Hawking radiation from point of view of a falling observer

This paper tells that Hawking claimed that the falling to a black hole observer will not detect any radiation. But only because the frequency of the Hawking radiation will be of the order $1/R_s$ so that the falling observer will not have a suitable detector.

This argument seems fallacious to me. First, the falling orserver can communicate to a distant observer who has a suitable detector. That observer will tell him that the BH indeed evaporates. So its radius will decrease for both falling and distant observer.

Before the falling observer will touch the hoizon, the radius of the BH will become smaller than his own dimentions so he will be able to detect the radiation. The smaller the BH will become, the higher frequency of the Hawking radiation will be. At some point the apparent temperature of the BH for a falling observer will reach infinity so that the BH will explode.

Am I right that this claim by Hawking is fallacious and in fact all ouside observers will detect Hawking radiation sooner or later?

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related question: physics.stackexchange.com/questions/22942/… –  Nathaniel Aug 4 '12 at 14:42

The analysis you made that the observer will see the black hole disappear before the observer falls in is false for large black holes. It is only true if the black hole is at the last stages of evaporation, so that the (necessarily microscopic) observer just can't get close to it in time, before it blows up.

For a standard black hole, the black hole horizon that the infalling observer sees disappears completely when the observer gets close. The reason is simply because to first order away from the horizon

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

$$f(r) = (1-{2m\over r})$$

defining a variable $r = 2M + u$, the metric turns into

$$ds^2 = C u dt^2 + {1\over C u} du^2 + (2M)^2 d\Omega^2$$

Where C is some constant. The sphere part is now constant to lowest order in u, and defining the variable $v$ by $v^2 \propto u$, you can reduce the time-radial part to standard Rindler form:

$$ds^2 = v^2 dt^2 + dv^2 + R^2 d\Omega^2$$

This is 4d Rindler space, if you use tangent space coordinates for the sphere. In these v,t coordinates, it is clear how to extend to the interior--- you just go on using the local x,t coordinates. This is equivalent to Kruskal or tortoise coodinates extension, since locally, near the horizon, Kruskal or any other extending coordinate system is just doing the u,v transformation.

For a static observer near the black hole, the Hawking thermal state is the same as the Unruh temperature the observer would see if accelerating in flat space. So switching off the rocket makes the radiation go away. It's the same phenomenon, whether the horizon is a black hole horizon or an Unruh acceleration horizon, because the two metrics are locally the same.

There is still the minor issue of whether the curvature of the black hole is important when you are close enough. For this question, it is important to note that the local metric defines the local temperature state (the period of t in imaginary time is the only relevant thing, and this is the periodicity in the polar form of the metric in the v coordinates above). The curvature of the geometry is only important at scales comparable to the size of the black hole. So if you want to see Hawking radiation, you have to tune your detector to find wavelengths comparable to the size of the entire black hole, and for a local detector, this requires a long time, to build up the coherent oscillations, and you don't have a long time, because you are going to fall into the black hole. This is Hawking's point. It is correct, but misleading. The reason the radiation vanishes is that it is Unruh radiation when you are close, the two geometries are the same.

This can be clarified using Unruh radiation. If an observer is accelerating in empty Minkowki space, this observer sees thermal particles coming out of the acceleration horizon and free-falling back in, so a thermal bath of particles from all directions. This thermal bath clearly disappears when the observer switches off the rocket and stops accelerating.

But the Hawking radiation from a black hole, for an observer sitting at constant r very near the horizon is just identical to the Unruh radiation the observer would expect at this acceleration. This means that you should identify this radiation with Unruh radiation, and it should have the property that if you switch off your rocket, and fall into the black hole, then the radiation goes away.

The argument is by analogy with Unruh radiation, and the analogy is often hidden using imaginary time methods. When you use imaginary time, the Unruh temperature is the inverse period of the Rindler time coordinate. This is also true when you use the imaginary time continuation of the black hole exterior. At any point on the horizon, the space looks like Unruh's, so the effects should be the same.

But this is not a mathematical proof, and many people, beginning with t'Hooft and now a collaboration including Polchinski, have tried to argue that there is something drastically wrong with this picture at the horizon. Without a full string theoretic calculation of what happens to an observer falling into a black hole, it is hard to say this is established with 100% certainty, although I find the arguments above fully convincing, and the counterarguments are no good.

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Why have the comments been deleted? By the way, +1 to cancel the unfair -1 –  drake Aug 7 '12 at 16:01
But if the free falling observer is far from the horizon, then he sees the radiation, or not ? And the radiation is blue shifted if he moves towards the black hole. –  jjcale Oct 21 '12 at 15:22