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; Problems 4.3, 4.4.

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.

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