# A Hawking and Unruh effects -- Differences and interpretation

Tags:
1. Aug 1, 2017

### 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?

2. Aug 1, 2017

### Staff: Mentor

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

3. Aug 1, 2017

### Staff: Mentor

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).

4. Aug 1, 2017

### Joker93

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)?

5. Aug 1, 2017

### Staff: Mentor

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.

6. Aug 1, 2017

### Staff: Mentor

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.

7. Aug 1, 2017

### Joker93

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]

8. Aug 1, 2017

### Staff: Mentor

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.

9. Aug 1, 2017

### Joker93

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.

10. Aug 1, 2017

### Staff: Mentor

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).

11. Aug 1, 2017

### Joker93

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