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