Page:Das Relativitätsprinzip und seine Anwendung.djvu/14

Fig. 6.

The second remark is based on the transformation equations for the electric moment ${\mathfrak {P}}$ (p. 84), which shows the impossibility (because the magnetization ${\mathfrak {M}}$ occurs in them) to clearly distinguish between polarization and magnetization electrons. In a magnetized body $({\mathfrak {M}}\neq 0)$, as seen from a reference system, it can rather be ${\mathfrak {P}}=0$, while ${\mathfrak {P}}'$ is different from zero in another reference system. This shall now be applied to a special case, where we confine ourselves to magnitudes of first order. The considered body (e.g. a steel magnet) shall contain only conduction electrons and such ones (when the body is at rest) which produce ${\mathfrak {M}}$, yet not ${\mathfrak {P}}$; it shall have the shape of an infinitely extended even plate, bounded by two planes $a,b$: the middle plane is made by us to the $yz$-plane (Fig. 6). When it is at rest, a constant magnetization ${\mathfrak {M}}_{y}$ may exist in the $y$-direction, while ${\mathfrak {P}}=0$. If the body acquires the velocity $v$ in the $z$-direction, then an observer not participating at the motion, is observing the electric polarization

${\mathfrak {P}}_{x}=-{\frac {v}{c}}{\mathfrak {M}}_{y}$

Now we imagine two conductors $c,d$ at both sides of the body, which together with it are forming two equal condensers, and they shall be short circuited by a wire (from $c$ to $d$). When in motion, $d$ charges will arise upon $c$ now, which can be calculated as follows. Since it is evidently impossible that a current exists in the $x$-direction, it is ${\mathfrak {E}}_{1x}=0$ or ${\mathfrak {E}}_{x}={\frac {v}{c}}{\mathfrak {B}}_{y}$. Since the process is stationary, it becomes ${\mathfrak {\dot {B}}}=0$; then the existence of a potential $\varphi$ follows from $\operatorname {rot} \ {\mathfrak {E}}=0$. If $\Delta$ is the thickness of the plate, then one has

$\varphi _{a}-\varphi _{b}={\frac {v}{c}}\Delta {\mathfrak {B}}_{y}.$

From the symmetry of the arrangement if evidently follows

$\varphi _{d}-\varphi _{a}=\varphi _{b}-\varphi _{c}{,}$

and because the plates $c,d$ are short circuited, it must be

$\varphi _{d}=\varphi _{c}$;

from that if follows

$\varphi _{d}-\varphi _{a}=-{\frac {v}{2c}}\Delta {\mathfrak {B}}_{y}.$

If $\gamma$ is the capacity of one of the two condensers, then the charge of the plate $d$ becomes equal to

$-{\frac {v}{2c}}\gamma \Delta {\mathfrak {B}}_{y}{,}$

and $c$ obtains the oppositely equal amount.