# Page:Grundgleichungen (Minkowski).djvu/22

 (V') $\mathfrak{e}'=\epsilon\mathfrak{E}',\ \mathfrak{M}'=\mu\mathfrak{m}',\ \mathfrak{s}'=\sigma\mathfrak{E}'$,

where $\epsilon,\ \mu,\ \sigma$ are the dielectric constant, magnetic permeability, and conductivity for the system x', y', z', t', i.e. in the space-time point x, y, z, t of matter.

Now let us return, by means of the reciprocal Lorentz-transformation to the original variables x, y, z, t, and the magnitudes $\mathfrak{w},\varrho,\mathfrak{s,e,m,E,M}$ and the equations, which we then obtain from the last mentioned, will be the fundamentil equations sought by us for the moving bodies.

Now from § 4, and § 6, it is to be seen that the equations A), as well as the equations B) are covariant for a Lorentz-transformation, i.e. the equations, which we obtain backwards from A') B'), must be exactly of the same form as the equations A) and B), as we take them for bodies at rest. We have therefore as the first result: —

The differential equations expressing the fundamental equations of electrodynamics for moving bodies, when written in $\varrho$ and the vectors $\mathfrak{s,\ e,\ m,\ E,\ M}$ are exactly of the same form as the equations for moving bodies. The velocity of matter does not enter in these equations. In the vectorial way of writing, we have

 (I) $\begin{array}{rcrl} (I) & \qquad & curl\ \mathfrak{m}-\frac{\partial e}{\partial t} & =\mathfrak{s},\\ \\(II) & & div\ \mathfrak{e} & =\varrho,\\ \\(III) & & curl\ \mathfrak{E}+\frac{\partial\mathfrak{M}}{\partial t} & =0,\\ \\(IV) & & div\ \mathfrak{M} & =0\end{array}$,

The velocity of matter occurs only in the auxilliary equations which characterise the influence of matter on the basis of their characteristic constants $\epsilon,\ \mu,\ \sigma$. Let us now transform these auxilliary equations x, y, z into the original co-ordinates x, y, z, and t.)

According to formula 15) in § 4, the component of $\mathfrak{e}'$ in the direction of the vector $\mathfrak{w}$ is the same us that of $\mathfrak{e}+[\mathfrak{wm}]$, the component of $\mathfrak{m}'$ is the same as that of $\mathfrak{m}-[\mathfrak{we}]$, but for the perpendicular direction $\mathfrak{\bar{w}}$, the components of $\mathfrak{e}'$ and $\mathfrak{m}'$