# Variation of the kinetic term wrt the metric in scalar field theory

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?

• 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". Commented Apr 22 at 16:19
• 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 Commented Apr 22 at 16:52
• @KP99, I didn't fully understand what you mean. Commented Apr 22 at 17:18

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.