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I am reading this paper on Black hole phase transition in AdS, but for the life of me I cannot get the signs right for the expression of the Action of a Black Hole in AdS (eq (2.9)).

Consider the AdS Black hole metric : $$ -Vdt^2+\frac{1}{V}dr^2+r^2 d\Omega_2\\ V = 1+\frac{r^2}{b^2}-\frac{2M}{r} $$

One can compute the Hawking temperature of such a black hole, and we find : $$ \beta = \frac{4\pi b^2 r_+}{b^2+3r_+} $$

Where $r_+$ is the schwarzschild radius, i.e. such that $V(r_+) = 0$. Note that this gives us the equation $M=\frac{1}{2}r_+(1+\frac{r_+^2}{b^2})$ which we will need. Now, we would like to compute the difference of the action of this Ads Black hole solution, minus the action of the pure AdS metric. We will do so in the Euclidean version, in other words we will Wick-rotate the metric to make it positive definite.

Consider the Einstein-Hilbert action with negative cosmological constant : $$ S = \frac{1}{16\pi}\int \sqrt{|g|}(R+\frac{6}{b^2}) $$

We don't include the Gibbons-Hawking term, because as claimed in the paper it should not contribute (although I am still unsure why, but this is a question for another day).

For the AdS BH metric, as well as pure AdS, we simply have $R = -\frac{12}{b^2}$. This means that computing the action merely involves computing the volume of the whole spacetime :

$$ S = \frac{-3}{2b^2}\int d\tau \int r^2dr $$ Where in the above equation we already computed the integral over $d\phi$ and $d\theta$ because it will contribute in the same way for the black hole and the pure AdS metric.

Now, as we know, in the case of the Black hole metric, to avoid the conical singularity at $r=r_+$, we should periodically identify $\tau \sim \tau+\beta$, where $\beta$ is the temperature of the Black Hole. Then, the full expression for its action should be :

$$ S_{bh}=\frac{-3}{2b^2}\int_0^\beta d\tau \int_{r_+}^{\rho}r^2 dr=-\frac{1}{2b^2}\beta(\rho^3-r_+^3) $$ Where we cutoff the integral at $r=\rho$, which is to be taken to $\infty$ later.

Now, we want to subtract the action for pure AdS, but we have to choose the period for $\tau$, in other words, choose the "temperature" of AdS. As seems to be hinted in the paper, and as it is done in other places, we take the temperature $\beta_{AdS}$ such that it matches locally at the boundary $r=\rho$ with the temperature of the Black hole. This is summarized as :

$$\beta_{AdS}\sqrt{1+\frac{\rho^2}{b^2}} =\beta \sqrt{1+\frac{\rho^2}{b^2}-\frac{2M}{\rho}}$$

So we are left with : $$ S_{BH}-S_{AdS} =-\frac{1}{2b^2}\beta(\rho^3-r_+^2-\frac{\beta_{AdS}}{\beta}\rho^3)\\ = -\frac{1}{2b^2}\beta(-r_+^2+\rho^3-\rho^3\left(1-\frac{M}{\rho+\rho^3/b^2}\right))\\ = -\frac{1}{2b^2}\beta(-r_+^2+Mb^2)\\ = \frac{1}{4b^2}\beta r_+(r_+^2-b^2) $$ Where we used the fact that $\rho\rightarrow \infty$

You can check that this gives exactly the right formula that Page and Hawking found (2.9), but with a flipped sign. I would like to know where this flipped sign comes from, as it is very important. Indeed, if we consider ourselves in the grand canonical ensemble at temperature $\beta$, the action with my sign yields an entropy which is negative, same goes for the Energy of the system.

I thought this might come from a missed sign in the Wick Rotation, but if our action is $S = \int L d^4x$, then our path integral gets a weight $e^{iS}$. By wick rotating, we have $\int L d^4x = i\int L_{eucl}d^4x$. So that the weight becomes $e^{-S_{eucl}}$. And for me, the action I have written down is precisely $S_{eucl}$ if we take the metric to be the wick rotated one...

So I am at a loss, I have no idea where this flipped sign could have come from.

EDIT :

Looking at this wikipedia article, I realize I do actually have the wrong sign for the Action after the Wick rotation, and it's from here I am missing the minus sign (so all the tedious computation can be ignored, as they should be right). However I still am unsure where in the Wick rotation I am going wrong.

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The Ads-Schwarzschild metric reads (I am working by setting $b\equiv 1$) $$ds_{AdS-Sch}^{2} = -\left(1-\frac{2M}{r} + r^{2}\right)dt^{2} + \left(1-\frac{2M}{r} + r^{2}\right)^{-1}dr^{2} + r^{2}d\Omega^{2}_{2}.$$

The gravitational action with cosmological constant $\Lambda=-d(d-1)/2b^{2}$ is (see equation $(4)$ here) $$I=-\frac{1}{2\kappa}\int_{\mathcal{M}}d^{d+1}x\sqrt{|g|}\ \left(R-\frac{d(d-1)}{b^{2}}\right) - \frac{1}{\kappa}\int_{\mathcal{\partial M}}d^{d}x\sqrt{|\gamma|}\Theta + \frac{1}{\kappa}I_{ct}\left(\gamma_{\mu\nu}\right),$$

where $\Theta$ is the extrinsic curvature tensor and $I_{ct}$ is the counter term.

The Einstein-Hilbert action reads (with $R=4\Lambda$) $$I_{AdS-Sch} = -\frac{1}{2\kappa}\int_{\mathcal{M}}d^{4}x\sqrt{|g|}\ (R-2\Lambda)=-\frac{2\Lambda}{16\pi G}\int_{-\beta}^{\beta}dt\int_{r_{H}}^{R}d^{3}x\sqrt{|g|}$$ $$=-\frac{\Lambda}{8\pi G}(2\beta)\int_{r_{H}}^{R}r^{2}dr\int d\Omega_{2}^{2} = \frac{8\pi \beta}{\kappa}\left(R^{3} - r_{H}^{3}\right),$$

where $\Lambda=-\frac{3}{b^{2}}$ (with $b\equiv1$). Now, the AdS action reads $$I_{AdS} = -\frac{2\Lambda}{2\kappa}\beta_{1}\int_{0}^{R}r^{2}dr\int d\Omega^{2} = -\frac{8\pi \Lambda\beta_{1}}{6\kappa}R^{3}.$$

Then, at the boundary $r=R$,

$$\beta_{1}\sqrt{1+R^{2}} = \beta\sqrt{1-\frac{2M}{R}+R^{2}},$$

and thus, $$I_{AdS}=-\frac{8\pi\Lambda R^{3}}{6\kappa}\beta\sqrt{\frac{1-\frac{2M}{R}+R^{2}}{1+R^{2}}}.$$

Now, the difference between the actions give us $$I_{Ads}-I_{AdS-Sch} \approx \frac{8\pi\Lambda T}{\kappa}\left(\frac{r_{H}^{3}}{6} - \frac{R^{3}}{6} + \frac{R^{3}}{6}\left(1-\frac{M}{R+R^{3}}\right)\right) = \frac{8\pi \beta}{2\kappa}\left(M - r_{H}^{3}\right),$$

which upon substituting for $M$ and $\beta$ yields (with $G\equiv 1$)

$$\Delta I = \frac{8\pi}{16\pi G}\left(\frac{4\pi r_{H}}{1+3r_{H}}\right)\left(\frac{r_{H}}{2} + \frac{r_{H}^{3}}{2} - r_{H}^{3}\right) = \frac{\pi r_{H}^{2}}{1+3r_{H}}\left(1-r_{H}^{2}\right),$$

which is the desired answer.

Edit:

The path integral transforms as follows under a Wick rotation $$Z=\int \mathcal{D}[g,\phi]e^{i I[\phi]}\rightarrow \int \mathcal{D}[g,\phi]e^{-I[\phi]},$$ where $I=I_{EH} = -\frac{1}{2\kappa}\int_{\mathcal{M}}d^{d}x\sqrt{|g|}(R-2\Lambda)$ is the Euclidean version of the Einstein-Hilbert action, which differs from the Lorentzian version by an overall minus sign. Now, the calculation for the AdS-Schwarzschild metric yields the Euclidean action $$I_{EAdS-Sch} = -\frac{\Lambda}{\kappa}\int_{0}^{\beta}\int_{r_{H}}^{R}r^{2}dr\int d\Omega^{2} = -\frac{\Lambda}{6G}\beta\left(R^{3} - r_{H}^{3} \right)$$

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  • $\begingroup$ In your first line of calculation you wrote : $\frac{1}{2\kappa}\int_{\mathcal{M}}d^{4}x\sqrt{|g|}\ (R-2\Lambda)=-\frac{2\Lambda}{16\pi G}\int_{-\beta}^{\beta}dt\int_{r_{H}}^{R}d^{3}x\sqrt{|g|}$. Since $R=4\Lambda$, you have the wrong sign in front, that's why you recover the desired answer, and not with a minus sign as I do. However, you made a mistake... I know now that the answer to the question lies in the wick rotation, the euclidean action really is : $I_{AdS-Sch} = -\frac{1}{2\kappa}\int_{\mathcal{M}}d^{4}x\sqrt{|g|}\ (R-2\Lambda)$. I will write an answer tomorrow ot explain why ! $\endgroup$ – Frotaur Mar 20 at 23:38
  • $\begingroup$ @Frotaur I have appended the sign, and obtain the correct answer. Please check. $\endgroup$ – Spoilt Milk Mar 21 at 7:14
  • $\begingroup$ The answer is corrected, but as is mine if I start with an inverted sign. My question was more, why do we have a minus sign in front of the euclidean action, when in the minkowskian action, there is no minus sign. (see my edit on the question) $\endgroup$ – Frotaur Mar 21 at 19:44

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