Physics Asked on June 18, 2021
Considering the action for induced gravity:
$$S=int d^4xsqrt{-g}left(epsilonphi^2 R-frac{1}{2}(partialphi)^2+V(phi)right).$$
I was trying to get the metric field equations by doing the usual procedure and for the $delta(phi^2 epsilon R)$ term, using the Leibniz rule twice, I got this:
$$ delta(phi^2epsilon R) = [epsilon phi^2 R_{munu} +g_{munu}square(epsilonphi^2) – nabla_{mu}nabla_{nu}(epsilonphi^2)] delta g^{munu} $$
Is this correct? Or do I have to do some type of chain rule for $phi^2$ in the Leibniz rules?
Yes it is correct. You can aply the covariant derivatives in $phi^2$:
$$nabla_{mu}nabla_{nu}phi^2 = nabla_{mu}(2phi nabla_{nu} phi) = 2 nabla_{mu}phinabla_{nu} phi + 2phinabla_{mu}nabla_{nu} phi$$
and the same for the box term, but your calculation is correct!
Correct answer by ApolloRa on June 18, 2021
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