UPDATE June 25, 2024: Did we refute thermodynamics, too? If we have a solid block of matter at absolute zero, and warm it up with random photons of a very long wavelength, it will later radiate away the heat as photons of a higher energy. The number of photons is reduced.
Why is entropy conserved? Maybe because the time at which the infrared photons leave, adds more entropy?
Or is it so that a short wavelength photon carries away more information because of the short wavelength?
Maybe it is best to check if a perpetuum mobile can be constructed with Hawking radiation. A perpetuum mobile is well defined, while "information", or entropy, is not.
Let us have a large vessel full of photon gas at a low temperature. We squeeze the vessel until the photon gas collapses into a black hole. After that, the hypothetical Hawking radiation will make the black hole to evaporate. The final stage of the evaporation releases photon gas in a violent explosion. Does this allow us to construct a perpetuum mobile?
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This is the question in the Bekenstein bound.
If we feed a black hole with photons whose wavelength is ~ 17 X the Schwarzschild radius, then the black hole will probably contain the Bekenstein bound worth of entropy.
Maximizing the number of individual particles and minimizing the frequency f
The way to increase entropy is to divide the mass M to the maximum possible number of particles, typically photons. The wavelength of the photons is restricted by
λ < 2 ∛ V,
if V is a cube.
The energy of a single photon is
h f,
where h is Planck's constant and f is the frequency. If we can make the speed of light c slower inside V, then each photon has less energy, even though its wavelength is the same as before.
We can attempt to increase the entropy inside V by slowing down light inside V. This is what gravity does.
Using phonons in a solid
Let us have two million low-energy particles. We build a solid material from one million particles. We assume a potential of a form k x² between those particles. The energy of the rest 1 million particles we insert into the solid as phonons.
The solid has to withstand an enormous energy which is put into phonons: the energy is as large as the mass-energy of the solid. We have to make k extremely large, which, in turn, means that the speed of sound is close to c.
Also, since the solid only has some 100³ particles, we cannot have short wavelengths in the phonons.
It does not look like that using a solid and phonons can increase the entropy, compared to a photon gas.
Feeding a black hole with extremely long wavelength photons: there is no Bekenstein bound
Is there a reason why a black hole would be unable to absorb photons whose wavelength is much larger than 17 times the Schwarzschild radius?
The probability of absorbing such a photon is very low, but not zero? What is the waveform like close to the event horizon of a Schwarzschild black hole?
The metric is:
where
rs = 2 G M / c²,
if M is the mass-energy of the black hole measured from far away.
The local proper time close to the event horizon runs at the rate
sqrt(1 - rs / r)
relative to the coordinate time, or the time of a distant observer.
The proper radial length is stretched by the factor
1 / sqrt(1 - rs / r)
relative to the radial coordinate r length. Thus, the coordinate speed of light, c', is only
c' = (1 - rs / r) c
close to the horizon.
If we send a wave of a length λ from a radial coordinate R toward the horizon, there will be
R
~ ∫ 1 / (1 - rs / r) * dr
rs
= ∞
cycles before the wave reaches the horizon.
If we have a very long wavelength far away in space, the wavelength will be tiny relative to the size of the black hole as the wave approaches the horizon. It is plausible that a photon of such a wavelength can be captured by the black hole.
We can feed the black hole with photons whose energy is arbitrarily small. This proves that the entropy can be arbitrarily large. There is no Bekenstein bound for black holes!
Feeding very small energy photons to a black hole through a harness around the black hole
Above we calculated that a very long wavelength photon has a reasonably short wavelength close to the horizon of a black hole.
Let us build a harness around the black hole such that it is relatively close to the horizon. Now we can efficiently feed photons of a very small energy into the black hole.
The information content, or entropy, is a vague concept
Above we proved that we can feed a black hole with photons of a very long wavelength. But is their "entropy" really very high?
We can heat up a block of solid material with very long wavelength radio waves. Later, the block will radiate away infrared photons which have a much shorter wavelength. There is no contradiction in this. The number of photons can decrease.
An attempt at constructing a perpetuum mobile which converts heat into mechanical energy
Let us have a vessel full of photon gas which is black body radiation for an extremely low temperature T. Let the gas be in a thermodynamic equilibrium with a black hole inside the vessel, assuming thst Hawking radiation exists.
A typical Hawking photon has a wavelength of 17X the Schwarzschild radius.
If we want to squeeze the photon into a "package" which can easily be dropped into the black hole, we have to make the wavelength 1/17 of the original. Thus, N might be 17X M. Can we harvest so much mechanical energy in a cycle?
As we lower the package closer to the event horizon, the photon is blueshifted. Does this allow us to squeeze the package smaller?
We could have an observer sitting close to the horizon. Now and then he will catch a very much blueshifted photon. He may even be able to make particles with a positive rest mass from the radiation hitting him from up and down.
The observer can then lower a particle with a positive rest mass closer to the horizon and harvest some mechanical energy.
Did we just prove that it is possible to harvest mechanical energy from a system which is in an equilibrium?
The same argument can be used for an ideal gas. Sometimes, a gas atom with a very high kinetic energy will hit an observer, and he can harvest some mechanical energy. But that is not considered a perpetuum mobile.
The hypothetical Hawking radiation causes radiation pressure. If we have a container of a volume V close to the horizon, the pressure from downward is larger than from upward. The gravity of the radiation inside the container probably makes up for the difference. It is like an atmosphere in an equilibrium. We cannot harvest energy by moving a container full of gas up or down.
Any gravitating body whose temperature is above absolute zero and which is inside a vessel in a thermodynamic equilibrium, possesses a similar atmosphere of infrared photons. This is not unique for a black hole.
The Hawking temperature
The temperature for a solar mass black hole is 60 nanokelvins. Wien's displacement law gives 50 km or 17 rs as the typical wavelength of the infrared radiation.
The Hawking formula says that the black hole is in a thermodynamic equilibrium with black body radiation whose typical wavelength can just slip through the "hole" in space.
Is this reasonable?
Let us imagine an extremely strong rigid shell which encloses the black hole, very close to the horizon. What is the equilibrium temperature of the shell?
We probably get the temperature from the blueshift and Wien's displacement law. The power of black body radiation is ~ T⁴. One factor of T is explained by the blueshift, the rest T³ is probably explained by the fact that the strong gravity field pulls most photons back to the surface of the shell. Only photons leaving approximately vertically can escape.
A collapse of a photon cloud
Let us make a black hole by letting dust to collapse. The lightest possible speck of dust is a photon. The "kinetic energy", or heat energy, of the a photon is its own energy. Under these assumptions, the lowest temperature that the cloud of dust of a certain size can have is roughly the Hawking temperature T.
If we let the photon cloud collapse, does the temperature for an observer far away appear the constant T regardless of the phase of the collapse? The Hawking radiation hypothesis would suggest this.
Let the surface of the cloud fall to a potential
U(r) << 1,
where f / U(r) is the frequency of a photon of a frequency f sent by a static observer at the radius r.
If the temperature T(r) of the surface goes as
~ 1 / U(r),
and the surface of cloud is static as it falls (it cannot be), then a faraway observer would see the temperature as the constant T.
But the surface of the photon cloud is falling down, eventually at a velocity which is almost the local coordinate velocity of light.
The Doppler effect will make the wavelength of the outgoing radiation much larger than in the case of a static cloud surface. A long wavelength wave has problems escaping from the small "hole" around the forming event horizon. Furthermore, the energy of escaping photons is reduced by the Doppler effect.
We conclude that the radiation that a faraway observer sees will have an intensity much less than the black body radiation of the temperature T. The main wavelength of the outgoing radiation will be much larger than 17 rs.
Here we have yet another strong argument against the existence of Hawking radiation. A collapsing photon cloud will radiate much less than what the Hawking radiation would be.
If the Hawking radiation exists, when will it start? Our outline of the process does not indicate any moment when Hawking radiation should start.
Conclusions
It is hard to define what is the information content, or entropy, for a given flux of radiation. One could claim that a single photon can code an infinite amount of information.
We then looked at a better defined problem: can one use Hawking radiation to construct a perpetuum mobile? It looks like a perpetuum mobile is not possible. Hawking radiation forms an "atmosphere" around a black hole, and one cannot make a perpetuum mobile from the atmosphere of a celestial body.
We then turned to studying a collapse of a photon cloud. We could as well study a collapse of particles with a non-zero rest mass. The surface of the collapsing clouds falls almost at the speed of light. The Doppler effect reduces the outgoing radiation intensity dramatically. This does not agree with the Hawking hypothesis at all. There should be a flux of black body radiation at a fixed temperature T.
In principle we could stop the collapse process if we had superstrong materials. If Hawking radiation would exist, we would have hard time explaining from where did it get its energy.
We obtained new strong arguments against the existence of Hawking radiation.
If the black hole would be a static system in a thermodynamic equilibrium with the surrounding space, then it looks likely that it would radiate at the temperature T. But a black hole, or a frozen star is not static, at least if we look at it as a classical object. Does quantum mechanics bring another aspect to this?
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