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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.

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  • $\begingroup$ 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

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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 $B (\delta A) B + \delta B = 0$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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