# Page:Über die Möglichkeit einer elektromagnetischen Begründung der Mechanik.djvu/8

Naturally, the energy ${\displaystyle {\mathfrak {E}}}$ of the resting ellipsoid, may not contain the velocity ${\displaystyle v}$.

Thus, since ${\displaystyle e}$ is invariable, ${\displaystyle a}$ is variable:

 ${\displaystyle 2a={\frac {e^{2}\arcsin Av{\sqrt {1-A^{2}v^{2}}}}{Av{\mathfrak {E}}}},}$ ${\displaystyle E={\mathfrak {E}}{\frac {Av\left(1+{\frac {1}{2}}A^{2}v^{2}\right)}{{\sqrt {1-A^{2}v^{2}}}\arcsin Av}}}$

or by series expansion

 (7) ${\displaystyle E={\mathfrak {E}}\left(1+{\frac {2}{3}}A^{2}v^{2}+{\frac {16}{45}}A^{4}v^{4}\dots \right).}$

The energy increase caused by motion, is thus in first approximation

${\displaystyle {\frac {2}{3}}{\mathfrak {E}}A^{2}v^{2}={\frac {m}{2}}v^{2},}$

thus the inertial mass ${\displaystyle m={\tfrac {4}{3}}{\mathfrak {E}}A^{2}}$.

By that, the mass defined by inertia would only be constant at small velocities, and would increase with increasing velocity. Since inertia as well as gravitation emerging from the body, is proportional to the number of quanta composing the body, it follows, that the mass defined by inertia must be proportional to the one specified by gravitation. If a body with mass ${\displaystyle m={\tfrac {4}{3}}{\mathfrak {E}}A^{2}}$, is attracted by a body of mass ${\displaystyle M}$ up to a distance of ${\displaystyle r}$, then the electromagnetic energy store of gravitation is diminished by the amount ${\displaystyle \epsilon {\tfrac {4}{3}}{\mathfrak {E}}A^{2}M/r}$, where ${\displaystyle \epsilon }$ denotes the gravitational constant.

This energy is transformed into kinetic energy to produce the velocity ${\displaystyle v}$. Thus we have

${\displaystyle {\frac {2}{3}}{\mathfrak {E}}A^{2}v^{2}\left(1+{\frac {8}{15}}A^{2}v^{2}\dots \right)=\epsilon {\frac {{\frac {4}{3}}{\mathfrak {E}}A^{2}M}{r}},}$

or, since ${\displaystyle v=dr/dt}$

 (8) ${\displaystyle {\frac {1}{2}}\left({\frac {dr}{dt}}\right)^{2}={\frac {\epsilon M}{r}}\left(1-{\frac {8}{15}}A^{2}\left({\frac {dr}{dt}}\right)^{2}\right).}$

Instead of it, we can write

 (9) ${\displaystyle {\frac {1}{2}}\left({\frac {dr}{dt}}\right)^{2}\left(1+{\frac {16}{15}}A^{2}{\frac {\epsilon M}{r}}\right)={\frac {\epsilon M}{r}}.}$