Are there quantum mechanical aspects to this?
In our previous blog post we remarked that a black hole, or a frozen star, is not "static", because matter keeps falling closer to the event horizon, though the coordinate speed of the fall is minuscule. We argued that the Doppler effect causes the radiation out from a black hole to be much less than what Hawking radiation predicts.
The distance of a classical particle to the horizon may quickly become less than, say, 10⁻⁷² m in the Schwarzschild coordinates. In quantum mechanics, such short distances are nonsensical. How does quantum mechanics handle this?
The proper distance to the horizon is larger, though.
Dropping a small photon to the event horizon of a Schwarzschild black hole
Let us drop a photon of a wavelength 1 km to a black hole of 1 solar mass. The mass-energy of such a photon is
h c / λ * 1 / c² = 2 * 10⁻⁴⁵ kg,
which corresponds to a Schwarzschild radius of
2 G M / c² = 10⁻⁷² m.
In quantum mechanics, the wave function of the photon is quickly squeezed into a volume which is 10⁻⁷² meters thick in the Schwarzschild coordinates and about 2 km wide.
What is the proper thickness of the volume?
The radial metric is stretched by a factor
1 / sqrt(1 - r / rs) = 10³⁸.
Therefore the proper thickness of the volume is
~ 10⁻³⁴ m.
Because of the blueshift, the wavelength of the originally 1000 m photon is only
~ 10⁻³⁵ m
down there.
We see that the wave function of the photon "fits" easily into the thin layer above the old horizon. That suggests that the photon does behave much like a classical particle whose horizontal location on the event horizon has a large uncertainty of 1 km.
However, the further development of the photon becomes obscure in a classical treatment, because the photon itself affects the location of the event horizon.
The coordinate speed of light c' at the position of the classical photon particle is ~ 10⁻⁶⁸ m/s. If we assume that the speed does not become greater than that, then the position of the photon is essentially unchanged for the next 10³⁰ years, as seen by a faraway observer.
If we assume that we can determine the quantum behavior of the system by summing probability amplitudes of different "paths" of the photon, like in the Feynman diagrams, then the quantum behavior of the photon is much like the classical behavior.
When would the Hawking radiation formula hold?
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| | |
| M T λ M T λ |
| | |
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"space" "black hole"
Let us have two immense vessels filled with black body radiation, each vessel containing a mass-energy M. The temperature T of the black body radiation is the Hawking temperature of the Schwarzschild black hole of the mass M.
We connect the vessels with a hole whose radius is a few times the Schwarzschild radius of M, or a little shorter than the main wavelength λ of the black body radiation.
The system is in a thermodynamic equilibrium. Some radiation seeps through the hole into both directions. We may imagine that the vessel on the left is the "space" and the vessel on the right is the "black hole", and they are in an equilibrium. This would be a model of Hawking radiation, as Jacob Bekenstein and Stephen Hawking might have envisioned it.
However, the model is wrong. The black hole on the right is just forming, and presumably, almost all photons move to the direction of the black hole center, away from the hole. There is no equilibrium.
Note that since the photons are predominantly moving to one direction, they are ordered and have less entropy than black body radiation which would come from every direction. Here we have another example where lower entropy implies less radiation out.
Hawking radiation would require that the radiation on the right side would quickly "thermalize" with the walls of the vessel, and start coming equally from every direction. This is not something that we expect about a black hole. The radiation falling in does not have time to collide at the center and form "thermalized" radiation.
The myth of the event horizon "capturing" one of a virtual pair of particles
In folklore, some people have tried to explain Hawking radiation by a mechanism where a virtual pair of particles is born and the event horizon captures the one with a negative energy. The black hole swallows the negative energy particle. The particle with positive energy is "freed" and escapes as a real particle. This real particle is supposed to be Hawking radiation.
Let us analyze this hypothesis.
1. In this blog we hold the view that virtual particles only exist as temporary states when real particles collide. The hypothesis breaks this principle because the virtual pair is born from space which only contains a macroscopic, static, gravity field.
2. In particle colliders we do not see any negative energy particles being produced. There is no empirical evidence that such particles can exist as independent particles.
3. Nature seems to have a "transaction" mechanism which ensures that energy and momentum is conserved in all interactions. In the hypothesis, if the positive energy particle carries a momentum p and an energy E, then the negative energy particle carries -p and -E, and moves to the same direction as the positive energy particle. Why would the black hole swallow just one of these particles? How do we satisfy conservation of momentum and energy?
What is a virtual particle close to the event horizon?
Let us imagine a particle collider placed just at the event horizon. In Feynman diagrams, there are lines which represent virtual particles. What happens to these lines?
The particles in the collision are falling freely. In freely falling coordinates, the collision should happen in much the same way as in zero gravity, because of an equivalence principle. However, a faraway observer sees the process to "freeze", first from the lower side, and then entirely. The collision lasts very long in coordinate time, maybe forever.
wave 1
-----------~~~~---------- field 1
| interaction
-----------~~-------------- field 2
wave 2
What is a frozen virtual particle? In this blog our view is that a virtual particle is not a free particle, but an interacting particle. In the wave interpretation, two or more waves are interacting.
The waveform of a virtual particle is not the ordinary sine wave, but can be distorted in many ways. This explains why a virtual particle can carry an arbitrary amount of energy and momentum.
Thus, a frozen virtual particle is simply a snapshot of the collision. The field is twisted to a form which only last for a very short proper time in the eyes of a co-falling observer. A faraway observer may measure that to last forever.
Could it happen that a negative energy virtual particle is frozen at the horizon and a positive energy real particle escapes?
Negative energy particles in a Feynman diagram
Above we have a vacuum polarization loop in a Feynman diagram. Time progresses from left to right, the wavy line is a photon, and the loop is a virtual electron-positron pair which is created and annihilated.
The positron and the electron may have arbitrary energies in the loop, as long as the sum of their energies is the energy of the photon.
Conclusions
Let us close this blog post. We will next look at Feynman diagrams, or other arguments, which might support the existence of the Schwinger effect or the existence of Hawking radiation. In both cases, the crucial question is if something like a "negative energy particle" can exist independently.
A negative energy particle is a formal calculation tool in a Feynman diagram. There is no empirical evidence that such a particle is a genuine natural phenomenon. It might just be a term in a calculation formula.
We have collected many arguments against the existence of Hawking radiation. The Schwinger effect brings a new aspect to this.
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