I'm reading Tong's notes on GR http://www.damtp.cam.ac.uk/user/tong/gr.html and i cannot understrand how he derived the equation that relates the varation of the metric with its inverse in page 141 under equation 4.2.

A different approach could be that $$g_{\mu\nu}g^{\mu\nu}=d \to \delta g_{\mu\nu}g^{\mu\nu} = -\delta g^{\mu\nu}g_{\mu\nu}$$ where $d$ is the spacetime dimension, but i cannot understand his calculation.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/295005/2451 , physics.stackexchange.com/q/703909/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 26, 2023 at 7:38
  • $\begingroup$ These answers did not helped. $\endgroup$
    – VladimirA
    Mar 26, 2023 at 7:43
  • $\begingroup$ Hint: since $g_{\rho\mu}g^{\mu\nu}=\delta_\rho^\nu$ is constant, $(\delta g_{\rho\mu})g^{\mu\nu}+g_{\rho\mu}\delta g^{\mu\nu}=0$. Multiply by $g^{\rho\sigma}$, relabel viz. $\mu\leftrightarrow\sigma$, then use the fact metric tensors are symmetric. $\endgroup$
    – J.G.
    Mar 26, 2023 at 9:16

1 Answer 1


Your approach won't work. How are you to isolate the $\delta g^{\mu \nu}$? You only have one scalar equation!

In the notes, we have schematically $AB = D$ where $D$ is a constant matrix and $B$ is the inverse to $A$. Now take the variation $(\delta A) B + A (\delta B) = 0$ (if you prefer think of this as equations for each component). Now we can isolate $\delta B$ by multiplying by the inverse of $A$ (which is $B$), so $B (\delta A) B + \delta B = 0$.

  • $\begingroup$ That $D$ is the identity matrix is important here, since your last step requires $BA=D$. $\endgroup$
    – J.G.
    Mar 26, 2023 at 9:09
  • $\begingroup$ My approach works fine, it is just a different approach as i clearly mention, by which we can relate the inverse variation with the normal variation. Of course, i do not expect to get a tensor equation from a scalar one. $\endgroup$
    – VladimirA
    Mar 26, 2023 at 9:10
  • $\begingroup$ @J.G. That $D$ is the identity matrix is not particularly important - its constancy is however. $\endgroup$ Mar 26, 2023 at 10:06
  • 1
    $\begingroup$ @VladimirA No your approach won't work, as I clearly mention. $\delta g^{\mu \nu}$ is a tensor with many components. The point is to solve for $\delta g^{\mu\nu}$, not have a scalar equation relating $\delta g_{\mu \nu}$ with $\delta g^{\mu\nu}$. $\endgroup$ Mar 26, 2023 at 10:06
  • $\begingroup$ I'm trying to obtain the variation of the metric determinand, my approach works for that, i'm sorry that i forgot to mention this $\endgroup$
    – VladimirA
    Mar 26, 2023 at 15:01

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.