Near-Extremal Throats and AdS2

An extremal charged black hole develops a long throat whose leading two-dimensional metric is AdS2, but its first departure from extremality is not a finite-energy excitation of rigid AdS2. The deformation changes the dilaton, shortens the throat, and reconnects it to the exterior. This page derives that statement in a fixed-charge ensemble, using four-dimensional Reissner–Nordström as a transparent example, and identifies the linear-in-temperature entropy that the JT description must reproduce.

Required background. Gauge Theory and String-Theory Preparation provides the gauge-charge normalization and low-energy string vocabulary used below.

Helpful background. Quantum Gravity supplies effective-gravity limits; Quantum Information supplies ensemble language; Semiclassical Gravity provides horizon thermodynamics; and Black-Hole Information supports the later spectral interpretation.

The double zero and the AdS2 throat

Take the Einstein–Maxwell solution

$$ds^2=-f(r)dt^2+\frac{dr^2}{f(r)}+r^2d\Omega_2^2, \qquad f(r)=\frac{(r-r_+)(r-r_-)}{r^2}.$$

Here the geometric charge parameter is defined by $r_+r_-=q^2$ and the mass by $r_++r_-=2GM$; this convention avoids hiding electromagnetic factors inside a symbol called $Q$. Extremality means $r_+=r_-=r_0=\lvert q\rvert$. Set $r=r_0+\epsilon\rho$ and $t=\tau/\epsilon$, then take $\epsilon\to0$ while holding $(\rho,\tau)$ fixed. The double zero of $f$ gives

$$ds^2\longrightarrow -\frac{\rho^2}{r_0^2}d\tau^2+ \frac{r_0^2}{\rho^2}d\rho^2+r_0^2d\Omega_2^2,$$

so both the AdS2 and sphere radii equal $r_0$. Here $\epsilon$ is dimensionless and $\rho$ has dimensions of length. This is a scaling region, not a replacement for the asymptotically flat solution. In a near-extremal family the overlap with the exterior obeys $r_+-r_-\ll r-r_0\ll r_0$; the temperature must be measured using the same time coordinate as the metric. The nearly constant dilaton and throat response are developed in Maldacena, Stanford, and Yang 2016, §§ 2.1–3, pp. 3–10.

Spherical reduction makes the missing degree of freedom visible. The two-dimensional dilaton is proportional to the transverse area, $\Phi\propto r^2/G$. Its extremal value supplies a topological entropy, while the fluctuation $\varphi\propto r^2-r_0^2$ records how the sphere opens toward the exterior. Freezing $\varphi$ would freeze precisely the gravitational response that carries nonzero energy.

The leading near-extremal deformation

At fixed $q$, write $r_+-r_-=\delta$ with $\delta/r_0\ll1$. The Hawking temperature and Bekenstein–Hawking entropy are

$$T=\frac{r_+-r_-}{4\pi r_+^2}, \qquad S_{\rm BH}=\frac{\pi r_+^2}{G}.$$

To leading order, $\delta=4\pi r_0^2T$ and $r_+=r_0+\delta/2+O(\delta^2/r_0)$. Therefore

$$S_{\rm BH}=S_0+\frac{4\pi^2r_0^3}{G}T+O(T^2), \qquad M-M_0=\frac{2\pi^2r_0^3}{G}T^2+O(T^3).$$

The second relation follows independently from the fixed-charge first law $dM=T\,dS$. It is an important normalization check: the entropy is linear in $T$, the energy is quadratic, and the heat capacity is positive in this near-extremal sector. To keep this $T$ normalized to the original asymptotic time, set $x=r-(r_++r_-)/2$ and $x_h=\delta/2$. To leading order in $x/r_0$ and $\delta/r_0$, the finite-temperature throat is

$$ds_2^2=-\frac{x^2-x_h^2}{r_0^2}dt^2+ \frac{r_0^2\,dx^2}{x^2-x_h^2}, \qquad T=\frac{x_h}{2\pi r_0^2},$$

with the last equality at this order. If instead $x=\epsilon\rho$, $x_h=\epsilon\rho_h$, and $t=\tau/\epsilon$, the finite throat temperature is $T_\tau=\rho_h/(2\pi r_0^2)$ and the asymptotic temperature is $T=\epsilon T_\tau$. The radial dilaton still distinguishes states that are locally related by AdS2 coordinate changes.

Why strict AdS2 is not a closed finite-energy system

The throat limit sends the exterior energy scale to infinity relative to the rescaled time, while a finite excitation backreacts strongly on the AdS2 boundary. In the two-dimensional equations, matter stress changes the dilaton and hence the location and extrinsic curvature of the regulated boundary. Almheiri and Polchinski 2015, §§3.2–3.3, pp. 8–9 exhibit this backreaction explicitly: rigid AdS2 boundary conditions do not admit the ordinary finite-energy excitation spectrum one might infer from a fixed-background field theory.

The controlled statement is consequently narrower. For temperatures well below the inverse throat scale and below the first Kaluza–Klein or matter gap, the leading gravitational response is universal and described by the dilaton boundary mode. Extra light fields, a short throat, changing charge, or temperatures approaching the matching scale invalidate that single-sector description. The exterior is not an optional correction; it fixes the boundary condition and the coefficient of the low-energy action. The next page packages this response as Jackiw–Teitelboim gravity.

Checks and limitations

Three checks keep the scaling honest. First, the limiting metric has $R_2=-2/r_0^2$, fixing the AdS2 radius. Second, the horizon area remains $4\pi r_0^2$, so the extremal entropy survives the limit. Third, integrating the linear entropy with the first law reproduces the quadratic energy above extremality. None of these checks establishes a microscopic degeneracy, and none says that every extremal throat has no additional light modes. Rotation, multiple charges, compactification moduli, and ensemble changes modify coefficients and sometimes the field content, although the double-zero logic remains useful.

There is also a useful order-of-limits check. One must first specify the fixed-charge near-extremal family, then take the low-temperature expansion, and only afterward discard modes above the throat gap. Taking an exact extremal limit first erases the thermal scale whose response is being measured; taking a decoupling limit without matching leaves the boundary normalization undetermined. These operations need not commute.

Exercises

Starting from the exact formulas for $T$ and $S_{\rm BH}$, derive the coefficient of the entropy linear in temperature at fixed $q$, and explain why holding $M$ rather than $q$ fixed would not be the same thermodynamic problem.

Solution

Let $r_\pm=r_0\pm\delta/2+O(\delta^2)$; fixed $q$ enforces $r_+r_-=r_0^2$ and only affects the expansion at second order. Substitution into the temperature gives $T=\delta/(4\pi r_0^2)+O(\delta^2)$. Expanding the area then yields $S_{\rm BH}-S_0=(2\pi r_0/G)(\delta/2)=4\pi^2r_0^3T/G$. Fixing the mass would allow the charge to vary and would add an electrostatic work term to the first law, so the coefficient would not test the same throat ensemble.

References

• Almheiri, Ahmed, and Joseph Polchinski. “Models of AdS2 Backreaction and Holography.” Journal of High Energy Physics 11 (2015) 014. arXiv:1402.6334.
• Maldacena, Juan, Douglas Stanford, and Zhenbin Yang. “Conformal Symmetry and Its Breaking in Two Dimensional Nearly Anti-de-Sitter Space.” Progress of Theoretical and Experimental Physics 2016 (2016) 12C104. arXiv:1606.01857.