A black body is classically defined as a perfect absorber of radiation. That seems to fit a black hole pretty well. The only remaining question, then, is does there exist a physical process to get radiation from the inside to the outside? In fact there is and there are a few ways to think about it.

a. What would normally be vacuum fluctuations of virtual particle/antiparticle pairs happen near the event horizon and are made real by one crossing over and the other not. This requires some energy to provide mass to the emitted particle and it comes at the expense of the mass of the black hole since the one that falls in has negative total energy. I always found this explanation difficult to follow.

b. We know from the uncertainty principle that particles with a long enough wavelength cannot be thought of as confined within the black hole. From this point of view it is just a quantum tunneling process and if you take a wavelength about equal to the Schwarzschild radius, you can derive the effective temperature of the emitted particles which is consistent with the thermal picture.

c. Go deep into the weeds of quantum field theory. This is the most accurate way to understand but also the hardest to absorb. You have to think of states of a quantum field as a superposition of arbitrarily long wavelength field modes. When a horizon exists, either due to a gravitational field or due to two observers being in relative states of acceleration, the modes that one observer defines will be different from the other observer. This means that what one observer perceives as a vacuum will be perceived as a gas of particles by the other observer. In other words, the definition of particle states is observer dependent.

The first two explanations should be thought of as figurative only since the whole lesson of quantum field theory is that there are no particles, there are only fields.

The Hawking radiation, which for distant observer takes the form of black-body radiation, is the consequence of quantum effects taking place in vicinity of black hole horizon (just outside the bh horizon). The strong curvature influences the particle-antiparticle creation in the vacuum close to the black hole horizon, effectively preempting the pair to anihilate. Roughly said, one particle is "lost" under horizon while the other travels to distant observer. Overall average of this process gives you Hawking radiation. For details see eg. [Wald, Robert M.,

Black holes and relativistic stars / edited by Robert M. Wald. Chicago : University of Chicago Press, 1998 ].

The Black Holes certainly don't emit black body radiation. Its the accretion disk around accreting black holes which can emit black body like radiation.

That's what the party line says. I don't think any distant observer has seen radiation coming from a black hole in the form of black body radiation. But what he has seen is a plethora of analogies and look-a-likes. In the expression for the hyperbolic velocity, which is the logarithm of the longitudinal Doppler shift, the hyperbolic velocity is replaced by a product of acceleration times proper time. The Doppler shift is then given by an exponential shift whose absolute square of the Fourier transform is said to represent the spectral density. What appears to be the Planck term becomes just that when the acceleration is set equal to the absolute temperature. But, what they forget to do is to check if the integral over all frequencies gives back the Stefan-Boltzmann law. The reason probably is that the integral diverges! Moreover, accelerations do not cause Doppler shifts: this is clearly seen by introducing the geometric mean of outward and return times of a light signal for the proper time in the expression for the hyperbolic velocity, while in the argument of the logarithm is the ratio of the return and outward times of a signal which gets reflected somewhere in between. This shows that the hyperbolic velocity cannot be set equal to a constant acceleration times the proper time.

Quantum Hawking radiation emitted by a non-rotating black hole is, in fact, grey-body, though thermal. Moreover, by studying its spectral deviation from the black-body radiation, it is possible, in principle, to prove that this radiation is emitted by a black hole with some definite mass, not by other body in thermal equilibrium. Furthermore, from the angular distribution of this radiation, the black hole location in space can be determined. So, this radiation brings us the maximum possible information which we could expect from a process occured and calculated in the Schwarzschild space-time metric.

Bernard,

You are right to be somewhat skeptical of the basis of the Hawking effect. However, arguing on very general grounds about black bodies vs. black holes is not going to get you anywhere. The devil here is in the details.

In fact, the result depends on assumptions which were not stated explicitly in Hawking's original papers and are often obscured or forgotten in more recent discussions. The crucial assumption is that the quantum state is chosen so that there is a finite and very small stress-tensor on the black hole horizon. The state that is often assumed or considered "preferred" is called the Unruh state based on:

"Notes on black hole evaporation"

W.G. Unruh, Phys.Rev. D14 (1976) 870

(This paper has 1673 citations).

If the quantum field is prepared in the Unruh state, then one can show --in free field theory--that a Hawking thermal flux results at late times--also ignoring any backreaction effect of the radiation which is presumed small on the black hole itself. This seems to many a "reasonable" state, because it agrees with the classical expectation that nothing particular happens at the horizon.

However, from the beginning it has been clear that there are other initial states in which the quantum field can be prepared for which this assumption (and it si mpure assumption) is violated. See e.g.

D. Boulware, Phys. Rev. D 11, 1404 (1975). (referred to in Unruh's paper) and

Phys. Rev. D 13, 2169 (1976); and

"Trace Anomalies and the Hawking Effect"

S.M. Christensen, S.A. Fulling, Phys.Rev. D15 (1977) 2088-2104.

These authors showed that there are states that produce large vacuum stresses on the future horizon and therefore violate the assumption needed to derive the Hawking effect. Rather than being pathological, these states can be argued to be more typical:

"Macroscopic Effects of the Quantum Trace Anomaly"

E. Mottola and R. Vaulin, Phys.Rev. D74 (2006) 064004 and a review in

"New Horizons in Gravity: The Trace Anomaly, Dark Energy and Condensate Stars"

E. Mottola, Acta Phys.Polon. B41 (2010) 2031-2162

Indeed, the Hawking Radiation emitted by Black Holes takes the form of a blackbody curve, i.e. the black hole also is a blackbody. The effect is explained nicely and graphically in Steven Hawkings Book "The Universe in a Nutshell". In simple terms, empty space is filled with quantum fluctuations, which can be imagined as so calle "virtual particle pairs": A pair of particle/antiparticle borrow some energy from the vacuum energy and start to exist together for a short period of time, before they annihilate again. This is the also the reason for Heisenberg's Uncertainty Principle.

In the vicinity of the black hole it can happen, that one of the partners of the virtual particle pair is swallowed by the black hole, while the other partner can leave the environment of the black hole. The net effect is the emission of one real particle from the black hole, which has to pay back half of the borrowed energy to the vacuum and thus has to shrink a little. This is the Hawking radiation.

Now there are three possibilities for such a virtual particle pair: (a) both of them annihilate before the black hole can affect them. (b) both of them are swallowed at once, and (c) only one is swallowed and the other one is radiated away as Hawking radiation. If you think about it, case c is most likely, if the wavelength of the virtual particle is comparable to the size of the black hole event horizon. If the particle is much smaller it is more likely that both are swallowed; if the wavelength is much bigger than the black hole, it is more likely that none of them is swallowed.

This means that the characteristic wavelength of the emitted particle (Hawking radiation) corresponds to the Schwarzschild Radius of the Black Hole and is close to the maximum of the Blackbody Radiation Curve of the Hawking Radiation.

Gunther,

There is confusion between the condition of adiabatic expansion of an ultrarelativistic gas RT=constant, and Wien's law, lambda_mT=constant, which determines the maximum frequency, or minimum wavlength, of the spectral frequency distribution of black-body radiation. Using the Schwarzschild radius would give a temperature of 10^{24} K, well above particle thresholds. If you just think about what an "emitted" particle has to do with equilibrium black-body radiation you will see that there are no conditions that would give rise to such a spectrum, e.g., the cavity walls heated to a constant temperature with a black-body absorbing and emitting radiation at all frequencies.

The vacuum has a thermal equation of state energy density= - pressure=constant. The latter is imposed by adiabaticity. How can energy be borrowed from a pure constant? Moreover, such an equation of state would render the vacuum unstable, which would tend to break up under the influence of negative pressure.

Alexei,

I have never heard of the angular distribution of black-body radiation. By its equilibrium nature it has to be uniform.

Emil,

Maybe the devil is in the details. How do you get from the quantum state on the black-hole horizon to black-body radiation? Black holes cannot evaporate using black-body radiation. The latter is an equilibrium process between the rates of absorption and emission of radiation. This is guaranteed by the Carnot-Clapeyron equation as Lord Rayleigh knew well back in 1902. I also wouldn't be impressed with citation numbers; science is not a democracy where the correct theory gets the majority of votes.

Regarding the relation between acceleration and absolute, or the lack thereof, please see: https://www.researchgate.net/post/How_can_absolute_temperature_ever_be_proportional_to_acceleration?ev=tp_feed_post_xview

A free falling observer will not experience acceleration? g=(G rho)^{1/2}v, where rho is the density and v, the velocity. The free fall time is just the inverse of the factor in front of it.

If there no density how can there be mass? This violates the equivalence between heat content and mass. The Schwarzschild solution requires the Ricci tensor to vanish, but then we learn there is matter present when the integration constant is set equal to 2m. That means the energy-momentum tensor is not zero. Then why aren't Einstein's equations satisfied with a non-vanishing Ricci tensor? Free falling with respect to what? Is there or isn't there gravitational acceleration?

Bernard,

Science is not a democracy, but it is a human activity. The point about the citation numbers is that this is where most people's view of the matter comes from, so it behooves you to read and understand this paper if you want to have a basis of discussion with others in the field. You may criticize or disagree, of course. But in order for this criticism to be worthwhile, or to the point at all, you need to at least recognize what has been done and what is the other point of view.

The answer to your question about how you get from quantum state to black body radiation is that regularity of the state on the horizon implies that its Green's function G(x,x') is periodic in imaginary time. This periodicity is a property of the Schwarzschild metric analytically continued to imaginary time. According to standard finite temperature field theory the imaginary time periodicity of G(x,x') is the inverse temperature beta= hbar/kT of the state.

Emil,

It is this type of "analytic continuation" has led to some of the most absurd results in physics. Absolute temperature cannot to be identified with acceleration, absolute temperature with surface gravity, etc. Temperature cannot increase as the mass decreases (it defies the second law), and if a body is not accelerating it has zero temperature? (temperature is the average kinetic energy of molecular motion) Black body radiation needs a thermal bath, and the horizon is not a thermal bath. No equilibrium can be established between the rate of absorption and emission of radiation on the horizon, or anywhere else in a black hole.

If we are talking about the same vacuum, it has an equation of state vacuum energy energy density=-pressure. If it is another vacuum, I have never heard of it. Mass has heat equivalent, as Planck showed in 1906: energy+stresses. I don't understand how energy can be attributed to curved space. Does positive curvature have one form of energy and negative curvature another? Then how can geometry be separated from energy-stress-mass, which you say does not contain matter? To my knowledge no one has discovered gravity waves.

A black body is classically defined as a perfect absorber of radiation. That seems to fit a black hole pretty well. The only remaining question, then, is does there exist a physical process to get radiation from the inside to the outside? In fact there is and there are a few ways to think about it.

a. What would normally be vacuum fluctuations of virtual particle/antiparticle pairs happen near the event horizon and are made real by one crossing over and the other not. This requires some energy to provide mass to the emitted particle and it comes at the expense of the mass of the black hole since the one that falls in has negative total energy. I always found this explanation difficult to follow.

b. We know from the uncertainty principle that particles with a long enough wavelength cannot be thought of as confined within the black hole. From this point of view it is just a quantum tunneling process and if you take a wavelength about equal to the Schwarzschild radius, you can derive the effective temperature of the emitted particles which is consistent with the thermal picture.

c. Go deep into the weeds of quantum field theory. This is the most accurate way to understand but also the hardest to absorb. You have to think of states of a quantum field as a superposition of arbitrarily long wavelength field modes. When a horizon exists, either due to a gravitational field or due to two observers being in relative states of acceleration, the modes that one observer defines will be different from the other observer. This means that what one observer perceives as a vacuum will be perceived as a gas of particles by the other observer. In other words, the definition of particle states is observer dependent.

The first two explanations should be thought of as figurative only since the whole lesson of quantum field theory is that there are no particles, there are only fields.

Bernard:

1. The Schwarzschild solution is an exterior solution which is why the energy tensor is zero. The "mass" in that solution is that which is measured by a distant observer.

2. Energy can be attributed to curved space but it is a tricky concept in general relativity. Abraham Pais' bio of Einstein has a careful development of the concept.

3. Temperature can be attributed to acceleration and to surface gravity but you have to be careful. First, the temperature attributed to acceleration is not the temperature of the accelerated observer but rather the temperature of the thermal bath in which he finds himself as a consequence of his acceleration. Also, the second law applies only to thermally closed systems and a radiating black hole is not closed unless you include the radiation as well. If you do that, the second law is satisfied perfectly well.

Bernard,

I actually agree with some of your criticisms. Analytic continuation is mathematics. It has to be physically justified. Please have a look at some of the papers I referenced in earlier ripostes. You may be interested also in:

P. O. Mazur and E. Mottola, e-print arXiv: gr-qc/0109035.

P. O. Mazur and E. Mottola, {\it Proc. Nat. Acad. Sci.} {\bf 101}, 9545

(2004).

@Werner, The space metrics used in GR, e.g., RW metrics, all have constant curvature, so it is all attractive or all repulsive. The difference between charges and mass is that two like charges repel and the energy needed to separate them comes from the field, so that the field has positive energy. [You don't have to worry about unlike charges since there is there is overall electrical neutrality.] But when you bring two masses together they attract and the energy needed to keep them apart comes from a negative field energy. It is this conclusion that made Maxwell drop any hope of deriving a theory of gravitation.

There is also no evidence that gravity propagates at the speed of light. No one has ever seen gravitational aberration. Moreover, there are big differences between nonlinear waves and linearized waves so that there is no reason why linear waves should be representative of nonlinear phenomena, e.g. solitons.

Your vacuum is worse than an aether, which has well-defined characteristics. No equation of state, no properties.

@Paul,

BB radiation depends on a dynamic equilibrium between rates of absorption and emission---not only absorption. In all the scenarios you give you refer to quantum considerations and it is impossible that these could balance the rate of absorption. I don't understand who is accelerating, the thermometer or the radiation. The second law does not apply only to isolated systems! It was Planck who used the second law to derive is expression for the spectral density of radiation which is clearly not an isolated system.

@Emil,

There is also a difference between mathematical analytic continuation and what physicists do. In the mathematical definition, the original variable is complex and the lack of analyticity is due the presence of a singular point. Analytic continuation refers to a function being continued analytically to points outside the original circle of convergence. There is no reason why a real variable like time should become complex, and even more be imaginary. This is a well-known trick of converting the Schrodinger equation into a diffusion equation, or a hyperbolic wave with its d'Alembertian into a Laplace-Beltrami operator, or a Fresnel integral into Wiener-Kac integral. Where is the justification for the physical interpretation?

Bernard,

The correspondence between the unitary Schrodinger time evolution operator and the statistical factor exp(-beta H) with beta imaginary time is a mathematical identity, so this requires no further justification. The assumption that is being made in Hawking's calculation is that this "natural" periodicity of the metric is also to be extended to all the Green's functions and quantum states defined on the Schwarzschild background. I quite agree that this is an assumption, which requires physical justification, and indeed may not be true of actual collapsed stars.

@Werner,

The RW metric does not coincide with any of the known metrics of Non-Euclidean geometry, and is therefore wrong. Even Robertson doesn't mention that metric in his book but gives the correct metrics of non-Euclidean geometry. The Schwarzschild metric for the inner solution is a metric of constant negative curvature but the outside solution is not. How can a metric change on going from the inner to the outer solution? On the outside the mass is constant, while on the inside its the density that is constant. The de Sitter solution is a cosmological constant different than zero. That is responsible for the negative pressure, which should have caused contraction not expansion. If gravity propagates at the speed of light then why aren't the potentials evaluated as retarded ones?

The correct metrics of constant curvature are given in Robertson-Noonan eqns (14.12), Relativity and Cosmology. Hyperbolic distance is arctanh, not arcsinh as in Landau and Lifshitz (111.12), Classical Theory of Fields. The latter is the circumference of a hyperbolic circle of radius r, i.e. 2 pi R sinh r/R, where R is the absolute constant of the geometry (31.5) Busemann and Kelly, Projective Geometry and Projective Metrics. It was even known to Einstein that the ratio of the circumference of a rotating disc to the radius was greater than 2 pi. In the RW metric at constant r, you would get the ratio equal to 2 pi for k=-1 (constant negative curvature). So RW is wrong and is not to be found in the Robertson-Noonan monograph except in the forward by Fowler.

The Schwarzschild solution is inconsistent. The outer solution has non-constant curvature while the inner solution has constant curvature. There is no matching condition on the boundary. It goes from M/r->rho r^2, i.e., from constant mass M to constant density rho. In Moller, eqn (79) of Ch. XI, the \Lambda term is negative, cf. (98) and the spatial geometry of the inner solution is given by (104) with 2m missing. This I interpret as going from one of constant mass to one of constant density. Moreover, the inner solution (104) does not agree with the Riemann metric (14-10) in Roberston-Noonan. The historical evolution is Riemann->Beltrami->Poincare. These are known metrics of constant curvature, and I showed that it is the Beltrami metric that characterizes a uniformly rotating disc. One of the major results is that you cannot have local observers at the (inertial) center and on the periphery communicating their time differences and thereby knowing their positions on the disc. This is what Einstein contended who did not realize that as you move about on the disc your rulers and clocks will shrink or enlarge with you so that you have no way of knowing where you are in reference to someone else. It will also take you an infinite time to reach the boundary of the disc. Any theory of relativity must reduce to these metrics in the constant curvature limit so that there was no reason for looking for new metrics that would be solutions to Einstein's equation in the constant curvature limit.

I was asked this question. As the discussion is already very engaged and that many interesting points of view are covered, I will answer as if the question had just been freshly asked.

Like Bernard, I attach the highest importance to the second principle of thermodynamics which establishes the irreversibility of physical phenomena, especially during heat exchange. I believe that this principle is as important as the first principle and that it is not a simple exception of the latter. It also leads to a cosmological arrow of an irreversible time.

That said, the possibility that black holes can radiate, while respecting the second principle is certainly not excluded. A cup of coffee could tell us why: As the hot water for a cup of coffee can evaporate when the surrounding air is cooler (heat can only go from hot to cold), black holes may evaporate only if their temperature is higher than the background radiation that bathes the entire universe.

Hawking suggested that the early universe was so dense that its space had collapsed into countless mini black holes with the Planck mass, which evaporated and were reborn at every 10-43 seconds. Their temperature was warmer than that of the ambient primeval universe. So they irradiated, emitted radiance and particles, going from hot to cold, following the thermodynamic arrow. Their temperature and their mass decreased. It respects the 2nd law of thermodynamics.

This condition is not fulfilled in the present universe which has a temperature of 3 degrees Kelvin after an evolution of about 15 billion years. Knowing that the temperature of the universe decreases with the expansion, it was deduced that a black hole of the mass of the Sun will begin its evaporation at t = 1020 years, when the temperature of the universe will be dropped to a ten-millionth of a degree Kelvin. A galactic black hole will have to wait until t = 1034 years and a hypergalactic black hole up to t = 1039 years before they begin to evaporate. The mass and the temperature of the black hole will decrease by irradiating when its temperature becomes warmer than the one of the ambient universe. This follows the 2nd law of thermodynamics.

Even if Einstein did not think about it (because he excluded the ether, which was the vacuum energy of his time), the flat space-time, without gravitation, of the special Relativity can be associated with a black body. If we take into account the quantum fluctuations in the space, we deduct that the acceleration (equivalent to a uniform gravitational field.) polarizes the space. If the observer transports with him a particle detector, he will measure a "quantum noise" in the form of a radiation of black body, from which the temperature is proportional to its acceleration. So when the acceleration increases, the «quantum mass» increases as well as the temperature.

Again, the second principle of the thermodynamics is respected. In all these cases, the temperature of a black hole is proportional to its ‘quantum mass’, which makes a link between black holes, thermodynamics and the second principle. Many would be entitled to think that this is Hawkinian lucubrations. After all, Einstein did not believe the existence of black holes and there is no experimental data on them.

The cherry on the black hole seems to have been when Emil, Paul, Werner…have suggested that the mass could decrease with the increment of temperature. Bernard claims that this is total absurdity. So much that they are likely to be right, in spite of common sense. Why? Because they use a conformal mass M and this mass evolves to the inverse of the quantum mass m. For instance, when the primordial mini black hole emits particles and radiation, its mass (quantum) and its temperature are decreasing. The temperature of the ambient universe, which forms a thermal bath, continues to decline with the expansion but the total mass (classic) of the universe increases.

Hawking expresses this puzzling link between gravitation and thermodynamics by postulating that the temperature of a black hole is inversely proportional to its classic mass. If we add some mass or energy to the black hole, it becomes heavier and therefore it cools. If the vacuum around a mini black hole, or a big black hole, is polarized by thermal or gravitational energy (Hawking’s mechanism), the temperature will become warmer and the mass will evaporate.

Paradox! Contradiction! We deal with two kinds of opposite masses, which reflect the tragi-comedy of physics since the early twentieth century: two major opposite theories, each of which has known great successes, and which remain more than ever irreconcilable

Russell, I don't know what quantum mass is---nor does thermodynamics. I also must admit I don't know what conformal mass is either.

The second law is universal, but has to be used with caution. First and foremost, you must define your thermodynamic system. No one seems to bother about this particular. To say that hot coffee thrown into a black hole increases its entropy is entirely meaningless since you don't define what thermodynamic system you are talking about. Whoever is throwing the coffee in is part of the thermodynamic system or not?

Then comes radiation, what radiation are we talking about? If it is thermal radiation the system must be isothermal. The walls of the cavity must be maintained at a constant temperature. Having said this, you can't make black holes evaporate through thermal radiation. If the walls of the cavity absorb more than they emit their temperature will rise and this contradicts the isothermal assumption. It is no longer thermal radiation so you can't equate evaporation with thermal radiation.

Since, as you mention, there is no experimental data on black holes, you (Hawking, or anyone else) can hypothesize what you want and won't be proven wrong. However, thermodynamics follows common sense, but the fabricated thermodynamics of black holes does not.

Hahaha! Bernard can you please explain clearly with proof why the black hole thermodynamics is wrong doesn't follow the common sense. I'm working on that in my project. Please suppose that I'm not a specialist of thermodynamics. Show please the incompatibility.

Crispin, Please go to arXiv:1110.5322. Note that there is a typo in Eq. 5. The Planck energy, E_{pl}, should be squared in the denominator. If you have any problems I'd be happy to explain.

Ok Bernard.

I think the other Paul is on the right track so I will just add my two cents.

You can think of a black hole, in particular the black hole's horizon, as a quantum amplifier that takes quantum vacuum fluctuations at the input and gives real particles at the output. The insight behind this reasoning comes from the parametric amplifier from quantum optics. The parametric amplifier uses a laser incident on a nonlinear crystal to amplify the fluctuations in the crystal into entangled photon pairs. The black holes does the same thing, and one can show that the mathematics underlying Hawking radiation is equivalent to a parametric amplifier (in this case non-degenerate). From this point you can follow the usual line or reasoning about one particle being trapped and the other one free. Or if you want you can think of it in the QFT way by relating positive and negative frequency field modes at the input and output (equivalent to different observers if you like to think that way), and calculate the resulting Bogolyubov transformation between the operators.

The next question one could ask is why is the radiation thermal? The answer for this comes from the way Hawking calculated this emission process. In Hawking's calculation, the mass of the black hole is assumed to be fixed and is left unquantized. He did this for good reasons; If the black hole is much larger than the Planck mass, the energy lost due to the Hawking effect is very small, and since we don't yet know how to do quantum gravity, it is best to leave the black hole as a classical entity. However in doing so, the black hole has effectively become a classical fixed-amplitude source of unlimited energy. In this case the only part of the system that is quantized is the Hawking radiation inside and outside of the black hole. This is a bipartite system that can readily be shown to always give a thermal spectrum. Again, the exact same physics occurs in the parametric amplifier. Even if you let the mass of the black hole change in time. As long as it remains unquantized, quantum optics models show the radiation is always thermal. In contrast, if you look a quantum optics models where the energy source (equivalent to black hole mass) is quantized, then the radiation starts out thermal but eventually deviates from thermal as the energy is reduced to near the Planck scale. This is probably what most people think happens.

I would also like to point out that since Hawking radiation is generated even when the space time is fixed indicates that this emission process actually has nothing to gravity itself! The only reason we need gravity here is to generate the horizon. The horizon is the thing actually responsible for this effect. However, many other systems, that have nothing to do with gravity, have also been shown to have horizons. This line of thinking is the framework behind analogue models of Hawking radiation.

Finally, one can also ask why is the temperature of the thermal radiation inversely proportional to the black hole's mass? Indeed, there is no blackbody that behaves this way. The best answer that I can give is that this comes from conformal symmetry near the black hole horizon. In this region, the space time is effectively reduced to (1+1)-dimensions. For a Schwarzschild black hole this means radius and time only. It has been shown using conformal field theory that one gets the inverse mass dependence directly from symmetry arguments. Several papers by Wilczek, Iso, and Umetsu discuss this. Note also that the conformal field theory is really only valid for large black holes. Exactly where the radiation is thermal.

The most convincing "mathematical derivation" I have seen to derive the Planck factor and which gives Hawking temperature shows that it is entirely wrong. It begins with Born's calculation of hyperbolic motion [Moller, Theory of Relativity, p. 75] and identifies the hyperbolic velocity as the product of the uniform acceleration and the proper time. Then using the definition of the hyperbolic velocity as the logarithm of the longitudinal Doppler effect, it gives an exponential Doppler shift that is supposedly caused by the acceleration. But, it fails to give the correct expression for time dilation (which is a second order Doppler effect) so if that doesn't exist then the Doppler shift caused by acceleration doesn't exist either. But, this is only a minor point.

Then in order to get the frequency spectrum, one [Padmanabhan, "Thermodynamical aspects of gravity", arXiv:0911.5004v2] take the Fourier transform of the exponential Doppler shift. What you want is the square of its absolute value, and if you do the calculation you come out with what looks like a Planck factor if you associate the absolute temperature with the acceleration. This give the mass proportional to the temperature, except when it is realized that the R^2 in the denominator is the Schwarzschild radius. Then the temperature becomes proportional to the inverse of the mass instead of the mass. So there is one dependency of mass on temperature say at the surface of the earth, and an inverse dependency for a black hole. Thermodynamics does not work that way.

A slight problem with the spectral density is that its integral over all frequencies diverges so you don't get back Stefan's law. You don't even get the correct number of oscillators in any given frequency range. Is this the mathematics you are talking about to show that there is thermal radiation both inside and outside of the horizon? "A source of unlimited energy" constitutes a perpetual motion machine.

they do not emit radiations but it appear so..New theory says Black holes aren't black