Page:CunninghamPrinciple.djvu/9

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

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

and α' 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 ξ₁ ξ₂ their coordinates in B at any time τ,

$x_{2}-x_{1}=\alpha '\left(\xi _{2}-\xi _{1}\right)$

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

$a'^{2}\left(1-{\frac {v^{2}}{c^{2}}}\right)=1$, i. e. $\alpha '=\left(1-{\frac {v^{2}}{c^{2}}}\right)^{-{\frac {1}{2}}}$.

Thus the transformation is finally

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

Now let points not on the axis of x be considered. Since the axes of x and ξ coincide at all time, y and z always vanish when η and ζ vanish.

Hence $y=\lambda \eta$ and $z=\mu \zeta$, and λ and μ 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

$\eta =y/\lambda ,\ \phi (-v)={\frac {1}{\lambda }}$, and therefore λ=1.

Similarly μ=1.

The general transformation between x y z t and ξ η ζ τ is therefore

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