It is known that any accelerated observer is subject to a heat bath due to Unruh radiation. The principle of equivalence suggests that any stationary observer on the surface of a massive body should also experience heat bath. Indeed by substituting surface gravity g into the formula for Unruh radiation one can find the black body temperature for a hypothetical hyper-cool planet:

$$T = \frac{\hbar g}{2\pi c k}$$

which is $3.9766×10^{-20}\,\rm{K}$ for Earth

one even can find the time which it will take for Earth to evaporate: $5.69×10^{50}$ years.

Since the heat in the super-cold Earth cannot come out of nothing one should assume that it will come from decay of particles due to a certain mechanism.

Sometimes I heared an argument that an event horizon is needed for Hawking radiation to exist. But this can be countered by assumption of possibility of decay due to quantum virtual black holes (which inevitably should appear due to uncertainty principle, and the more massive and dense body is the greather concentration of virtual black holes inside it will be, eventually becoming similar to the concentration of bobbles inside a body of boiling water). Or just suggest that any massive body due to uncertainty principle can quantum tunnel into a black hole state so to emit Hawking radiation.

So what is the conclusion here?

  • Can we say that all massive bodies are surrounded by the atmosphere of heated vacuum?

This is a weaker preposition: thermal state of surrounding vacuum does not mean energy transfer if the system is in thermodynamic equilibrium.

  • Any body gradually evaporates, i.e. transfers its energy to the surrounding vacuum until completely vanishes?

This is a stronger preposition and suggests emission of radiation al loss of mass.

  • $\begingroup$ I would warn you against extrapolating black hole temperature to estimating evaporation times. the back reaction problem (quantum effects create matter, which then perturbs the spacetime, which then change the quantum effects) in semi-classical gravity is extremely non-trivial, and once the star has radiated a large portion of its mass, the back-reaction effects will not be negligible. The premise of this problem, where the quantum radiation interacts with a matter distribution, is even more complicated. $\endgroup$ – Jerry Schirmer Jan 11 '11 at 2:08
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    $\begingroup$ Actually this estimation is mostly inaccurate because the temperature will decrease as the body evaporates (due to decrease of surface acceleration), unlike a BH which evaporates at accelerating rate. Thus one should better estimate the "half-evaporation" time for Earth rather than the expected time of complete evaporation which is infinite. The figure in the question is just for illustration, it is calculated out of presumption that the rate of evaporation does not change. One can better think of it as of an approximate time of evaporation of Earth mass from Saturn. $\endgroup$ – Anixx Aug 20 '11 at 8:41

There's a simple argument that massive bodies that are not black holes cannot emit Hawking radiation. Consider a single proton in its ground state. It does not emit radiation because if it were, it would have to decay to some lower-energy state. This eventually may happen when the proton decays, but the radiation is not black-body radiation, so it can't be considered Hawking radiation.

Now, let's consider a very large crystal of diamond1 in its lowest-energy state. Similarly, it cannot emit any radiation until one of its protons decays2, in which case it will probably emit some high-energy gamma rays. So any radiation it is emitting is not black-body Hawking radiation.

1Note for pedants: The comments point out that diamond is only stable at high pressures. So feel free to replace it with a silicon crystal.

2Note: Proton decay isn't part of the Standard Model, but lots of physicists think quantum gravity is going to require proton decay exists, because in many theories of gravity, black holes don't preserve baryon and lepton number. If this is the case, Planck-scale interactions should violate baryon number conservation. (Also, many Grand Unified Theories lead to proton decay, although for some of these, the experimental limits on proton decay rule them out.)

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    $\begingroup$ This answer neglects insanely improbable p+e to two photons which is gravitationally allowed. $\endgroup$ – Ron Maimon Jan 1 '12 at 17:08
  • $\begingroup$ @Ron: you're absolutely right. That's possibly more probable than proton decay, and needless to say it also will not produce black-body Hawking radiation. $\endgroup$ – Peter Shor Jan 1 '12 at 17:13
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    $\begingroup$ This answer only considers a black-body radiation but if a proton decays due to virtual black hole formation, the resulting photons will not belong to a black body spectrum, but still technically this can be considered Hawking radiation. That is if the black hole is too small its Hawking radiation is not necessarily following the black body distribution. $\endgroup$ – Anixx Jan 1 '12 at 17:42
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    $\begingroup$ A crystal of diamond immersed in vacuum, even a “very large” one, will unlikely have such thing as “lowest-energy state” because diamond is thermodynamically unstable at zero pressure. $\endgroup$ – Incnis Mrsi Oct 25 '14 at 11:45
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    $\begingroup$ One should make clear that proton decay and @RonMaimon improbable channel only occur within speculative still GUTS type models, not the standard model., as people can misunderstand physics.stackexchange.com/questions/462181/… $\endgroup$ – anna v Feb 22 '19 at 5:51

The answer is clearly no, but it is interesting to see what goes wrong with the argument. I think the problems lies in the distinction between Unruh’s effect and Hawking’s. In Unruh’s case one has to be careful what is and what is not implied: an accelerated detector in empty flat space will behave as if immersed in a heat bath, in that it detects particles distributed thermally. However, this does not imply there is radiation, in the sense of energy flow from one place to another, empty space is empty, even in Rindler coordinates. It is only for a real black hole that you have real radiation emanating from the horizon.

(I vaguely remember a decent discussion of this in Birrell and Davies).

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    $\begingroup$ Also presuming that black hole radiates and a body with identical mass but not a black hole does not, suggests that there are two different kinds or vacuum around them, an absurd idea. $\endgroup$ – Anixx Jan 11 '11 at 7:01
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    $\begingroup$ 1. Hawking formula has to do with the horizon, e.g the temperature has to do with the surface gravity, etc. It does not apply to objects with no horizons. $\endgroup$ – user566 Jan 11 '11 at 7:39
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    $\begingroup$ 2. Body as dense as a black hole is a black hole. Body with identical mass but much larger, is a different object which does not radiate. $\endgroup$ – user566 Jan 11 '11 at 7:40
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    $\begingroup$ 3. Virtual black holes are an hypothetical process, which if exists has some interesting consequence, but I don’t think that radiation from any massive object whatsoever is one of them. Placing the vacuum in a gravitational gradient is not sufficient to create Hawking radiation. $\endgroup$ – user566 Jan 11 '11 at 7:46
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    $\begingroup$ To repeat the answer to the “two kinds of vacuum” comment - the details of a body consists of more than its mass, there are many many possible mass distributions with identical total mass, and the vacuum around them, as well as many other details, depend on the mass distribution. There is nothing absurd by saying that Hawking radiation depends on more than that one number, the total mass. It manifestly does, as the radiation is zero without an horizon. $\endgroup$ – user566 Jan 11 '11 at 15:50

The answer is no, because there is no horizon. A time-independent gravitational field emits no particles, because energy is conserved, just like a stationary charge distribution does not radiate. This is consistent with the equivalence principle, as explained below.

In the Unruh effect, the particle emmission is from the "black wall" acceleration horizon behind the observer. If you place a refrigerated baffle between the black wall and the observer, the radiation will be absorbed by the baffle, and will not be seen by the observer. An observer standing on a planet is indistinguishable from an accelerating observer with a planet between him and the acceleration horizon, and such an observer sees no emissions from the horizon, because the horizon is invisible.

You might object that the horizon is infinite, and the planet is finite, so shouldn't you be able to see the horizon far enough out? But if you go far enough out, you notice the curvature caused by the planet, and this will push the horizon out to infinity, because there is no horizon after all, you're just standing on a planet.

  • $\begingroup$ It seems you did not read the question. Your stand that "no horizon=no radiation" has been already addressed by a conjecture that the horizon can emerge virtually. $\endgroup$ – Anixx Aug 15 '11 at 12:02
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    $\begingroup$ I didn't say no horizon means no radiation--- I said if there is stuff between you and the horizon, you see the stuff, not the emissions from the far-away virtual horizon. The virtual black hole business is silly--- a virtual black holes can pop in and pop out, but it conserves energy and doesn't lead to any detectible particle production unless there is a horizon visible. Every static solution has a conserved energy. $\endgroup$ – Ron Maimon Aug 18 '11 at 6:12
  • $\begingroup$ "a virtual black holes can pop in and pop out, but it conserves energy" - why it should conserve energy? why matter cannot decay through VBH fourmation? $\endgroup$ – Anixx Aug 19 '11 at 7:29
  • $\begingroup$ @Anixx: you conflate two different things: conservation of energy (a consequence of the time translation invariance), and conservations of so named “numbers” (except for charges, see “gauge theories”, are not thought to have a deep substantiation). $\endgroup$ – Incnis Mrsi Oct 25 '14 at 12:37
  • $\begingroup$ This thread is long dead, but I think it's necessary to remind everyone that Hawking radiation is not produced at the horizon. In fact, it is produced over an extended area many times larger than the horizon. It is a non-local, quantum mechanical effect. $\endgroup$ – Al Nejati Oct 23 '18 at 22:32

My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the equivalence principle, expect such a charge to radiate. I believe the answer is that it does not, but that the analysis is subtle. You could start with the article D. Boulware, "Radiation from a uniformly accelerated charge", Annals of Physics 124 (1980), 169-187 and work your way forward from there by looking at citations of that paper.

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    $\begingroup$ OK Then imagine a star which consists only of electrons. It becomes a black hole, it evaporates... $\endgroup$ – Anixx Jan 11 '11 at 0:56
  • $\begingroup$ Actually whether protons decay is an open question (en.wikipedia.org/wiki/…) $\endgroup$ – Sklivvz Jan 11 '11 at 1:05
  • $\begingroup$ @ Jeff Harvey You did not understand my question. My question was not whether electron is stable (it probably is), but whether an observer on the surface of a massive planet will observe Hawking radiation. $\endgroup$ – Anixx Jan 11 '11 at 1:20
  • $\begingroup$ I rewrote the question. But if a planet consists only of cold hydrogen at near-zero temperature, one cannot suggest other source of energy other than proton decay. $\endgroup$ – Anixx Jan 11 '11 at 1:44
  • $\begingroup$ @Anixx: gravitational contraction of the planet would be another possible source of energy. $\endgroup$ – Jerry Schirmer Jan 11 '11 at 2:05

I would think that you can have some evaporation for massive objects without horizons. The usual way to think of black hole evaporation (see wikipedia) is that you get spontaneous particle pair creation near the horizon. Then particle '1' get sucked in and particle '2' escapes the black hole. Energy in total is conserved, so the black hole loses mass.

Now think of this planet. We have no horizon, but we do have a gravitational field. So when a spontaneous pair gets created (say outside the atmosphere), there is a (hyper small) chance that one will get sucked into the planet, while the other flies free, lowering the mass of the earth. I think that this is what you are talking about with your 'virtual black holes'.

So what would be the rate? Perhaps less than one particle per lifetime of the universe. The type of particle involved may also need to have special properties, e.g.: be extremely heavy/light and be charge free.

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    $\begingroup$ This nonsense evidently stems from a hyper-simplistic Hawking radiation “explained” for children, at Wikipedia or not. $\endgroup$ – Incnis Mrsi Oct 25 '14 at 12:02
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    $\begingroup$ Please visit Incnis Mrsi's profile page. $\endgroup$ – Tom Andersen Jan 2 '15 at 15:16
  • $\begingroup$ I don't think your answer is bad at all. I'm unfortunately not smart enough to say it's "Correct", but what you describe is how I picture Hawking radiation working in my tiny brain. On earth, it would almost never happen. On a Neutron star it might happen - occasionally-maybe. But, that's my assumption on how it works, I'm not certain. (voted you up cause I don't think you deserved a vote down). $\endgroup$ – userLTK Aug 28 '15 at 0:00
  • $\begingroup$ Isn't true what you are saying. Your explanation of Hawking radiation is very far from the real picture. $\endgroup$ – Nogueira Jan 6 '16 at 7:45

The Unruh effect can be calculated by a lot of ways. But all this calculations make the same assumption (should be), but in different ways: some of this assumptions can be very unclear by some calculations. One of those assumptions that are hidden by the usual calculations are this: the vacuum state is a non-local state in the sense that any local observable fluctuates after a measurement. In other words, the projector $|0\rangle\langle 0|$ do not commute with $\phi(x)$.

Note that the derivation of the Unruh effect are based on QFT on flat Minkowski space-time. Accelerated observers in a Minkowski space-time are described by the Rindler coordinates. And you can see that in these coordinates we have a horizon. Then this observer can't see an entirely spacelike hypersurface. In other words, he can't access the entire Hilbert space of the system. This is why he describe the vacuum state of the inertial frame as mixed state described by a proper density matrix $\rho$.

To get a picture of "what is going on":enter image description here

The Unruh effect is due the fact that the vacuum state correlates all the space-time $\langle\phi(x)\phi(y)\rangle\neq 0$, and this interval $(x,\,y)$ can cross the horizons. Integrating out the field at points that do not belongs to the Rindler region create an thermal radiation on this region.

Now let answer your question: No. Thermal radiation only show up when the horizon show up too. Explain this is simple: Unruh effect happens because the horizon. The traditional calculation with Bogoliubov transformation don't make this clear but is there, hidden in the middle of the calculations. So the Unruh effect is non-local because depend on the assumption about the existence of an horizon.

In a star or in a black hole the Riddler coordinates are good only locally, is a polynomial expansion over the radius. We can do this expansion through all the shells around the object. When we make this expansion more and more close to the object the acceleration increase and the horizon get closer. After you pass through to the interior of the star the acceleration start to fall down make the horizon far and far until you reach the center of the star. In the black hole we don't have any interior to save us from the horizon because the horizon came first of any possible interior.

For more about the Unruh effect see this article.

A. Higuchi, G. E. A. Matsas, and D. Sudarsky (1992) Bremsstrahlung and zero-energy Rindler photons

  • $\begingroup$ It seems you did not read the question and comments to other answers. Your stand that "no horizon=no radiation" has been already addressed by a conjecture that the horizon can emerge virtually. $\endgroup$ – Anixx Jan 6 '16 at 10:49

As the question is currently phrased, the answer is a straightforward one. Hawking radiation does not occur with any object except one with a horizon, i.e. a black hole. The arguments leading to Hawking radiation are subtle, but, conceptually, the idea (via Susskind in The Black Hole War) is that quantum jitters become thermal jitters at the event horizon. (Susskind's explanation is much better, of course.)

To be a bit more complete and concrete, consider a standard quantum fluctuation of energy arising from the uncertainty relation between energy and time. Fluctuations on sufficiently small time scales will be large enough to result in particle-antiparticle pair creation. At the horizon of a black hole, the pair can be separated so that one particle falls into the black hole, while the other is shot out. This transforms a virtual particle (pair) into a real particle, which is then able to interact with other matter.

Now, one could say that this process is possible in any system with non-zero mass. However, the detailed and subtle mathematics of the process require an event horizon in order to occur. As a thought experiment, imagine that any mass could produce this effect. It would result in a wide-scale violation of the Uncertainty Principle because we would see changes in energy that exceeded those allowable over the associated time frame.

One other objection is that black hole temperature, which is tied to Hawking radiation, increases as the mass of the black hole decreases. If "normal" mass were to behave in a similar way, elementary particles would have tremendous "black-hole-like" temperatures.

  • $\begingroup$ "It would result in a wide-scale violation of the Uncertainty Principle because we would see changes in energy that exceeded those allowable over the associated time frame." - can you please elaborate on this? It is known that particles decay due to energy fluctuations indeed occurs (with nuclear forces, for example), even if total energy of the potential barrier. It is not evident why this process cannot happen with a massive object. $\endgroup$ – Anixx Jan 12 '11 at 2:30
  • $\begingroup$ Regarding your second comment that this would lead to any particle to have very large temperature, you are wrong. Yes, the temperature of a black hole rises as its radius decreases, but only because the surface gravitational acceleration also increases. The Hawking temperature is proportional to surface gravity. As any particle has very tiny mass, its surface gravity is also very small and as such it cannot emit much Hawking radiation. That said, the radius does not matter, what matters is the surface gravity. $\endgroup$ – Anixx Jan 12 '11 at 2:34
  • $\begingroup$ @Anixx My reference to violation of the Uncertainty Principle is not with regard to particle decay, but particle-antiparticle pair creation. If pairs did not annihilate and return their borrowed energy to the vacuum, we would see quantum fluctuations in energy that are not allowable. $\endgroup$ – Mitchell Jan 12 '11 at 2:45
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    $\begingroup$ @Anixx: how much quantitative general relativity or quantum field theory do you know? $\endgroup$ – Jerry Schirmer Jan 12 '11 at 3:30
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    $\begingroup$ The qualitative answer depends on quantitative physics. $\endgroup$ – Jerry Schirmer Jan 12 '11 at 18:10

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