Page:Elektrische und Optische Erscheinungen (Lorentz) 037.jpg



In addition, the mutually corresponding volume elements, and therefore also the ions, shall have the same charges in $$S_{1}$$ and $$S_{2}$$.

If we apply to all magnitudes, which are related to the second system, a prime so they can be distinguished, then

$$\rho'=\rho\sqrt{1-\frac{\mathfrak{p}^{2}}{V^{2}}}{,}$$

and

$$\frac{\partial^{2}\omega'}{\partial x'^{2}}+\frac{\partial^{2}\omega'}{\partial y'^{2}}+\frac{\partial^{2}\omega'}{\partial z'^{2}}=\rho'=\rho\sqrt{1-\frac{\mathfrak{p}^{2}}{V^{2}}}.$$

Then the equation (23) can be written in the form

$$\frac{\partial^{2}\omega}{\partial x'^{2}}+\frac{\partial^{2}\omega}{\partial y'^{2}}+\frac{\partial^{2}\omega}{\partial z'^{2}}=\rho$$

then

$$\omega=\frac{\omega'}{\sqrt{1-\frac{\mathfrak{p}^{2}}{V^{2}}}}{,}$$

and since in the second system

$$\mathfrak{E'}_{x}=4\pi V^{2}\frac{\partial\omega'}{\partial x'}\text{, etc.}$$

thus

$$\mathfrak{E}_{x}=\mathfrak{E'}_{x},\ \mathfrak{E}_{y}=\sqrt{1-\frac{\mathfrak{p}^{2}}{V^{2}}}\mathfrak{E'}_{y},\ \mathfrak{E}_{z}=\sqrt{1-\frac{\mathfrak{p}^{2}}{V^{2}}}\mathfrak{E'}_{z}.$$

The same relations, as they exist between the components of $$\mathfrak{E}$$ and $$\mathfrak{E}'$$, also exist, since the charges in $$S_{1}$$ and $$S_{2}$$ are equal, between the force components acting on an ion.

If in the second system at certain places $$\mathfrak{E}'=0$$, then $$\mathfrak{E}$$ vanishes at the corresponding points of the first system.

§ 24. Several implications of this theorem are obvious. From ordinary electrostatics, we know for example that an excess of positive (or negative) ions can be distributed over a