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?


1 Answer 1


$g_{\mu\lambda}g^{\lambda\nu}=\delta_\mu^\nu$ 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.

  • 1
    $\begingroup$ Identity tensor, btw +1 $\endgroup$
    – basics
    Commented Nov 10, 2022 at 23:54
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.