# Page:AbrahamMinkowski1.djvu/16

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