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Chapter IV.
Differentiation of Quaternions.

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 differentiation of a quaternion function of any number of scalar independent variables. In fact, this process is identical with ordinary differentiation.

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 r = F(q) be a function of a quaternion q,

dr = dFq = L∞n {F(q + dq/n) − F(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

f(q, dq),
p.95 where f is a new function, depending on the form of F ; homogeneous and of the first degree in dq ; but not, in general, capable of being put in the form
f(q) dq.

138. To make more clear these last remarks, we may observe that the function

f(q, dq),
thus derived as the differential of F(q), is distributive with respect to dq. That is
f(q, r + s) = f(q, r) + f(q, s),
r and s being any quaternions.

For

f(q, r + s) = L∞n {F(q + (r + s)/n) − F(q)}
= L∞n {F(q + r/n + s/n) − F(q + s/n)} + F(q + s/n) − Fq}
= L∞n {F(q + s/n + r/n) − F(q + s/n)} + L∞n {F(q + s/n) − Fq}
= f(q, r) + f(q, s).
And, as a particular case, it is obvious that if x be any scalar,
f(q, xr) = xf(q, r).

139. And if we define in the same way

dF(q, r, s ......)
as being the value of
= L∞n {F( q + dq/n, r + dr/n, s + ds/n,......) − F(q, r, s,......)},
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 f 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 quaternions, each factor is to be differentiated as if it alone were variable ; and the terms corresponding to these are to be added for the complete differential. This differs from the ordinary process of scalar differentiation solely in the fact that, on account of the non-commutative property of quaternion multiplication, each factor must in general be differentiated in situ. Thus
d(qr) = dq . r + qdr, but not generally = rdq + qdr.
p.96

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 transformations.

(1) (Tρ)2 = − ρ2.

The differential of the left-hand side is simply, since Tρ is a scalar,

2Tρ dTρ.

That of ρ2 is

L∞n {(ρ + dρ/n)2 − ρ2}
= L∞n ((2/n)Sρdρ + (dρ)2/n2) (§ 104)
= 2Sρdρ.

Hence TρdTρ = −Sρdρ,

or dTρ = −S . Uρdρ = Sdρ/Uρ,
or dTρ/Tρ = Sdρ/ρ.
(2) Again, ρ = Tρ Uρ
dρ = dTρ . Uρ + TρdUρ,
whence dρ/ρ = dTρ/Tρ + dUρ/Uρ
= S dρ/ρ + dUρ/Uρ by (1).
Hence dUρ/Uρ = Vdρ/ρ.
This may be transformed into Vdρ . ρ/ρ2 or Vρdρ/Tρ2, &c.
(3) (Tq)2 = qKq
2TqdTq = d(qKq) = L∞n[ (q + dq/n)K(q + dq/n) − qKq],
= L∞n ( (qKdq + dqKq)/n + (1/n2)dqKdq),
= qKdq + dqKq,
= qKdq + K(qKdq) (§ 55),
= 2S . qKdq = 2S . dqKq.
p.97
Hence dTq = S . dqUKq = S . dqUq−1 = TqS(dq/q) ,
since Tq = TKq,   and   UKq = Uq−1.

[If q = ρ, a vector, Kq = Kρ = −ρ, and the formula becomes

dTρ = −S . Uρdρ, as in (1).]

Again,

dq = TqdUq + UqdTq,
which gives dq/q = dTq/Tq + dUq/Uq ;
whence, as S(dq/q) = dTq/Tq,
we have V(dq/q) = dUq/Uq.
(4) d(q2) = L∞n{(q + dq/n)2 − q2}
= qdq + dq . q
= 2S . qdq + 2 Sq . Vdq + 2Sdq . Vq.

If q be a vector, as ρ, Sq and Sdq vanish, and we have

d(ρ2) = 2Sρdρ, as in (1).
(5) Let q = r1/2.
This gives dr1/2 = dq. But
dr = d(q2) = qdq + dq . q.
This, multiplied by q and into Kq, gives the two equations
qdr = q2dq + qdq . q,
and drKq = dqTq2 + qdq . Kq.
Adding, we have
qdr + dr . Kq = (q2 + Tq2 + 2Sq .q)dq = 4Sq . qdq ;
whence dq, i.e. dr1/2 , 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) qq−1 = 1,
qdq−1 + dq . q−1 = 0
∴ dq−1 = −q−1dq . q−1.

If q is a vector, = ρ suppose,

dq−1 = dρ−1 = −ρ−1dρ . ρ−1
T. Q. I. 7 p.98 = dρ/ρ2 −(2/ρ)S(dρ/ρ)
= (dρ/ρ −2Sdρ/ρ)/ρ
= − K(dρ/ρ)/ρ.
(7) q = Sq + Vq,
dq = dSq + dVq.
But dq = Sdq + Vdq.

Comparing, we have

dSq = Sdq,   dVq = Vdq.
Since Kq = Sq − Vq, we find by a similar process
dKq = Kdq.
(8) In the expression qaq−1 where a is any constant quaternion, q may be regarded as a mere versor, so that
(Tq)2 = 1 = qKq = qq−1.
Thus S . dqKq = 0 ;
and hence dqq−1 = −qdq−1,
as well as
q−1dq = −dq−1q,
are vectors. But, if a = a + α, where a is a scalar, qaq−1 = a, i.e. constant, so that we are concerned only with d(qαq−1).

Hence d(qαq −1) = dqαq−1 − qαq−1dqq−1,

= dqq−1 . qαq−1 −qαq−1 . dqq−1,
= 2V . dqq−1qαq−1 = −2V . qdq−1qαq−1.
(9) With the restriction in (8) above we may write
q = cos u + θ sin u,
where Tθ = 1;   Sθdθ = 0.
Hence q−1 = cos u − θ sin u ;
−q−1dq = dq−1q = {− (sin u + θ cos u) du − dθ sin u} (cos u + θ sin u)
= −θ du − dθ sin u (cos u + θ sin u) ;
−qdq−1 = dqq−1 = θdu + dθ sin u (cos u − θ sin u).

Both forms are represented as linear functions of the rectangular system of vectors

θ, dθ, θdθ.
p.99

If the plane of q be fixed, θ is a constant unit vector, and

dqq−1 = −dq−1q = θdu.
(10) The equation (belonging to a family of spheres)
T( (ρ + α)/(ρ − α) ) = e
gives Sdρ {(ρ + α) − e2(ρ − α)} = 0;
or, by elimination of e,
Sdρ {(ρ + α)−1 − (ρ − α)−1} = 0;
whose geometrical interpretation gives a well-known theorem.

If we confine our attention to a plane section through the vector α, viz.

S . γαρ = 0, S . γαdρ = 0;   or   S . βρ = 0, S . βdρ = 0,
where β||Vγα||Vαρ ;
we have
dρ||V.β{(ρ + α)−1 − (ρ − α)−1} or V.dρVβ{(ρ + α)−1 − (ρ − α)−1} = 0.
(11) Again, from
SU (ρ + α)/(ρ − α) = e
(which is the equation of the family of tores produced by the rotation of a group of circles about their common chord) we have
SU . (ρ + α)(ρ − α) = −e.
Also this gives   VU . (ρ + α)(ρ − α) = β = √(1 − e2) U . Vαρ.

We obtain from the first of these, by differentiation,

S {(V(dρ/(ρ + α)) . U(ρ + α) U(ρ − α) + U(ρ + α) V(dρ/(ρ − α)) U(ρ − α)} = 0.
or S . β {(ρ + α)−1 − (ρ − α)−1} = 0.

If we consider β 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 β is perpendicular to the plane of α, ρ (and, of course, dρ),

V . dρ {(ρ + α)−1 − (ρ − α)−1} = 0.
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.

7—2
p.100
(12) As a final instance, take the equation
ρ−1 ρ′ρ = α,
where ρ′ stands for dρ/ds, s being the arc of a curve.

By § 38, α is a unit vector, and the expression shews by its form that it belongs to a plane curve. Let β be a vector in its plane, and perpendicular to α. Operate by S . β and we get

(2Sβρ/ρ2) Sρρ′ − Sβρ′ = 0,
whose integral is
ρ2 − Sβρ = 0,
the tensor of β being the constant of integration.

141. Successive differentiation of course presents no new difficulty.

Thus, we have seen that

d(q2) = dq . q + qdq.

Differentiating again, we have

d2(q2) = d2q . q + 2(dq)2 + qd2q,
and so on for higher orders.

If q be a vector, as ρ, we have, § 140 (1),

d(ρ2) = 2Sρdρ.
Hence d2(ρ2) = 2 (dρ)2 + 2 Sρd2ρ, and so on.
Similarly d2Uρ = − d( (Uρ/Tρ2)Vρdρ).

But d(1/Tρ2) = − d(2dTρ/Tρ3) = d(2Sρdρ/Tρ4)

and d . Vρdρ = V . ρd2 ρ.
Hence d2Uρ = (Uρ/Tρ4)(Vρdρ)2
− (UρVρd2ρ)/(Tρ2) − (2UρVρdρSρdρ)/(Tρ4)
= (Uρ/Tρ4) {(Vρdρ)2 + ρ2Vρd2ρ − 2VρdρSρdρ }*.
* 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
(Uρ/Tρ4)V . ρ {(dρVρdρ + d2ρ . ρ2 − 2 dρSρdρ}.

142. If the first differential of q be considered as a constant quaternion, we have, of course,

d2q = 0,   d3q = 0, &c.
and the preceding formulæ become considerably simplified. p.101

Hamilton has shewn that in this case Taylor’s Theorem admits of an easy extension to quaternions. That is, we may write

f(q + xdq) = f(q) + xdf (q) + (x2 / 1.2) d2f(q) + ......
if d2q = 0 ; subject, of course, to particular exceptions and limitations as in the ordinary applications to functions of scalar variables. Thus, let
f(q) = q3 , and we have
df(q) = q2dq + qdq . q + dq . q2,
d2f(q) = 2dq . qdq + 2q(dq)2 + 2 (dq)2q,
d3f(q) = 6 (dq)3,
and it is easy to verify by multiplication that we have rigorously
(q + xdq)3 = q3 + x (q2dq + qdq . q + dq.q2) +
x2(dq . qdq + q (dq)2 + (dq)2q) + x3 (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 ρ, the vector of any point in space.

Let

F(ρ) = C,
where F is a scalar function and C an arbitrary constant, be the equation of a series of surfaces. Its differential,
f(ρ, dρ) = 0,
is, of course, a scalar function : and, being homogeneous and linear in dρ, § 137, may be thus written,
Sνdρ = 0,
where ν is a vector, in general a function of ρ.

This vector, ν, is shewn, by the last written equation, to have the direction of the normal to the given surface at the extremity of ρ. It is, in fact, perpendicular to every tangent line dρ ; §§ 36, 98.

p.102

145. This leads us directly to one of the most remarkable operators peculiar to the Quaternion Calculus ; viz.

  ∇ = i d/dx + j d/dy + k d/dz
 
 (1),
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 x, y, z are Cartesian co-ordinates referred to these as axes, it is implied in (1) that ∇ is an Invariant. This will presently be justified. Meanwhile it is easy to see that if ρ be the vector of any point in space, so that
ρ = ix + jy + kz,
  ∇ρ = −3
 
 (2),
  ∇Tρ = ∇√(x2 + y2 + z2) = (ix + jy + kz) /√(x2 + y2 + z2) = ρ/Tρ = Uρ
 
 (3),
  ∇Tρn = n(Tρ)n − 1 ∇Tρ = n(Tρ)n − 2 ρ
 
 (4),
of which the most important case is ∇(1/Tρ) = − ρ/(Tρ)3 = − ρ/(Tρ2)
 
 (5).
A second application gives ∇2(1/Tρ) = − ∇ρ/(Tρ)3 − ∇(1/(Tρ3)) . ρ
 
 (6).

Again

∇ρ = −3 = ∇Tρ . Uρ + Tρ . ∇Uρ = −1 + Tρ . ∇Uρ,
so that ∇ Uρ = −2/Tρ
 
 (7).

By the definition (1) we see that

  ∇2 = − {(d/dx)2 + (d/dy)2 + (d/dz)2}
 
 (8),
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), α being any constant vector,

Sα∇ . ρ − ∇Sαρ = V . αV∇ρ = 0,
from which
∇Vαρ + Vα∇ . ρ = (Sα∇ . ρ − αS∇ρ) + (αS∇ρ − ∇Sαρ ) = 0.
p.103

[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 ∇ is always to the left of (though not necessarily contiguous to) its subject, ρ.]

146. By the help of these elementary results, of which (3) and (7) are specially noteworthy, we easily find the effect of ∇ upon more complex functions.

For instance, taking different modes of operating, we have with

α = ia +jb + kc
  Sα∇ . ρ = ∇Sαρ = −∇(ax + by + cz) = − (ia + jb + kc) = −α
 
 (1),
or thus ∇Sαρ = iSαi + jSαj + kSαk = −α ;
while −Vα∇ . ρ = ∇Vαρ = −∇Vρα = −∇(ρα − Sρα)
  = 3α − α = 2α
 
 (2),
∇Vαρ = iVαi + jVαj + kVαk = 2α.

From the latter of these we have

∇(Vαρ/Tρ3) = (2α/Tρ3) − (3ρVαρ/Tρ5) = − (2αρ2 − 3ρVαρ)/Tρ5)
  = (αρ2 − 3ρSαρ)/(Tρ5)
 
 (3),
[where note that the first of these values is obtained thus,
∇(Vαρ/Tρ3) = (∇Vαρ/Tρ3) + ∇(1/Tρ3) . Vαρ
The order is of vital importance.]

This, in its turn, gives

  S . δρ∇(Vαρ/Tρ3) = −Sαδρ/(Tρ3) − 3SαρSρδρ/(Tρ5) = −δSαρ/(Tρ3)
 
 (4).
where δ 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 form
  S . δρ∇(Vαρ/Tρ3) = δSα∇ . (1/Tρ) = Sα∇ . δ(1/Tρ)
 
 (4′).
And it is easy to see that S may be substituted for V in the left-hand member. [The reason for this may be traced in the result of § 145(6).]

As an addition to these examples, note that (by (2) of § 148, below)

S δρ∇ . (Vαρ/Tρ3) = − δ(Vαρ/Tρ3),
p.104 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

  ∇ΣαSβρ = −Σβα
 
 (5),
a result which will be found useful in next Chapter.

147. Still more important are the results obtained from the operator ∇ when it is applied to

  σ = iξ + jη + kζ
 
 (1),
and functions of this vector. (Here , ξ, η, ζ are functions of x, y, z, so that σ is a vector whose value is definite for each point of space.)

We have at once

∇ρ = −(dξ/dx + dη/dy + dζ/dz) + i(dζ/dy − dη/dz)
  + j(dξ/dz − dζ/dx) + k(dη/dx − dξ/dy)
 
 (2).

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 ρ, it is clear that

  S∇σ = −(dξ/dx + dη/dy + dζ/dz)
 
 (3),
represents the consequent condensation of the group of points (say particles of a fluid) originally in the neighbourhood of ρ, while
  V∇σ = i(dζ/dy − dη/dz) + j(dξ/dz − dζ/dx) + k(dη/dx − dξ/dy)
 
 (4),
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 applications. They are introduced here in their semi-Cartesian form merely to shew the importance of the operator ∇.

148. Let us recur to the equation of § 144, viz.

  F(ρ) = C
 
 (1).
Ordinary complete differentiation gives
dF = (dF/dx) dx + (dF/dy) dy + (dF/dz) dz,
p.105 or, what is obviously the same,
  dF = −Sdρ∇F
 
 (2),
which we may write, if we please, as
−Sdρ∇ . F.

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

−Sdρ∇
is equivalent to total differentiation as involved in the passage from ρ to ρ + dρ. Hence, of course, as in § 144
dF(ρ) = 0 = Sνdρ = −Sdρ∇F,
and thus (as dρ may have any of an infinite number of values)
  ν = −∇F
 
 (3).

If we pass from one surface of the series (1) to a consecutive one by the vector δρ, we have

−Sνδρ = δC,
Hence −ν−1δC is a value of δρ ; so that the tensor of ν 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, ν, as given by (3), represents the vector force at the point ρ ; if (1) be a set of isothermals, ν (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, α, and adding three such expressions together. Thus if

σ = αF + βF1 + γF2,
we obtain at once
  dσ = −S(dρ∇)σ = −Sdρ∇ . σ
 
 (4).
But here the brackets, or the point, should (at first, at least) be employed ; otherwise we might confound the expression with
dσ = −S . dρ∇σ
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 = −Sdρ∇ . q
 
 (5).
p.106

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 S(dρ∇) are employed. 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.

Examples to Chapter IV.

1. Shew that
(a) d . SUq = S . UqV(dq/q) = −S(dq/qUVq)TVUq,
(b) d . VUq = V . Uq−1V(dq . q−1),
(c) d . TVUq = S(dUq/UVq) = S(dq/qUVq)SUq,
(d) d . αx = (π/2)αx + 1dx,
(e) d2 . Tq = {S2 . dqq−1 −S . (dqq−1)2}Tq = −TqV2(dq/q).
2. If Fρ = Σ . SαρSβρ + (1/2)gρ2
give dFρ = Sνdρ,
shew that ν = ΣV . αρβ + (g + ΣSαβ)ρ.
3. Find the maximum and minimum values of Tρ, when
(a) (ρ − α)2 = −α2 ;
(b)  (ρ − α)2 = −α2, }
Sβρ = 0;
(c) ρ2 − SαρSβρ = −α2 ;
(d)  ρ2 − SαρSβρ = −α2 , }
Sγρ = 0 ;

Point out the differences, geometrical and analytical, between (a), (c) on the one hand, and (b), (d) on the other.

State each of the problems in words.

p.107 4. With ∇ as in § 145, shew that
S∇ (qαq−1) = 2S . ∇q αq−1 = 2S . αq−1∇q.
V∇ (qαq−1) = 2qαq−1S . ∇qq−1 − 2S(qαq−1∇)q . q−1
where q is a function of ρ, and α any constant vector.
5. Shew that, if α, β, γ be a constant rectangular unit-vector system,
qαq−1d . qαq−1 + qβq−1d . qβq−1 + qγq−1d . qβq−1 = 4 dqq−1.
6. Integrate the differential equations :—
(a) ρ˙ + αSβρ = γ,
(b) q˙ + αq = b,
(c) θ˙ = Vαθ. Proc. R. S. E. 1870.
7. Shew that
(a) ∫VαdρSβρ = (1/2)V . α (ρSβρ + V . β∫Vρdρ).
(b) ∫dρVαρ = (1/2)(ρVαρ − V . α∫Vρdρ) + S . α∫Vρdρ.
(c) ∫V . VαdρVβρ = (1/2)(αS . β∫Vρdρ + βS . α∫Vρdρ − ρS . αβρ),
(d) ∫S . VαdρVβρ = (1/2)(SαρSβρ − ρ2Sαβ − S . αβ∫Vρdρ).

When these integrals are taken round a closed plane curve we have

∫Vρdρ = 2Aγ,
where A is the area, and γ a unit vector perpendicular to its plane. In this case
∫dρVαρ = AVγα + 2ASγα,
∫V . VαdρVβρ = A(αSβγ + βSαγ),
∫S . VαdρVβρ = AS . αβγ.
8. State, in words, the geometrical theorems involved in the equations of § 140, (10), (11), (12).
9. Shew (by means of § 91) that
∇1S . σ1∇σ = S . ∇σ∇1 . σ1,
where ∇, ∇1, operate respectively on σ, σ1 ; but, after the operations are performed, we put
∇1 = ∇, σ1 = σ.