Graduate Solving Almheriri's Dilaton-Gravity Model in AdS##_2##

[Fgard]

I am going through Almheriri's article about " Models of AdS$_2$ and I am stuck on a derivation. I think they make some kind of assumption which I don't understand.

What I am trying to do, is to compute the equation of motion by varying the action with respect to the metric. Unfortunately I am stuck; I know what the answer is, but it is not what I get. I'll describe first what I have done, firstly I have computed the Ricci scalar from the conformal gauge
$$ ds^2=-e^{2\omega}dx^-dx^+$$
which is $R=8e^{-2\omega}\partial_+\partial_-\omega$. The action is
$$I=\int d^2x\sqrt{-h}(\Phi^2R+\lambda(\partial\Phi)^2-U(\Phi^2/d^2)) $$
where, $\Phi^2$ is the dilation, and $U$ is an arbitrary potential. To highlight the metric dependence of the action I have written as
$$I=\int d^2x\sqrt{-h}(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2)). $$
Then I have the variation is
$$ \delta I=\int d^2x\left[\delta(\sqrt{-h})(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2))+\sqrt{-h}(\Phi^2R_{\alpha\beta}\delta(h^{\alpha\beta})+\lambda \partial_{\alpha}\Phi\partial_{\beta}\Phi\delta(h^{\alpha\beta})) \right]$$
where I have used that the boundary is static, and I have that the variation of $\delta(\sqrt{-h})=-\frac{1}{2}\sqrt{-h}h_{\alpha\beta}\delta(h^{\alpha\beta}).$
Plugging that in,
$$-\frac{1}{2}h_{\alpha\beta}(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2))+\Phi^2R_{\alpha\beta}+\lambda \partial_{\alpha}\Phi\partial_{\beta}\Phi=0, $$
then I could replace $h^{\alpha\beta}h_{\alpha\beta}=\delta^{\alpha}_{\beta}$. But the problem is that I know that the answer is supposed to be just
$$ -e^{2\omega}\partial_+\left(e^{-2\omega}\partial_+\Phi^2 \right)=0,$$
and I don't see how I should get that. Especially, I don't understand how they get rid of the potential term.

 

[craigthone]

I am going through Almheriri's article about " Models of AdS$_2$ and I am stuck on a derivation. I think they make some kind of assumption which I don't understand.

What I am trying to do, is to compute the equation of motion by varying the action with respect to the metric. Unfortunately I am stuck; I know what the answer is, but it is not what I get. I'll describe first what I have done, firstly I have computed the Ricci scalar from the conformal gauge
$$ ds^2=-e^{2\omega}dx^-dx^+$$
which is $R=8e^{-2\omega}\partial_+\partial_-\omega$. The action is
$$I=\int d^2x\sqrt{-h}(\Phi^2R+\lambda(\partial\Phi)^2-U(\Phi^2/d^2)) $$
where, $\Phi^2$ is the dilation, and $U$ is an arbitrary potential. To highlight the metric dependence of the action I have written as
$$I=\int d^2x\sqrt{-h}(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2)). $$
Then I have the variation is
$$ \delta I=\int d^2x\left[\delta(\sqrt{-h})(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2))+\sqrt{-h}(\Phi^2R_{\alpha\beta}\delta(h^{\alpha\beta})+\lambda \partial_{\alpha}\Phi\partial_{\beta}\Phi\delta(h^{\alpha\beta})) \right]$$
where I have used that the boundary is static, and I have that the variation of $\delta(\sqrt{-h})=-\frac{1}{2}\sqrt{-h}h_{\alpha\beta}\delta(h^{\alpha\beta}).$
Plugging that in,
$$-\frac{1}{2}h_{\alpha\beta}(\Phi^2h^{\alpha\beta}R_{\alpha\beta}+\lambda h^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi-U(\Phi^2/d^2))+\Phi^2R_{\alpha\beta}+\lambda \partial_{\alpha}\Phi\partial_{\beta}\Phi=0, $$
then I could replace $h^{\alpha\beta}h_{\alpha\beta}=\delta^{\alpha}_{\beta}$. But the problem is that I know that the answer is supposed to be just
$$ -e^{2\omega}\partial_+\left(e^{-2\omega}\partial_+\Phi^2 \right)=0,$$
and I don't see how I should get that. Especially, I don't understand how they get rid of the potential term.

I guess you missed some terms in $\delta R$
$$\delta R=R_{\alpha\beta} \delta g^{\alpha\beta}+g_{\alpha\beta} \delta R^{\alpha\beta}=R_{\alpha\beta} \delta g^{\alpha\beta}+g_{\alpha\beta}\nabla^2 \delta g^{\alpha\beta}-\nabla_\alpha \nabla_\beta \delta g^{\alpha\beta}$$

 

[haushofer]

A quick remark: you have double indices, so your expression in the OP becomes probably

I am going through Almheriri's article about " Models of AdS$_2$ and I am stuck on a derivation. I think they make some kind of assumption which I don't understand.

Plugging that in,
$$-\frac{1}{2}h_{\alpha\beta}(\Phi^2h^{\mu\nu}R_{\mu\nu}+\lambda h^{\mu\nu}\partial_{\mu}\Phi\partial_{\nu}\Phi-U(\Phi^2/d^2))+\Phi^2R_{\alpha\beta}+\lambda \partial_{\alpha}\Phi\partial_{\beta}\Phi=0, $$
then I could replace $h^{\alpha\beta}h_{\alpha\beta}=\delta^{\alpha}_{\beta}$. But the problem is that I know that the answer is supposed to be just
$$ -e^{2\omega}\partial_+\left(e^{-2\omega}\partial_+\Phi^2 \right)=0,$$
and I don't see how I should get that. Especially, I don't understand how they get rid of the potential term.

Your remark about the contraction between your metric and inverse metric then becomes wrong (which is wrong anyway; a total contraction becomes 2, the trace of the identity matrix in 2 dimensions).