Page:Grundgleichungen (Minkowski).djvu/42

and have the value given on the right-hand side of (79). Therefore the general relations

(80)	$S_{h1}S_{1k}+S_{h2}S_{2k}+S_{h3}S_{3k}+S_{h4}S_{4k}=0$

h, k being unequal indices in the series 1, 2, 3, 4, and

(81)	$S_{h1}S_{1h}+S_{h2}S_{2h}+S_{h3}S_{3h}+S_{h4}S_{4h}=L^{2}-Det^{\frac {1}{2}}fDet^{\frac {1}{2}}F$

for h = 1,2,3,4.

Now if instead of F and f in the combinations (72) and (73), we introduce the electrical rest-force $\Phi$, the magnetic rest-force $\Psi$ and the rest-ray $\Omega$ [(55), (56) and (57)], we can pass over to the expressions, —

(82)	$L=-{\frac {1}{2}}\epsilon \Phi {\bar {\Phi }}+{\frac {1}{2}}\mu \Psi {\bar {\Psi }}$,

(83)	$S_{hk}=-{\frac {1}{2}}\epsilon \Phi {\bar {\Phi }}e_{hk}-{\frac {1}{2}}\mu \Psi {\bar {\Psi }}e_{hk}$ $+\epsilon (\Phi _{h}\Phi _{k}-\Phi {\bar {\Phi }}w_{h}w_{k})+\mu (\Psi _{h}\Psi _{k}-\Psi {\bar {\Psi }}w_{h}w_{k})$ $-\Omega _{h}w_{k}-\epsilon \mu w_{h}\Omega _{k}\qquad \qquad (h,k=1,2,3,4)$

Here we have

$\Phi {\bar {\Phi }}=\Phi _{1}^{2}+\Phi _{2}^{2}+\Phi _{3}^{2}+\Phi _{4}^{2},\ \Psi {\bar {\Psi }}=\Psi _{1}^{2}+\Psi _{2}^{2}+\Psi _{3}^{2}+\Psi _{4}^{2}$, $e_{hh}=1,\ e_{hk}=0(h\gtrless k)$.

The right side of (82) as well as L is an invariant in a Lorentz transformation, and the 4✕4 element on the right side of (83) as well as $S_{hk}$, represent a space time vector of the second kind. Remembering this fact, it suffices, for establishing the theorems (82) and (83) generally, to prove it for the special case $w_{1}=0,\ w_{2}=0,\ w_{3}=0,\ w_{4}=i$. But for this case ${\mathfrak {w}}=0$, we immediately arrive at the equations (82) and (83) by means (45), (51), (60) on the one hand, and ${\mathfrak {e}}=\epsilon {\mathfrak {E}},\ {\mathfrak {M}}=\mu {\mathfrak {m}}$ on the other hand.

The expression on the right-hand side of (81), which equals

$=\left({\frac {1}{2}}({\mathfrak {mM}}-{\mathfrak {eE}})\right)^{2}+({\mathfrak {em}})({\mathfrak {EM}})$

is $\geqq 0$, because $({\mathfrak {em}})=\epsilon \Phi {\bar {\Psi }},\ ({\mathfrak {EM}})=\mu \Phi {\bar {\Psi }}$, now referring back to 79), we can denote the positive square root of this expression as $Det^{\frac {1}{2}}S$.

Since ${\bar {f}}=-f,\ {\bar {F}}=-F$, we obtain for ${\bar {S}}$, the transposed matrix of S, the following relations from (78),

(84)	$Ff={\bar {S}}-L,\ f^{*}F^{*}=-{\bar {S}}-L$.

Then is