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

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

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

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