Page:AbrahamMinkowski1.djvu/16

 If one inserts '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)

where the vector

is determined from

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

then the components of $$\mathfrak{W}$$ become

and it follows from (40a)

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

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: