# Page:Über die Möglichkeit einer elektromagnetischen Begründung der Mechanik.djvu/11

These expressions satisfy Maxwell's equations, when

${\displaystyle {\frac {d}{dt}}=-v{\frac {\partial }{\partial x}}}$

${\displaystyle \left(1-A^{2}v^{2}\right){\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}=0.}$

But if ${\displaystyle v}$ is depending on ${\displaystyle t}$, we have

${\displaystyle {\frac {d}{dt}}={\frac {\partial }{\partial t}}-v{\frac {\partial }{\partial x}}.}$

If our value shall generally hold for ${\displaystyle x}$, it also must be

${\displaystyle {\frac {\partial X}{\partial t}}}$ small against ${\displaystyle v{\frac {\partial X}{\partial x}}}$.

Now it is

${\displaystyle {\frac {\partial X}{\partial t}}={\frac {\partial ^{2}}{\partial x\partial t}}U\left(1-A^{2}v^{2}\right),}$

thus it must be

${\displaystyle {\frac {\partial }{\partial t}}\left[{\frac {\partial U}{\partial x}}\left(1-A^{2}v^{2}\right)\right]}$ small against ${\displaystyle v{\frac {\partial ^{2}U}{\partial x^{2}}}\left(1-A^{2}v^{2}\right)}$,

or

${\displaystyle A^{2}x{\frac {\partial v}{\partial t}}}$ small against ${\displaystyle 1-A^{2}v^{2}}$,

Also the values of ${\displaystyle Y,Z}$ and ${\displaystyle M,N}$, give

${\displaystyle \left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}{\frac {\partial v}{\partial t}}}$ is small against ${\displaystyle 3x\left(1-A^{2}v^{2}\right)}$

and

 ${\displaystyle \left(1-A^{2}v^{2}\right)\left[\left(x^{2}+\left(1-A^{2}v^{2}\right)\varrho ^{2}\right)\right]{\frac {\partial v}{\partial t}}}$ ${\displaystyle -\left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}v^{2}{\frac {\partial v}{\partial t}}}$

must be small against ${\displaystyle 3x\left(1-A^{2}v^{2}\right)v^{2}}$.

This condition is fulfilled, when the dimensions of the space, in which the energy comes essentially into consideration, are sufficiently small. Because the terms to be neglected all contain the linear dimensions in a higher power. Though ${\displaystyle dv/dt}$ may not be too great and the absolute velocity ${\displaystyle v}$ not too small.

When this neglect is allowed, then we can put for the change of kinetic energy

${\displaystyle {\frac {d}{dt}}\left({\frac {m}{2}}v^{2}\right)=mv{\frac {dv}{dt}}=K{\frac {dr}{dt}}dt=m{\frac {dr}{dt}}{\frac {d^{2}r}{dt^{2}}},}$