I. Kinematical Part.Show the German source

II. Electrodynamical Part.Show the German source

§ 9. Transformation of the Maxwell-Hertz Equations with Consideration of Convection Currents.Show the German source

We start from the equations:Show the German source $$\begin{aligned} \frac{1}{V} \left\{ u_x \varrho + \frac{\partial X}{\partial t} \right\} &= \frac{\partial N}{\partial y} - \frac{\partial M}{\partial z} , &\qquad \frac{1}{V} \frac{\partial L}{\partial t} &= \frac{\partial Y}{\partial z} - \frac{\partial Z}{\partial y} , \\ \frac{1}{V} \left\{ u_y \varrho + \frac{\partial Y}{\partial t} \right\} &= \frac{\partial L}{\partial z} - \frac{\partial N}{\partial x} , &\qquad \frac{1}{V} \frac{\partial M}{\partial t} &= \frac{\partial Z}{\partial x} - \frac{\partial X}{\partial z} , \\ \frac{1}{V} \left\{ u_z \varrho + \frac{\partial Z}{\partial t} \right\} &= \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} , &\qquad \frac{1}{V} \frac{\partial N}{\partial t} &= \frac{\partial X}{\partial y} - \frac{\partial Y}{\partial x} , \end{aligned}$$$V$Speed of light$u_x$Charge velocity (stationary system)$\varrho$Charge density, stationary system$X$Electric field, stationary system$t$Time in the stationary system$N$Magnetic field, stationary system$y$Position in the stationary system$M$Magnetic field, stationary system$z$Position in the stationary system$L$Magnetic field, stationary system$Y$Electric field, stationary system$Z$Electric field, stationary system$u_y$Charge velocity (stationary system)$x$Position in the stationary system$u_z$Charge velocity (stationary system)Explain this equationShow the German source whereShow the German source $$\varrho = \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z}$$$\varrho$Charge density, stationary system$X$Electric field, stationary system$x$Position in the stationary system$Y$Electric field, stationary system$y$Position in the stationary system$Z$Electric field, stationary system$z$Position in the stationary systemExplain this equationShow the German source denotes $4\pi$ times the density of electricity, and $(u_x, u_y, u_z)$ the velocity vector of electricity.Show the German source If we imagine the electric masses to be invariably bound to small rigid bodies (ions, electrons), these equations are the electromagnetic foundation of Lorentz’s electrodynamics and optics of moving bodies.Show the German source

If these equations, which are to hold in the system $K$, are transformed to the system $k$ with the help of the transformation equations of § 3 and § 6, we obtain the equations:Show the German source $$\begin{aligned} \frac{1}{V} \left\{ u_\xi \varrho' + \frac{\partial X'}{\partial \tau} \right\} &= \frac{\partial N'}{\partial \eta} - \frac{\partial M'}{\partial \zeta} , &\qquad \frac{\partial L'}{\partial \tau} &= \frac{\partial Y'}{\partial \zeta} - \frac{\partial Z'}{\partial \eta} , \\ \frac{1}{V} \left\{ u_\eta \varrho' + \frac{\partial Y'}{\partial \tau} \right\} &= \frac{\partial L'}{\partial \zeta} - \frac{\partial N'}{\partial \xi} , &\qquad \frac{\partial M'}{\partial \tau} &= \frac{\partial Z'}{\partial \xi} - \frac{\partial X'}{\partial \zeta} , \\ \frac{1}{V} \left\{ u_\zeta \varrho' + \frac{\partial Z'}{\partial \tau} \right\} &= \frac{\partial M'}{\partial \xi} - \frac{\partial L'}{\partial \eta} , &\qquad \frac{\partial N'}{\partial \tau} &= \frac{\partial X'}{\partial \eta} - \frac{\partial Y'}{\partial \xi} , \end{aligned}$$$V$Speed of light$u_{\xi }$Charge velocity (moving system)${\varrho }^{\prime }$Charge density, moving system$X^{\prime }$Electric field, moving system$\tau$Time in the moving system$N^{\prime }$Magnetic field, moving system$\eta$Position in the moving system$M^{\prime }$Magnetic field, moving system$\zeta$Position in the moving system$L^{\prime }$Magnetic field, moving system$Y^{\prime }$Electric field, moving system$Z^{\prime }$Electric field, moving system$u_{\eta }$Charge velocity (moving system)$\xi$Position in the moving system$u_{\zeta }$Charge velocity (moving system)Explain this equationShow the German source whereShow the German source $$\begin{aligned} \frac{u_x - v}{1 - \frac{u_x v}{V^2}} &= u_\xi , \\ \frac{u_y}{\beta \left( 1 - \frac{u_x v}{V^2} \right)} &= u_\eta , &\qquad \varrho' = \frac{\partial X'}{\partial \xi} + \frac{\partial Y'}{\partial \eta} + \frac{\partial Z'}{\partial \zeta} = \beta \left( 1 - \frac{v u_x}{V^2} \right) \varrho . \\ \frac{u_z}{\beta \left( 1 - \frac{u_x v}{V^2} \right)} &= u_\zeta . \end{aligned}$$$u_x$Charge velocity (stationary system)$v$Speed of the moving system$V$Speed of light$u_{\xi }$Charge velocity (moving system)$u_y$Charge velocity (stationary system)$\beta$Lorentz factor$u_{\eta }$Charge velocity (moving system)${\varrho }^{\prime }$Charge density, moving system$X^{\prime }$Electric field, moving system$\xi$Position in the moving system$Y^{\prime }$Electric field, moving system$\eta$Position in the moving system$Z^{\prime }$Electric field, moving system$\zeta$Position in the moving system$\varrho$Charge density, stationary system$u_z$Charge velocity (stationary system)$u_{\zeta }$Charge velocity (moving system)Explain this equationShow the German source Since — as follows from the addition theorem of velocities (§ 5) — the vector $(u_\xi, u_\eta, u_\zeta)$ is nothing other than the velocity of the electric masses measured in the system $k$, it is thereby shown that, on the basis of our kinematical principles, the electrodynamic foundation of Lorentz’s theory of the electrodynamics of moving bodies is in accordance with the principle of relativity.Show the German source

It may also be briefly remarked that the following important theorem can easily be deduced from the equations developed: If an electrically charged body moves in any manner in space and its charge, regarded from a coordinate system moving with the body, does not change in the process, then its charge also remains — regarded from the “stationary” system $K$ — constant.Show the German source

In conclusion I remark that in working on the problem treated here my friend and colleague M. Besso stood loyally by my side, and that I owe him several valuable suggestions.Show the German source

Bern, June 1905.Show the German source

(Received 30 June 1905.)Show the German source