Page:The Meaning of Relativity - Albert Einstein (1922).djvu/33

time $t$ has the co-ordinates $x_{\nu }$ If we express this acceleration by partial differential coefficients, we obtain, after dividing by $\rho$,

${\frac {\delta u_{\nu }}{\delta t}}+{\frac {\delta u_{\nu }}{\delta x_{\sigma }}}u_{\sigma }=-{\frac {1}{\rho }}{\frac {\delta p_{\nu \sigma }}{\delta x_{\sigma }}}+X_{\nu }$

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We must show that this equation holds independently of the special choice of the Cartesian system of co-ordinates. $(u_{\nu })$ is a vector, and therefore ${\frac {\delta u_{\nu }}{\delta t}}$ is also a vector, ${\frac {\delta u_{\nu }}{\delta x_{\sigma }}}$ is a tensor of rank 2, ${\frac {\delta u_{\nu }}{\delta x_{\sigma }}}u_{\tau }$ is a tensor of rank 3. The second term on the left results from contraction in the indices $\sigma ,\tau$. The vector character of the second term on the right is obvious. In order that the first term on the right may also be a vector it is necessary for $p_{\nu \sigma }$ to be a tensor. Then by differentiation and contraction ${\frac {\delta p_{\nu \sigma }}{\delta x_{\sigma }}}$ results, and is therefore a vector, as it also is after multiplication by the reciprocal scalar ${\frac {1}{\rho }}$. That $p_{\nu \sigma }$ is a tensor, and therefore transforms according to the equation

$p_{\mu \nu }'=b_{\mu \alpha }b_{\nu \beta }p_{\alpha \beta },$

is proved in mechanics by integrating this equation over an infinitely small tetrahedron. It is also proved there by application of the theorem of moments to an infinitely small parallelopipedon, that$p_{\nu \sigma }=p_{\sigma \nu }$, and hence that the tensor of the stress is a symmetrical tensor. From what has been said it follows that, with the aid of the rules