1
$\begingroup$

Using QFT in curved spacetime, Hawking was able to show that black holes evaporate. Whilst this has never been observered, the commonly excepted statement is that black holes continually radiate until some point at which they explode, leaving no trace of the black hole behind.

My question is then to what extent can the singularity be removed given that there is no smooth mapping between a manifold that has a singularity and a manifold that does not? What happens to the singularity when a black hole evaporates?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ Why would there need to be a smooth mapping? We have no universally accepted theory of quantum gravity, it is not clear which theory should model the actual "moment of evaporation" at all, let alone that this theory would require a smooth transition. $\endgroup$
    – ACuriousMind ♦
    Commented Apr 12, 2020 at 10:45
  • $\begingroup$ @ACuriousMind, this is an interesting point. Theories like canonical loop quantum gravity argue for the dismissal of a smooth pseudo-reimannian manifold at the Plank scale in favor of a discrete nature of spacetime. I had not considered this - in which case as you say the assumption of a smooth mapping need not hold. How then the singularity will evolve as the black hole radiates will depend on the details of a quantum theory of gravity. Thankyou for this point. $\endgroup$
    – Jack Hughes
    Commented Apr 12, 2020 at 10:50
  • 2
    $\begingroup$ Can we assume that the singularity has formed in the first place? Hawking radiation assumes a steady state solution of Einstein's equation in which there is already a singularity, but from the point of view of an exterior observer, it takes an infinite time for singularities to form. $\endgroup$
    – Charles Francis
    Commented Apr 12, 2020 at 10:59
  • 1
    $\begingroup$ Even ignoring quantum gravity problems, can there be said to be a singularity at all if the mass-energy there is zero ? There's nothing to maintain a singularity. Note we presume the singularity was "switched on" at some point, so why is there a problem thinking it could be "switched off" ? $\endgroup$
    – StephenG - Help Ukraine
    Commented Apr 12, 2020 at 11:28
  • 1
    $\begingroup$ There are two issues with your question. One is that singularities don’t exist in reality, but only in some outdated relativity textbooks misguiding many people. The other is that there is no Hawking radiation In General Relativity. The fundamental principle of General Relativity is that physics does not depend on our view (frame of reference). In contrast, the Hawking radiation is based on the Unruh principle that the reality is different for different observers. These two principles are in a direct contradiction with each other. $\endgroup$
    – safesphere
    Commented Apr 12, 2020 at 19:06

1 Answer 1

Reset to default
1
$\begingroup$

Hawking radiation is predicted using the steady state solution for a black hole which has already formed (this is discussed in some textbooks, such as MTW's Gravitation), but I think it is first important to discuss whether such a steady state solution can appear in our universe.

Because time appears to stop at the Schwarzschild radius, an issue is raised as to whether a singularity can actually form. In 1939 Julius Robert Oppenheimer (one of those known as the “father of the atomic bomb” for their role in the Manhattan Project) and one his students, Hartland Snyder, published the first calculation of gravitational collapse They concluded that, from the point of view of an exterior observer,

“it is impossible for a singularity to form in a finite time.”

The Schwarzschild radius was interpreted as a boundary at which time stopped. Black holes were called frozen stars, never actually becoming singular. They calculated that an observer on the surface of a collapsing star would see a different result:

“The total time of collapse for an observer comoving with the stellar matter is finite, and for this idealized case and typical stellar masses, of the order of a day.”

Thus, a collapsing star will create a singularity in finite proper time for an infalling observer, but infinite time for an outside observer. The interior solution, even if real to an infalling observer, exists only in our infinite future, and cannot exist at all in a universe with a Big Crunch.

Improve this answer
$\endgroup$
7
  • $\begingroup$ In fact the interior solution is not even defined as a part of our universe, but only as a mathematical extension to it. Pre-collapse, there is no interior solution and even post collapse there is nothing to say that the event horizon is not a singular point, which only looks like sphere in particular coordinates -- the argument used by MTW to extend beyond the event horizon fails because you cannot impose the general principle of relativity on a boundary, or on a singularity. $\endgroup$
    – Charles Francis
    Commented Apr 12, 2020 at 17:13
  • $\begingroup$ It is inaccurate to state that the black hole interior exists only in the infinite future of outside observers. This suggests that the interior region is always in the causal future of any outside observer. However, of any outside observer there will be parts of the interior region (including parts of the singularity) that are spacelike separated from the observer. $\endgroup$
    – TimRias
    Commented Apr 13, 2020 at 11:22
  • $\begingroup$ "Hawking radiation is predicted using the steady state solution for a black hole which has already formed" This is false. Hawking radiation can be derived using an "atrophysical" spacetime of a collapsing star. At no point does the derivation require the presence of a singularity. $\endgroup$
    – TimRias
    Commented Apr 13, 2020 at 11:27
  • $\begingroup$ @mmeent, agreed, this was a very informal way of expressing that the interior is never in the causal past of the exterior observer. It uses the outside observer's notion of determined by Einstein synchroneity, as in Schwarzschild coordinates. I sacrificed some accuracy for the sake of simplicity. $\endgroup$
    – Charles Francis
    Commented Apr 13, 2020 at 11:28
  • 1
    $\begingroup$ @mmeent, The Schwarzschild time coordinate is the natural time coordinate for an external observer, and expresses the causal relationship correctly from that point of view. Ingoing coordinates cease to have empirical meaning at the event horizon. Extending beyond the event horizon remains an unverifiable hypothesis. $\endgroup$
    – Charles Francis
    Commented Apr 13, 2020 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.