Page:CunninghamPrinciple.djvu/9

546 Thus

$$\begin{array}{l} \xi=a'(x-vt),\\ \\\tau=a'\left\{ -\frac{vx}{c^{2}}+t\right\} .\end{array}$$.

If this transformation be reversed we have

$\begin{array}{l} x=\alpha'(\xi+v\tau)\\ \\t=\alpha'\left(+\frac{v\xi}{c^{2}}+\tau\right)\ \mathrm{where}\ \alpha'=\frac{1}{a'\left(1-\frac{v^{2}}{c^{2}}\right)}\end{array}$|undefined

and &alpha;' will be the same Function of (-v) that a' is of v.

But the transformation shows that if x1 x2 be two points fixed relative to A and &xi;1 &xi;2 their coordinates in B at any time &tau;,

i. e. a line of length l as seen by A appears to be of length $$\tfrac{l}{\alpha'}$$, as seen by B moving relatively to it. But this will be the same whichever be the direction of B's motion along the axis of x, so that if $$\alpha'=f(v),\ f(v)=f(-v)$$, i.e. $$a'=\alpha'$$.

Hence

Thus the transformation is finally

$\begin{array}{l} \xi=\beta(x-vt),\\ \\\tau=\beta\left(-\frac{vx}{c^{2}}-t\right)\ \mathrm{where}\ \beta=\left(1-\frac{v^{2}}{c^{2}}\right)^{-\frac{1}{2}}\end{array}$|undefined

Now let points not on the axis of x be considered. Since the axes of x and &xi; coincide at all time, y and z always vanish when &eta; and &zeta; vanish.

Hence $$y=\lambda\eta$$ and $$z=\mu\zeta$$, and &lambda; and &mu; will not change if the velocity of motion of B be changed from v to -v ; thus if $$\lambda=\phi(v)$$; $$\phi(v)=\phi(-v)$$.

But since by reversing the transformation

Similarly &mu;=1.

The general transformation between x y z t and &xi; &eta; &zeta; &tau; is therefore

$$\begin{array}{c} y=\eta,\ z=\zeta,\\ \\\xi=\beta(x-vt)+c_{1}y+d_{1}z\,\\ \\\tau=\beta\left(-\frac{vx}{c^{2}}+t\right)+c_{2}y+d_{2}z.\end{array}$$.