Page:Grundgleichungen (Minkowski).djvu/31

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(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.