Proving Lagrangian EOM with Non-minimal Coupling

[kalish]

Hi, I am currently in training and while deriving the EOM for a specific lagrangian I am having difficulties to prove that$$g^{\mu \nu} \delta R_{\mu \nu} B(\phi) = (\nabla_{\mu} \nabla_\nu - \square B g_\mu_\nu)$$ I am ashamed it might be a simple calculus but I don't see how. If you had just hints to help me that would be fair.

Moreover I would like to check wether $$\square = \frac{\partial_\mu (\sqrt{-g}g^\mu^\nu)\partial_\nu}{\sqrt{-g}}$$ as I found or $$\square = \frac{\partial_\mu\sqrt{-g}g^\mu^\nu\partial_\nu}{\sqrt{-g}}$$ as I read into one reference.


Thanks.

 

[kalish]

well latex doesn't semm to work pretty well with me...

 

[Bill_K]

◻Φ = (1/√-g)(∂_μ(√-g g^μν ∂_νΦ))

 

[kalish]

thanks.
But the more important was how I can prove the first equation about the riemann tensor.

it lacks a$$\delta g{\mu \nu}$$ on the right member of the first equation. But however I think I can manage now.

and you should read $$(\nabla_\mu \nabla_\nu -g_{\mu\nu}\square)B$$

but I found it