together $a + ib + jc$ and
$x + iy + jz$. It was natural to assume that the product
$$= ax - by - cz + i(ay + bx) + j(az + cx) + ij(bz + cy);$$
but what are we to do with $ij$? Shall it be of the form
$\alpha + \beta i + \gamma j$? Its square would seem to be $=
1$, because $i^2 = j^2 = -1$; and this might tempt us to take
$ij = 1$ or $ij = -1$; but with neither assumption shall we have
the sum of the squares of the coefficients of $1$, $i$, and $j$
in the product $=$ to the product of the corresponding sums of
squares in the factors. Take the simplest case of a product,
namely, the case where it becomes a square; we have
$a^2 - b^2 - c^2 + 2iab + 2jac +2ijbc$
\dottriangle\
and
$(a^2 - b^2 - c^2)^2 + (2ab)^2 + (2ac)^2 = (a^2 + b^2 + c^2)^2$;
the condition respecting the {\it moduli} is \dottriangle\
fulfilled, if we suppress the term involving $ij$ altogether, and
what is more $a^2 - b^2 - c^2$, $2ab$, $2ac$ are precisely the
coordinates of the {\it square-point}, so to speak, deduced from
the point $a$, $b$, $c$, in space, by a slight extension of
Mr.~Warren's rule for points in a plane. [It is a long time
since I read this book, but the grand features of his view cannot
be forgotten.] In fact, if we double, in its own plane, the
rotation from the positive semiaxis of $x$ to the radius vector
of the point $a$, $b$, $c$, we attain the direction of the radius
vector drawn to $a^2 - b^2 - c^2$, $2ab$, $2ac$.
Behold me therefore tempted for a moment to fancy that $ij = 0$.
But this seemed odd and uncomfortable, and I perceived that the
same suppression of the term which was {\it de trop} might be
attained by assuming what seemed to me less harsh, namely that
$ji = - ij$. I made therefore $ij = k$, $ji = -k$, reserving to
myself to inquire whether $k$ was $=0$ or not. For this purpose,
I next multiplied $a + ib + jc$, as a multiplier, into
$x + ib + jc$, as a multiplicand, keeping still, as you see, the
two factor lines in one common plane with the unit line. The
result was
$$ax - b^2 - c^2 + i(a + x)b + j(a + x)c + k(bc - bc),$$
in which the coefficient of $k$ still vanishes; and
$ax - b^2 - c^2$, $(a + x)b$, $(a + x)c$ are easily found to be
the correct coordinates of the {\it product-point}, in the sense
that the rotation from the unit line to the radius vector of
$a$, $b$, $c$, being added in its own plane to the rotation from
the same unit-line to the radius vector of the other factor-point
$x$, $b$, $c$, conducts to the radius vector of the lately
mentioned product-point; and that this latter radius vector is in
length the product of the two former. Confirmation of $ij = -
ji$; but no information yet of the value of $k$. Try boldly then
the general product of two triplets, and seek whether the law of
the moduli is satisfied when we suppress the $k$ altogether. Is
$$(a^2 + b^2 + c^2) (x^2 + y^2 + z^2)
= (ax - by -cz)^2 + (ay + bx)^2 + (az + cx)^2 ?$$
No, the first member exceeds the second by $(bz - cy)^2$. But
this is just the square of the coefficient of $k$, in the
development of the product
$(a + ib + jc)(x + iy + jz)$, if we grant that $ij = k$,
$ji = -k$, as before. And here there dawned on me the notion
that we must admit, in some sense, {\it a fourth
dimension}\footnote{${}^3$}{The writer has this moment been informed
(in a letter from a friend) that in the Cambridge Mathematical
Journal for May last a paper on Analytical Geometry of $n$
dimensions has been published by Mr.~Cayley, but regrets that he
does not yet know how far Mr.~Cayley's views and his own may
resemble or differ from each other.}
of space for the purpose of calculating with triplets; or
transferring the paradox to algebra, must admit a {\it third}
distinct imaginary symbol $k$, not to be confounded with either
$i$ or $j$, but equal to the product of the first as multiplier,
and the second as multiplicand; and therefore was led to
introduce {\it quaternions}, such as $a + ib + jc + kd$, or
$(a,b,c,d)$.
I saw that we had probably $ik = -j$, because $ik = iij$ and
$i^2 = -1$; and that in like manner we might expect to find
$kj = ijj = -i$; from which I thought it likely that $ki = j$,
$jk = i$, because it seemed likely that if $ji = -ij$, we should
have also $kj = - jk$, $ik = -ki$. And since {\it the order of
multiplication of these imaginaries is not indifferent}, we
cannot infer that $k^2$, or $ijij$, is $+1$, because
$i^2 \times j^2 = -1 \times -1 =+1$. It is more likely that
$k^2 = ijij = - iijj = -1$. And in fact this last assumption is
necessary, if we would conform the multiplication of quaternions
to the law of the multiplication of moduli. For multiplying
$a + ib + jc + kd$ as multiplier into
$a' + i b' + j c' + k d'$ as multiplicand, and granting that
$k^2$ is real, we find for the real part of the product,
$aa' - bb' - xx' + k^2 dd'$; and for the coefficient of $k$,
$ad' + da' +$ terms arising from the multiplication of $i$ and $j$
together; in order then that $2a a' d d'$ may disappear in the
expression of the square of the modulus of the product, by being
added to $2k^2 aa' dd'$, it is necessary that we should have $k^2
= -1$.
My assumptions were now completed, namely,
$$\left\{ \eqalign{
i^2 &= j^2 = k^2 = -1; \cr
jk &= -kj = i; \cr}\quad
\eqalign{
ij &= -ji = k; \cr
ki &= -ik = j: \cr}
\right.
\leqno {\rm (A.)}$$
and with these I was obliged to assume that if
$$(a, b, c, d) (a', b', c', d') = (a'', b'', c'', d''),$$
then the four following equations of multiplication hold good:
$$\left\{ \eqalign{
a'' &= aa' - bb' - cc' - dd';\cr
b'' &= ab' + ba' + cd' - dc';\cr
c'' &= ac' + ca' + db' - bd';\cr
d'' &= ad' + da' + bc' - cb'.\cr}
\right.
\leqno {\rm (B.)}$$
But I considered it essential to try whether these equations were
consistent with the law of moduli, namely
$$a''^2 + b''^2 + c''^2 + d''^2
= (a^2 + b^2 + c^2 + d^2) (a'^2 + b'^2 + c'^2 + d'^2),$$
without which consistence being verified, I should have regarded
the whole speculation as a failure. Judge then of my pleasure
when, after a careful examination, I found that the whole
twenty-four products not involving squares in the development of
the sum of the squares of the four quadrinomials (B.)
destroyed each other; and that the sixteen products involving
squares were just those which arise otherwise from the
multiplication of $a^2 + b^2 + c^2 + d^2$ and
$a'^2 + b'^2 + c'^2 + d'^2$.\footnote{${}^4$}{Mr.~Graves has since
pointed out to the writer that this theorem of ordinary algebra is
not new, and has very elegantly extended it.}
We have, then, this first law for the {\it multiplication of two
quaternions} together, that the modulus of the product is equal
to the product of the moduli of the factors.
Division of quaternions is easy. The equations (B.) of
multiplication give
$$\left\{ \eqalign{
a' &= (a^2 + b^2 + c^2 + d^2)^{-1}
(aa'' + bb'' + cc'' + d d'');\cr
b' &= (a^2 + b^2 + c^2 + d^2)^{-1}
(-ba'' + ab'' + dc'' - cd'');\cr
c' &= (a^2 + b^2 + c^2 + d^2)^{-1}
(-ca'' + ac'' + bd'' - db'');\cr
d' &= (a^2 + b^2 + c^2 + d^2)^{-1}
(-da'' + ad'' + cb'' - bc'').\cr}
\right.
\leqno {\rm (C.)}$$
The modulus of the quotient is the quotient of the moduli. A
quaternion divided by itself gives for quotient
$(1,0,0,0) = 1$.
Addition and subtraction require no notice.
Making
$$\left\{ \eqalign{
a = \mu \cos\rho, \quad
b = \mu \sin \rho \cos \phi, \quad
c &= \mu \sin \rho \sin \phi \cos \psi, \cr
d &= \mu \sin \rho \sin \phi \sin \psi,\cr}
\right.
\leqno {\rm (D.)}$$
I would call $\rho$ the {\it amplitude} of the quaternion,
$\phi$ its {\it colatitude}; and $\psi$ its {\it longitude}.
The {\it modulus} is $\mu$. Representing the three coefficients
$b$, $c$, $d$ of the imaginary triplet $ib + jc + kd$, by the
rectangular coordinates of a point in space; $\mu \sin \rho$, as
being the radius vector of this point, might be called the
{\it radius}, or perhaps the {\it length}, of the quaternion.
We may speak of the {\it inclination} of one quaternion to
another, and its cosine is
$$\cos \phi \cos \phi'
+ \sin \phi \sin \phi' \cos (\psi' - \psi)
= {bb' + cc' + dd' \over \mu \mu' \sin \rho \sin \rho'}.$$
If with the amplitudes $\rho$, $\rho'$, $\rho''$ of any two
factors and their product, as sides, we construct a spherical
triangle, the angle opposite to the amplitude of the product will
be the supplement of the inclination of the factors to each
other; and the angle opposite to the amplitude of either factor
will be the inclination of the other factor to the
product.\footnote{${}^5$}{An improved form of this theorem of the
spherical triangle was communicated by the writer to the Royal
Irish Academy in November 1843, and has been published in the
Philosophical Magazine for July 1844.}
This {\it theorem of the spherical triangle}, combined with the
{\it law of the moduli}, requires only besides the following
{\it rule of rotation}, to decide on which side of the plane of
the two factor-lines the product-line is found, in order to
complete the geometrical construction for the equations
(B.) of multiplication of quaternions. In whichever
direction, to the right or to the left, the positive semiaxis of
$i$ must rotate round that of $k$ in order to approach to that of
$j$; in the same direction must the multiplicand line turn round
the multiplier-line in order to approach the product-line (or the
plane containing the product-line and the multiplier-line).
If the factor-lines coincide with each other in direction, the
product-line coincides with each of them; and its amplitude is
the sum of theirs; because the spherical triangle degenerates
into a single arc, two angles vanishing and the third becoming
equal to two right angles. If the amplitudes of the factors are
also supplementary to each other, the amplitude of the product is
$\pi$, and its sine vanishes; the product has therefore in this
{\it no length}, or radius, but becomes a {\it pure real}, and
at the same time a negative quantity; because
$\mu'' \sin \rho'' = 0$, $\mu'' \cos \rho'' = - \mu''$. The
factors are in this case of the forms
$$( \mu \cos \rho,\enspace
\mu \sin \rho \cos \phi,\enspace
\mu \sin \rho \sin \phi \cos \psi,\enspace
\mu \sin \rho \sin \phi \sin \psi ).$$
and
$$( -\mu' \cos \rho,\enspace
\mu' \sin \rho \cos \phi,\enspace
\mu' \sin \rho \sin \phi \cos \psi,\enspace
\mu' \sin \rho \sin \phi \sin \psi );$$
and their product is
$$(-\mu \mu',0,0,0) = - \mu \mu'.$$
Making the factors equal,
$\displaystyle \rho' = {\pi \over 2},\quad \mu' = \mu$,
we find that
$$(0,\enspace
\mu \cos \phi,\enspace
\mu \sin \phi \cos \psi,\enspace
\mu \sin \phi \sin \psi )^2 = - \mu^2.$$
The square of a pure imaginary is real and negative: the square
root of a real negative is a pure imaginary, having a determined
length, but a wholly undetermined direction. The square root of
$-1$ is any one of the infinitely many pure imaginaries, of the
form
$$\sqrt{-1} = i \cos \phi + j \sin \phi \cos \psi
+ k \sin \phi \sin \psi.$$
In general, however, the square root of a quaternion has only two
values, which differ only in sign: for if
$$(a'', b'', c'', d'') = (a, b, c, d)^2,$$
then
$$a'' = a^2 - b^2 - c^2 - d^2,\quad
b'' = 2 ab,\quad
c'' = 2 ac,\quad
d'' = 2 ad;$$
$$\sqrt{a''^2 + b''^2 + c''^2 + d''^2}
= a^2 + b^2 + c^2 + d^2,$$
$$a^2 = {\textstyle {1 \over 2}} (a''
+ \sqrt{a''^2 + b''^2 + c''^2 + d''^2});$$
and $a^2$ cannot vanish, except in the case of the pure negative
square, $b'' = c'' = d'' = 0$, $a'' < 0$.
Multiplication will be easy if we are familiar with the rules for
the product of two pure imaginaries. This product is, by
(B.),
$$(0, b, c, d) (0, b', c', d')
= ( - bb' - cc' - dd',\, cd' - dc',\, db' - bd',\, bc' - cb');$$
the product-line is perpendicular to the plane of the factors; its
length is the product of their lengths multiplied by the sine of
the angle between them: and the real part of the product, with
its sign changed, is the same product of the lengths of the
factors multiplied by the cosine of their inclination.
Finally we may always decompose the latter problem into these two
others; to multiply two pure imaginaries which agree in
direction, and to multiply two which are at right angles with
each other. In the first case the product is a pure negative,
equal to the product of the lengths or moduli with its sign
changed. In the second case, the product is a pure imaginary
of which the length is the product of the lengths of the factors,
and which is perpendicular to both of them. The distinction
between one such perpendicular and its opposite may be made by
the rule of rotation already stated.
There seems to me to be something analogous to {\it polarized
intensity} in the pure imaginary part; and to {\it unpolarized
energy} (indifferent to direction) in the real part of a
quaternion: and thus we have some slight glimpse of a future
Calculus of Polarities. This is certainly very
vague,\footnote{${}^6$}{A pen has been drawn across a clause or
two in the foregoing copy of this letter, as having reference
only to this guess of a future physical application. A
postscript has also been suppressed, as being comparatively
private. In other respects, it has been thought that no change
ought to be made, if the letter were to be printed at all; and
therefore the indulgence of the reader is requested for some
expressions which have the air of attaching to the subject an
importance greater than it may deserve.}
but I hope that most of what I have said above is clear and
mathematical. Hoping that you may be tempted to pursue the view
which has thus opened, I remain, with best wishes,
\vskip12pt
\line{\hfill Your sincere friend,}
\line{\hfill {\sc William R. Hamilton}}
\bye