In general relativity one can derive the Einstein Field Equations by the principle of least action through variations with respect to the inverse of the metric tensor. In some modified theories of gravity, such as the Brans-Dicke Theory, a scalar field is added to the Einstein Hilbert Action and the gravitational constant is replaced by a function of the scalar field. I am not quite sure how to derive the field equations from this action, more specifically the part where the scalar field is attached to the Ricci scalar $\phi R$.

The Brans-Dicke Action is $$S_{BD} = \int d^4x \sqrt{-g}\left[ \frac{1}{16\pi}\left(\phi R - \frac{\omega }{\phi}g^{ab} \partial _a\phi\partial _b \phi \right) +L_M \right].$$

The resulting field equation is $$G_{ab} = \frac{8\pi}{\phi}T_{ab}+\frac{\omega}{\phi^2} (\partial_a\phi\partial_b\phi-\frac{1}{2}g_{ab}\partial_c\phi\partial^c\phi) +\frac{1}{\phi}(\nabla_a\nabla_b\phi-g_{ab}\Box\phi).$$

I also want to derive a new field equation for practice. So my questions are:

  1. How does one derive the equations of motion?

  2. How to perform the variation of the following action? $$S=\int d^4x \sqrt{-g} \left[ \frac{1}{16\pi G} R - \phi( \nabla_{\mu} g_{ab} \nabla_{\nu}g_{ab}) - 2\Lambda+L_M) \right] $$

The Ricci scalar, the cosmological constant, and the matter Lagrangian will variate simply like the Einstein Hilbert Action to: $$\delta S = \int d^4x \sqrt{-g} \left[ \frac{1}{\kappa} \left( R_{ab}-\frac{1}{2}Rg_{ab}+\Lambda g_{ab} \right) -T_{ab} \right]\delta g^{ab}.$$ What about the extra term? Would one simply variate with respect to $\phi$, or is the variation of the covariant derivative of the metric tensor also required? If the latter is true, then would the variation of this extra term be $$\frac{\partial L}{\partial g_{ab}}-\partial _\mu\frac{\partial L}{\partial (\nabla_{\mu}g_{ab})}=0.$$ Any help would be appreciated. By the way, is $\nabla_{\mu}g_{ab}\nabla_{\nu}g_{ab}$ an expression that shows the rate of change (derivative) of the metric tensor with respect to a coordinate $(t, x, y, z)$?

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    $\begingroup$ Usually, without torsion, you choose the (unique) connection such as $\nabla_\mu g_{ab} = 0$, see this PSE question $\endgroup$ – Trimok Jan 10 '14 at 19:36
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    $\begingroup$ The basic intuition underpinning the Brans-Dicke theory should be "What if we replace Newton's constant $G$ by a scalar field $\phi$? (Or, depending on your religion, $\phi^{-1}$?)" ... everything else follows from that. $\endgroup$ – Alex Nelson Jan 10 '14 at 23:05
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    $\begingroup$ Even with torsion you still get $\nabla_\mu g_{ab} = 0$. You'd need the non-metricity tensor as well to make it something else, which is pretty scarcely used. $\endgroup$ – Slereah Jan 10 '14 at 23:53

Find the answer to question 1 below. Question 2. is weird since $\nabla_\mu g_{\alpha\beta} = 0$ (when the connection is metric compatible) as mentioned by @Trimok. In any case, the variation of the action can be derived using the method described below.

We start with the BD action $$ S = \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + S_M $$ where $S_M$ is the matter action. To determine Einstein's field equations, we vary the action w.r.t to the metric. We will use the formulae (ref. wikipedia) \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} + \nabla_\sigma \left( g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} - g^{\mu\sigma} \delta \Gamma^\rho_{\rho\mu} \right) \end{split} \end{equation} The variation of the Christoffel tensor is \begin{equation} \begin{split} \delta \Gamma^\lambda_{\mu\nu} & = \delta g^{\lambda\rho} g_{\rho\alpha} \Gamma^\alpha_{\mu\nu} + \frac{1}{2} g^{\lambda\rho} \left( \partial_\mu \delta g_{\nu\rho} + \partial_\nu \delta g_{\mu\rho} - \partial_\rho \delta g_{\mu\nu} \right) \\ & = \frac{1}{2} g^{\lambda\rho} \left( \nabla_\mu \delta g_{\nu\rho} + \nabla_\nu \delta g_{\mu\rho} - \nabla_\rho \delta g_{\mu\nu} \right) \\ & = - \frac{1}{2} \left( g_{\nu\alpha} \nabla_\mu \delta g^{\alpha\lambda} + g_{\mu\alpha} \nabla_\nu \delta g^{\alpha\lambda} - g_{\mu\alpha} g_{\nu\beta} \nabla^\lambda \delta g^{\alpha\beta} \right) \end{split} \end{equation} where we used $\delta g_{\mu\nu} = - g_{\mu\alpha} g_{\nu\beta} \delta g^{\alpha\beta}$. This implies \begin{equation} \begin{split} g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} &= -\nabla_\alpha \delta g^{\alpha\sigma} + \frac{1}{2} g_{\alpha\beta}\nabla^\sigma\delta g^{\alpha\beta} \\ g^{\mu\sigma} \delta \Gamma^\lambda_{\lambda\mu} &= - \frac{1}{2}g_{\alpha\beta} \nabla^\sigma \delta g^{\alpha\beta} \end{split} \end{equation} which implies \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} - \nabla_\mu \nabla_\nu \delta g^{\mu\nu} + g_{\mu\nu} \nabla^2 \delta g^{\mu\nu} \end{split} \end{equation} Finally, from 1, we also have $$ \delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} $$ Finally, we are ready to compute the variation of the action. We have \begin{equation} \begin{split} \delta S &= \frac{1}{16\pi}\int d^4 x \delta \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi \delta R - \frac{\omega}{\phi} \delta g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + \delta S_M \\ &= - \frac{1}{32\pi}\int d^4 x \sqrt{-g} g_{\mu\nu} \left[ \phi R - \frac{\omega}{\phi} g^{\alpha\beta} \partial_\alpha \phi \partial_\beta \phi \right] \delta g^{\mu\nu} + \int d^4 x \frac{ \delta S_M}{\delta g^{\mu\nu}} \delta g^{\mu\nu} \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \left( \phi R_{\mu\nu} - \nabla_\mu \nabla_\nu \phi + g_{\mu\nu} \nabla^2 \phi \right) - \frac{\omega}{\phi} \partial_\mu \phi \partial_\nu \phi \right] \delta g^{\mu\nu} \end{split} \end{equation} Requiring that the variation of the action vanish (to leading order in $\delta g^{\mu\nu}$) gives \begin{equation} \begin{split} G_{\mu\nu} &= - \frac{16\pi}{\phi \sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} Recall that the stress-tensor is defined as \begin{equation} \begin{split} T_{\mu\nu} = - \frac{2}{\sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} \end{split} \end{equation} Thus \begin{equation} \begin{split} G_{\mu\nu} &= \frac{8\pi}{\phi} T_{\mu\nu} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} which is the Brans-Dicke equation.

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    $\begingroup$ I see how to do it now. In regards to my second question, does the extra term simply become zero as the covariant derivative of the metric tensor is zero? So the action now becomes the familiar Einstein-Hilbert Action? $\endgroup$ – user33941 Jan 11 '14 at 0:57
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    $\begingroup$ That's correct. $\endgroup$ – Prahar Mitra Jan 11 '14 at 1:00
  • $\begingroup$ How do you prove $\delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu}$ ? I cannot follow the wikipedia entry... $$ \delta \sqrt{-g} = - \frac{1}{2 \sqrt{-g} } \delta g = - \frac{1}{2 \sqrt{-g} } g g^{\mu\nu} \delta g_{\mu\nu} = - \frac{1}{2 \sqrt{-g} } \sqrt{-g} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} = - \frac{1}{2 } \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu} \neq \frac{1}{2 } \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu}$$ What about the sign-error? $\endgroup$ – Breaking Mad Sep 5 '16 at 12:22
  • $\begingroup$ @BreakingM_a_t Note that by definition $\delta g = \det ( g + \delta g) - \det g = \det g \ [ \det ( 1 + g^{-1} \delta g) - 1 \ ] $. Now, to compute $\det ( 1 + g^{-1} \delta g)$ to leading order in $\delta g$, we use the identity $\log \det M = \text{tr} \log M$. Then, we have $\det ( 1 + M ) = \exp \log \det ( 1 + M ) = \exp \text{tr} \log ( 1 + M ) = \exp [ \text{tr} ( M + {\cal O}(M^2) ) ] = \exp(\text{tr} M) + {\cal O}(M^2) = 1 + \text{tr} M + {\cal O}(M^2)$. $\endgroup$ – Prahar Mitra Feb 26 '17 at 20:39
  • $\begingroup$ @BreakingM_a_t This implies $\delta g = \det g \text{tr} ( g^{-1} \delta g ) = \det g g^{\mu\nu} \delta g_{\mu\nu}$. Thus, we find that $\delta \sqrt{ - g } = - \frac{1}{2 \sqrt{-g} } \delta g = - \frac{1}{ 2 \sqrt{ - g } } g g^{\mu\nu} \delta g_{\mu\nu} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu}$. $\endgroup$ – Prahar Mitra Feb 26 '17 at 20:48

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