# How to obtain the field equations in Brans-Dicke theory from the action?

The action for the Brans-Dicke-Jordan theory of gravity is $$\\S =\int d^4x\sqrt{-g} \; \left(\frac{\phi R - \omega\frac{\partial_a\phi\partial^a\phi}{\phi}}{16\pi} + \mathcal{L}_\mathrm{M}\right).$$ And the field equations of the gravitation field are $$G_{ab} = \frac{8\pi}{\phi}T_{ab}+\frac{\omega}{\phi^2} (\partial_a\phi\partial_b\phi-\frac{1}{2}g_{ab}\partial_c\phi\partial^c\phi) +\frac{1}{\phi}(\nabla_a\nabla_b\phi-g_{ab}\Box\phi).$$ I tried to vary this action w.r.t $g_{ab}$ but failed. How can I obtain these from the action? Where can I get the detailed derivation? Thx!

• What have you tried so far? Are you familiar with how to vary the Einstein Hilbert action (see eg: en.wikipedia.org/wiki/…)? If you can handle the Einstein Hilbert action, this is a relatively simply extension. To vary the Brans Dicke action wrt $g^{ab}$ you only need to know how to vary $\sqrt{-g}$, $R$, and $g^{ab}$, all of which appear in varying Einstein Hilbert. (Technical aside: it's usually easier to vary wrt $g^{ab}$ than $g_{ab}$). – Andrew Dec 10 '13 at 2:41
• Also if you look at the eoms, $G_{ab}$ and $T_{ab}$ make sense, they appear after varying Einstein Hilbert. The term proportional to $\omega$ is the stress energy for a scalar field, so that makes sense. The tricky term is the last one. There are some integrations by parts you have to do in varying the Einstein Hilbert term. Since $\phi$ multiples $R$, you will pick up some extra derivatives on $\phi$ when you vary $\sqrt{-g} \phi R$ that were not present after you vary $\sqrt{-g} R$. Again, if you go through varying the Einstein Hilbert action carefully, you can see how this term arises. – Andrew Dec 10 '13 at 2:52
• I can derive the Einstein equations from the Einstein Hilbert action, but when I vary $\sqrt{-g}\phi R$ I cannot get the extra derivatives on $\phi$. Where can I find some details about this derivation? Thx. – DHA Dec 10 '13 at 4:44
• Hm, I don't know a source that does it in detail. But here's basically how it works. On wikipedia you can see that $\delta R^\mu_{\ \nu\rho\sigma}\sim\nabla \delta \Gamma$. In $\sqrt{-g}R$, this leads to a term $\sqrt{-g} \nabla(\delta \Gamma)$, which is a total derivative. However in Brans Dicke you have $\sqrt{-g} \phi \nabla (\delta \Gamma)$, which is not a total derivative. You need to re express $\delta \Gamma$ in terms of $\delta g$, then integrate the $\nabla$ onto $\phi$. There is a handy trick: in the locally inertial frame $\nabla \delta \Gamma = \partial \delta \Gamma$. – Andrew Dec 10 '13 at 5:08
• Maybe this paper will interest you, while treating a more general case. The general Euler-Lagrange equations are given by equations $2,3$. Standard Brans-Dicke corresponds to $f= -\omega \frac{\partial_\mu \phi \partial^\mu \phi}{\phi}$, take also $J=\Lambda=0$), see equations $12,13$. Be careful that you have coupled equations, for instance $\square\phi$ depends on $T$ and $\omega$, so you may have different (but equivalent) representations for the Euler-Lagrange equations. – Trimok Dec 10 '13 at 12:31

I have been in your situation. I agree with Andrew and I've found this helpful http://arxiv.org/abs/1002.0617v4. Go directly to appendix B, There you will find the answer of an identical problem and you shall use appendix A, as well.

• Hi maha, we usually expect answers to stand on their own. Can you include a brief summary of the solution you link to or quote some of the useful content? – Brandon Enright Jan 31 '14 at 19:43
• @Brandon Well,I'm actually a beginner in the field and as you can see I avoided to use any symbol because I can't use the TeX language yet.Also I have no reputation yet to try guiding DHA via a comment.So I had to use what I'm allowed to which was answering! However I shall edit my answer as soon as possible.Thanks Brandon. – maha Feb 1 '14 at 7:20

The action which describes Brans-Dicke theory is given by,

$$S=\frac{1}{16\pi G}\int d^4x \, \sqrt{|g|} \left( -\Phi R + \frac{\omega}{\Phi}\partial_\mu \Phi \partial^\mu \Phi \right)$$

which features a scalar field $\Phi$ coupling to gravity through the Ricci scalar, and with its own kinetic term. To obtain the equations of motion, we vary our action with respect to the scalar and metric, like so,

$$\delta S = \frac{1}{16\pi G} \int d^4x \, \delta \Phi \left( -R - \frac{2\omega}{\Phi} \square \Phi + \frac{\omega}{\Phi^2} \partial_\mu \Phi \partial^\mu\Phi\right)$$ $$-\delta g^{\mu\nu} \left(\Phi G_{\mu\nu}-\frac{\omega}{\Phi} \partial_\mu \Phi \partial_\nu \Phi + \frac{1}{2}g_{\mu\nu}\frac{\omega}{\Phi}\partial_\lambda \Phi \partial^\lambda \Phi\right) + \Phi (\nabla_\mu\nabla_\nu \delta g^{\mu\nu}-\square g_{\mu\nu}\delta g^{\mu\nu})$$ where we have already performed an integration by parts. From the variation, we may deduce, $$\Phi G_{\mu\nu} - \nabla_\mu \nabla_\nu \Phi + g_{\mu\nu} \square \Phi - \frac{\omega}{\Phi} \left( \partial_\mu \Phi \partial_\nu \Phi - \frac{1}{2}g_{\mu\nu} (\nabla\Phi)^2\right) = 8\pi T_{\mu\nu}$$

for some background matter with stress-energy tensor $T_{\mu\nu}$. There is an additional equation of motion due to the scalar field, namely,

$$\Phi R + 2\omega \square \Phi - \frac{\omega}{\Phi} (\nabla\Phi)^2 = 0$$

which is zero providing the scalar field does not couple to the background matter. We can now take a trace with respect to the metric of the first equation, obtaining,

$$-\Phi R+3\square \Phi + \frac{\omega}{\Phi}(\nabla \Phi)^2 = 8\pi T$$

presuming $d=4$, where $T \equiv T^\mu_\mu$. Adding this equation to the previous, we find,

$$(3+2\omega) \square \Phi = 8\pi T.$$

The parameter $\omega$ measures how strongly $\Phi$ couples to matter content. We can rewrite the 'Einstein' field equations as,

$$R_{\mu\nu}-\frac{1}{\Phi}\nabla_\mu \nabla_\nu \Phi + \frac{1}{\Phi}g_{\mu\nu} \square \Phi - \frac{\omega}{\Phi^2}\partial_\mu \Phi \partial_\nu \Phi = \frac{8\pi}{\Phi}T_{\mu\nu} - g_{\mu\nu}\frac{\omega}{\Phi} \Phi \square \Phi$$

by expanding the Einstein tensor and substituting the relation between the Ricci scalar and field. We can now write a relation between the Ricci tensor, field and stress-energy tensor, namely,

$$R_{\mu\nu}-\frac{1}{\Phi}\nabla_\mu \nabla_\nu \Phi - \frac{\omega}{\Phi^2}\partial_\mu \Phi \partial_\nu \Phi = \frac{8\pi}{\Phi} \left( T_{\mu\nu}-\frac{(\omega+1)}{(3+2\omega)}T g_{\mu\nu} \right)$$

In "The Scalar-Tensor Theory of Gravitation", of Yasunori Fujii and Kei-ichi Maeda you can find explicitly the solution, in Appendix C (pag. 195). Personally, I really didn't like this book and even this demonstration it's very difficult to follow.

So I did it in another way. Use the usual theory for the GR part, and isolate this term:

$\int d^4 x \sqrt{-g}\Phi \delta R_{\mu\nu}g^{\mu\nu}$

Then you can use the Palatini Identity: $$\delta R_{\mu\nu}=\left( \delta \Gamma^{\alpha}_{\mu\nu} \right)_{;\alpha}-\left( \delta \Gamma^{\alpha}_{\mu\alpha} \right)_{;\nu}$$

So:

$$\begin{split} \delta R_{\mu\nu}g^{\mu\nu} &=g^{\mu\nu}\left[ \left( \delta \Gamma^{\alpha}_{\mu\nu} \right)_{;\alpha}-\left( \delta \Gamma^{\alpha}_{\mu\alpha} \right)_{;\nu} \right]= \dots \\ & \dots =\left(g^{\mu\nu} \delta \Gamma^{\alpha}_{\mu\nu} \right)_{;\alpha}-\left(g^{\mu\nu} \delta \Gamma^{\alpha}_{\mu\alpha} \right)_{;\nu}=\left(g^{\mu\nu} \delta \Gamma^{\beta}_{\mu\nu} -g^{\mu\beta} \delta \Gamma^{\alpha}_{\mu\alpha} \right)_{;\beta} \end{split}$$

In GR you don't have $\Phi$, so this term simply goes to zero thanks to Gauss's Theorem. Now you need to integrate by parts twice. The second integration by parts comes from the explicit expression of $\delta \Gamma^{\alpha}_{\mu\nu}$. In order to simplify, go into a Locally inertial frame, where:

$$\delta \Gamma^{\beta}_{\mu\nu}=\frac{1}{2}g^{\rho\beta}[ \left(\partial_\nu \delta g_{\rho\mu}\right)+\left(\partial_\mu \delta g_{\rho\nu} \right)-\left(\partial_\rho \delta g_{\nu\mu} \right) ]$$

At the end go to a general frame, so $\partial \rightarrow \nabla$. These are all the difficult steps, there are some simple calculations to be done in the middle.