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I have trouble with obtaining d'Alambert equation for scalar field in Brans-Dicke gravity (http://www.scholarpedia.org/article/Jordan-Brans-Dicke_Theory).

B-D gravity langrangian density is given by: $$\mathcal{L}=\phi R + 16\pi\mathcal{L}_{matter} -\frac{\omega(\phi_{,\alpha}\phi^{,\alpha} )}{\phi},$$ with corresponding variational principle:

$$\delta\int (-g)^{1/2}[\phi R + 16\pi\mathcal{L}_{matter} -\frac{\omega(\phi_{,\alpha}\phi^{,\alpha} )}{\phi}]d^4x = 0.$$

When i'm using standard Euler-Langrange equation for $\phi$:

$\frac {\partial\mathcal{L}}{\partial\phi}-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0$

I get:

$\frac {\partial\mathcal{L}}{\partial\phi}=R + (-\omega\frac {\partial{\frac{\phi_{,\alpha}\phi^{,\alpha}}{\phi}}}{\partial\phi})=R-(-\frac{\omega(\phi_{,\alpha}\phi^{,\alpha} )}{\phi^2})=R + \frac{\omega(\phi_{,\alpha}\phi^{,\alpha} )}{\phi^2}$

and:

$\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=-\frac{\omega}{\phi} \partial_\mu(\frac{\partial\mathcal{\frac{\phi_{,\alpha}\phi^{,\alpha}}{\phi}}}{\partial(\partial_\mu\phi)})=-\frac{\omega}{\phi} g^{\alpha\beta}(\partial_\mu(\frac{\partial_\alpha\phi\partial_\beta\phi}{\partial(\partial_\mu\phi)}))=-\frac{\omega}{\phi} g^{\alpha\beta}\partial_\mu(\delta^\mu_\alpha\partial_\beta\phi+\delta^\mu_\beta\partial_\alpha\phi)=-\omega \partial_\mu[g^{\mu\beta}\partial_\beta\phi+g^{\alpha\mu}\partial_\alpha\phi]=-\frac{\omega}{\phi} \partial_\mu[g^{\beta\mu}\partial_\beta\phi+g^{\alpha\mu}\partial_\alpha\phi] =-\frac{\omega}{\phi} \partial_\mu2g^{\mu\beta}\partial_\beta\phi=-2\frac{\omega}{\phi} g^{-1/2} \partial_\mu[g^{1/2}g^{\mu\beta}\partial_\beta\phi]=-2\frac{\omega}{\phi} \Box\phi$

Which after plugging into E-L equation gives:

$0=R+\omega\frac{\phi_{,\alpha}\phi^{,\alpha}}{\phi^2}+(2\omega\Box\phi)/\phi$

After multiplying by $\phi/2\omega$ i arrived at:

$0=\frac{\phi R}{2\omega}+\frac{\phi_{,\alpha}\phi^{,\alpha}}{2\phi}+\Box\phi$

However correct equation should be (from article linked above):

$0=\frac{\phi R}{2\omega}-\frac{\phi_{,\alpha}\phi^{,\alpha}}{2\phi}+\Box\phi$

How to do it correctly?

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There is a mistake in the beginning, you should use E-L equations in curved spacetime. Here is a detailed calculation

The action is given by, $$ S^{(BD)} = \int d^4 x \sqrt{|g|} \left[ \phi R - \frac{\omega}{\phi} g^{\mu \nu} \, \nabla_\mu \phi \nabla_\nu \phi - V(\phi) \right] $$

EoM for BD scalar field

Euler - Lagrange Equations in curved spacetime \begin{align} \frac{\partial\mathcal{L}}{\partial\phi}=\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] \end{align} OR \begin{align} \frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] \end{align} Let us evaluate the $\phi$ differentiation, \begin{align} \frac{\partial\mathcal{L}}{\partial\phi} = R + \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \nabla_\nu \phi - \frac{\partial V}{\partial \phi} \end{align} and evaluating RHS yields, \begin{align} \begin{aligned} \nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] &= \nabla_{\mu} \left[ - \frac{\omega}{\phi} g^{\mu \nu} \left(\nabla_\nu \phi + \underbrace{\nabla_{\mu} \phi \, \delta^\mu{}_\nu}_{\nabla_\nu \phi} \right)\right] \\[1ex] &= \nabla_{\mu} \left[ - \frac{\omega}{\phi} g^{\mu \nu} \left(2\nabla_\nu \phi \right)\right]\\[1ex] &= -2 \omega \nabla_{\mu} \left( \frac{1}{\phi} g^{\mu \nu} \nabla_{\nu} \phi \right) \\[1ex] &= -2 \omega \nabla_{\mu} \left( \frac{1}{\phi} \nabla^{\mu} \phi \right) \\[1ex] &= -2 \omega \left[ \left(\nabla_{\mu} \frac{1}{\phi}\right) \nabla^\mu \phi + \frac{1}{\phi} \underbrace{\nabla_\mu \nabla^\mu \phi}_{\Box \phi} \right] \\[1ex] &= -2 \omega \left[ \left(-\frac{1}{\phi^2} \nabla_{\mu} \phi \right) \nabla^\mu \phi + \frac{1}{\phi} \Box \phi \right]\\[1ex] &= \frac{2 \omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi - \frac{2\omega}{\phi} \Box \phi \end{aligned} \end{align}

Therefore, \begin{align} \begin{aligned} \frac{\partial\mathcal{L}}{\partial\phi} - \nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] = R + \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi - \frac{\partial V}{\partial \phi} + \frac{2 \omega}{\phi} \Box \phi \end{aligned} \end{align} \begin{align} \boxed{R - \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi - \frac{\partial V}{\partial \phi} + \frac{2 \omega}{\phi} \Box \phi = 0} \end{align}

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