Inrtroduction

In this note we describe the Almheiri-Polchinski (AP) model \cite{Almheiri:2014cka} which is a particular Jackiw-Teitelboim (JT) model \cite{Jackiw:1984je,Teitelboim:1983ux} of dilaton gravity in two dimensions with striking similarities to the Sachdev–Ye–Kitaev (SYK) model \cite{Kitaev:2015, Sachdev-Ye:1993} (everything is encoded in the boundary theory which is given by a Schwarzian action).

In here we will mainly follow \cite{Almheiri:2014cka, Almheiri:2019psf} but also \cite{Engelsoy:2016xyb,Maldacena:2016upp}.

Eternal ${\rm AdS}^2$ Black hole

The APJT model is a dynamical theory of gravity in two dimensions with no local excitations, i.e. the value of the metric tensor $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$, which is a dynamical variable here (as opposed to pure gravity in two dimensions),  and the value of the dilaton field $\phi$, which plays a crucial role in the existence of stable black hole configurations in two dimensions,  is uniquely fixed in terms of the stress-energy-momentum tensor of the matter field $f$ (through the equations of motion). We will mostly assume that the matter theory is given by a conformal field theory which can also be holographic. We will also assume that the matter field $f$ interacts with dilaton field $\Phi$ only through the metric tensor and not directly. The action reads explicitly

\begin{eqnarray} S[g,\Phi,f]&=& S_{JT}[g,\Phi]+D_{\rm CFT}[g,f]\nonumber\\ &=&\frac{1}{16\pi G}\int d^2x\sqrt{-g}(\Phi^2 R-V(\Phi))+S_{\rm CFT}[g,f]. \label{action}\end{eqnarray}

The AP potential $V$ is given explicitly by

\begin{eqnarray} V(\Phi)=2-2\Phi^2.
\end{eqnarray}

This potential guarantees that spacetime has a constant negative scalar curvature (integrate out $\phi=\Phi^2$ to obtain the delta function $\delta (R+2))$, i.e. we must have $R=-2$. In other words, spacetime is precisely  anti-de Sitter spacetime  ${\bf AdS}^2$. In fact this potential also guarantees that the matter action does not depend directly on the dilaton field but depends on it only indirectly through the metric tensor. The dilaton filed itself is the crucial ingredient allowing  the existence of black hole configurations on this two-dimensional ${\rm AdS}$ background.

We check these facts more explicitly as follows. First, we write the metric in the conformal gauge as follows

\begin{eqnarray}ds^2=-e^{2\omega(u,v)}du dv\end{eqnarray}

The light-cone coordinates $u$ and $v$ will also be denoted as $u=x^+$ and $v=x^-$ where $x^{\pm}=t\pm z$.

The equations of motion (by varying the dilaton field and the metric tensor) will then read

\begin{eqnarray}4\partial_u\partial_v\omega+e^{2\omega}=0.\label{eom1}\end{eqnarray}

\begin{eqnarray}-e^{2\omega}\partial_u(e^{-2\omega}\partial_u\Phi^2)=8\pi G T_{uu}.\label{eom2}\end{eqnarray}

\begin{eqnarray}-e^{2\omega}\partial_v(e^{-2\omega}\partial_v\Phi^2)=8\pi G T_{vv}.\label{eom3}\end{eqnarray}

\begin{eqnarray}2\partial_u \partial_v \Phi^2+e^{2\omega}(\Phi^2-1)=16\pi G T_{uv}.\label{eom4}\end{eqnarray}

The stress-energy-momentum tensor of the matter field is given by the usual formula

\begin{eqnarray}T_{ab}=-\frac{2}{\sqrt{-g}}\frac{\delta S_{\rm CFT}[g,f]}{\delta g^{ab}}.\end{eqnarray}

For conformal matter, such as a massless free scalar field $f$ given by the Klein-Gordon action $S_{\rm CFT}=\frac{1}{32\pi G}\int d^2x \sqrt{-g}(\nabla f)^2$, the off-diagonal component $T_{uv}$ of the stress-energy-momentum tensor vanishes classically.  We also recall that in this case $T_{uu}=T_{++}=\frac{1}{16\pi G}\partial_+f\partial_+f$ and $T_{vv}=T_{--}=\frac{1}{16\pi G}\partial_-f\partial_-f$.

The most general solution of (\ref{eom1}) is the ${\rm AdS}^2$ metric, i.e. $e^{2\omega}=4/(u-v)^2$.  We have explicitly the metric

\begin{eqnarray}ds^2=-\frac{4}{(x^+-x^-)^2}dx^+dx^-=\frac{1}{z^2}(-dt^2+dz^2)~,~x^{\pm}=t\pm z.\label{sol1}\end{eqnarray}

The gravitational sector will be treated semi-classically. In other words, we replace the stree-energy-momentum tensor by its expectation value, viz

\begin{eqnarray}T_{ab}=\langle T_{ab}\rangle.\end{eqnarray}

For $\langle T_{ab}\rangle=0$ the most general solution of the equations of motion (\ref{eom2}), (\ref{eom3}) and (\ref{eom4}) is given by the dilaton field

\begin{eqnarray}\Phi^2=1+\frac{a-\mu x^+x^-}{x^+-x^-}.\label{sol2}\end{eqnarray}

This dilaton field represents an eternal black hole with two asymptotic boundaries.

The case $a=0$ represents pure ${\rm AdS}^2$ or more precisely the Poincaré patch of ${\rm AdS}^2$ whereas the solution with $a>0$ (and $\mu>0$) represents an ${\rm AdS}^2$ black hole  (it is the Rindler wedge of the Poincaré patch).

The most general solution of the equations of motion is a conformal transformation $x=f(y)$ of (\ref{sol1}) and (\ref{sol2}), viz

\begin{eqnarray}ds^2=-\frac{4f^{\prime}(y^+)f^{\prime}(y^-)dy^+dy^-}{(f(y^+)-f(y^-))^2}~,~\Phi^2=1+\frac{a-f(y^+)f(y^-)}{f(y^+)-f(y^-)}.\end{eqnarray}

A static form of this black hole configuration is obtained by means of the conformal transformation

\begin{eqnarray}x=f(y)=\frac{1}{\sqrt{\mu}}\tanh \sqrt{\mu} y.\label{diffeo}\end{eqnarray}

The black hole solution reads then (we set $a=1$)

\begin{eqnarray}ds^2=-\frac{4\mu dy^+dy^-}{\sinh^2\sqrt{\mu}(y^+-y^-)}~,~\Phi^2=1+\sqrt{\mu}\coth\sqrt{\mu}(y⁺-y^-).\end{eqnarray}

The $x$ coordinates cover the whole geometry of the spacetime manifold whereas the $y$ coordinates cover only the exterior of the black hole.  At high temperature ($\mu \longrightarrow 0$) we can make the identification

\begin{eqnarray}z=\frac{y^+-y^-}{2}.\end{eqnarray}

Thus, the boundary $z=0$ of ${\rm AdS}^2$ is  located in the $y$ coordinates at $y^+-y^-=0$. On the other hand, the horizon $z\longrightarrow \infty$ of ${\rm AdS}^2$ corresponds in the $y$ coordinates either to the future horizon $y^+\longrightarrow +\infty$ (or equivalently $x^+\longrightarrow 1/\sqrt{\mu}$) or to the past horizon  $y^-\longrightarrow -\infty$ (or equivalently $x^+\longrightarrow -1/\sqrt{\mu}$).

This black hole configuration after Euclidean rotation becomes periodic in the variable $y^+-y^-$ with period $\beta_0=1/T_0$ given by

\begin{eqnarray}2\sqrt{\mu}=\frac{2\pi}{\beta_0}\Rightarrow T_0=\frac{\sqrt{\mu}}{\pi}.\end{eqnarray}

This is Hawking temperature. This can also be checked in the Schwarzschild coordinates defined by

\begin{eqnarray}\rho=\sqrt{\mu}\coth \sqrt{\mu}(y^+-y^-)~,~T=\frac{y^++y^-}{2}.\end{eqnarray}

The metric and the dilaton take then the form

\begin{eqnarray}ds^2=-4(\rho^2-\mu)dT^2+\frac{d\rho^2}{\rho^2-\mu}~,~\Phi^2=1+\rho.\end{eqnarray}

The Hawking temperature is then given by

\begin{eqnarray}T_0&=&\frac{1}{4\pi}\partial_{\rho}\sqrt{-\frac{g_{TT}}{g_{\rho\rho}}}|_{\rho=\sqrt{\mu}}\nonumber\\&=&\frac{\sqrt{\mu}}{\pi}.\label{T_Hawking}\end{eqnarray}

${\rm AdS}^2$ Black hole formed from gravitational collapse

We look at the equations of motion more carefully. First, we trivially check that $\exp(2\omega)=4/(x^+-x^-)^2$ solves the equation of motion (\ref{eom1}). Next, in terms of the ansatz $\Phi^2=M/(x^+-x^-)$ we write the constraints (\ref{eom2}) and (\ref{eom3}) in the form

\begin{eqnarray}\partial_+^2M=-8\pi G(x^+-x^-)T_{++}(x^+)~,~ \partial_-^2M=-8\pi G(x^+-x^-)T_{--}(x^-).\label{eom5}\end{eqnarray}

The final equation of motion (\ref{eom4}) reads

\begin{eqnarray}\partial_+\partial_-M+(\partial_+M-\partial_-M-2)/(x^+-x^-)=8\pi G(x^+-x^-)T_{+-}.\label{eom6}\end{eqnarray}

The most general solution of (\ref{eom5}) is

\begin{eqnarray}M=a+bx^++cx^-+dx^{+}x^--I^++I^-.\end{eqnarray}

\begin{eqnarray}I^{\pm}=8\pi G\int_{x_0^{\pm}}^{x^{\pm}}dx^{\prime\pm}(x^{\prime \pm}-x^{\mp})(x^{\prime\pm}-x^{\pm})T_{\pm\pm}(x^{\prime\pm}).\end{eqnarray}

For conformal theory we have $T_{+-}=0$. The solution of (\ref{eom6}) is given by the requirement

\begin{eqnarray}b-c=2.\end{eqnarray}

The solution is then (with $b=b^{\prime}+1$)

\begin{eqnarray}\Phi^2=1+\frac{a+b^{\prime}(x^++x^-)+dx^+x^-}{x^+-x^-}-\frac{I^+-I^-}{x^+-x^-}.\label{res0}\end{eqnarray}

For $T_{uv}=\langle T_{uv}\rangle =0$ we get the solution

\begin{eqnarray}\Phi^2=\frac{a+b^{\prime}(x^++x^-)+dx^+x^-}{x^+-x^-}.\label{eom7}\end{eqnarray}

By an ${\rm SL}(2,R)$ symmetry which acts as $t\longrightarrow t^{\prime}=(at+b)/(ct+d)$ with $ad-cb=1$ we can bring this solution to the form

\begin{eqnarray}\Phi^2=1+\frac{a-\mu x^+x^-}{x^+-x^-}.\label{eom8}\end{eqnarray}

This is the dilaton profile corresponding to an eternal ${\rm AdS}^2$ black hole.

Next, we would like to derive the black hole solution formed from gravitational collapse. We consider then the effect of an infalling matter pulse into the black hole, i.e. the effect of a schokwave of energy $E_S$ traveling on the null curve $x^-=0$ starting from the boundary $z=0$ at time $t=0$. The stress-energy-momentum tensor is given by

\begin{eqnarray}T_{--}=\langle T_{--}\rangle =E_S\delta (x^-).\end{eqnarray}

We compute immediately $I^+=0$, $I^-=8\pi G E_S x^+x^-$. The dilaton profile becomes then

\begin{eqnarray}\Phi^2=\frac{a+b^{\prime}(x^++x^-)+(d+8\pi GE_S) x^+x^-}{x^+-x^-}.\end{eqnarray}

By an ${\rm SL}(2,R)$ symmetry we can bring this solution to the form

\begin{eqnarray}\Phi^2=\frac{a-(\mu+8\pi GE_S) x^+x^-}{x^+-x^-}.\label{res2}\end{eqnarray}

We can deduce from this formula the relationship between the mass of the black hole and its temperature. If we start from a pure ${\rm AdS}^2$ space we can set $\mu=0$ and thus we obtain

\begin{eqnarray}\Phi^2=\frac{a-8\pi GE_S x^+x^-}{x^+-x^-}.\label{eom9}\end{eqnarray}

By comparing (\ref{eom9}) and (\ref{eom8}) and using (\ref{T_Hawking}) we deduce the relationship between the energy and temperature as

\begin{eqnarray}8\pi GE_S=\mu_S=(\pi T_S)^2.\end{eqnarray}

Thus for the eternal ${\rm AdS}^2$ black hole we obtain the relationship

\begin{eqnarray}8\pi GE_0=\mu_0=(\pi T_0)^2.\end{eqnarray}

By considering now the  effect of an infalling matter pulse into the black hole (black hole formed by gravitational collapse) we obtain

\begin{eqnarray}8\pi GE_1=\mu_1=(\pi T_1)^2.\end{eqnarray}

The energy $E_1$ is simply given by the energy $E_0$ of the eternal black hole plus the energy $E_S$ of the infalling matter, viz $E_1=E_0+E_S$. Thus, the relationship between the the new temperature $T_1$, the old temperature $T_0$ and the energy of the pulse $E_S$ is given by

\begin{eqnarray}(\pi T_1)^2=(\pi T_0)^2+8\pi G E_S.\end{eqnarray}

Quantum correction and coupling to a heat bath

The next step is to add quantum corrections which in the case of a conformal theory are encoded in the conformal anomaly.  The  Almheiri-Polchinski (AP) model is really characterized by transparent boundary conditions at the boundary as opposed to the reflecting boundary conditions characterizing the usual Jackiw-Teitelboim (JT) model. In other words, the (right) boundary of ${\rm AdS}^2$ is coupled to a heat bath at zero temperature into which Hawking radiation can escape and hence we have a simulated black hole evaporation process with the associated Hawking radiation and the consequent black hole information loss problem.

The matter sector which is independent of the dilaton field (and only interacts with it through the constraints) is treated as a conformal field theory on a fixed ${\rm AdS}^2$ background in the coordinates $x$. The fields are subjected to reflecting boundary conditions. In the external heat bath ${\rm M}^2$ we have the same conformal field theory in the coordinates $y$. The coupling between the ${\rm AdS}^2$ space and the heat bath occurs at $t=0$, i.e. at $x^-=0$ which results in a shokwave of energy $E_S$ infalling from the boundary into the black hole. We have the following metrics

The boundary conditions of the metric $ds^2_{\rm AdS^2}=(-dt^2+dz^2)/z^2$ and the dilaton $\Phi^2=1+(1-\mu(t^2-z^2))/2z$, i.e. their values at the boundary $z=\epsilon$  are given by (where $u$ is the time variable on the boundary)

\begin{eqnarray}g_{tt}|_{\rm boundary}=\frac{1}{\epsilon^2}=\frac{-t^{\prime 2}+z^{\prime 2}}{z^2}~,~\Phi^2|_{\rm boundary}=\frac{1}{2\epsilon}.\end{eqnarray}

The diffeomorphism $x=f(y)$ is chosen such that the boundary is simple at constant value in the $y$ coordinates, viz

\begin{eqnarray}\epsilon=\frac{y^+-y^-}{2}.\end{eqnarray}

Of course before the coupling between the ${\rm AdS}^2$ and the heat bath ${\rm M}^2$ is turned on at $t=0$ this diffeomorphsim is given by  (\ref{diffeo}) which corresponds to the static form of the eternal black hole solution.

Approaching the future/past horizon $y^{\pm}\longrightarrow\pm\infty$ on the boundary $y^+-y^-=0$ means that $u=T|_{\rm boundary}=\frac{y^++y^-}{2}|_{\rm boundary}\longrightarrow \pm\infty$ which in the $x$ coordinates is equivalent to spending the times $\pm t_{\infty}={\rm lim}_{u\longrightarrow\pm\infty}f(u)$. Before the coupling between ${\rm AdS}^2$ and the heat bath these times $\pm t_{\infty}$ are precisely the future/past horizon times $\pm t_{\infty}=f(\pm \infty)=x^{\pm}=\pm 1/\sqrt{\mu_0}=\pm 1/\pi T_0$. After the coupling the temperature changes to $T_1$ and in order for the wormhole to remain not traversable  the new event horizon is required to lie outside the original horizon and hence it can be reached in less time, i.e. $t_{\infty}$ after the coupling must satisfy $t_{\infty}\lt 1/\pi T_0$. The idea is that the so-called Averaged Null Energy Condition (ANEC) on the horizon must always be satisfied in order to maintain boundary causality \cite{Maldacena:2018lmt,Galloway:2018dak}.

Let us now introduce the Euclidean time $\tau=i t$ and the Euclidean (complex) coordinates $x$ and $\bar{x}$ by

\begin{eqnarray}x^+=\bar{x}=t+z~,~x^-=-x=t-z.\end{eqnarray}

Thus, $t=(x^++x^-)/2=(\bar{x}-x)/2$ and $z=(x^+-x^-)/2=(\bar{x}+x)/2$. The ${\rm AdS}^2$ boundary is at $z=0$ whereas the bulk is $z>0$. The initial state at $t=0$ is therefore the Hartle-Hawking state on ${\rm AdS}^2$ which is given by the vacuum on the half-line $z>0$.  The heat bath is another half-line $z<0$ with the same CFT prepared in the same vacuum state. The physical time (time on the boundary and in the heat bath) is $u$ and not $t$ (which is the bulk time). They are related by a diffeomorphism $t=f(u)$ (we choose $0=f(0)$).

We must therefore go from the coordinates $x$ (defined on ${\rm AdS}^2$) to the coordinates $y$ (defined on the heat bath) by the diffeomorphism $f$, viz $x=f(y)$ and $\bar{x}=f(\bar{y})$. The boundary is simple in the $y$ coordinates located at $y+\bar{y}=0$ while the physical time is $T=(\bar{y}-y)/2$ (we choose $f(y)=-f(-y)$).

At the initial time $t=u=0$ we have $y=\bar{y}=f^{-1}(z)$ and ${\rm AdS}^2$ corresponds to the right half-line $y>0$ whereas the heat bath corresponds to the left half-line $y<0$ . In general, ${\rm AdS}^2$ corresponds to the right half-plane $y+\bar{y}>0$ whereas the heat bath corresponds to the left half-plane $y+\bar{y}<0$.

The initial quantum state of the heat bath is therefore given by the Euclidean path integral on the left lower half-plane (the half-line vacuum). On the other hand, the initial quantum state of ${\rm AdS}^2$ is given by an Euclidean path integral on the right lower half-plane with a deformed boundary (a Virasoro descendant of the usual CFT vacuum on the half-line). In conclusion, we have in the $y$ coordinates a combined coupled system evolving in the physical time by the usual Hamiltonian of conformal field theory on the line.

The initial state is then time-reflection symmetric given by an Euclidean path integral over a simply connected space with a single boundary. Therefore it is a descendent of the half-line vacuum. The Cauchy surface at $t=0$ can thus be mapped by a means of an appropriate diffeomorphism to a half-line parameterized by a coordinate $w\in [0,\infty[$, i.e. we can map our initial quantum state to the half-line vacuum.

The goal is to derive the diffeomorphism $w=w(x)$ (in the ${\rm AdS}^2$ region $x>0$), the diffeomorphism $w=w(y)$ (in the heat bath ${\rm M}^2$ region $y<0$), the stress-energy-momentum tensor of the conformal matter in both regions and the energy $E_S$ of the initial shockwave due to the turning on of the coupling between ${\rm AdS}^2$ and ${\rm M}^2$ at time $t=0$. This will allow us to determine the quantum corrections to the Hawking temperature.

First, at $t=0$ the stress-energy-momentum tensor of ${\rm AdS}^2$ is zero in the physical coordinates $y$, and also zero in the Poincare coordinates $x$ since the Weyl anomaly between $x$ and $y$ is zero ($x=x(y)$ is an $SL(2,R)$ transformation). Similarly, the heat bath ${\rm M}^2$ is at zero tempertaure and thus the corresponding stress-energy-momentum tensor is also zero. We have then

\begin{eqnarray}\langle T_{xx}(x)\rangle=0~(x>0)~,~\langle T_{yy}(y)\rangle=0~ (y<0)~,~t=0\end{eqnarray}

At later times we use the transformation law of the stress-energy-momentum tensor under the diffeomorphism $w$, viz

\begin{eqnarray}(\frac{dw}{dx})^2\langle T_{ww}(w)\rangle=\langle T_{xx}(x)\rangle-\frac{c}{24\pi}S(w,x).\end{eqnarray}

\begin{eqnarray}(\frac{dw}{dy})^2\langle T_{ww}(w)\rangle=\langle T_{yy}(y)\rangle+\frac{c}{24\pi }S(w,y).\end{eqnarray}

The number $c$ is the central charge of the conformal field theory and $S$ is the so-called Schwarzian which is defined by

\begin{eqnarray}S(w,x)=\{w,x\}&=&\frac{w^{\prime\prime\prime}(x)}{w^{\prime}(x)}-\frac{3}{2}\frac{w^{\prime\prime 2}(x)}{w^{\prime 2}(x)}\nonumber\\&=&(\frac{w^{\prime\prime}}{w^{\prime}})^{\prime}-\frac{1}{2}(\frac{w^{\prime\prime}}{w^{\prime}})^2.\end{eqnarray}

We take the diffeomorphism $w$ to be an $SL(2,R)$ transformation of the relevant coordinates, i.e.  a Mobius map which guarantees the vanishing of the stress-energy-momentum tensor  in the $w$ coordinates. Hence, we obtain the energy-momentum tensor

\begin{eqnarray}\langle T_{xx}(x)\rangle=-\frac{c}{24\pi}S(w,x).\end{eqnarray}

\begin{eqnarray}\langle T_{yy}(y)\rangle=-\frac{c}{24\pi }S(w,y).\end{eqnarray}

We go back to the initial time $t=0$. The diffeomorphism (or conformal transformation) $w$ will map the ${\rm AdS}^2$ region $x>0$ to the interval $[0,w_0]$ while it will map the heat bath ${\rm M}^2$ region $y<0$ to the interval $[w_0,\infty[$ and it is given explicitly by \cite{Almheiri:2019psf}

\begin{eqnarray}w(x)=\frac{w_0^2}{w_0+x}~,~x>0.\end{eqnarray}

\begin{eqnarray}w(y)=w_0+f^{-1}(-x)~,~x<0.\end{eqnarray}

We write this

\begin{eqnarray}w(x)=\frac{w_0^2}{w_0+x}\theta(x)+(w_0-y)\theta(-x).\end{eqnarray}

We compute (using $f(0)=0$)

\begin{eqnarray}w^{\prime}(x)=-\frac{w_0^2}{(w_0+x)^2}\theta(x)-y^{\prime}(x)\theta(-x).\end{eqnarray}

Then we compute (using $y^{\prime}(x)=1/f^{\prime}(y)$ and $f^{\prime}(0)=1$ where primes denote derivatives with respect to the appropriate variable)

\begin{eqnarray}w^{\prime\prime}(x)=\frac{2w_0^2}{(w_0+x)^3}\theta(x)-y^{\prime\prime}(x)\theta(-x).\end{eqnarray}

Hence

\begin{eqnarray}\frac{w^{\prime\prime}(x)}{w^{\prime}(x)}=-\frac{2}{w_0+x}\theta(x)+\frac{y^{\prime\prime}(x)}{y^{\prime}(x)}\theta(-x).\end{eqnarray}

As a consequence we have (using $y^{\prime\prime}/y^{\prime}=-f^{\prime\prime}/f^{\prime 2}$ and hence $\frac{y^{\prime\prime}}{y^{\prime}}|_{x=0}=\frac{y^{\prime\prime}}{y^{\prime}}|_{y=0}=-f^{\prime\prime}(0)$)

\begin{eqnarray}(\frac{w^{\prime\prime}(x)}{w^{\prime}(x)})^{\prime}=\frac{2}{(w_0+x)^2}\theta(x)+(\frac{y^{\prime\prime}(x)}{y^{\prime}(x)})^{\prime}\theta(-x)-\frac{2}{w_0}\delta(x)+f^{\prime\prime}(0)\delta(x).\end{eqnarray}

We get then the Schwarzian and the energy-momentum tensor

\begin{eqnarray}S(w,x)=\{w,x\}=\{y,x\}\theta(-x)-(\frac{2}{w_0}-f^{\prime\prime}(0))\delta(x).\end{eqnarray}

\begin{eqnarray}\langle T_{xx}(x)\rangle=-\frac{c}{24\pi}\{y,x\}\theta(-x)+\frac{c}{24\pi}(\frac{2}{w_0}-f^{\prime\prime}(0))\delta(x).\label{res1}\end{eqnarray}

Thus, we can make the identification

\begin{eqnarray}E_S= \frac{c}{24\pi}(\frac{2}{w_0}-f^{\prime\prime}(0)).\end{eqnarray}

However,  ${\rm AdS}^2$ is dual to a conformal quantum mechanics at the boundary and thus it is more natural to map the ${\rm AdS}^2$ region to a single point. In other words, we must take the limit $w_0\longrightarrow 0$ and hence $E_S\longrightarrow \infty$. As it turns out, this is indeed the physically sensible limit in order to avoid acausal correlations \cite{Almheiri:2019psf}. The diffeomorphism $w$ becomes a mapping to the upper-half plane given by

\begin{eqnarray}w(x)=(\frac{12\pi E_S}{c})^{-1}\frac{1}{x}\theta(x)+f^{-1}(-x)\theta(-x).\end{eqnarray}

We also write the result (\ref{res1}) in the form

\begin{eqnarray}\langle T_{x^-x^-}(x⁻)\rangle=-\frac{c}{24\pi}\{y^-,x^-\}\theta(x^-)+E_S\delta(x^-).\end{eqnarray}

We use this result in (\ref{res0}). We compute $I^+=0$ as before but now

\begin{eqnarray}I^-=8\pi GE_S x^+x^--\frac{k}{2}\int_0^{x^-}dt(x^+-t)(x^--t)\{u,t\}~,~t=f(u)~,~k=\frac{c.G}{3}.\end{eqnarray}

The dilaton field, including quantum corrections, becomes (compare with equation (\ref{res2}) and subsequent equations)

\begin{eqnarray}\Phi^2&=&\frac{1-(\mu_0+8\pi GE_S)x^+x^-+\frac{k}{2}\int_0^{x^-}dt(x^+-t)(x^--t)\{u,t\}}{x^+-x^-}\nonumber\\&=&\frac{1-(8\pi T_1)^2x^+x^-+\frac{k}{2}I(x^+,x^-)}{x^+-x^-}.\end{eqnarray}

The boundary theory

The  Schwarzian plays a crucial role in this problem since the underlying dynamics is one-dimensional on the boundary. Indeed, the space ${\rm AdS}^2$ is characterized by a boundary and the action should be enhanced by a boundary term, viz

\begin{eqnarray} S[g,\Phi,f]\longrightarrow S[g,\Phi,f]&=& S_{JT}[g,\Phi]+D_{\rm CFT}[g,f]+S_{b}[g,\Phi]\nonumber\\ &=&\frac{1}{16\pi G}\int_{\cal M} d^2x\sqrt{-g}(\Phi^2 R-V(\Phi))+S_{\rm CFT}[g,f]+S_{b}[g,\Phi]. \end{eqnarray}

The boundary term is given by

\begin{eqnarray}S_{b}[g,\Phi]=\frac{1}{8\pi G}\int_{\partial\cal M} du\sqrt{-\gamma}\Phi^2K.\end{eqnarray}

$K$ is the scalar extrinsic curvature. More precisely, $K=g^{\mu\nu}K_{\mu\nu}=\gamma^{\mu\nu}K_{\mu\nu}$ where $\gamma_{\mu\nu}$ is the induced metric at the boundary and the extrinsic curvature tensor $K_{\mu\nu}$ describes how the boundary ${\partial \cal M}$ is curved with respect to the manifold ${\cal M}={\rm AdS}^2$ in which it is embedded.

Since the equation of motion of the dilaton enforces the ${\rm AdS}^2$ geometry the bulk action is zero and we only need to focus on the boundary term.

Let us consider the Euclidean metric $ds^2=(dt^2+dz^2)/z^2$. The boundary conditions become

\begin{eqnarray}g_{tt}|_{\rm boundary}=\frac{1}{\epsilon^2}=\frac{t^{\prime 2}+z^{\prime 2}}{z^2}~,~\Phi^2|_{\rm boundary}=\frac{1}{2\epsilon}\Rightarrow \epsilon=\frac{z}{\sqrt{t^{\prime 2}+z^{\prime 2}}}.\end{eqnarray}

We can solve explicitly in powers of the cutoff $\epsilon$ to find

\begin{eqnarray}z=\epsilon t^{\prime}+O(\epsilon^2).\end{eqnarray}

We compute then

\begin{eqnarray}z^{\prime}=\epsilon t^{\prime\prime}+O(\epsilon^2)~,~z^{\prime\prime}=\epsilon t^{\prime\prime\prime}+O(\epsilon^2).\end{eqnarray}

\begin{eqnarray}\epsilon^{\prime}=\frac{z^{\prime}}{\sqrt{t^{\prime 2}+z^{\prime 2}}}-\frac{z(t^{\prime}t^{\prime\prime}+z^{\prime}z^{\prime\prime})}{(t^{\prime 2}+z^{\prime 2})^{3/2}}=0+O(\epsilon^2).\end{eqnarray}

The primes are derivatives with respect to $u$ which is the time parameter on the boundary. The tangent vector at the boundary is $e^{\mu}=\partial_{u}x^{\mu}=(t^{\prime},z^{\prime})$ whereas the normal vector is $n^{\mu}=\epsilon(z^{\prime},-t^{\prime})$. By construction these two vectors are orthogonal, i.e. $n_{\mu}e^{\mu}=0$ and furthermore $n^{\mu}$ is normalized, i.e. $n_{\mu}n^{\mu}=1$.

We compute the scalar curvature by the formula

\begin{eqnarray}K=\nabla_{\mu}n^{\mu}&=&\partial_{\mu}n^{\mu}+\Gamma_{\mu\nu}^{\mu}n^{\nu}\nonumber\\&=&\partial_{\mu}n^{\mu}+\frac{1}{2}g^{\mu\beta}\partial_{\nu}g_{\beta\mu} n^{\nu}\nonumber\\&=&-\frac{t^{\prime 2}-z^{\prime 2}}{t^{\prime}\sqrt{t^{\prime 2}+z^{\prime 2}}}+\frac{2z}{(t^{\prime 2}+z^{\prime 2})^{3/2}}(z^{\prime\prime}t^{\prime}-t^{\prime\prime}z^{\prime})+2\frac{t^{\prime 2}-z^{\prime 2}}{t^{\prime}\sqrt{t^{\prime 2}+z^{\prime 2}}}\nonumber\\&=&-\frac{z}{\epsilon t^{\prime}}+\frac{2}{(t^{\prime 2}+z^{\prime 2})^{3/2}}(t^{\prime}(zz^{\prime\prime}+t^{\prime 2}+z^{\prime 2})-z z^{\prime}t^{\prime\prime})\nonumber\\&=&1+2\epsilon^2\{t,u\}.\end{eqnarray}

So the extrinsic curvature to leading order in $\epsilon^2$ is equal to the Schwarzian. Note, that in going from the second line to the third line we have replaced in both terms the derivatives $\partial_t$ and $\partial_z$ with $\partial_u/t^{\prime}$ and $\partial_u/z^{\prime}$ respectively.

The boundary term becomes

\begin{eqnarray}S_{b}[g,\Phi]&=&(-1)(\frac{1}{2})\frac{1}{8\pi G}\int_{\partial\cal M} du\frac{1}{\epsilon}\frac{1}{2\epsilon}. 2\epsilon^2\{t,u\}\nonumber\\&=&-\frac{1}{16\pi G}\int du \{t,u\}.\end{eqnarray}

The minus sign is due to the Euclidean signature whereas the factor $1/2$ is due to the fact that the boundary of ${\rm AdS}^2$ is constituted of two identical disconnected segments.

Thus, we have spontaneous symmetry breaking of conformal symmetry along the boundary down to the $SL(2,R)$ Mobius transformations $t\longrightarrow (at+b)/(ct+d)$ with $ad-cb=1$. Indeed, the Schwarzian is only invariant under these transformations, viz

\begin{eqnarray}\{\frac{at+b}{ct+d},u\}=\{t,u\}.\end{eqnarray}

The field $t=t(u)$ acts as the corresponding pseudo Nambu Goldstone modes associated with this spontaneous breaking \cite{Maldacena:2016upp}.

The ADM energy associated with boundary translations $u\longrightarrow u+\delta u$ is immeidately given from the above action by the Schwarzian, viz

This can be obatined by varying the boundary metric and computing the corresponding stress-energy-momentum tensor \cite{Almheiri:2014cka}.

We can check that for $u<0$, where the diffeomorphism $t=f(u)=\frac{1}{\pi T_0}\tanh (\pi T_0 u)$, this ADM energy $E(u)$ is precisely equal to the energy of the eternal black hole $E_0=\pi T_0^2/8G$.

Exercises

Exercise $1$: Calculate the equations of motion (\ref{eom1}), (\ref{eom2}), (\ref{eom3}) and (\ref{eom4}). Determine the constraints.

Exercise $2$: Show by using an ${\rm SL}(2,R)$ symmetry that we can bring the solution (\ref{eom7}) to the form (\ref{eom8}).

Exercise $3$: By varying the boundary metric compute the boundary stress-energy-momentum tensor and show that the ADM energy is given by equation (\ref{ADM}).

Exercise $4$: Show that for $u<0$ the ADM energy $E(u)$ is precisely equal to the energy of the eternal black hole $E_0=\pi T_0^2/8G$.

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