Page:SahaElectrodynamics.djvu/31

 the electron is at the origin of co-ordinates, and moves with the velocity v along the X-axis of the system. It is clear that at this moment ($$t=0$$) the electron is at rest relative to the system k, which moves parallel to the X-axis with the constant velocity v.

From the suppositions made above, in combination with the principle of relativity, it is clear that regarded from the system k, the electron moves according to the equations

$$m\frac{d^{2}\xi}{dt^{2}}=eX',\ m\frac{d^{2}\eta}{dt^{2}}=eY',\ m\frac{d^{2}\zeta}{dt^{2}}=eZ'$$,

in the time immediately following the moment, where the symbols ($$\xi, \eta, \zeta, \tau, X', Y', Z'$$) refer to the system k. If we now fix, that for $$t=v=y=z=0$$, $$\tau = \xi = \eta = \zeta=0$$, then the equations of transformation given in § 3 (and 6) hold, and we have :

$$\left. \begin{array}{c}\tau=\beta\left(t-\frac{v}{c^{2}}x\right),\ \xi=\beta(x-vt),\ \eta=y,\ \zeta=z,\\ \\X'=X,\ Y'=\beta\left(Y-\frac{v}{c}N\right),\ Z'=\beta\left(Z+\frac{v}{c}M\right) \end{array} \right\}$$

With the aid of these equations, we can transform the above equations of motion from the system k to the system K, and obtain :—

{{MathForm1|(A)|$$\left. \begin{array}{c} \frac{d^{2}x}{dt^{2}}=\frac{e}{m}\frac{1}{\beta^{3}}X,\ \frac{d^{2}y}{dt^{2}}=\frac{e}{m}\frac{1}{\beta}\left(Y-\frac{v}{c}N\right),\ \\ \\\frac{d^{2}z}{dt^{2}}=\frac{e}{m}\frac{1}{\beta}\left(Z+\frac{v}{c}M\right)\end{array} \right\}$$}}