# 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

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