Page:Das Relativitätsprinzip und seine Anwendung.djvu/14

 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 $$(\mathfrak{M}\neq0)$$, 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.