Undergrad Ricci tensor from this action

[ergospherical]

Here is an action for a theory which couples gravity to a field in this way:$$S = \int d^4 x \ \sqrt{-g} e^{\Phi} (R + g^{ab} \Phi_{;a} \Phi_{;b})$$I determine\begin{align*}
\frac{\partial L}{\partial \phi} &= \sqrt{-g} e^{\Phi} (R + g^{ab} \Phi_{;a} \Phi_{;b}) \\
\nabla_a \frac{\partial L}{\partial(\nabla_a \Phi)} &= 2\sqrt{-g} e^{\Phi} g^{ab} (\Phi_{;a} \Phi_{;b} + \Phi_{;ba})
\end{align*}giving $R = g^{ab} (\Phi_{;a} \Phi_{;b} + 2\Phi_{;ba})$. Now vary the action with respect to the metric,\begin{align*}
\frac{\delta S}{\delta g^{ab}} &= -\frac{1}{2}\sqrt{-g} e^{\Phi} g_{ab} (R + g^{cd} \Phi_{;c} \Phi_{;d}) + \sqrt{-g} e^{\Phi}(\frac{\delta R}{\delta g^{ab}} + \Phi_{;a} \Phi_{;b}) \end{align*}Put $\delta R_{ab} / \delta g^{ab} = R_{ab}$ and insert the previous equation for $R$. Zero the variation and cancel the common factor $\sqrt{-g} e^{\Phi}$,$$0 = -g_{ab} g^{cd} (\Phi_{;c} \Phi_{;d} + \Phi_{;dc}) + R_{ab} + \Phi_{;a} \Phi_{;b}$$This should give $R_{ab} = \Phi_{;ba}$ but it doesn't work out that way because the indices are mangled. Can somebody see the error?

 

[PeterDonis]

Here is an action for a theory

Is there a particular source from which you got this? It looks like it might be the general action for one kind of scalar-tensor theory of gravity.

 

[ergospherical]

It's just a very old exam. I presumed it was just something arbitrary the examiner thought up, but may have some application.

 

[OlderWannabeNewton]

Not sure if you eventually managed to figure it out but your error is you're overloading indices when taking the variations. Generally speaking, it's better to give separate indices for the field you're taking variations with respect to, compared to the indices in the quantity you're computing variations of.

We have $$\frac{\delta S}{\delta \nabla_c \Phi} = 2\sqrt{-g}e^{\Phi}\nabla^c \Phi,$$ hence $$\nabla_c\frac{\delta S}{\delta \nabla_c \Phi} = 2\sqrt{-g}e^{\Phi}\nabla^c \nabla_c \Phi.$$ Thus $$R = 2\nabla^a \nabla_a \Phi - \nabla^a \Phi \nabla_a \Phi.$$

We can now compute $\frac{\delta S}{\delta g^{ab}}$. The general expression you calculated is indeed correct: $$R_{cd} = \frac{\delta S}{\delta g^{cd}} = \sqrt{-g}e^{\Phi}\left(\frac{\delta R}{\delta g^{cd}} - \frac{1}{2}g_{cd}R + \nabla_c \Phi \nabla_d \Phi - \frac{1}{2}g_{cd}\nabla^e \Phi \nabla_e \Phi\right).$$

We have $$\frac{\delta R}{\delta g^{cd}} = 2\nabla_c \nabla_d \Phi - \nabla_c \Phi \nabla_d \Phi.$$ Plugging in for $R$ we find the desired result $$R_{ab} = \nabla_a \nabla_b \Phi.$$

Hope that helps!

 

[ergospherical]

Yeah, it was a mistake in $\nabla_a \frac{\partial L}{\partial(\nabla_a \Phi)}$. Hard for me to figure out why it happened. :)