# How does the temperature evolve in a star collapsing into a black hole?

The temperature of a massive star is very high, and for neutron stars it is typically above $10^5\ \mathrm{K}$. (https://en.wikipedia.org/wiki/Neutron_star#Mass_and_temperature)

However, the model based on Hawking radiation gives the temperature of a black hole as $\hbar c^3/8\pi GMk_B$ (https://en.wikipedia.org/wiki/Hawking_radiation)

which gives very cold temperatures for black holes of stellar masses.

How do the two meet? Do we know of a phenomenon that causes the star to cool down as it collapses? or is the Hawking temperature only applicable to the event horizon? in which case, what is the temperature of the singularity, and its thermic conductivity?

• Not being an expert in black hole collapse, I would presume that the temperature of material falling toward (but not yet having reached) the singularity is very high. When the material gets close to the singularity, this question becomes 'how does temperature behave in a quantum gravity setting?' To external observers, the only physical quantities that an observer can have knowledge of are mass, angular momentum, and electric charge. Since mass for a Schwartzchild black hole is $\propto T^{-1}$, observers outside the black hole will see a very low temperature. – Bob Aug 22 '17 at 18:24

Matter falling into a black hole is accelerated to relativistic speeds, i.e., speeds comparable to the speed of light. Friction converts some of this kinetic energy to heat, so the matter gets extremely hot. This is why a black hole's accretion disk commonly emits x-rays: hotter matter emits electromagnetic waves at higher frequencies.

Once the matter passes through the event horizon, it only exists for a matter of milliseconds (for a solar-mass black hole) before it impacts the singularity. In our current theories, this means that it ceases to exist. In any case, we can't observe this part of the process from the outside, since information-carrying radiation can't get out through the event horizon.

However, the model based on Hawking radiation gives the temperature of a black hole as ℏc3/8πGMkB [...] which gives very cold temperatures for black holes of stellar masses.

This temperature is purely hypothetical, since it will not be until the very distant future that any black hole finds itself not exposed to an incoming radiation background that is many orders of magnitude hotter than the Hawking temperature. (We also have no particular reason to trust predictions of Hawking radiation, since semiclassical gravity is inconsistent and has never been tested against observation.) That temperature, supposing that it eventually occurs, is not the temperature of the matter that gravitationally collapsed. It's the temperature of the radiation in the empty space surrounding the black hole.

• How can empty space have a temperature? – descheleschilder Mar 21 '19 at 3:31
• @descheleschilder This is a separate question and should not be a comment. But a quick answer is that heat may be transferred through empty space by e.g. radiation. – Codename 47 Mar 21 '19 at 10:09
• @descheleschilder: Thanks for the comment, I'll edit to clarify. – user4552 Mar 21 '19 at 13:40

To figure this out, you need to have a model of the collapse. One simple model used for this purpose is the Oppenheimer-Snyder model, which describes the symmetric collapse of a uniform sphere of matter under relativistic gravity.

In this model, the interior of the sphere - i.e. the region of space where the matter is - behaves basically like a small patch of the Universe undergoing the Big Bang in reverse, i.e. a Big Crunch. If this is assumed to in any way approximate the reality, then one can reasonably surmise the answer would be that the temperature and density will rise similarly to just such a reverse of the Big Bang, so you would also expect the various phenomena that occurred there to happen as well, until (presumably, at least), they reach on the order of the Planck temperature, about $$10^{32}\ \mathrm{K}$$, and density, around $$10^{96}\ \mathrm{kg/m^3}$$, and at that point, quantum-gravitational effects will take over and that's where our evidence-backed theories run out. It may also be possible they run out even sooner than this - we just don't know.

Insofar as "reconciling" this with the apparently cold temperature predicted by the Hawking formula is concerned, the answer is that they are discussing the results of measurements done at two very different points in space-time. One of these is at the very center the collapsing star that was forming the black hole as the collapse is taking place, the other measurement is being done from very far away long, long after the supernova of that star has faded out.