Hawking and Unruh effects -- Differences and interpretations

[Joker93]

Hello,
I am a bit confused on the relation between the Hawking effect(radiation) and the Unruh effect.
What I understood with my little knowledge is that the Hawking temperature is the temperature that is emitted at the event horizon of a black hole as measured by an observer at infinite spatial distance from the black hole. So, since the Unruh effect is observer-dependent, I see an analogy of the Hawking-Unruh effects with the effect of gravitational time dilation.
In gravitational time dilation, the proper time(an invariant) between two events is different from the coordinate time(observer-dependent) in the same way that the Hawking temperature is observer independent (as this paper explains: https://link.springer.com/content/pdf/10.1007/JHEP10(2016)161.pdf ) and the Unruh effect is observer-dependent. (it's like the Hawking temperature is the "proper temperature")

So, is this right?
Also, how are the two effects(Hawking and Unruh) exactly related?
[the wikipedia article might help: https://en.wikipedia.org/wiki/Hawking_radiation --go to the section "emission process"--I don't really get the idea completely though]

Extra: If my explanation of the Hawking temperature is correct(i.e. the temperature emitted at the horizon as measured by an observer at infinity) and since it is finite, wouldn't the background temperature that we should measure be huge since we are far away from any black hole and there is a huge number of black holes in the Universe?

Thanks in advance.

 

[PeterDonis]

the Hawking temperature is the temperature that is emitted at the event horizon of a black hole as measured by an observer at infinite spatial distance from the black hole.

Heuristically, yes, this is a reasonable way to view it. (There are a lot of subtleties lurking in the math, though...)

since the Unruh effect is observer-dependent, I see an analogy of the Hawking-Unruh effects with the effect of gravitational time dilation.

I don't think this analogy is valid. The observer dependence of the Unruh effect is also present in the Hawking effect: an observer free-falling into the black hole does not see any Hawking radiation, just as an inertial observer in flat spacetime does not see any Unruh radiation.

(There are various speculative proposals related to resolving the black hole information paradox that claim that an observer free-falling into a black hole would see Hawking radiation, or at least something similar to it--but those proposals break the correspondence between the Hawking effect and the Unruh effect that you are describing.)

how are the two effects(Hawking and Unruh) exactly related?

The exact relationship is difficult to describe without math which I am not an expert on. But heuristically, both effects are manifestations of the fact that which state of the quantum field is the "vacuum state" (state of lowest energy) depends on the state of motion of the observer; a free-falling observer sees a different field state as the "vacuum" from an accelerated observer. So, for example, a state that looks like the vacuum to an inertial observer will look like a thermal state at some finite temperature to an accelerated observer (where the temperature is proportional to the acceleration). The basic math describing this is the same for both the Unruh effect and the Hawking effect; the only differences are the geometry of the spacetime (Minkowski for Unruh, Schwarzschild for Hawking) and the state the quantum field is assumed to be in (the Minkowski vacuum for Unruh, and IIRC a state called the Hartle-Hawking vacuum for Hawking).

A discussion that might be helpful is on the Usenet Physics FAQ site here:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

 

[PeterDonis]

If my explanation of the Hawking temperature is correct(i.e. the temperature emitted at the horizon as measured by an observer at infinity) and since it is finite, wouldn't the background temperature that we should measure be huge since we are far away from any black hole and there is a huge number of black holes in the Universe?

There are a huge number of black holes, but they all have a very, very, very low temperature. And temperatures don't add; if our sky were filled with a uniform distribution of black holes all at temperature $T$, that would just mean our whole sky was at temperature $T$. Try computing what $T$ will be assuming a black hole of one solar mass, which is a good enough lower bound on the mass of black holes that might be in our sky (the Hawking temperature goes down as the mass goes up, so this is an upper bound on the temperature $T$ of our sky due to Hawking radiation from black holes).

 

[Joker93]

Heuristically, yes, this is a reasonable way to view it. (There are a lot of subtleties lurking in the math, though...)
I don't think this analogy is valid. The observer dependence of the Unruh effect is also present in the Hawking effect: an observer free-falling into the black hole does not see any Hawking radiation, just as an inertial observer in flat spacetime does not see any Unruh radiation.

(There are various speculative proposals related to resolving the black hole information paradox that claim that an observer free-falling into a black hole would see Hawking radiation, or at least something similar to it--but those proposals break the correspondence between the Hawking effect and the Unruh effect that you are describing.)
The exact relationship is difficult to describe without math which I am not an expert on. But heuristically, both effects are manifestations of the fact that which state of the quantum field is the "vacuum state" (state of lowest energy) depends on the state of motion of the observer; a free-falling observer sees a different field state as the "vacuum" from an accelerated observer. So, for example, a state that looks like the vacuum to an inertial observer will look like a thermal state at some finite temperature to an accelerated observer (where the temperature is proportional to the acceleration). The basic math describing this is the same for both the Unruh effect and the Hawking effect; the only differences are the geometry of the spacetime (Minkowski for Unruh, Schwarzschild for Hawking) and the state the quantum field is assumed to be in (the Minkowski vacuum for Unruh, and IIRC a state called the Hartle-Hawking vacuum for Hawking).

A discussion that might be helpful is on the Usenet Physics FAQ site here:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

Thanks for the reply! I appreciate it. I have a couple of questions though:

So, when a book mentions the Hawking temperature and says that it is equal to 1/(8πGM), it actually wants to say that it is the temperature of the event horizon as seen by an observer at infinity after it is redshifted?
Also, you say that the Hawking effect is observer-dependent. But, most books state that the Hawking temperature is the one given above: a constant. So, how is this observer dependent?

Lastly, Carroll might shade some light through a section of his book (Spacetime and Geometry: An introduction to General Relativity) on p.413.
If you have access to it, you can see his last paragraph (After eq.9.173) where he explains it clearly enough.
Less clear for me is the following: for equation 9.172 he basically red-shifts the temperature as seen by an observer that has an acceleration α at point r (see equation 9.170). So, he redshifts the Unruh effect's temperature to give the Hawking temperature. So, is he saying that no matter what the acceleration of the Unruh observer is, the corresponding temperature will translate to the Hawking temperature 1/(8πGM) for the static observer at infinity(which means that for that distant, static observer, the acceleration of the Unruh observer is irrelevant)?

 

[PeterDonis]

So, when a book mentions the Hawking temperature and says that it is equal to 1/(8πGM), it actually wants to say that it is the temperature of the event horizon as seen by an observer at infinity after it is redshifted?

Heuristically, yes.

most books state that the Hawking temperature is the one given above: a constant. So, how is this observer dependent?

Because it's specific to the observer at infinity. There are two dependencies on the observer involved:

(1) Observers who are "hovering" at some finite altitude above the hole's horizon will see a different temperature than the observer at infinity; how different depends on their altitude. As they get very close to the horizon, hovering observers will see temperatures that increase without bound.

(2) As I mentioned before, observers who are free-falling into the hole will see no Hawking radiation at all. To them the hole is a vacuum.

 

[PeterDonis]

is he saying that no matter what the acceleration of the Unruh observer is, the corresponding temperature will translate to the Hawking temperature 1/(8πGM) for the static observer at infinity(which means that for that distant, static observer, the acceleration of the Unruh observer is irrelevant)?

Basically, yes. The only caveat is that the "Unruh observer" picture for a black hole is only really valid very close to the horizon, because the dependence of the acceleration (and hence of the observed temperature) on distance from the horizon is different in flat spacetime (which is the proper setting for the Unruh effect) vs. the curved spacetime around a black hole.

 

[Joker93]

Heuristically, yes.
Because it's specific to the observer at infinity. There are two dependencies on the observer involved:

(1) Observers who are "hovering" at some finite altitude above the hole's horizon will see a different temperature than the observer at infinity; how different depends on their altitude. As they get very close to the horizon, hovering observers will see temperatures that increase without bound.

(2) As I mentioned before, observers who are free-falling into the hole will see no Hawking radiation at all. To them the hole is a vacuum.

So, the Hawking temperature is much more general than just 1/(8πGM)? Is it also independent of the Unruh effect(which does not seem so since Carroll uses the Unruh temperature--he redshifts it--to derive the 1/(8πGM) Hawking temperature)? Thus far, it seems to me that the Unruh effect has to do with the acceleration of the observer while the Hawking effect has to do with the spatial distance from the black hole but I think that it's not so simple.

[Sorry for the many questions; it's just that some resources I found really confused me]

 

[PeterDonis]

So, the Hawking temperature is much more general than just 1/(8πGM)?

The Hawking effect is much more general; all "hovering" observers in the spacetime around a black hole see Hawking radiation, not just the observer at infinity. The Hawking temperature is defined by the expression you gave, which is the temperature seen by an observer at infinity. This is mainly for convenience, so people can talk about things like the dependence of the temperature on the mass of the hole, the implications for evaporation of the hole, etc., without getting bogged down in discussions of how things vary from observer to observer for a given hole.

Is it also independent of the Unruh effect

I'm not sure what you mean. The two effects are different effects, physically, although there are similarities between them.

Carroll uses the Unruh temperature--he redshifts it--to derive the 1/(8πGM) Hawking temperature

Carroll's "derivation" is really a heuristic one, making use of the similarity of a "hovering" observer very close to a black hole's horizon to an accelerated observer in flat spacetime. It's not meant (as far as I can tell) to be a rigorous claim that the two effects are the same thing.

it seems to me that the Unruh effect has to do with the acceleration of the observer while the Hawking effect has to do with the spatial distance from the black hole

The Hawking effect has to do with the mass of the black hole. Spatial distance doesn't come into it unless you want to look at the variation in the observed temperature with the altitude of a "hovering" observer above the horizon. But that also happens for the Unruh effect: if you consider a family of accelerated observers in flat spacetime that are maintaining a constant distance from each other (these are called "Rindler observers" in the literature), so that they all have the same Rindler horizon, the "temperature" of Unruh radiation they observe will vary from observer to observer.

 

[Joker93]

The Hawking effect is much more general; all "hovering" observers in the spacetime around a black hole see Hawking radiation, not just the observer at infinity. The Hawking temperature is defined by the expression you gave, which is the temperature seen by an observer at infinity. This is mainly for convenience, so people can talk about things like the dependence of the temperature on the mass of the hole, the implications for evaporation of the hole, etc., without getting bogged down in discussions of how things vary from observer to observer for a given hole.
I'm not sure what you mean. The two effects are different effects, physically, although there are similarities between them.
Carroll's "derivation" is really a heuristic one, making use of the similarity of a "hovering" observer very close to a black hole's horizon to an accelerated observer in flat spacetime. It's not meant (as far as I can tell) to be a rigorous claim that the two effects are the same thing.
The Hawking effect has to do with the mass of the black hole. Spatial distance doesn't come into it unless you want to look at the variation in the observed temperature with the altitude of a "hovering" observer above the horizon. But that also happens for the Unruh effect: if you consider a family of accelerated observers in flat spacetime that are maintaining a constant distance from each other (these are called "Rindler observers" in the literature), so that they all have the same Rindler horizon, the "temperature" of Unruh radiation they observe will vary from observer to observer.

Thanks for the replies! They are really helpful.

So, a "hovering" observer sees a temperature which is the sum of the temperature from the Hawking effect(not 1/(8πGM) in general) and from the Unruh effect?

Lastly, do you know of any good resource to learn these effects and their differences? While your answers are great, I still feel a little confused.

 

[PeterDonis]

So, a "hovering" observer sees a temperature which is the sum of the temperature from the Hawking effect(not 1/(8πGM) in general) and from the Unruh effect?

No. There is only one thing going on. If the spacetime geometry is Minkowski, it's the Unruh effect. If the spacetime geometry is Schwarzschild, it's the Hawking effect. If the spacetime geometry is Schwarzschild but you just look at a small patch of spacetime close to the horizon, you can heuristically understand what an accelerated observer sees by using the Unruh effect as an approximation (because you can view the small patch of spacetime as flat and the hole's horizon as the Rindler horizon of the accelerated observer); but it's just an approximation to the full math of the Hawking effect.

do you know of any good resource to learn these effects and their differences? While your answers are great, I still feel a little confused.

This is an "A" level topic, so you'll need a lot of background knowledge to be able to make sense of the references. (One of the reasons it's so hard to find good sources on this topic is that it's so hard to explain it without bringing in all that background knowledge.) The best one I know of is Wald's monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. It discusses both the Unruh effect and the Hawking effect (and in fact this is where I first read about both of them).

 

[Joker93]

No. There is only one thing going on. If the spacetime geometry is Minkowski, it's the Unruh effect. If the spacetime geometry is Schwarzschild, it's the Hawking effect. If the spacetime geometry is Schwarzschild but you just look at a small patch of spacetime close to the horizon, you can heuristically understand what an accelerated observer sees by using the Unruh effect as an approximation (because you can view the small patch of spacetime as flat and the hole's horizon as the Rindler horizon of the accelerated observer); but it's just an approximation to the full math of the Hawking effect.
This is an "A" level topic, so you'll need a lot of background knowledge to be able to make sense of the references. (One of the reasons it's so hard to find good sources on this topic is that it's so hard to explain it without bringing in all that background knowledge.) The best one I know of is Wald's monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. It discusses both the Unruh effect and the Hawking effect (and in fact this is where I first read about both of them).

Thanks a lot! I truly appreciate it!
The last bit clarified some stuff. I will continue on with the book that you mentioned.

 

[Swamp Thing]

In the Hawking effect, the energy that escapes ultimately comes from stuff that fell in, which is why we expect that it would eventually deplete the hole.

What about the Unruh effect in flat spacetime? If the observer absorbs a lot of Unruh photons, presumably they will experience a corresponding retarding force. (They would have to throttle up their rocket just a little bit in order to maintain the same acceleration in the face of the Unruh photon pressure). To me, this suggests that the photon energy that they absorbed came from their own rocket or jet pack or whatever was trying to accelerate them in the first place.

Applying this logic back to the black hole, if a hovering (jet pack equipped) probe absorbs a lot of photons just outside the horizon, would there be a net radiation force acting against its rocket thrust, tryng to push it towards the hole? If so, the probe would have to throttle up a bit in order to continue hovering in the same "place". In this case, doesn't the absorbed energy come from the jet pack, rather than from the black hole -- analogous to the previous paragraph?

 

[PeterDonis]

What about the Unruh effect in flat spacetime? If the observer absorbs a lot of Unruh photons, presumably they will experience a corresponding retarding force.

No, they won't, because the Unruh photons should appear to be coming from "below" the observer. At least, that's what would be expected since the Rindler horizon of the accelerating observer is "below" the observer, and the Unruh radiation is supposed to appear to be coming from the horizon. However, I don't know if any treatments of the Unruh effect have actually analyzed the directionality of the Unruh radiation (i.e., whether it appears to be coming from a particular direction, and if so, what direction).

 

[Swamp Thing]

I don't know if any treatments of the Unruh effect have actually analyzed the directionality of the Unruh radiation (i.e., whether it appears to be coming from a particular direction, and if so, what direction).

Maybe we can attempt this as follows:
Imagine two spherical radiation probes with a spring between them. When the spring is released, they begin to accelerate away from each other. They begin to absorb Unruh photons over their surfaces. Now if the Unruh radiation field is isotropic, it cannot exert any net force on either probe. This means that the acceleration is the same as it would be without the radiation, and the final kinetic energy likewise. We have thus heated up the two surfaces without sacrificing any energy from the spring. This seems like extracting usable energy from the vacuum, which doen't seem a likely possiblility.

The only way to avoid this is if the Unruh radiation is non-isotropic such that the kinetic energy of the probes would be reduced.

 

[PeterDonis]

if the Unruh radiation field is isotropic, it cannot exert any net force on either probe. This means that the acceleration is the same as it would be without the radiation

No, it doesn't, because heating up the probes increases their rest masses, which decreases their acceleration given that the force exerted by the spring is constant.

 

[Swamp Thing]

No, it doesn't, because heating up the probes increases their rest masses, which decreases their acceleration given that the force exerted by the spring is constant.

True.

Can we sum up by saying that all of the extra energy (kinetic + thermal) came from the spring, not from the vaccum, but the istotropic Unruh field has thermalized some of the orderly motion imparted by the spring?

 

[PeterDonis]

Can we sum up by saying that all of the extra energy (kinetic + thermal) came from the spring, not from the vaccum, but the istotropic Unruh field has thermalized some of the orderly motion imparted by the spring?

I think that would be a reasonable description, yes. Note that the "orderly motion imparted by the spring" in your scenario is isotropic, so there is no violation of conservation of momentum if some of the spring's energy ends up as heat instead of kinetic energy of the probes.