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136.) To differentiate a function of a function of a quaternion we proceed as with scalar variables, attending to the peculiarities already pointed out.

137.] A case of considerable importance in geometrical applications of quaternions is the differentiation of a scalar function of 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, $ 130, may be thus written,

Svdp = 0, where v is a vector, in general a function of p.

This vector, v, is easily seen to have the direction of the normal to the given surface at the extremity of p; being, in fact, perpendicular to every tangent line dp, $$ 36, 98. Its length, when F is a surface of the second degree, is as the reciprocal of the distance of the tangent-plane from the origin. And we will shew, later, that if

prix + jy + kz, then

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dz

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EXAMPLES TO CHAPTER IV.

dq

(a.) d.SUq = S.Ugv dg

UVq

1. Shew that

dg

S. 9

TVU,

qUVO (6.) d.VUq=V.Uq-1V (dq.q-1),

dUq

dq (c.) d.TVUq=S

S SUq,

qura
(d.) d.a" = a@+idx,

2
(e.) do.Tq = {82.dqq-2–8.(d99-1)2}Tq=-Tqvadq.

9
2. If
Fp= 2.Sapsßp+ 3gp?

give dFp= Sydp,
shew that ν= ΣΥ.αρβ+(9+ Σ Καβ)ρ.

T

CHAPTER V.

THE SOLUTION OF EQUATIONS OF THE FIRST DEGREE.

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138.] WE have seen that the differentiation of any function whatever of a quaternion, q, leads to an equation of the form

dr = f (q, dq), where f 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 f is easily seen to be

dr =f(q, dq) = Vadqb+S.cdq, where a, b, and c may be any quaternion functions of q whatever.

. Every possible term of a linear and homogeneous function is reducible to this form, as the reader may easily see by writing down all the forms he can devise. Taking the scalars of both sides, we have

S.cdq = SdqSc + S.V

dqVc. But we have also, by taking the vector parts,

Vdr = EV.adyb = Sdq. Tab+ EV.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.BVdq, by the help of formulæ like those in $$ 90, 91. Solving this, we have Vdq, and Sdq is then found from the preceding equation.

Sdr =

=

=

or

139.] The problem may now be stated thus. Find the value of p from the equation

P

aspp + a,s3.pt ... = 2.aspp =y, where a, b, a,, B1, ... y are given vectors. (It will be shewn later

[ that the most general form requires but three terms, i. e. six vector constants a, b, a,, B1, 22, B, in all.] If we write, with Hamilton,

φρ = Σ.α.βρ, the given equation may be written

φρ =γ,

P = 4-ly, and the object of our investigation is to find the value of the inverse function 0-1.

140.] We have seen that any vector whatever may be expressed in terms of any three non-coplanar vectors. . Hence, we should ex

, pect à priori that a vector such as poop, or $3p, for instance, should be capable of expression in terms of p, pp, and pap. [This is, of course, on the supposition that p, bp, and $p are not generally coplanar. But it may easily be seen to extend to this case also. For if these vectors be generally coplanar, so are op, pop, and 0%p, since they may be written o, pr, and 0%. And thus, of course, 0%p can be expressed as above. If in a particular case, we should have, for some definite vector p, op=gp where g is a scalar, we shall obviously have pap=gʻp and 0%p=gop, 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, op, op are coplanar. For then we have an equation of the form

$ap = Ap + Bør, which gives

φρ = Αφρ + Βφορ

ABP+(A + B2)op.
So that o'p is in the same plane.]
If, then, we write
-0%p & Xp+yop +204p, .

(1) it is evident that x, y, z are quantities independent of the vector P, and we can determine them at once by processes such as those in $$ 91, 92.

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

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the right side of the equation, and the solution is complete. For by putting 0-1p for p and transposing, the equation becomes

P

-«p-1 = yp + zop+p; that is, the unknown inverse function is expressed in terms of direct operations. If x vanish, while y remains finite, we substitute 0-2p for p, and have

-yg-p= zp+op, and if x and y both vanish

- 20-1p=p. 141.] To illustrate this process by a simple example we shall take the very important case in which o belongs to a central surface of the second order; suppose an ellipsoid ; in which case it will be shewn (in Chap. VIII.) that we may write

øp = -a-isip-62jSjp-cokSkp. Here we have

фі = а2 і,
=
$2i = a^i,

031 = à®i,
øj = 62;, фj = 1+ј, $3j = 66j,

pk = cak, pok=c4k, pk = ck. Hence, putting separately i, j, k for p in the equation (1) of last section, we have

-a6 = x+ya+ zao,
-76 = x+y62 +264,

-C6 = + y c2 + zc4.
Hence ao, 82, c2 are the roots of the cubic

$3 +262 + y $ + x = 0, which involves the conditions

z=-(a2 +62 +ca),
y=a2b2 +62c2 + c a?,

x = -a262ca.
Thus, with the above value of $, we have

0%p = a62c4p-(a_62 + b2c2 +c%a2) pp+ (a+62 +c2)%p. 142.] Putting p-10 in place of p (which is any vector whatever)

$ and changing the order of the terms, we have the desired inversion of the function in the form

a%b2c0-10 (a262 +6202 +c4a4) 0–(a2 +62 +ca) po +020, where the inverse function is expressed in terms of the direct function. For this particular case the solution we have given is complete, and satisfactory; and it has the advantage of preparing the reader to expect a similar form of solution in more complex cases.

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V.pop

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143.] It may also be useful as a preparation for what follows, if we put the equation of $ 141 in the form 0 = 0 (0) = 0%p-(a? +62 +c%) o'p +(a2b2 + b2c2 + c2a2) op-a62c2p

{$3–(a + b +62) $+ (a2b2 + 62c2 +ca) $-a-62c2} p {(0-a?)(0–62)(°-c)} p.

(2) This last transformation is permitted because o is commutative with scalars like a?, i.e. Q(a_p) = a-op. Here we remark that (by § 140) the equation

(.ρφρ = 0,
, or op

= 9P where 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 a2, 62, c.

144.] 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

φρ = ΣΥ.qρη, 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.o where o is any vector whatever, Soop = ESOT.qpr = ESp.roq = Spo'o,

(3) if we agree to write φ'σ = ΣΥ.rση, and remember the proposition of $ 88. The functions and $ are thus conjugate to one another, and on this property the whole investigation depends. 145.] Let , u be any two vectors, such that M

,

φρ = Υλμ. Operating by S.A and Sou we have

Sa p = 0, Sμφρ = 0.
But, introducing the conjugate function $', these become

Spør = 0, Spo'u = 0, and give p in the form mp=1$'10'u, where m is a scalar which, as we shall presently see, is independent of , H, and p.

M
But our original assumption gives

ρ = φ'Υλμ; hence we have -Υλμ = Vφλφ'μ,

(4) and the problem of inverting is solved.

=0.

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