Page:Grundgleichungen (Minkowski).djvu/31

(37), we have the identity $Det^{\frac {1}{2}}({\mathsf {\overline {A}}}f{\mathsf {A}})=Det\ {\mathsf {A}}\ Det^{\frac {1}{2}}f$. Therefore $Det^{\frac {1}{2}}f$ becomes an invariant in the case of a Lorentz transformation [see eq. (26) Sec. § 5].

Looking back to (36), we have for the dual matrix

$({\mathsf {A}}^{-1}f^{*}{\mathsf {A}})({\mathsf {A}}^{-1}f{\mathsf {A}})={\mathsf {A}}^{-1}f^{*}f{\mathsf {A}}=Det^{\frac {1}{2}}f.{\mathsf {A}}^{-1}{\mathsf {A}}=Det^{\frac {1}{2}}f$,

from which it is to be seen that the dual matrix $f^{*}$ behaves exactly like the primary matrix f, and is therefore a space time vector of the II kind; $f^{*}$ is therefore known as the dual space-time vector of f with components $f_{14},\ f_{24},\ f_{34},\ f_{23},\ f_{31},\ f_{12}$.

6°.If w and s are two space-time vectors of the 1st kind then by $w{\bar {s}}$ (as well as by $s{\bar {w}})$) will be understood the combination

(43)	$w_{1}s_{1}+w_{2}s_{2}+w_{3}s_{3}+w_{4}s_{4}$

In case of a Lorentz transformation ${\mathsf {A}}$, since $(w{\mathsf {A}})({\mathsf {\bar {A}}}{\bar {s}})=w{\bar {s}}$ this expression is invariant. — If $w{\bar {s}}=0$, then w and s are perpendicular to each other.

Two space-time rectors of the first kind w, s gives us a 2✕4 series matrix

$\left|{\begin{array}{cccc}w_{1},&w_{2},&w_{3},&w_{4}\\s_{1},&s_{2},&s_{3},&s_{4}\end{array}}\right|$

Then it follows immediately that the system of six magnitudes

(44)	$w_{2}s_{3}-w_{3}s_{2},\ w_{3}s_{1}-w_{1}s_{3},\ w_{1}s_{2}-w_{2}s_{1},\ w_{1}s_{4}-w_{4}s_{1},\ w_{2}s_{4}-w_{4}s_{2},\ w_{3}s_{4}-w_{4}s_{3}$

behaves in case of a Lorentz-transformation as a space-time vector of the II. kind. The vector of the second kind with the components (44) are denoted by [w,s]. We see easily that $Det^{\frac {1}{2}}[w,s]=0$. The dual vector of [w,s] shall be written as [w,s]*.

If w is a space-time vector of the 1st kind, f of the second kind, wf signifies a 1✕4 series matrix. In case of a Lorentz-transformation ${\mathsf {A}}$, w is changed into $w'=w{\mathsf {A}}$, f into $f'={\mathsf {A}}^{-1}f{\mathsf {A}}$, therefore $w'f'=w{\mathsf {A}}\ {\mathsf {A}}^{-1}f{\mathsf {A}}=(wf){\mathsf {A}}$, i.e., wf is transformed as a space-time vector of the 1st kind. We can verify, when w is a space-time vector of the 1st kind, f of the 2nd kind, the important identity

(45)	$[w,wf]+[w,wf^{*}]^{*}=(w{\bar {w}})f$.