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The Minkowski metric tensor have the relation $\eta_{ij} \eta^{jk}=\delta_i {^k}$. That is the inverse of the Minkowski matrix is the matrix itself.
Analogously, is it true that $g_{ij} g^{jk}=\delta_i {^k}$, where $g_{ij}$ is the metric tensor in a curved space? If yes how to prove this? I came up with the confusion while finding the chrischoffel symbol I came up with a equation $$2 \Gamma^{\gamma}_{\alpha j} g_{i \gamma}=g_{ij,\alpha}+g_{\alpha i,j}-g_{j\alpha,i} .$$ To eliminate $g_{i\gamma}$ I have to find the inverse of the metric tensor in tensor notation. Can anyone suggest how would I solve this problem?

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3 Answers 3

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The matrix $(g^{-1})^{\mu\nu}$ of components of the inverse metric tensor field$^1$ is not necessarily equal to the matrix $g_{\mu\nu}$ of components of the metric tensor field, if that's what you're asking.

However, there is a standard notational shorthand convention to write $(g^{-1})^{\mu\nu}$ as $g^{\mu\nu}$ because the upper indices already indicate that we're talking about the inverse metric.

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$^1$ The metric tensor field is a covariant symmetric $(0,2)$ tensor field, while the inverse metric tensor field is a contravariant symmetric $(2,0)$ tensor field.

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    $\begingroup$ 1. $g^{j m}$ in the equation $\Gamma$: $\Gamma^k_{li}=\frac{1}{2}g^{jm}(\frac{\partial g_{ij}}{\partial q^l}+\frac{\partial g_{lj}}{\partial q^i}-\frac{\partial g_{il}}{\partial q^j})$ of Christoffel symbol is the component of the inverse of the matrix representing the metric tensor. Is it right? 2. But it is it a covariant metric tensor too? $\endgroup$
    – walber97
    Mar 23, 2019 at 14:14
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    $\begingroup$ 1. Yes. 2. No, it is instead contravariant. $\endgroup$
    – Qmechanic
    Mar 23, 2019 at 17:03
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    $\begingroup$ What does (0, 2) tensor field and (2, 0) tensor field mean? $\endgroup$ Jan 26, 2021 at 13:54
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    $\begingroup$ Hi @HelloGoodbye: Plainly speaking, it counts the number of upper and lower indices of the tensor components. $\endgroup$
    – Qmechanic
    Jan 26, 2021 at 13:57
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    $\begingroup$ @HelloGoodbye: E.g. $g_{\mu\nu}=2\eta_{\mu\nu}$. $\endgroup$
    – Qmechanic
    Feb 17, 2021 at 23:15
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Even in flat spacetime, the metric does not have to have the form $\operatorname{diag}(1,-1,-1,-1)$. For example, if you want to do physics in SI units rather than natural relativistic units where $c=1$, then you want something like $g=\operatorname{diag}(c,-1,-1,-1)$. This is not the same as its own inverse.

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  • $\begingroup$ And if you use anything else than Cartesian coordinates, the diagonal may look different, or the off-diagonal elements may even be non-zero if the grid axes are not orthogonal. I don't know why you would use such a coordinate system, especially in flat spacetime; I'm just saying that you can, and if you do it will cause the metric tensor to look different. $\endgroup$ Jan 26, 2021 at 14:00
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I'm surprised this hasn't been mentioned before, but$$g_{ij}g^{jk}=g_{i\mu}g^{\mu k}-g_{i0}g^{0k}=\delta_i^k-g_{i0}g^{0k}.$$

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