Page:LorentzGravitation1916.djvu/16

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draw a line perpendicular to \sigma_{2} and \sigma_{1}. Let B_1 be the point, where it cuts thus last, plane, the "base", and A_{1} the point where this plane is encountered by the generating line through A_{2}. If then \angle A_{1}A_{2}B_{1}=\vartheta, we have

\overline{A_{2}B_{1}}=\overline{A_{2}A_{1}}\cos\vartheta (13)

The strokes over the letters indicate the absolute values of the distances A_{2}B_{1} and A_{2}A_{1}.

It can be shown (§ 8) that, all quantities being expressed in natural units, the "volume" of the prism P is found by taking the product of the numerical values of the base \sigma_{1} and the "height" A_{2}B_{1}.

Let now linear three-dimensional extensions perpendicular to A_{1}A_{2} be made to pass through A_1 and A_2. From these extensions the lateral boundary of the prism cuts the parts \sigma'_{1} and \sigma'_{2} and these parts, together with the lateral surface, enclose a new prism P', the volume of which is equal to that of P. As now the volume of P' is given by the product of \overline{A_{2}A_{1}} and \sigma'_{1}, we have with regard to (13)

\sigma'_{1}=\sigma{}_{1}\cos\vartheta

If now we remember that, if a vector perpendicular to \sigma_{1} is projected on the generating line, the ratio between the projection and the vector itself (viz. between their absolute values) is given by \cos\vartheta and that a connexion similar to that which was found above between a normal section \sigma'_{1} of the prism and \sigma_{1}, also exists between \sigma'_{1} and any other oblique section, we easily find the following theorem:

Let \sigma and \bar{\sigma} be two arbitrarily chosen linear three-dimensional sections of the prism, \mathrm{N} and \bar{\mathrm{N}} two vectors, perpendicular to \sigma and \bar{\sigma} resp. and of the same length, S and \bar{S} the absolute values of the projections of \mathrm{N} and \bar{\mathrm{N}} on a generating line. Then we have

S\sigma=\bar{S}\bar{\sigma} (14)


§ 19. After these preliminaries we can show that the left hand side of (10) is equal to 0, if the numbers g_{ab} are constants and if moreover both the rotation \mathrm{R}_{e} and the rotation \mathrm{R}_{h} are everywhere the same. For the two parts of the integral the proof may be given in the same way, so that it suffices to consider the expression

\int\left[\mathrm{R}_{e}\cdot\mathrm{N}\right]_{x}d\sigma (15)

Let X_{1},\dots X_{4} be the components of the vector \mathrm{N}, expressed in x-units. From the distributive property of the vector product it then follows that each of the four components of