Page:CunninghamPrinciple.djvu/8

Rh transformation must be of the form

Now &xi; will not in general be infinite unless x is infinite, and also when x and t are zero &xi; and &tau; are also zero. Hence the transformation must be of the simpler form

$\begin{array}{llll} \xi=a'x+b't & & \frac{x=b''\xi-b'\tau}{\triangle}\\ & \mathrm{or} & & \triangle=(a'b-b'a)\\ \tau=ax+bt & & t=\frac{-a''\xi+a'\tau}{\triangle}\end{array}$

the coefficients a', b', a", b" being functions of the relative velocity v.

Now if a point starts from A at time t=0 and travels with B, its coordinate &xi; is always zero by virtue of the relation x=vt.

Hence

$$b'=-a'v\,$$

i.e.

$$\xi=a'(x-vt)\,$$

Now consider what is involved in saying that if a point moves along the axis of x relative to A with the velocity v of light, it also moves with velocity c relative to B. If a point moves from the position x at time t to the position x+&delta;x at time t+&delta;t let the corresponding changes in &xi; and &tau; be &delta;&xi;, &delta;&tau;.

Then

$$\delta\xi=a'\delta x+b'\delta t,\,$$

$$\delta\tau=a\delta x+b\delta t\,$$

Hence

$$\frac{\delta\xi}{\delta\tau}=\frac{a'\delta x+b'\delta t}{a\delta x+b\delta t}$$.

Hence if the point has velocity n in A's system of coordinates and $$\nu$$ in that of B

$$\nu=\frac{a'n+b'}{an+b}$$

In particular if $$n=\pm c$$, $$\nu=\pm c$$,

$$ac^{2}\pm c(b-a')-b'=0$$,

so that

$$b=a'\,$$ and $$a=\frac{b'}{c^{2}}=-\frac{va'}{c^{2}}$$.