# 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 ${\displaystyle c^{2}}$.

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

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

where the vector

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

is determined from

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

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

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

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

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

and it follows from (40a)

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

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

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

Since one has

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

then it becomes with respect to (40a)

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

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

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

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

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

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