Page:SahaElectrodynamics.djvu/13

two-fold application of the transformation-equations, we obtain

$t'=\phi (-v)\beta (-v)\left\{\tau +{\frac {v}{c^{2}}}\xi \right\}=\phi (v)\phi (-v)t$,

$x'=\phi (v)\beta (v)(\xi +v\tau )=\phi (v)\phi (-v)x$, etc.

Since the relations between (x', y', z', t'), and (x, y, z, t) do not contain time explicitly, therefore K and k' are relatively at rest.

It appears that the systems K and k' are identical.

$\therefore \phi (v)\ \phi (-v)=1$,

Let us now turn our attention to the part of the y-axis between ($\xi =0,\eta =0,\zeta =0$), and ($\xi =0,\eta =1,\zeta =0$). Let this piece of the y-axis be covered with a rod moving with the velocity v relative to the system K and perpendicular to its axis ;—the ends of the rod having therefore the co-ordinates

$\left.{\begin{array}{lll}x_{1}=vt,&y={\frac {l}{\phi (v)}},&z_{1}=0\\x_{2}=vt,&y_{2}={\frac {l}{\phi (v)}},&z_{2}=0\end{array}}\right\}$

Therefore the length of the rod measured in the system K is ${\frac {l}{\phi (v)}}$. For the system moving with velocity (-v), we have on grounds of symmetry,

${\frac {l}{\phi (v)}}={\frac {l}{\phi (-v)}}$

$\therefore \phi (v)=\phi (-v),\ \therefore \phi (v)=1$