Page:The Meaning of Relativity - Albert Einstein (1922).djvu/70

must hold. We shall now show that this equation leads to the same law of motion of a material particle as that already obtained. Let us imagine the matter to be of infinitely small extent in space, that is, a four-dimensional thread; then by integration over the whole thread with respect to the space co-ordinates $x_{1},x_{2},x_{3}$, we obtain

${\begin{aligned}\int K_{1}dx_{1}dx_{2}dx_{3}=\int {\frac {\delta T_{14}}{\delta x_{4}}}dx_{1}dx_{2}&dx_{3}=\\&-\mathbf {i} {\frac {d}{dl}}\left\{\int \sigma _{0}{\frac {dx_{1}}{d\tau }}{\frac {dx_{4}}{d\tau }}dx_{1}dx_{2}dx_{3}\right\}.\end{aligned}}$

Now $\int dx_{1}dx_{2}dx_{3}dx_{4}$ is an invariant, as is, therefore, also $\int \sigma _{0}dx_{1}dx_{2}dx_{3}dx_{4}$. We shall calculate this integral, first with respect to the inertial system which we have chosen, and second, with respect to a system relatively to which the matter has the velocity zero. The integration is to be extended over a filament of the thread for which $\sigma _{0}$ may be regarded as constant over the whole section. If the space volumes of the filament referred to the two systems are $dV$ and $dV_{0}$ respectively, then we have

$\int \sigma _{0}dVdl=\int \sigma _{0}dV_{0}d\tau$

and therefore also

$\int \sigma _{0}dV=\int \sigma _{0}dV_{0}{\frac {d\tau }{dl}}=\int dm\mathbf {i} {\frac {d\tau }{dx_{4}}}.$

If we substitute the right-hand side for the left-hand side in the former integral, and put ${\frac {dx_{1}}{d\tau }}$ outside the sign