(58) |
$|\Phi \Psi ]=i[w,\Omega ]^{*}$, |

*i,e.*

$\Phi _{1}\Psi _{2}-\Phi _{2}\Psi _{1}=i(w_{3}\Omega _{4}-w_{4}\Omega _{3})$, etc.
.

The vector $\Omega$ fulfills the relation

(59) |
$(w{\bar {\Omega }})=w_{1}\Omega _{1}+w_{2}\Omega _{2}+w_{3}\Omega _{3}+w_{4}\Omega _{4}=0$, |

which we can write as

$\Omega _{4}=i({\mathfrak {w}}_{x}\Omega _{1}+{\mathfrak {w}}_{y}\Omega _{2}+{\mathfrak {w}}_{z}\Omega _{3})$
and $\Omega$ is also *normal* to *w*. In case ${\mathfrak {w}}=0$, we have $\Phi _{4}=0,\ \Psi _{4}=0,\ \Omega _{4}=0$, and

(60) |
$\Omega _{1}=\Phi _{2}\Psi _{3}-\Phi _{3}\Psi _{2},\ \Omega _{2}=\Phi _{3}\Psi _{1}-\Phi _{1}\Psi _{3},\ \Omega _{3}=\Phi _{1}\Psi _{2}-\Phi _{2}\Psi _{1}$, |

I shall call $\Omega$, which is a space-time vector 1st kind the *Rest-Ray*.

As for the relation E), which introduces the conductivity $\sigma$, we have

$-w{\bar {s}}=-(w_{1}s_{1}+w_{2}s_{2}+w_{3}s_{3}+w_{4}s_{4})={\frac {-\left|{\mathfrak {w}}\right|s_{\mathfrak {w}}+\varrho }{\sqrt {1-{\mathfrak {w}}^{2}}}}=\varrho '$
This expression gives us the *rest-density* of electricity (see §8 and §4). Then

(61) |
$s+(w{\bar {s}})w$ |

represents a space-time vector of the 1st kind, which since $w{\bar {w}}=1$, is *normal* to *w*, and which I may call the *rest-current*. Let us now conceive of the first three component of this vector as the *x-, y-, z* co-ordinates of the space-vector, then the component in the direction of ${\mathfrak {w}}$ is

${\mathfrak {s_{w}}}-{\frac {\left|{\mathfrak {w}}\right|\varrho '}{\sqrt {1-{\mathfrak {w}}^{2}}}}={\frac {{\mathfrak {s_{w}}}-\left|{\mathfrak {w}}\right|\varrho }{\sqrt {1-{\mathfrak {w}}^{2}}}}={\frac {\mathfrak {F_{w}}}{1-{\mathfrak {w}}^{2}}}$
and the component in a perpendicular direction is ${\mathfrak {s_{\bar {w}}}}={\mathfrak {F_{\bar {w}}}}$.

This space-vector is connected with the space-vector ${\mathfrak {F}}={\mathfrak {s}}-\varrho {\mathfrak {w}}$, which we denoted in § 8 as the conduction-current.

Now by comparing with $\Phi =-wF$, the relation (E) can be brought into the form

(E) |
$s+(w{\bar {s}})w=-\sigma wF$. |