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,

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

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