Page:The Meaning of Relativity - Albert Einstein (1922).djvu/92

is also a vector. Since this is the case for an arbitrary choice of the $dx_{\sigma }$, it follows that

$A^{\mu };\sigma ={\frac {\delta A^{\mu }}{\delta x_{\sigma }}}+\Gamma _{\sigma \alpha }^{\mu }A^{\alpha }$

(71)

is a tensor, which we designate as the co-variant derivative of the tensor of the first rank (vector). Contracting this tensor, we obtain the divergence of the contra-variant tensor $A^{\mu }$. In this we must observe that according to (70),

$\Gamma _{\mu \sigma }^{\sigma }={\frac {1}{2}}g^{\sigma \alpha }{\frac {\delta g_{\sigma \alpha }}{\delta x_{\mu }}}={\frac {1}{\sqrt {g}}}{\frac {\delta {\sqrt {g}}}{\delta x_{\mu }}}.$

(72)

If we put, further,

$A^{\mu }{\sqrt {g}}={\mathfrak {A}}^{\mu }$

(73)

a quantity designated by Weyl as the contra-variant tensor density ^[1] of the first rank, it follows that,

${\mathfrak {A}}={\frac {\delta {\mathfrak {A}}^{\mu }}{\delta x_{\mu }}}$

(74)

is a scalar density.

We get the law of parallel displacement for the co-variant vector $B_{\mu }$ by stipulating that the parallel displacement shall be effected in such a way that the scalar

$\phi =A^{\mu }B_{\mu }$

remains unchanged, and that therefore

$A^{\mu }\delta B_{\mu }+B_{\mu }\delta A^{\mu }$

1. This expression is justified, in that $A^{\mu }{\sqrt {g}}dx={\mathfrak {A}}^{\mu }dx$ has a tensor character. Every tensor, when multiplied by ${\sqrt {g}}$, changes into a tensor density. We employ capital Gothic letters for tensor densities.