What happens to a star's heat when it gets torn apart by a black hole?

[mitosis]

I'm sure this is a really basic question for the black hole experts here. Consider a situation where a star approaches and then crosses the event horizon of a super massive black hole, the star gets torn apart eventually undergoing spaghettification as it approaches the singularity. My question is, since black holes are really cold and the star begins this process as really hot, what happens to the star's heat?

 

[Ibix]

What do you mean by the star's heat? Do you mean the thermal energy stored in it? That just counts as mass, so the star has slightly more mass than it would do cold, and the black hole has slightly more mass than it would if it swallowed the same number of cold hydrogen atoms as made up the star.

Or do you mean why does the star radiate and the black hole doesn't? Because the black hole doesn't let heat or light escape, except a tiny amount of Hawking radiation. In that sense nothing happens to the star's heat. It's just that it can't radiate to the outside universe once it's crossed the event horizon.

 

[mitosis]

What do you mean by the star's heat? Do you mean the thermal energy stored in it? That just counts as mass, so the star has slightly more mass than it would do cold, and the black hole has slightly more mass than it would if it swallowed the same number of cold hydrogen atoms as made up the star.

Or do you mean why does the star radiate and the black hole doesn't? Because the black hole doesn't let heat or light escape, except a tiny amount of Hawking radiation. In that sense nothing happens to the star's heat. It's just that it can't radiate to the outside universe once it's crossed the event horizon.

Yes, I meant the thermal energy of the star. I have read that black holes are extremely cold, much colder than the 2-3 Kelvin of space, and the larger the black hole the colder it is. So does the thermal heat get transformed into mass as it enters the black hole?

 

[Ibix]

So does the thermal heat get transformed into mass as it enters the black hole?

It already contributes to the mass of the star. It contributes the same amount to the mass of the hole.

The point is that the "surface" of a black hole isn't really a physical thing. The matter of the star falls through and on to the singularity. So physical properties of the star like its temperature become irrelevant - apart from any contribution to mass, charge or angular momentum, they can't be measured. The "temperature" of the black hole isn't much to do with its contents (the concept doesn't even necessarily make sense, bizarrely enough). It is just a measure of its emission of Hawking radiation, which depends on the size of the hole and nothing else.

 

[mitosis]

Thank you Ibix, helpful explanations.

 

[Ibix]

You're welcome. Black holes are places here ordinary physical intuitions really do not apply, and descriptions of them in anything other than maths tend to lack a certain precision. But they are fascinating.

 

[mitosis]

So, were it not for Harking radiation, the event horizon would seem to an outside observer to have a temperature of absolute zero?

 

[Elroch]

Well, in the limit. You could never receive a signal from very close to the event horizon because such a signal would take a time tending to infinity to reach the remote observer (and clearly it has not had more than 13.8 billion years).
Perhaps better to say that anything very near the event horizon is so strongly gravitational red-shifted that it would appear to be almost at absolute zero if it could be seen. To be complete, this assumes it is not moving very rapidly towards the remove observer.

 

[Ibix]

So, were it not for Harking radiation, the event horizon would seem to an outside observer to have a temperature of absolute zero?

I'd be inclined to think in terms of temperature not being definable for a black hole. I'm not sure. I know that difficulties defining thermodynamic quantities like entropy were what led Hawking to his work on black holes, but I don't know that much about it yet. Others here do.

 

[PeterDonis]

I'd be inclined to think in terms of temperature not being definable for a black hole.

That's correct; classically, there is no way to define a temperature for a black hole, because there is no way for it to be in thermal equilibrium with anything else.

 

[Elroch]

An interesting relevant question is whether you can measure the temperature of a cold object without being in equilibrium with it. I believe this can be achieved by observing the flow of radiant energy from a warm object that is absorbed by the cold object, the "cooling effect" of the cold object (this is similar to the opposite done with a hot object out of equilibrium, where one may measure the radiation emitted by the object. This is routinely done without of equilibrium warm objects, including in work I have been involved in in the past). In simple terms, the surface of something pointing towards a cold object is lowered in temperature, because there is a net flow of radiant energy away from the warmer object from the surface. This lowering in temperature can at least in principle be measured and the temperature of the cold object inferred. (Note that since a lens that is warmer than the cold object would wreck the measurement, the cold object needs to be very large in the field of view for this to have a chance of working unless you can cool a lens sufficiently).

[EDIT: thinking a little more on this, it is clear the correct way to do it would be to actively cool the detector surface as much as possible. This increases the signal to noise ratio. This is a technique that is already important in astronomy for observations in longer wavelengths].

As a result, I would respectfully take the position that the classical black hole model does have a temperature and that temperature is absolute zero. If a classical black hole could exist, this is the temperature that would be measured in isolation by a version of the above procedure. Given that the Hawking temperature of a 1 stellar mass black hole is 0.00000006 Kelvin, a very delicate experiment would be needed to detect the difference, even if you could find a black hole in perfect isolation. Such an experiment might sound impractical, but so did observing gravitational waves to Einstein.

 

[PeterDonis]

An interesting relevant question is whether you can measure the temperature of a cold object without being in equilibrium with it.

Two objects exchanging heat in order to come to thermal equilibrium with each other is certainly a vaild procedure and can be used to measure temperature. The problem with a classical black hole is that you can't do this: you can shine radiant energy into it all you want and it won't come to thermal equilibrium with anything because it will never radiate anything back; it will just continue gaining mass.

I would respectfully take the position that the classical black hole model does have a temperature and that temperature is absolute zero.

The problem with this is that the entropy of a classical black hole would also have to be zero, but that would mean you could violate the second law of thermodynamics by the process you describe--just keep on dumping heat from some other object into the black hole, continuously cooling the other object indefinitely (approaching but never reaching absolute zero). And if you are violating any of the laws of thermodynamics, you can't depend on concepts like temperature having a well-defined meaning any more.

Given that the Hawking temperature of a 1 stellar mass black hole is 0.00000006 Kelvin, a very delicate experiment would be needed to detect the difference

Yes, that's true; in a practical sense, the temperature of any black hole of this size or larger can be taken to be absolute zero for almost all purposes. But theoretically, the ability of a black hole to emit Hawking radiation when quantum effects are taken into account makes a big difference when trying to apply the laws of thermodynamics to a system containing a black hole.