## Metric tensor under Lorentz transformation

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; 3,4.

The invariant interval in special relativity can be written as

$\displaystyle ds^{2}=\eta_{ij}dx^{i}dx^{j}$

where ${\eta_{ij}}$ is the metric tensor in flat space, with components ${\eta_{00}=-1}$, ${\eta_{ii}=+1}$ for ${i=1,2,3}$ and zero otherwise. Thus this relation is the same as

$\displaystyle ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}$

Under a Lorentz transformation, we get

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle \eta_{ij}dx^{\prime i}dx^{\prime j}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{ij}\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}dx^{a}dx^{b}$

Since the interval is invariant, we get

 $\displaystyle \eta_{ij}\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}dx^{a}dx^{b}$ $\displaystyle =$ $\displaystyle \eta_{ab}dx^{a}dx^{b}$ $\displaystyle \left(\eta_{ij}\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}-\eta_{ab}\right)dx^{a}dx^{b}$ $\displaystyle =$ $\displaystyle 0$

Since the last equation must be true for an infintesimal interval, the quantity in parentheses must be zero, so

$\displaystyle \eta_{ab}=\eta_{ij}\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}$

That is, if we apply a Lorentz transformation (the same transformation!) to each index in the metric tensor, we get the same tensor back again.

We can multiply this equation by an inverse transformation to get

$\displaystyle \left(\Lambda^{-1}\right)_{\;\; k}^{a}\eta_{ab}=\eta_{ij}\Lambda_{\;\; a}^{i}\Lambda_{\;\; b}^{j}\left(\Lambda^{-1}\right)_{\;\; k}^{a}$

Multiplying a transformation by its inverse gives the identity matrix:

$\displaystyle \Lambda_{\;\; a}^{i}\left(\Lambda^{-1}\right)_{\;\; k}^{a}=\delta_{\;\; k}^{i}$

So we get

 $\displaystyle \left(\Lambda^{-1}\right)_{\;\; k}^{a}\eta_{ab}$ $\displaystyle =$ $\displaystyle \eta_{ij}\delta_{\;\; k}^{i}\Lambda_{\;\; b}^{j}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{kj}\Lambda_{\;\; b}^{j}$

Repeating the process, we get

 $\displaystyle \left(\Lambda^{-1}\right)_{\;\; l}^{b}\left(\Lambda^{-1}\right)_{\;\; k}^{a}\eta_{ab}$ $\displaystyle =$ $\displaystyle \eta_{kj}\Lambda_{\;\; b}^{j}\left(\Lambda^{-1}\right)_{\;\; l}^{b}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{kj}\delta_{\;\; l}^{j}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{kl}$

Thus, not surprisingly, if we multiply the metric tensor by two inverse Lorentz transformations, we get the same tensor back.