Page:Über die Möglichkeit einer elektromagnetischen Begründung der Mechanik.djvu/5

H. A. Lorentz^[1] has alluded to the fact, that the length of a body in the direction of Earth's motion is contracted by the velocity $v$ of this motion in the ratio ${\sqrt {1-A^{2}v^{2}}}$, if the molecular forces could be replaced by electrostatic forces.

By that, Michelson's result would be explained, when one can neglect molecular motions. As to how far this is true, must be shown by gas-theoretical investigations.

For the explanation of gravitation, following Lorentz, we must assume two different kinds of electric polarizations. Any of them satisfies for itself Maxwell's equations. Additionally, with respect to static fields it is given

$X=-{\frac {\partial \phi }{\partial x}},\quad Y=-{\frac {\partial \phi }{\partial y}},\quad Z=-{\frac {\partial \phi }{\partial z}}$

and the energy

${\frac {1}{8\pi }}\int \int \int dx\ dy\ dz\ \left({\frac {\partial \phi }{\partial x}}\right)^{2}+\left({\frac {\partial \phi }{\partial y}}\right)^{2}+\left({\frac {\partial \phi }{\partial z}}\right)^{2}$ $={\frac {1}{8\pi }}\int dS{\frac {\partial \phi }{\partial n}}\phi -{\frac {1}{8\pi }}\int \int \int dx\ dy\ dz\ \phi \triangle \phi .$

If $\varphi$ or $\partial \varphi /\partial$ vanishes at the surface of space, the energy is

$=-{\frac {1}{8\pi }}\int \int \int dx\ dy\ dz\ \phi \triangle \phi .$

Now, according to (2)

$\triangle \phi =-4\pi \varsigma ,\quad \phi =\int \int \int {\frac {dx\ dy\ dz\ \varsigma }{r}},$

thus the integral

$={\frac {1}{2}}\int \int \int {\frac {dx\ dy\ dz\ \varsigma }{r}}\int \int \int dx'\ dy'\ dz'\ \varsigma '$ $=\int \int \int \int \int \int {\frac {\varsigma \varsigma 'dx\ dy\ dz\ dx'\ dy'\ dz'}{r}}$

1. H. A. Lorentz, Attempt of a Theory of Electrical and Optical Phenomena in Moving Bodies, Leiden 1895.