Page:CunninghamPrinciple.djvu/8

transformation must be of the form

$\xi ={\frac {a'x+b't+c'}{ax+bt+c}}$, $\tau ={\frac {a''x+b''t+c''}{ax+bt+c}}$,

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

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

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 ξ 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+δx at time t+δt let the corresponding changes in ξ and τ be δξ, δτ.

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

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

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

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

so that

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