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Varying $\partial_\lambda\phi\,\partial^\lambda\phi$ wrt the metric tensor $g_{\mu\nu}$ in two different ways gives me different results. Obviously I'm doing something wrong. Where am I going wrong?

Method 1: \begin{align} (\delta g_{\mu\nu})\,\partial^\mu\phi\,\partial^\nu\phi \end{align}

Method 2: \begin{align} &(\delta g^{\mu\nu})\,\partial_\mu\phi\,\partial_\nu\phi \nonumber \\ =&(-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma})\,\partial_\mu\phi\,\partial_\nu\phi \quad(\because \delta g^{\mu\nu}=-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma} \,\,\text{as can be checked by varying the identity}\,\, g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu) \nonumber\\ =&-(\delta g_{\rho\sigma})\,\partial^\rho\phi\,\partial^\sigma\phi \end{align} The second result differs from the first one by a minus sign. What's going wrong?

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    $\begingroup$ Method 1 is wrong because $\partial^\mu\phi=g^{\mu\nu}\partial_\nu\phi$ itself depends on the metric. Only the lower-indexed derivatives are "natural". $\endgroup$ Commented Apr 22 at 16:19
  • $\begingroup$ Check Method 2. Hint : $\partial_{\mu}\phi=\delta^{\sigma}_{\mu}\partial_{\sigma}\phi$. Now Kronecker Delta is a constant matrix so maybe try to slide it inside the variation of inverse metric $\endgroup$
    – paul230_x
    Commented Apr 22 at 16:52
  • $\begingroup$ @KP99, I didn't fully understand what you mean. $\endgroup$
    – vyali
    Commented Apr 22 at 17:18

1 Answer 1

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As mentioned in the comments, the first method is incorrect. $\partial_\mu \phi$ is something that is always well-defined in a manifold (with respect to some coordinate system) and does not depend on the metric at all. $\partial^\mu \phi$, however, is defined as $g^{\mu\nu} \partial_\nu \phi$, which clearly has a dependence on the metric. Therefore, if you want to use the first method, you also need to consider the variations in $\partial^\mu\phi$. In practice, I guess this will end up being more laborious than just performing the second method.

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