Page:LorentzGravitation1916.djvu/16

 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

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

§ 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

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