Metric tensor

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

$d{s}^2=g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }$

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

$d{s}^2=dc{t}^2-(d{x}^2+d{y}^2+d{z}^2)$

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

$||{\eta }_{\mu \nu }||=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\end{array}\right]$

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

$||g_{\mu \lambda }{g}^{\lambda \nu }||=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

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

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