If you have an extreme black hole (say with maximum charge), then it has temperature 0 and doesn't radiate. However, it seems to me that a neutral (uncharged) observer hovering above the horizon should still see Unruh radiation because he is undergoing a high acceleration.

Does this show that Hawking radiation and Unruh radiation are really different things? If not, how does one resolve this discrepancy?

And either way, doesn't the observer hovering over the horizon see the Unruh radiation escaping from the black hole? If he sees this, why is he wrong?

• I seem to remember that the derivation of Hawking radiation as Unruh radiation considers an observer at infinity (and there's a technical part involving the presence of the horizon that shows this isn't zero), so you shouldn't consider an observer "hovering above the horizon". I'll try to locate the relevant paper. – ACuriousMind Jun 2 '16 at 11:42
• @ACuriousMind: Actually, from Lubos's answer, it seems that an observer hovering just above the horizon is exactly what I should be looking at. I'd like to see the relevant paper. – Peter Shor Jun 2 '16 at 17:22
• I think it was this one. And, actually, Lubos says much the same: "So that's why the Unruh radiation is seen as a real, Hawking radiation by the observer at infinity " Hawking radiation is Unruh radiation at spatial infinity (although you derive it from the Unruh radiation from observers at finite distance). – ACuriousMind Jun 2 '16 at 18:33

First, the Unruh and Hawking radiation aren't quite "the same thing". They have a similar origin and the Unruh radiation may be considered a flat space (large black hole) limit of the Hawking radiation.

Now, the near-horizon metric of an extremal black hole is $AdS_2\times S^2$ while for a non-extremal one, the $AdS_2$ is replaced by the Rindler space.

This $AdS_2$ (two-dimensional anti de Sitter space) has consequences. First, in the static coordinates, the proper distance of any observer from the event horizon diverges. On the Wikipedia page I linked to, the formula $$ds^2=-\frac{r^2}{M^2}\,dt^2+\frac{M^2}{r^2}\,dr^2+M^2\,\big(d\theta^2+\sin^2\theta\,d\phi^2 \big)$$ implies that near $r=0$ (which corresponded to the horizon $r=M=Q$ in the original coordinates), $s =\int ds$ is proportional to the integral of $M/r$, and therefore logarithmically diverges.

The metric in the displayed formula above is locally $AdS_2$ – the curvature is constant and it's a maximally symmetric space etc. – but it is only a part of the $AdS_2$ space. The coordinates we got were the so-called "Poincaré coordinate" and they only covered a part of the $AdS_2$ space, the so-called Poincaré patch.

The Poincaré patch covers the green portion of the "global $AdS_2$" on the right part of the picture above. The observer sitting at the horizon however moves along the upper 45° tilted boundary of the green triangle and his trajectory inevitably is a geodesic. So he experiences no local acceleration – and no Unruh radiation. This is actually related to the fact that the near-horizon metric $AdS_2\times S^2$ with the appropriate electric or magnetic flux is a solution to Einstein's equations by itself – while the non-extremal near-horizon metric isn't a solution by itself.

Because the local curvature of these trajectories vanishes, the Unruh temperature vanishes, as also expected from the fact that when the "two horizons" coincide, the gravitational acceleration at the horizon vanishes.

So because the acceleration and temperature near this horizon is zero, there is no Unruh or Hawking radiation seen by this observer.

In the non-extremal case, there is a radiation that the observer keeping himself a bit above the horizon sees. Locally, it may be interpreted as the Unruh radiation, and the Unruh radiation could be undone in the flat space by using the non-accelerating reference frame. However, in a finite non-extremal black hole spacetime, things are different. The static Schwarzschild coordinates behave at $r=\infty$ as non-accelerating coordinate in the Minkowski space, but near $r=r_0$, they behave as the locally accelerating frame where the Unruh radiation exists. With the Schwarzschild choice of the time and the corresponding energy, we know that the fields aren't in the ground state of this Schwarzschild $H$ near $r=r_0$ because there's the Unruh radiation. Because $H$ is a symmetry of the background, it must be true after some time, too. At $r\to\infty$, these excitations must be still there, even though the curvature may already be neglected at $r\to\infty$.

So that's why the Unruh radiation is seen as a real, Hawking radiation by the observer at infinity (where the attraction by the black hole becomes negligible).

• You're saying that a neutral observer hovering near the horizon of an extremal black hole experiences no acceleration? Does the mass in the black hole not attract him then? – Peter Shor Jun 2 '16 at 12:09
• Yes. The limit of the rest-frame acceleration for the observer approaching the event horizon of an extremal black hole is zero. In general relativity, whether gravity "attracts" someone depends on coordinates. Look at the green diagrams. If an observer sits close to the RN event horizon but not quite at it, it moves along the non-null trajectory on the right, and that line simply is intrinsically straight, as clear from the right AdS2 picture where it's the vertical boundary of the diagram. – Luboš Motl Jun 2 '16 at 12:16
• Let me try to explain this thing - which disagrees with the Newtonian intuition - using a different analogy. Something is falling to the black hole, OK? So a force $F$ acts on it. And a force $F$ over distance $s$ makes work $Fs$, OK? But the distance is infinite while the work that the black hole does on the object can't go to infinity, so the force must $F\to 0$. None of these considerations have good analogies in the Newtonian gravity in the flat space because the proper distances from and near a surface of a localized object cannot diverge. – Luboš Motl Jun 2 '16 at 12:23
• One more way to see why the acceleration is zero. In the Rindler space, the low-radius highly curved hyperbola experiences a big intrinsic acceleration $a=1/r$ OK? But here the hyperbola goes near the left vertex of the green triangle in either part of the picture, and that's near the global AdS boundary. The AdS boundary naturally repels matter - oscillating curves inside the AdS cylinder are geodesics - so this curved hyperbola is exactly the kind of geodesics that AdS considers natural. The whole difference from the non-extremal case is the divergent profile of g_rr. – Luboš Motl Jun 2 '16 at 14:05
• The answer above isn't correct: a neutral observer hovering just above an extremal black hole horizon does experience a proper acceleration. Unlike for a non-extremal black hole, however, this acceleration is finite in the limit that the horizon is approached. The surface gravity $\kappa$ is equal to the proper acceleration $a$ times the redshift factor, which is the norm of the horizon generating Killing vector, $|\xi|$. At the horizon $|\xi|=0$, but $a$ is finite for an extremal black hole, so $\kappa=0$. – Ted Jacobson Aug 13 '18 at 21:44