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I am trying to derive equation

$$ G_{\mu\nu}=2\nabla_{\mu}\nabla_{\nu}\phi-2\nabla_{\mu}\phi\nabla_{\nu}\phi+3g_{\mu\nu}(\nabla\phi)^2-2g_{\mu\nu}\Box\phi-\lambda^2g_{\mu\nu}e^{2\phi} $$

from these notes.

The given metric is $$ ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}+\frac{1}{\lambda^2}e^{-2\phi}d\Omega^2 $$

I tried calculating the Christoffel symbols and derive the equation using $$ G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} $$ but I'm not sure on how to handle the $d\Omega^2$ components of the metric.

Edit: The action for the given equations of motion is $$ \mathcal{S}=\frac{1}{2\pi}\int{d^2x\sqrt{-g} e^{-2\phi}\left(R+2\nabla_{\mu}\phi\cdot\nabla^{\mu}\phi+2\lambda^2e^{2\phi}\right)} $$

My Answer: Special thanks to ApolloRa I litterally used multiple answers from you so thanks!

Firstly, to obtain the Einstein tensor we have to vary the action and from the equations of motion solve for $R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$. Now the action reads

$$ \mathcal{S}=\frac{1}{2\pi}\int{d^2x \sqrt{-g}e^{-2\phi}\left(R+2g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi+2\lambda^2e^{2\phi}\right)} $$ the variation w.r.t to $g_{\mu\nu}$ is $$ \delta\mathcal{S}=\frac{1}{2\pi}\int{d^2x}\left(\delta\sqrt{-g}e^{-2\phi}\left(R+2g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi+2\lambda^2e^{2\phi}\right)+\sqrt{-g}\left(\delta(e^{-2\phi}R)+2\delta(e^{-2\phi}g_{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi\right)\right) $$ Now I show $\delta(e^{-2\phi}R)=\delta(e^{-2\phi}g^{\mu\nu}R_{\mu\nu})$ where $$ \delta(e^{-2\phi}g^{\mu\nu}R_{\mu\nu}) = e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu} + g^{\mu\nu}e^{-2\phi}\delta R_{\mu\nu} $$

Then we use the Palatini identity where we have $$ e^{-\phi}\delta R_{\mu\nu} g^{\mu\nu} = e^{-2\phi}(g_{\mu\nu}\Box - \nabla_\mu\nabla_\nu)\delta g^{\mu\nu} $$ Now for these 2 terms we use the Leibniz rule as many times as needed to get the $\delta g_{\mu\nu}$ outside the derivative.

When we do that we get $$ e^{-2\phi} g_{\mu\nu}\Box \delta g^{\mu\nu} = \text{total derivative} -2e^{-2\phi}g_{\mu\nu}\delta g^{\mu\nu}\Box \phi + 4e^{-2\phi}g_{\mu\nu}\delta g^{\mu\nu}(\nabla\phi)^2 $$ and

$$ e^{-2\phi}\nabla_\mu\nabla_\nu\delta g^{\mu\nu} = \text{total derivative} +\nabla_\mu\nabla_\nu(e^{-2\phi}) \delta g^{\mu\nu} $$

Now combining these together we get $$ \delta(e^{-2\phi}R) = \text{total derivatives} +e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu}-2e^{-2\phi}g_{\mu\nu}\Box\phi\delta g^{\mu\nu}+4e^{-2\phi}g_{\mu\nu}(\nabla\phi)^2\delta g^{\mu\nu}+2\nabla_\mu\nabla_\nu e^{-2\phi}\delta g^{\mu\nu}-4\nabla_\mu\phi\nabla_\nu\phi e^{-2\phi}\delta g^{\mu\nu} $$ Now the variation of the action becomes $$ \delta\mathcal{S} = \frac{1}{2\phi}\int{d^2x\left(-\frac{1}{2}\sqrt{-g}\delta g^{\mu\nu}g_{\mu\nu}e^{-2\phi}(R+2(\nabla\phi)^2+2\lambda^2e^{2\phi})+\sqrt{-g}\left(\text{total derivatives} + e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu}-2e^{-2\phi}g_{\mu\nu}\Box\phi\delta g^{\mu\nu}+4e^{-2\phi}g_{\mu\nu}(\nabla\phi)^2\delta g^{\mu\nu}+2\nabla_\mu\nabla_\nu\phi e^{-2\phi}\delta g^{ \mu\nu}-2\nabla_\mu\phi\nabla_\nu\phi e^{-2\phi}\delta g^{\mu\nu}\right)+\sqrt{-g} 2 e^{-2\phi}\nabla_\mu\phi\nabla_\nu\phi\delta g^{\mu\nu}\right)} $$ Imposing $\delta\mathcal{S}=0$ and that the total derivatives vanish we get $$ -\frac{1}{2}g_{\mu\nu}(R+2(\nabla\phi)^2+2\lambda^2e^{2\phi})+ R_{\mu\nu}-2g_{\mu\nu}\Box\phi+4g_{\mu\nu}(\nabla\phi)^2+2\nabla_\mu\nabla_\nu\phi - 4\nabla_\mu\phi\nabla_\nu\phi+2\nabla_\mu\phi\nabla_\nu\phi = 0 $$ Now solving for $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=G_{\mu\nu}$ I get $$ G_{\mu\nu}=-2\nabla_\mu\nabla_\nu\phi +2\nabla_\mu\nabla\nu\phi-3g_{\mu\nu}(\nabla\phi)^2+2g_{\mu\nu}\Box\phi+\lambda^2g_{\mu\nu}e^{2\phi} $$ which is almost correct if you ignore the minus sign. I have been searching for the sign error for too long now. Where is the error?

PS:ApolloRa thank you again!

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    $\begingroup$ You haven’t mentioned some crucial details: $\mu$ and $\nu$ range over only $t$ and $r$, and $g_{\mu\nu}$ and $\phi$ are functions only of $t$ and $r$. $\endgroup$
    – G. Smith
    Commented Feb 28, 2021 at 20:35

2 Answers 2

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If you want to derive the Einstein equation coming from the action the author gives you have to vary with respect to the metric tensor everywhere it appears and cancel total divergence terms everywhere they appear.

The box and the double covariant derivative terms come from the $e^{-2\phi} R$ term in the action, see for example my answer here: Metric field equations for the Jordan-Brans-Dicke action

In the field equation $\phi$ represents the scalar field and not the angle see author's comment in parenthesis.

Comment on version after the sixth edit:

The term $-\nabla_{a}\nabla_{b}\delta g^{ab} e^{-2\phi}$ comes with the minus sign. I obtained the variation of this term as:

$$(4\nabla_{a}\phi \nabla_{b}\phi e^{-2\phi} - 2e^{-2\phi} \nabla_{a}\nabla_{b}\phi) \delta g^{ab}$$

It seems to me that you forgot the initial minus sign.

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  • $\begingroup$ Does that mean that the variation of the terms with no R or $\sqrt{-g}$ is zero? Also what about the $\nabla_{\mu}\phi\nabla^{\mu}\phi$ term? $\endgroup$ Commented Mar 5, 2021 at 21:34
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    $\begingroup$ First of all for the nature of your question, you're trying to obtain a field equation not a component of the field equation. Plugging the metric, christoffels etc into the Einstein tensor will not give you what you're looking fot (the first equation in your post). $\endgroup$
    – Noone
    Commented Mar 5, 2021 at 21:39
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    $\begingroup$ Variation with respect to the metric will yield the equation you want. All terms are at least minimally coupled to gravity via $\sqrt{-g}$, so all terms contribute. Unfortunately all terms need to be varied in order to obtain the equation. $\endgroup$
    – Noone
    Commented Mar 5, 2021 at 21:44
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    $\begingroup$ Yes, when you vary with respect to the metric, $\phi$ is fixed: $\delta \phi =0$. $\endgroup$
    – Noone
    Commented Mar 5, 2021 at 21:44
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    $\begingroup$ The kinetic term of the scalar field also contains the metric: $$\delta (\sqrt{-g}(g^{ab}\nabla_{a}\phi\nabla_{b}\phi))$$ $\endgroup$
    – Noone
    Commented Mar 5, 2021 at 21:47
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Written out explicitly, you have $$\mathrm ds^2 = g_{00} (\mathrm dx^0)^2 + (g_{01}+g_{10}) \mathrm dx^0 \mathrm dx^1 + g_{11} (\mathrm dx^1)^2+ \frac{1}{\lambda^2} e^{-2\phi}\big((\mathrm dx^2)^2 + \sin^2(x^2) (\mathrm dx^3)^2\big)$$

or in more suggestive notation, $$\mathrm ds^2=g_{tt} \mathrm dt^2 + (g_{tr}+g_{rt}) \mathrm dt \mathrm dr + g_{rr} \mathrm dr^2 + \frac{1}{\lambda^2} e^{-2\phi}(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\varphi^2)$$

In matrix form, your metric would look like

$$g_{\mu\nu} = \pmatrix{g_{00} & g_{01} & 0 & 0 \\ g_{10} & g_{11} & 0 & 0 \\ 0 & 0 & \frac{1}{\lambda^2}e^{-2\phi} & 0 \\ 0 & 0 & 0 & \frac{1}{\lambda^2} e^{-2\phi} \sin^2(\theta)}$$

Using that, you should be able to compute the required quantities. Note that the assumption of spherical symmetry implies that $g_{\mu\nu}, \mu,\nu=0,1$ and $\phi$ are functions of $t$ and $r$ only.

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