Page:The Meaning of Relativity - Albert Einstein (1922).djvu/90

character, that they must depend linearly and homogeneously upon the $dx_{\nu }$ and the $A^{\nu }$. We therefore put

$\delta A^{\nu }=-\Gamma _{\alpha \beta }^{\nu }A^{\alpha }dx_{\beta }.$

(67)

In addition, we can state that the $\Gamma _{\alpha \beta }^{\nu }$ must be symmetrical with respect to the indices $\alpha$ and $\beta$. For we can assume from a representation by the aid of a Euclidean system of local co-ordinates that the same parallelogram will be described by the displacement of an element $d^{(1)}x_{\nu }$ along a second element $d^{(2)}x_{\nu }$ as by a displacement of $d^{(2)}x_{\nu }$ along $d^{(1)}x_{\nu }$. We must therefore have

${\begin{aligned}d^{(2)}x_{\nu }+(d^{(1)}x_{\nu }-\Gamma _{\alpha \beta }^{\nu }d^{(1)}x_{\alpha }d^{(2)}&x_{\beta })\\&=d^{(1)}x_{\nu }+(d^{(2)}x_{\nu }-\Gamma _{\alpha \beta }^{\nu }d^{(2)}x_{\alpha }d^{(1)}x_{\beta }).\end{aligned}}$

The statement made above follows from this, after interchanging the indices of summation, $\alpha$ and $\beta$, on the right-hand side.

Since the quantities $g_{\mu \nu }$ determine all the metrical properties of the continuum, they must also determine the $\Gamma _{\alpha \beta }^{\nu }$. If we consider the invariant of the vector $A^{\nu }$, that is, the square of its magnitude,

$g_{\mu \nu }A^{\mu }A^{\nu }$

which is an invariant, this cannot change in a parallel displacement. We therefore have

$0=\delta (g_{\mu \nu }A^{\mu }A^{\nu })={\frac {\delta g_{\mu \nu }}{\delta x_{\alpha }}}A^{\mu }A^{\nu }dx_{\alpha }+g_{\mu \nu }A^{\mu }\delta A^{\nu }+g_{\mu \nu }A^{\nu }\delta A^{\mu }$

or, by (67),

$\left({\frac {\delta g_{\mu \nu }}{\delta x_{\alpha }}}-g_{\mu \beta }\Gamma _{\nu \alpha }^{\beta }-g_{\nu \beta }\Gamma _{\mu \alpha }^{\beta }\right)A^{\mu }A^{\nu }dx_{\alpha }=0.$