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