Metric tensor

The metric tensor's elements are the coefficients read off of the line element

$ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }$

For special relativity rectilinear coordinate inertial frames are used which given

$x^{0}=ct$

$x^{1}=x$

$x^{2}=y$

$x^{3}=z$

the metric tensor will be designated $\eta _{\mu \nu }$ and the line element will be

$ds^{2}=dct^{2}-\left(dx^{2}+dy^{2}+dz^{2}\right)$

and the Minkowski metric tensor elements given by

$\eta _{00}=1$

$\eta _{11}=\eta _{22}=\eta _{33}=-1$

All other elements are 0.

Written as a matrix this is

$||\eta _{\mu \nu }||={\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}$

The metric tensor acts a an index raising

$T^{\mu }=g^{\mu \nu }T_{\nu }$

and lowering

$T_{\mu }=g_{\mu \nu }T^{\nu }$

opperator. And as an inner product operator in 4d spacetime

$U^{\mu }T_{\mu }=g_{\lambda \mu }U^{\mu }T^{\lambda }$

There is an inverse relationship between the contravariant and covariant metric tensor elements

$g_{\mu \lambda }g^{\lambda \nu }=\delta _{\mu }^{\nu }$

which can be expressed as the matrix

$||g_{\mu \lambda }g^{\lambda \nu }||={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}}$

So solving for the contravariant metric tensor elements given the covariant ones and vica-versa can be done by simple matrix inversion.

The covariant derivative of the metric with respect to any coordinate is zero

$g_{\mu \nu };_{\lambda }=0$

$g^{\mu \nu };_{\lambda }=0$

where the covariant derivative is done with the use of Christoffel symbols. And so of course the covariant divergence of the metric is also zero

$g^{\mu \nu };_{\nu }=0$