Page:AbrahamMinkowski1.djvu/16

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If one inserts Minkowski's connecting equations between the electromagnetic vectors into our system, then the momentum density in the moving body becomes equal to the energy current divided by c^{2}.

From (40) and (21) it follows, with respect to (37)

(40a) c\mathfrak{g=[EH]-q(qW]}

where the vector

(40b) \mathfrak{W=[DB]}-c\mathfrak{g}

is determined from

(40c) \mathfrak{W-q(qW)=[DB]-[EH]}

Let us direct the x-axis into the direction of \mathfrak{q}, and let us set

(40d) k^{2}=1-|\mathfrak{q}|^{2}

then the components of \mathfrak{W} become

(41) \begin{cases}
\mathfrak{W}_{x}=k^{-2}\left\{ [\mathfrak{DB}]_{x}-[\mathfrak{EH}]_{x}\right\} ,\\
\mathfrak{W}_{y}=[\mathfrak{DB}]_{y}-[\mathfrak{EH}]_{y},\\
\mathfrak{W}_{z}=[\mathfrak{DB}]_{z}-[\mathfrak{EH}]_{z},\end{cases}

and it follows from (40a)

(42) \begin{cases}
c\mathfrak{g}_{x}=\frac{\mathfrak{S}_{x}}{c}=k^{-2}[\mathfrak{EH}]_{x}-|\mathfrak{q}|^{2}k^{-2}[\mathfrak{DB}]_{x},\\
\\c\mathfrak{g}_{y}=\frac{\mathfrak{S}_{y}}{c}=[\mathfrak{EH}]_{y},\\
\\c\mathfrak{g}_{z}=\frac{\mathfrak{S}_{z}}{c}=[\mathfrak{EH}]_{z}.\end{cases}

The previous derivation has a gap; the proof is missing that equations (39) (assumed as being valid) are really satisfied. In order to prove this, we calculate the vector

\begin{array}{ll}
\mathfrak{R}' & =\mathfrak{[DE']+[BH']=\left[E'[qH]\right]-\left[H'[qE]\right]}\\
 & =\mathfrak{q(E'H)-q(EH')+E(qH')-H(qE')}\end{array}

Since one has

\begin{array}{c}
\mathfrak{E'H-EH'=q\left\{ [DE']+[BH']\right\} =(qR')},\\
\mathfrak{E(qH')-H(qE')=E(qH)-H(qE)=\left[q[EH]\right]},\end{array}

then it becomes with respect to (40a)

\mathfrak{R'-q(qR')}=[\mathfrak{q}c\mathfrak{g}]

One can – because according to the things said, the component of \mathfrak{R}' coinciding with the direction of vector \mathfrak{q}, is equal to zero – also write

(43) \mathfrak{R'=[DE']+[BH'}]=[\mathfrak{q}c\mathfrak{g}]

By that, condition (18a) is shown to be valid, and at the same time the gap in the previous derivation of the value of \mathfrak{g} is closed.

From (19) the value of the energy density follows:

(44) \psi=\frac{1}{2}\mathfrak{E'D}+\frac{1}{2}\mathfrak{H'B}+\mathfrak{q}c\mathfrak{g}