# Metric field equations for the Jordan-Brans-Dicke action

Considering the Jordan-Brans-Dicke action:

$$S=\int d^4x\sqrt{-g}\left(\phi R+\frac\omega\phi(\partial\phi)^2+\mathfrak{L_{m}}(\psi)\right).$$

I was trying to get the metric field equations by varying the metric and got this:

$$-\frac{1}{2}g_{\mu\nu}R+R_{\mu\nu}+\frac{\omega}{\phi^2}[-\frac{1}{2}g_{\mu\nu}(\partial\phi)^2+\partial_\mu\phi\partial_\nu\phi]-\frac{1}{2\phi}g_{\mu\nu}\mathfrak{L_{m}}(\psi)=0$$

I varied the terms $$\sqrt{-g}$$, $$R_{\mu\nu}$$ , $$g^{\mu\nu}$$ and $$\partial_\mu \phi \partial_\nu \phi g^{\mu\nu}$$. If we are only conserned for the equations of the metric field then this is it right? If I wanted the equations for the gravitational field we would have to vary w.r.t. the metric and the field $$\phi$$ right?

EDIT: On the 2nd Leibniz rule I considered:

$$-\nabla^{\alpha}\nabla_{\alpha}(g_{\mu\nu}\phi\delta g^{\mu\nu}) = -g_{\mu\nu}\nabla^{\alpha}\nabla_{\alpha}(\phi) \delta g^{\mu\nu}-g_{\mu\nu}\nabla^{\alpha} (\phi)\nabla_{\alpha}(\delta g^{\mu\nu})-g_{\mu\nu}\nabla_{\alpha} (\phi)\nabla^{\alpha}( \delta g^{\mu\nu})-g_{\mu\nu} \phi \nabla^{\alpha}\nabla_{\alpha}(\delta g^{\mu\nu})$$

I pulled out the metric so I dont have to deal with 6 terms. The ones we want are only the first and second in the RHS of this equation

• What happened to /phi R term?? Commented Oct 25, 2020 at 20:02
• I divided the whole equation by $\phi$ and that one would be the first term that appears Commented Oct 25, 2020 at 20:33
• Τhats wrong. You have to integrate by parts. Commented Oct 25, 2020 at 20:38
• You have already applied the first covariant derivative. You have to do it once again with the other covariant derivative. Commented Oct 26, 2020 at 14:23
• What do you mean? I applied both of them already. What am I confusing where? Commented Oct 26, 2020 at 14:48

The $$\delta(\phi R)$$ term will be:

$$\delta(\phi R) = \delta(\phi g^{\mu\nu}R_{\mu\nu}) = \phi\delta g^{\mu\nu}R_{\mu\nu} +\phi\delta R_{\mu\nu}g^{\mu\nu}$$

The term: $$\phi\delta g^{\mu\nu}R_{\mu\nu}$$ is ready, here the variation of the inverse metric tensor is already a multiplying factor. Now the second term is:

$$\phi\delta R_{\mu\nu}g^{\mu\nu} = \phi (g_{\mu\nu}\Box - \nabla_{\mu}\nabla_{\nu})\delta g^{\mu\nu}$$

where i've used the Palatini Identity. Now we have for example for the box term:

$$\phi g_{\mu\nu}\Box\delta g^{\mu\nu} = \phi g_{\mu\nu}\nabla^{\alpha}\nabla_{\alpha}\delta g^{\mu\nu} =\nabla^{\alpha}(\phi g_{\mu\nu}\nabla_{\alpha}\delta g^{\mu\nu}) -\nabla^{\alpha}\phi g_{\mu\nu}\nabla_{\alpha}\delta g^{\mu\nu}$$

The first term is a total derivative. We will ignore it as a boundary term. Now we use Leibniz rule once again:

$$-\nabla^{\alpha}\phi g_{\mu\nu}\nabla_{\alpha}\delta g^{\mu\nu} = -\nabla^{\alpha}\nabla_{\alpha}(g_{\mu\nu}\phi\delta g^{\mu\nu}) + g_{\mu\nu}\delta g^{\mu\nu}\nabla^{\alpha}\nabla_{\alpha}(\phi)$$

where i've used metric compatibillity. So we have:

$$\phi g_{\mu\nu}\Box\delta g^{\mu\nu} = g_{\mu\nu}\delta g^{\mu\nu}\nabla^{\alpha}\nabla_{\alpha}(\phi) = g_{\mu\nu}\delta g^{\mu\nu} \Box \phi$$ One has to do the same procedure for the two covariant derivatives. The other terms seem correct.

The problem here is that the Ricci Scalar is coupled with $$\phi$$. When i first came across such coupling terms i had the same problem. In the context of General relativity, the action is:

$$S = \int d^4x \sqrt{-g}R.$$

The variation gives rise to the term $$g^{\mu\nu}\delta R_{\mu\nu}$$. We can show that this term is a total derivative term and cancel it. In the context of Brans Dicke (or other geometric modifications to Einstein Gravity, $$f(R)$$ for example, Horndeski, or matter fields non-minimally coupled to gravity) this term is no longer a total divergence. Here, this term is : $$\phi\delta R_{\mu\nu}g^{\mu\nu}$$. $$\phi$$ makes things tricky, we cannot now discard this term as it is, it is not a total derivative term. Thus, we follow the procedure i described above.

Regarding the second part of the question, yes you have to vary also with respect to $$\phi$$. Here $$\phi$$ is not a matter field, it is a geometric quantity.

• Thank you! In all the action variations I´ve done so far R wasn´t coupled with any other field so I just do it fast by claiming the Palatini identity and Gauss theorem and say it´s zero. Commented Oct 25, 2020 at 22:04
• I understand your situation 100%. Been there, done exactly the same thing. Commented Oct 25, 2020 at 22:14
• Just one thing, isn´t there a $g_{\mu\nu}$ missing in the 4th equation you wrote on the 1st term of the RHS? Commented Oct 25, 2020 at 22:54
• Of course!! Thank you for pointing it out! Commented Oct 25, 2020 at 22:56
• Sorry could you elaborate in how you did the second Leibniz rule? I get 4 terms when I distribute the two covariant derivatives for this term: $\nabla^{\alpha} \nabla_{\alpha}(g_{\mu\nu} \phi \delta g^{\mu\nu})$ considering the metric compatibility I took the metric out directly and got the other 4 terms, do 2 of them cancel each other out? Commented Oct 26, 2020 at 12:16