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In Weinberg's Gravtation and Cosmology Part four chapter 12 section 2. There is a equation I don't quiet understand which is shown below.

$$0=\delta (g_{\mu \nu}g^{\nu \lambda})=g_{\mu \nu}\delta (g^{\nu \lambda})+g^{\nu \lambda}\delta (g_{\mu \nu})$$

I think $g_{\mu \nu}g^{\nu \lambda}$ is not constant, why the darivation of it is zero?

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1 Answer 1

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$g_{\mu\lambda}g^{\lambda\nu}=\delta_\mu^\nu$ is constant, so its variation is zero.

You can also write this equation as $g g^{-1} = 1$, where $1$ is the identity tensor. The variation of $g^{-1}$ is defined such that $gg^{-1}=1$ remains true after the variation.

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    $\begingroup$ Identity tensor, btw +1 $\endgroup$
    – basics
    Commented Nov 10, 2022 at 23:54
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    $\begingroup$ @basics An up-down 2-index tensor is a matrix :) $\endgroup$
    – Andrew
    Commented Nov 11, 2022 at 0:03
  • $\begingroup$ @Andrew An up-down 2-index tensor is a linear transformation from a vector space to itself. It becomes a matrix the moment you settle on a coordinate grid and use said coordinate grid to describe the tensor as a two-dimensional array of numbers. $\endgroup$
    – Arthur
    Commented Nov 11, 2022 at 11:39
  • $\begingroup$ @Andrew I have to disagree, firmly, since this is one of the easiest thing to believe to mess everything up with tensors. I'd agree if you say "the components of a $2^{nd}$-order tensor can be collected in a matrix" $\endgroup$
    – basics
    Commented Nov 11, 2022 at 13:08

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