Page:LorentzGravitation1916.djvu/17

$\left[\mathrm {R} _{e}\cdot \mathrm {N} \right]_{x}$

is a homogeneous linear function of $X_{1},\dots X_{4}$. Under the special assumptions specified at the beginning of this § these are every where, the same functions. Let us thus consider a definite component of (15) e.g. that which corresponds to the direction of the coordinate $x_{a}$. We can represent it by an expression of the form

$\int \left(\alpha _{1}X_{1}+\dots +\alpha _{4}X_{4}\right)d\sigma$

where $\alpha _{1},\dots \alpha _{4}$ are constants. It will therefore be sufficient to prove that the four integrals

$\int X_{1}d\sigma \dots \int X_{4}d\sigma$

(16)

vanish.

In order to calculate $\int X_{1}d\sigma$ we consider an infinitely small prism, the edges of which have the direction $x_{1}$. This prism cuts from the boundary surface $\sigma$ two elements $d\sigma$ and ${\overline {d\sigma }}$. Proceeding along a generating line in the direction of the positive $x_{1}$ we shall enter the extension $\Omega$ bounded by $\sigma$ through one of these elements and leave it through the other. Now the vectors perpendicular to $\sigma$, which occur in (15) and which we shall denote by $\mathrm {N}$ and ${\bar {\mathrm {N} }}$ for the two elements, have the same value.^[1] If, therefore, $S$ and ${\bar {S}}$ are the absolute values of the projections of $\mathrm {N}$ and ${\bar {\mathrm {N} }}$ on a line in the direction $x_{1}$, we have according to (14)

$Sd\sigma ={\bar {S}}{\overline {d\sigma }}$

(17)

Let first the four directions of coordinates be perpendicular to one another. Then the components of the vector obtained by projecting $\mathrm {N}$ on the above mentioned line are $X_{1},0,0,0$ and similarly those of the projection of ${\bar {\mathrm {N} }}:{\bar {X}}_{1},0,0,0$. But as, proceeding in the direction of $x_{1}$ we enter $\Omega$ through one element and leave it through the other, while $\mathrm {N}$ and ${\bar {\mathrm {N} }}$ are both directed outward, $X_{1}$ and ${\overline {X_{1}}}$, must have opposite signs. So we have

$S:{\bar {S}}=X_{1}:-{\bar {X}}_{1}$

and because of (17) we may now conclude that the elements $X_{1}d\sigma$

1. From § 10 it follows that if the length of a vector $\mathrm {A}$ that is represented by a line (§ 17) coincides with a radius-vector of the conjugate indicatrix, it is always represented by an imaginary number. We may however obtain a vector which in natural units is represented by a real number e.g. by 1 (§ 13) if we multiply the vector $\mathrm {A}$ by an imaginary factor, which means that its components and also those of a vector product in which it occurs are multiplied by that factor.