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This question is slightly related to this one Do all massive bodies emit Hawking radiation?, which I think was poorly posed and so didn't get very useful answers. There are several questions in this post so I hope people who answer will follow the question marks and give their opinion on each one.

Let's look at flat spacetime first. Inertial observers detect no particles in empty space. Non-inertial observers do detect particles because there is a non-trivial Bogolyubov transformation that mixes positive and negative frequency modes of the matter field. And finally, if the non-inertial observers see a horizon as in the Unruh case, the spectrum of detected particles is precisely thermal.

Now look at a black hole spacetime. Free falling observers will not detect any particles but distant/stationary observers will detect a flux of radiation from the black hole outward to infinity. Q1: Why is there a flux in this case and none in the Unruh case? Shouldn't the two situations be nearly identical for the case of a very large black hole or very near to the horizon? How would you physically distinguish between the two anyway? (Maybe this is equivalent to asking why the black hole should lose mass and evaporate)

Q2: Is it natural to expect that an observer who is not in free fall but is not stationary either (and is on some weird trajectory) will detect particles but that they will not be in a thermal state?

For massive compact objects without a horizon, free fallers should see no particles but other non-inertial observers should detect particles because of the Bogolyubov transformation as in the Unruh case. This probably depends a lot on the answer to Q1 but Q3: suppose you bring such an object arbitrarily close to the point of forming a black hole with a horizon (by adding mass) -- what would a stationary observer see? Would radiation and evaporation be observed only once the horizon is formed?

Q4: In the Unruh case, the energy from the radiation is accounted for by the agency that accelerates the detector. In the black hole case, it is accounted for by the evaporation of the black hole. How is it accounted for in the case of a stationary detector outside a massive compact object without a horizon (which doesn't evaporate)?

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My simple minded observation comes from the "aha" moment when learning of Hawking radiation from black holes. The vacuum continually creates and annihilates pairs of particles, the vacuum sea, everywhere. The finality of the horizon of a black hole is what turns a simple pair creation/annihilation virtual event into radiation: one part of the pair can escape while the other falls in the well for ever. This can happen only when there is a "no return" path for one of the pair. –  anna v May 2 '11 at 4:24
    
continued: massive bodies do not supply a "no return " path, nor accelerations etc. –  anna v May 2 '11 at 4:29
    
* ... the spectrum of detected particles is precisely thermal ...*, @dbrane keep in mind that this property only emerges when we integrate over the entire worldline of our observer. During any finite interval an observer will not see a thermal distribution. [This was pointed out to me by one of Paddy's student in IUCCA. Thanks dude. I forget your name.] –  user346 May 2 '11 at 6:42
    
Exactly, Anna. It's the event horizon - a surface that can only be crossed in one direction - that splits the virtual pairs and turns the exterior member of the pair to a real particle of the Hawking radiation. It's the event horizon that allows the physical particles to go inside which is why the lowest-energy state - local vacuum - remains the ground state of the freely falling observers. –  Luboš Motl May 2 '11 at 9:12
    
@Deepak, what you write about the non-thermality of the Unruh radiation is very bizarre. The Unruh radiation looks thermal at any moment of time. Of course, particles need to be accumulated to check the thermality, but at any timescale substantially longer than $O(1/a)$ in the $\hbar=c=1$ units, one always gets enough data to show the thermality. ... There are other issues that because of the preservation of information, the Hawking radiation is not exactly thermal in principle. But it's thermal for all practically measurable purposes. –  Luboš Motl May 2 '11 at 9:14
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3 Answers 3

Q1 - The black hole spacetime sees actual particles in the "static" frame because the diminishing curvature across the black hole spacetime continuously connects the static frame at infinity - which is really static and corresponds to freely falling, static observers - with the static observers who otherwise live in a strong gravitational field near the black hole event horizon.

So the Hamiltonian corresponding to the static time in the black hole case looks like the "static frame" of the Unruh case at infinity while it looks like the "accelerating frame" of the Unruh case near the event horizon. However, the natural Hamiltonian near the event horizon is linked to the freely falling frame - because the particles can continue to fall after they cross the event horizon.

So that's the frame in which there are no particles near the event horizon. Because the static frame is accelerating relatively to it, it will see a particle flux near the event horizon. And because the static coordinates connect this place with infinity, the same particle flux must be seen at infinity, too.

In the Unruh case, you can't turn the radiation to real particles in a flat space because every frame is either inertial or accelerating and we know very well that only the inertial frame has a flat spacetime and it sees no particles. Note that Unruh's argument is only valid locally - you may show that if one observer sees nothing, a nearby accelerating observer does see particles. You need to propagate the particles through a general curved spacetime to figure out how these particles will manifest themselves at a completely different location - and that's where the nontrivial curvature of the black hole spacetime kicks in.

But, as Anna suggested, it may be more pedagogic to link the creation of Hawking particles to the split virtual pairs near the event horizon. Or to the quantum tunneling from the black hole - the uncertainty principle prevents particles from being completely safely hidden inside the black hole.

Q2 - An observer in a general state of acceleration will detect a radiation that will generalize the radiation or no radiation seen in the two frames and it will be approximately thermal, too. There is always a frame in which the radiation may be argued to vanish - and all other frames in the same place will see Unruh-like radiation corresponding to their acceleration.

Q3 - Before the horizon gets formed, the situation is not translationally symmetric in time. So the nice Hamiltonian associated with the Killing vector field - showing that the spacetime is static - doesn't exist yet. At that moment, it's not possible to say whether some particles are "real". However, quite generally, freely falling observers define the frames in which there are no particles produced by the future black hole (there can be other "real" particles participating in the collapse, of course). That's not surprising because these freely falling particles are analogous to those that cross the event horizon of the completed black hole. All observers that accelerate relatively to them - in the same place - will see an Unruh-like radiation. It's important to realize that it's hard to detect the "exact radiation" during the formative moments of the black hole. The reason is that the creation of the black hole approximately takes time $t = R/c$ or so, and because the typical energy of the Hawking quanta is just $E = \hbar / t$ or so, we're near the saturation point of the uncertainty principle, so we can't measure the energy of the Hawking quanta too accurately. And the density of the Hawking particles is of order one per this $R^3\times t$ region of spacetime, too. Also note that once it's decided that the black hole is going to form, you can't slow the process down. A much longer time than the time of the black hole birth is needed to be sure about the precise spectrum of the Hawking radiation.

Q4 - The static observer sitting on a heavy star doesn't see any radiation! It's because there is no event horizon. So the accounting is zero equals zero. There is not even an outgoing flux equal to an incoming flux - after all, one couldn't guarantee any permanent incoming flux from an empty space. The only sensible frame - relatively to which the actual state of the empty space is approximately the ground state - is the static frame. And there are no particles. The presence of the rock solid star's surface guarantees that the freely falling frame is no good to define the vacuum state.

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The "who otherwise live in a strong gravitational field near the black hole event horizon" part is wrong - the field near the event horizon can be as weak as desired. A trillion solar mass black hole has about Earth's gravity at the horizon for instance. –  Tom Andersen Jan 1 '12 at 18:12
    
"The presence of the rock solid star's surface guarantees that the freely falling frame is no good to define the vacuum state" - can you explain this? How presence of star surface affects the vacuum properties? –  Anixx Jan 3 '12 at 6:16
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I agree with your Q1: How could an observer tell the difference between sitting on the event horizon of a large black hole (billions of solar mass black holes have really tame gravitational fields at the horizon) from being accelerated?

Is an event horizon really needed - what it amounts to is a 'one sided membrane'. Particle pair creation near this membrane occurs, one gets sucked in, the other escapes. The stronger the field, the larger percentage of pairs that get separated. Will that not happen - even around the earth? The gravitational stress at the horizon of a large enough black hole is the same as gravity right here on earth.

I wonder if you do the calculation for the radiation per surface area of a supermassive black hole with a 9.81 m/s**2 at the horizon, then take that rate and apply it to the earths surface area - what would be the rate? So I would answer Q1 with yes - the earth emits this kind of radiation.

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If the Earth emits this radiation, where is the energy coming from? –  Peter Shor Jan 1 '12 at 22:09
    
Dear @Tom, as Peter sketched, objects without event horizons can't emit the Hawking radiation because that would violate the energy conservation law. The horizon is a "one-sided membrane", as you correctly wrote, but the Earth's surface isn't asymmetric in this way. Things can go up and down. The region beneath the horizon is hidden so the mass it stored may change, thus rendering the energy conservation law irrelevant. But the Earth would have to get the energy somewhere. The situations aren't the same: black holes are different from stars and planets. –  Luboš Motl Jan 14 '12 at 18:55
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Here's my educated guess about the answer to Q1: Hawking radiation appears to emanate from the event horizon. Unruh radiation seems like it should be equivalent, in which case an accelerating observer would see a mass of thermal photons "popping out" of the event horizon behind her. However, in the observer's reference frame these particles are continually being accelerated toward the event horizon (i.e. they lose momentum at a constant rate, so their frequency decreases and they change direction), so they must eventually arc around and fall back into it. Since the flux of particles falling towards the event horizon must balance the flux of particles popping out of it, the overall flux is zero. The black hole case is different because the curvature of space-time means the event horizon is a closed surface and the thermal photons can escape to infinity, so the two fluxes don't cancel.

If this is right then it suggests a partial answer to Q2, because we need to assume the acceleration is constant in order to say the two fluxes are equal. An observer on a non-constantly accelerating trajectory would therefore potentially see a varying overall flux of particles, though I don't know whether they would be thermally distributed.

The "particles popping out of the event horizon" picture also suggests an answer to Q3. An observer on the surface of (say) the Earth doesn't see an event horizon, so there's nothing for the particles to pop out of. The difference between an observer on the surface of the Earth and an observer in empty space accelerating at 1g is again the curvature of space - in this case the flat version has an event horizon whereas the curved one doesn't. As I understand it, the "non-trivial Bogolyubov transformation that mixes positive and negative frequency modes of the matter field" arises in the derivation of the Unruh effect because of the need to work in Rindler coordinates, which cover only the part of space-time that can be observed. I would guess that if there's no event horizon then this won't happen.

If that reasoning is correct then the answer to the "do all massive bodies emit Hawing radiation" question is no, and we don't need an answer to Q4 any more because there's no energy that needs to be accounted for.

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Another possible answer to Q2 is that a nonuniformly accelerating observer who starts and ends her trajectory travelling inertially will never see an event horizon because no part of space-time ever becomes causally inacessible to her. Then according to the reasoning above she won't experience any Unruh-type photons at all, though she will still see red-shifted radiation from any normal matter in locations that would be behind the event horizon if she were accelerating constantly. This makes me skeptical about whether these phenomena can really exist in a universe of finite duration. –  Nathaniel Jan 2 '12 at 10:24
    
Actually the heat bath will be observed even in an accelerating box, well shielded from the apparent horizon. That is it is incorrect to suppose that the particles come directly from the horizon. –  Anixx Jan 3 '12 at 6:24
    
@Anixx, how do you reach that conclusion? I don't think I believe it. I can see that the Bogliubov transformation must predict the same level of EM radiation inside the box as everywhere else, but then because the box is sealed that just means there's a finite amount of thermal radiation just bouncing around inside it. Once you've removed that (by detecting it, for example) it's gone. Now the un-transformed EM field is no longer in the vacuum state so there's no reason to expect it to come back again, and you're left with an empty box. –  Nathaniel Jan 3 '12 at 10:25
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