Page:The Meaning of Relativity - Albert Einstein (1922).djvu/93

vanishes for every value assigned to $(A^{\mu })$. We therefore get

$\delta B_{\mu }=\Gamma _{\mu \sigma }^{\alpha }A_{\alpha }dx_{\sigma }$

(75)

From this we arrive at the co-variant derivative of the co-variant vector by the same process as that which led to (71),

$B_{\mu };\sigma ={\frac {\delta B_{\mu }}{\delta x_{\sigma }}}-\Gamma _{\mu \sigma }^{\alpha }B_{\alpha }.$

(76)

By interchanging the indices $\mu$ and $\sigma$, and subtracting, we get the skew-symmetrical tensor,

$\phi _{\mu \sigma }={\frac {\delta B_{\mu }}{\delta x_{\sigma }}}-{\frac {\delta B_{\sigma }}{\delta x_{\mu }}}.$

(77)

For the co-variant differentiation of tensors of the second and higher ranks we may use the process by which (75) was deduced. Let, for example, $(A_{\sigma \tau }$ be a co-variant tensor of the second rank. Then $A_{\sigma \tau }E^{\sigma }F^{\tau }$ is a scalar, if $E$ and $F$ are vectors. This expression must not be changed by the $\delta$-displacement; expressing this by a formula, we get, using (67), $\delta A_{\sigma \tau }$ whence we get the desired co-variant derivative,

$A_{\sigma \tau ;\rho }={\frac {\delta A_{\sigma \tau }}{\delta x_{\rho }}}-\Gamma _{\sigma \rho }^{\alpha }A_{\alpha \tau }-\Gamma _{\tau \rho }^{\alpha }A_{\sigma \alpha }.$

(78)

In order that the general law of co-variant differentiation of tensors may be clearly seen, we shall write down two co-variant derivatives deduced in an analogous way:

$A_{\sigma ;\rho }^{\tau }={\frac {\delta A_{\sigma }^{\tau }}{\delta x_{\rho }}}-\Gamma _{\sigma \rho }^{\alpha }A_{\alpha }^{\tau }+\Gamma _{\alpha \rho }^{\tau }A_{\sigma }^{\alpha }.$

(79)

$A_{;\rho }^{\sigma \tau }={\frac {\delta A^{\sigma \tau }}{\delta x_{\rho }}}+\Gamma _{\alpha \rho }^{\sigma }A^{\alpha \tau }+\Gamma _{\alpha \rho }^{\tau }A^{\sigma \alpha }.$

(80)

6