In this link, one of the answers contains the statement

If you examine the space-time near a finite area quantum black hole, you will see an approximate AdS space.

Presumably "approximate" means this is only true to some order in the distance from the horizon ? Could someone outline where the result comes from, or provide a reference.


2 Answers 2


The answer you quoted was invalid in most respects, including this one. The quantum character of a black hole has nothing to do with the AdS geometry; the AdS geometry is the near-horizon geometry of a classical black hole (or black brane). Which black hole? It has to be an extremal black hole, i.e. it must have the maximum value of a charge or angular momentum that is allowed for the given mass (or mass density, in the case of black branes).

For a black $p$-brane which is extended in time as well as $p$ additional spatial dimensions, one obtains $AdS_{p+2}$. For example, one may get an $AdS_2\times S^{d-2}$ from the near-horizon geometry of extremal black holes in $d=4$. See the derivations and comments e.g. around pages 57 and 104 in


or search "near-horizon [geometry]" in this paper or another introduction to AdS/CFT. Black holes that are non-extremal have a finite volume of the region near the horizon and quantum phenomena don't change anything about this fact. The extremality is needed for the infinite volume – and AdS spaces have an infinite volume. The derivation of the near-horizon limit of a black hole metric is a purely classical, geometric procedure: one neglects some subleading terms in the metric tensor and only keeps the leading ones (which dominate for a small difference $\Delta R$ from the horizon, relatively to the black hole radius), just like you would expect.

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    $\begingroup$ -1: Lubos, the answer is not invalid in any way. The near horizon geometry is AdS, and the only reason I said "quantum" for the black hole is to say I am considering a model string black hole, as you very well know. $\endgroup$
    – Ron Maimon
    Jan 8, 2012 at 5:20
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    $\begingroup$ No, sorry, this is not a possible interpretation of your flawed answer. The question you unsuccessfully tried to answer was one about decoherence – an important process in quantum mechanics – and the sentence made it very clear that you thought that the AdS near-horizon geometry depended on the discussion of quantum mechanics (or even decoherence) which it doesn't. As you can see, twistor understood it in the same way which led to his confusing question with the word "quantum" which has nothing whatsoever to do with the statement that AdS is the near-horizon geometry. $\endgroup$ Jan 8, 2012 at 7:24
  • $\begingroup$ By the way, you will have to give me over 15,000 fraudulent downvotes similar to this one to catch up with me, good luck. $\endgroup$ Jan 8, 2012 at 7:25
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    $\begingroup$ Many thanks for both answers guys. When I wrote the question I wasn't aware of the significance or otherwise of the quantum nature of the BH in the statement. Although it may we well known to you I didn't know about the near-horizon AdS asymptotics. I haven't done any physics for 28 years and am trying to catch up with what's been happening. In particular strings/holography look very interesting. I find that trying to follow the discussions here is a good way to do this. $\endgroup$
    – twistor59
    Jan 8, 2012 at 9:17
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    $\begingroup$ @Lubos: I am not competing with you. I upvoted you many times too (much more than the downvotes), when you merit it. It's just that I know exactly what you are insinuating, and it's totally wrong! I know why the AdS is "approximate", it's just near-horizon geometry! It is ridiculous to claim that this is misstated in my answer. The quantum situation is the reason why it is interesting, and to explain that I am considering a cold black hole, i.e. an extremal one, so that it has reversible dynamics. This is stuff everybody knows (or should know) from the 1990s, it's no longer current research! $\endgroup$
    – Ron Maimon
    Jan 8, 2012 at 15:03

The approximation is within purely classical gravity, and the parameter in which it is approximate is distance from the horizon. It is the expansion of a spacetime near the horizon of an extremal black hole to leading order in the ratio of the distance from the horizon to the radial parameter of the black hole. The reason for saying "quantum" in the answer is because the near-horizon description is most useful for cases where the near horizon behavior is the same as the behavior of bound strings, which are described by a pure gauge theory living on the branes which make up the quantum degrees of freedom of the model black hole. This case is called AdS/CFT.

When a black hole is not extremal, like a normal Schwarzschild black hole, the horizon is locally flat. So if you zoom in close to a patch of the horizon, you find ordinary Minkowski space, except that the external time coordinate becomes a Minkowski angle coordinate, just like Rindler space.

To see this formally, consider a Schwarzschild metric in a radial coordinate parametrized by u so that the usual radial coordinate r is given by $r=2M+u^2$

$$ -{u^2\over 2M + u^2 } dt^2 + {2M+u^2\over u^2} dr^2 + (2M+u^2)^2 d\Omega^2$$

Replace $dr$ by $2u du$ and consider u small, so that the metric becomes to leading order

$$ -{u^2\over 2M} dt^2 + 8M du^2 + (2M)^2 d\Omega^2 $$

The result is the Rindler space metric in the t,u coordinates, copied over a 2d sphere. Since the r,t coordinates are perpendicular to the sphere, expanding the sphere in locally flat coordinates, you recover 4d Minkowski space in Rindler form, at every local patch of the sphere. The $d\Omega^2$ part becomes $dy^2 + dz^2$ near a given point on the sphere (they are all the same), and then the usual coordinate change from Rindler space to Minkowski coordinates (with an additional rescaling to get rid of the 2M factor) works to show that the spacetime is locally Minkowski, i.e. that the horizon is not a special singular place.

The easiest extremal limit is the Reissner Nordstrom black hole:

$$ds^2 = - f(r) dt^2 + {1\over f(r)} dr^2 + r^2 d\Omega^2 $$

With $f(r) = 1-{2M\over r} + {Q^2\over r^2}$. In this case, extremality is Q=M, so that $f(r) = (1-{Q\over r})^2$. Now the horizon is at r=Q, and the integral of the distance ds from r=Q out gives infinity. This is the "infinite volume" business Lubos mentions--- it is just the infinite distance to the horizon, times the area of the sphere. It does not imply that objects cannot cross, because the horizon goes out noncausally to meet finite mass objects in finite time. Only infinitesimal test particles take forever to fall in, finite objects fall in in finite time (and I believe also fall back out in finite time, but that's another story).

Expanding r=Q+u, $f(r) = u^2/Q^2$ to leading order, and you get

$$ ds^2 = - {u^2\over Q^2} dt^2 + {Q^2\over u^2} du^2 + Q^2 d\Omega^2 $$

The first part is now a locally curved 2d spacetime, and the second part is the metric of a sphere of radius Q. The curvature of the first part must be constant across the sphere by symmetry, and it must be constant in u, since you are zooming into a small u limit, where the curvature of the radial sphere becomes constant, so that the curvature of the u part must be the negative of this by Einstein equation property of vanishing Ricci curvature. So it must be an AdS space.

But it is good to verify this explicitly. Using the logarithmic coordinate $x = Q log u$, and rescaling t by Q, the radial metric becomes

$$ ds^2 = - e^{-{2\over Q} x} dt^2 + dx^2 $$

This is now obviously homogeneous because a shift in t does nothing, while a shift in x is compensated by a rescaling of t. The curvature may be calculated quickly usinf the quick-and-dirty method described here: Ricci scalar for a diagonal metric tensor

$$ g_{\mu\nu} = - e^{- 2\alpha x} l_{00} + l_{11} $$

Where $\alpha = 1/Q$, differentiating

$$ \Gamma_{\xi,\mu\nu} = -\alpha e^{-2\alpha x} (l_{001}+l_{010}+l_{100})$$

$$ \Gamma^\xi_{\mu\nu} = -\alpha (l^0_{10} + l^0_{10} ) + \alpha e^{-2\alpha x} l^1_{00}$$

Which gives the differntiated part of the curvature

$$ \Gamma_, = 2\alpha^2 e^{-2\alpha\nu} l_{00} $$

and the product part

$$ \Gamma\Gamma = -\alpha^2 e^{-2\alpha x } l_{00} - \alpha^2 l_{11} $$

So that together they give the Ricci

$$ R_{\mu\nu} = -\alpha^2 g_{\mu\nu} $$

This is a constant negative curvature space, an AdS space.

Although I used the most trivial example of the 4d Reissner Nordstom solution, the argument is completely generic, it applies for all extremal black holes, where you have a nonvanishing curvature in the near-horizon limit. All these objects have an infinite distance to the horizon, and they all reduce to a homogeneous sphere or sphere-quotient times an AdS space in the transverse coordinates. The generic near-horizon behavior of black hole solutions is a foundation stone of AdS/CFT--- the near horizon degrees of freedom are the stuff that is stuck to the black hole, which are described in the weak coupling string theory by those strings that are stuck to the brane. The low energy limit of strings stuck to branes are necessarily described at low energy by the appropriate gauge theory.

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    $\begingroup$ Thanks for spelling it out so explicitly. As Lubos suggested, I did a search on "near horizon geometry" and turned up this reference which seems to give a pretty comprehensive list of the common cases. $\endgroup$
    – twistor59
    Jan 8, 2012 at 9:34
  • $\begingroup$ Why is this being downvoted? It is very mysterious to get a downvote on a correct answer with no explanation. $\endgroup$
    – Ron Maimon
    Jan 9, 2012 at 14:37
  • $\begingroup$ @Ron Maimon This seems to be a sign that You have some "friends" ... I thought voting should be used to tell apart correct/useful answers from wrong/not so useful answers. Using (anonymous and commentless) votes to express "friendship" makes this site less useful for everybody! $\endgroup$
    – Dilaton
    Jan 9, 2012 at 15:53
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    $\begingroup$ @Nemo: While you are probably right, there was a stupid factor of 4 error and a missing square error that might have precipitated the downvote. But it would be nice to point it out, so that it can be fixed. $\endgroup$
    – Ron Maimon
    Jan 9, 2012 at 16:39

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