and have the value given on the righthand 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 restforce $\Phi$, the magnetic restforce $\Psi$ and the restray $\Omega$ [(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 righthand 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