Page:Das Prinzip der Relativität und die Grundgleichungen der Mechanik.djvu/4

in this frame the equations of motion in the simple form 2) apply to it, and for the moving force the product of the electric charge e and the electric field strength has to be used. Now we transform the equations of motion to a second reference frame whose x-axis again coincides with the direction of velocity q, but which is now at rest in frame (x, y, z, t). For this we use on the one hand as regards the acceleration components, the relations 1), on the other hand as regards to the force components, the relations 4) by setting everywhere q in place of v. Finally, by a simple rotation of the coordinate axes we can enter the system (x, y, z, t), and by performing all these elementary calculations, we obtain the equations of motion in the form:

${\begin{cases}{\frac {m{\ddot {x}}}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}}}=e{\mathfrak {E}}_{x}-{\frac {e{\dot {x}}}{c^{2}}}({\dot {x}}{\mathfrak {E}}_{x}+{\dot {y}}{\mathfrak {E}}_{y}+{\dot {z}}{\mathfrak {E}}_{z})\\\\+{\frac {e}{c}}\left({\dot {y}}{\mathfrak {H}}_{z}-{\dot {z}}{\mathfrak {H}}_{y}\right)\ {\mathsf {etc}}.\end{cases}}$

5)

From the general admissibility of these three equations one can convince himself afterwards directly by the consideration that the equations have to remain true as regards the principle of relativity, if the primed quantities instead of the unprimed quantities are written throughout, and the constants c, e and m remain the same. In fact, this is confirmed in a very general way in consequence of the result of relations 1) and 4), for any value of v.

Now we want to bring the equations of motion to its simplest form. If we multiply them respectively by ${\dot {x}},{\dot {y}},{\dot {z}}$ and sum up, then it follows:

$e({\dot {x}}{\mathfrak {E}}_{x}+{\dot {y}}{\mathfrak {E}}_{y}+{\dot {z}}{\mathfrak {E}}_{z})={\frac {m({\dot {x}}{\ddot {x}}+{\dot {y}}{\ddot {y}}+{\dot {z}}{\ddot {z}})}{\left(1-{\frac {q^{2}}{c^{2}}}\right)^{\frac {3}{2}}}},$

and this results substituted in 5) gives, if one also puts:

$e{\mathfrak {E}}_{x}+{\frac {e}{c}}({\dot {y}}{\mathfrak {H}}_{z}-{\dot {z}}{\mathfrak {H}}_{y})=X,\ {\mathsf {etc}}.$ ${\frac {d}{dt}}\left\{{\frac {m{\dot {x}}}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}}}\right\}=X,\ {\mathsf {etc}}.$

6)