From the perspective of a far-away observer, matter falling into a black hole never crosses the boundary. Why doesn't a basic symmetry argument prove that Hawking radiation is therefore also frozen on the boundary, and therefore not observable? Wouldn't the hawking radiation have to have started its journey before the formation of the black hole? Furthermore, wouldn't the radiation be infinitely red-shifted?

  • $\begingroup$ I thought the never-cross-the-horizon argument only applied to (massless) test particles. If the particle has mass, then strictly speaking you have to compute it's own (admittedly small) contribution, and in that case the matter truly does cross the horizon, even for a distant observer. $\endgroup$
    – Warrick
    Aug 24, 2012 at 8:38

2 Answers 2


Classically, this is true. Something exiting from a classical static black hole would have had to have started before the universe was created. The frozen-on-the-horizon view of a black hole is the view you get under classical general relativity. When you add quantum mechanics, this view is no longer quite valid, and you get Hawking radiation, which from a far-away observer's viewpoint, interacts with the infalling matter.


Evaporating black holes behave qualitatively differently from static ones if you allow an infinite amount of time to pass. In particular, there is no event horizon for an evaporating BH, only an apparent horizon. Furthermore, if you stack the apparent horizons of a shrinking black hole plus time, the resultant surface is two-way transversible.

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    $\begingroup$ do you have any references for that? two-way transversible mean that light can traverse it, so far observers can see behind it, including the singularity $\endgroup$
    – lurscher
    Aug 23, 2012 at 19:55
  • $\begingroup$ @lurscher: You can look at any of Ashtekar's papers on Dynamic horizons. It's easy enough to see if you imagine the null ray that is trying to escape the black hole, sitting stationary on the boundary of the hole. If the black hole shrinks, then this null ray is now on the outside of the black hole, and will escape to infinity. $\endgroup$ Aug 23, 2012 at 21:25
  • $\begingroup$ A more technical approach will say that if you assume cosmic censorship, then you can prove the area theorem. Shrinking black holes explicitly contradict the area theorem. Also, if the black hole evaporates completely, then there should be no singularity at future timelike infinity. $\endgroup$ Aug 23, 2012 at 21:26
  • $\begingroup$ reference: relativity.livingreviews.org/Articles/lrr-2004-10 $\endgroup$ Aug 23, 2012 at 21:28
  • $\begingroup$ good, but the information taken by the ray waiting for the black hole to shrink to exit to outside needs to be erased at that boundary, or swamped in thermal noise, to keep thermodynamics law consistent, i think $\endgroup$
    – lurscher
    Aug 23, 2012 at 21:29

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