Page:The Meaning of Relativity - Albert Einstein (1922).djvu/88

But we can complement dx so that it becomes an invariant. If we form the determinant of the quantities

$g_{\mu \nu }'={\frac {\delta x_{\alpha }}{\delta x_{\mu }'}}{\frac {\delta x_{\beta }}{\delta x_{\nu }'}}g_{\alpha \beta }$

we obtain, by a double application of the theorem of multiplication of determinants,

$g'=\left|g_{\mu \nu }'\right|=\left|{\frac {\delta x_{\nu }}{\delta x_{\mu }'}}\right|^{2}\cdot \left|g_{\mu \nu }\right|=\left|{\frac {\delta _{\mu }'}{\delta x_{\nu }}}\right|^{-2}g.$

We therefore get the invariant,

${\sqrt {g'}}dx'={\sqrt {g}}dx.$

Formation of Tensors by Differentiation. Although the algebraic operations of tensor formation have proved to be as simple as in the special case of invariance with respect to linear orthogonal transformations, nevertheless in the general case, the invariant differential operations are, unfortunately, considerably more complicated. The reason for this is as follows. If $A^{\mu }$ is a contra-variant vector, the coefficients of its transformation, ${\frac {\delta x_{\mu }'}{\delta x_{\nu }}}$, are independent of position only if the transformation is a linear one. For then the vector components, $A^{\mu }+{\frac {\delta A^{\mu }}{\delta x_{\alpha }}}dx_{\alpha }$, at a neighbouring point transform in the same way as the $A^{\mu }$, from which follows the vector character of the vector differentials, and the tensor character of ${\frac {\delta A^{\mu }}{\delta x_{\alpha }}}$. But if the ${\frac {\delta x_{\mu }'}{\delta x_{\nu }}}$ are variable this is no longer true.