Page:The Meaning of Relativity - Albert Einstein (1922).djvu/71

of integration, we obtain,

$\mathbf {K} _{x}={\frac {d}{dl}}\left(m{\frac {dx_{1}}{d\tau }}\right)={\frac {d}{dl}}\left({\frac {m\mathbf {q} _{x}}{\sqrt {1-q^{2}}}}\right).$

We see, therefore, that the generalized conception of the energy tensor is in agreement with our former result.

The Eulerian Equations for Perfect Fluids. In order to get nearer to the behaviour of real matter we must add to the energy tensor a term which corresponds to the pressures. The simplest case is that of a perfect fluid in which the pressure is determined by a scalar $p$. Since the tangential stresses $p_{xy}$, etc., vanish in this case, the contribution to the energy tensor must be of the form $p\delta _{\nu \mu }$. We must therefore put

$T_{\mu \nu }=\sigma u_{\mu }u_{\nu }+p\delta _{\mu \nu }$

(51)

At rest, the density of the matter, or the energy per unit volume, is in this case, not $\sigma$ but $\sigma -p$. For

$-T_{44}=-\sigma {\frac {dx_{4}}{d\tau }}{\frac {dx_{4}}{d\tau }}-p\delta _{44}=\sigma -p.$

In the absence of any force, we have

${\frac {\delta T_{\mu \nu }}{\delta x_{\nu }}}=\sigma u_{\nu }{\frac {\delta u_{\mu }}{\delta x_{\nu }}}+u_{\mu }{\frac {\delta (\sigma u_{\nu })}{\delta x_{\nu }}}+{\frac {\delta p}{\delta x_{\mu }}}=0.$

If we multiply this equation by $u_{\mu }\left(={\frac {dx_{\mu }}{d\tau }}\right)$ and sum for the $\mu$'s we obtain, using (40),

$-{\frac {\delta (\sigma u_{\nu })}{\delta x_{\nu }}}+{\frac {dp}{d\tau }}=0$

(52)