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