Page:The Meaning of Relativity - Albert Einstein (1922).djvu/34

given above, the equation is co-variant with respect to orthogonal transformations in space (rotational transformations); and the rules according to which the quantities in the equation must be transformed in order that the equation may be co-variant also become evident.

The co-variance of the equation of continuity,

${\frac {\delta \rho }{\delta t}}+{\frac {\delta (\rho u_{\nu })}{\delta x_{\nu }}}=0$

(17)

requires, from the foregoing, no particular discussion.

We shall also test for co-variance the equations which express the dependence of the stress components upon the properties of the matter, and set up these equations for the case of a compressible viscous fluid with the aid of the conditions of co-variance. If we neglect the viscosity, the pressure, $p$, will be a scalar, and will depend only upon the density and the temperature of the fluid. The contribution to the stress tensor is then evidently

$p\delta _{\mu \nu }$

in which $\delta _{\mu \nu }$ is the special symmetrical tensor. This term will also be present in the case of a viscous fluid. But in this case there will also be pressure terms, which depend upon the space derivatives of the $u_{\nu }$. We shall assume that this dependence is a linear one. Since these terms must be symmetrical tensors, the only ones which enter will be

$\alpha \left({\frac {\delta u_{\mu }}{\delta x_{\nu }}}+{\frac {\delta u_{\nu }}{\delta x_{\mu }}}\right)+\beta \delta _{\mu \nu }{\frac {\delta u_{\alpha }}{\delta x_{\alpha }}}$

(for ${\frac {\delta u_{\alpha }}{\delta x_{\alpha }}}$ is a scalar). For physical reasons (no slipping)