# 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

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