Page:Grundgleichungen (Minkowski).djvu/35

From Wikisource
Jump to: navigation, search
This page has been proofread, but needs to be validated.

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


\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


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


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


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.