Page:Grundgleichungen (Minkowski).djvu/52

In consequence of this condition, the integral (7) taken over the whole range of the sickle, varies on account of the displacement as a definite function ${\mathsf {N}}+\delta {\mathsf {N}}$ of $\vartheta$, and we may call this function ${\mathsf {N}}+\delta {\mathsf {N}}$ as the mass action of the virtual displacement.

If we now introduce the method of writing with indices, we shall have

(9)	$d(x_{h}+\delta x_{h})=dx_{h}+{\underset {k}{\sum }}{\frac {\partial \delta x_{h}}{\partial x_{k}}}dx_{k}+{\frac {\partial \delta x_{h}}{\partial \vartheta }}d\vartheta \qquad \left({\begin{array}{c}k=1,2,3,4\\h=1,2,3,4\end{array}}\right)$.

Now on the basis of the remarks already made, it is clear that the value of ${\mathsf {N}}+\delta {\mathsf {N}}$, when the value of the parameter is $\vartheta$, will be: —

(10)	${\mathsf {N}}+\delta {\mathsf {N}}=\int \int \int \int v{\frac {d(\tau +\delta \tau )}{d\tau }}dx\ dy\ dz\ dt$,

the integration extending over the whole sickle $d(\tau +\delta \tau )$, where $d(\tau +\delta \tau )$ denotes the magnitude, which is deduced from

${\sqrt {-(dx_{1}+d\delta x_{1})^{2}-(dx_{2}+d\delta x_{2})^{2}-(dx_{3}+d\delta x_{3})^{2}-(dx_{4}+d\delta x_{4})^{2}}}$

by means of (9) and

$dx_{1}=w_{1}d\tau ,\ dx_{2}=w_{2}d\tau ,\ dx_{3}=w_{3}d\tau ,\ dx_{4}=w_{4}d\tau ,\ d\vartheta =0$

therefore: —

(11)	${\frac {d(\tau +\delta \tau )}{d\tau }}={\sqrt {-{\underset {h}{\sum }}\left(w_{h}+{\underset {k}{\sum }}{\frac {\partial \delta x_{h}}{\partial x_{k}}}w_{k}\right)^{2}}}\left({\begin{array}{c}k=1,2,3,4\\h=1,2,3,4\end{array}}\right)$.

We shall now subject the value of the differential quotient

(12)	$\left({\frac {d({\mathsf {N}}+\delta {\mathsf {N}})}{d\vartheta }}\right)_{(\vartheta =0)}$

to a transformation. Since each $\delta x_{h}$, as a function of $x_{1},\ x_{2},\ x_{3},\ x_{4},\ \vartheta$ vanishes for the zero-value of the paramater $\vartheta$, so in general ${\frac {\partial \delta x_{h}}{\partial x_{k}}}=0$ for $\vartheta =0$.

Let us now put

(13)	$\left({\frac {\partial \delta x_{h}}{\partial \vartheta }}\right)_{\vartheta =0}=\xi _{h}\qquad (h=1,2,3,4)$,

then on the basis of (10) and (11), we have the expression (12):

$-\int \int \int \int \ \nu {\underset {h}{\sum }}w_{h}\left({\frac {\partial \xi _{h}}{\partial x_{1}}}w_{1}+{\frac {\partial \xi _{h}}{\partial x_{2}}}w_{2}+{\frac {\partial \xi _{h}}{\partial x_{3}}}w_{3}+{\frac {\partial \xi _{h}}{\partial x_{4}}}w_{4}\right)dx\ dy\ dz\ dt$.

for the system $x_{1},\ x_{2},\ x_{3},\ x_{4}$ on the boundary of the sickle, $\delta x_{1},\ \delta x_{2},\ \delta x_{3},\ \delta x_{4}$ shall vanish for every value of $\vartheta$