Can photon-photon interactions alone produce black body radiation?

In summary, the answer to the question is yes, photon-photon interactions will thermalize a photon gas that is out of equilibrium.
  • #1
Rap
827
10
Suppose you have a container with perfectly reflecting walls, containing electromagnetic radiation that is not in equilibrium (i.e. does not have a Planck distribution of energies.) Will photon-photon interactions (QED and/or gravitational) produce a Planckian distribution after a sufficiently long time?

I don't care how long - fifty gazillion times the age of the universe, I don't care.
 
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  • #2
I suspect that bouncing off the walls would lead to thermal equilibrium very quickly. Photon-photon interactions (except right after the big bang) are extremely rare.
 
  • #3
mathman said:
I suspect that bouncing off the walls would lead to thermal equilibrium very quickly. Photon-photon interactions (except right after the big bang) are extremely rare.

No, bouncing off the walls does not change the energy of a photon. If the energy distribution were out of equilibrium, it would remain out.
 
  • #4
Sure it will. It can even change the number of photons.
 
  • #5
Vanadium 50 said:
Sure it will. It can even change the number of photons.

I don't see how - the energy has to be conserved, so if one photon hits, two photons reflect, then they are both of lower energy. That's not perfectly reflecting.
 
  • #6
Rap said:
No, bouncing off the walls does not change the energy of a photon. If the energy distribution were out of equilibrium, it would remain out.
Bouncing off the wall means an interaction takes place with the material in the wall. The wall is at a temperature - hot photons hitting the wall will tend to cool off, while cold photons tend to heat up.
 
  • #7
mathman said:
Bouncing off the wall means an interaction takes place with the material in the wall. The wall is at a temperature - hot photons hitting the wall will tend to cool off, while cold photons tend to heat up.

But this is counter to the assumption of perfectly reflecting walls. You are changing the subject.
 
  • #8
Photon-photon interactions (except right after the big bang) are extremely rare.
But they do take place. And he gave us fifty gazillion times the age of the universe, so one scatter every 13.7 billion years is still fifty gazillion collisions.
 
  • #9
Rap said:
But this is counter to the assumption of perfectly reflecting walls. You are changing the subject.

There is no such thing. The walls have to be made of something. That something will have a temperature.
 
  • #10
mathman said:
Bouncing off the wall means an interaction takes place with the material in the wall. The wall is at a temperature - hot photons hitting the wall will tend to cool off, while cold photons tend to heat up.

What is a "hot photon" and a "cold photon"?
 
  • #11
mathman said:
There is no such thing. The walls have to be made of something. That something will have a temperature.

You are missing the point of the question. The reflecting walls are a way of bounding the system in order to ask a question. They are not the point of the question. The question is whether photon-photon interactions will thermalize a photon gas that is out of equilibrium.

Assume a photon gas of practically infinite extent so that boundary conditions are practically negligible - will photon-photon interactions thermalize that photon gas?

Assume a closed universe composed only of photons, so that there are no boundary conditions - will photon-photon interactions thermalize that photon gas?

Do you see what I am asking?
 
  • #12
Rap said:
You are missing the point of the question. The reflecting walls are a way of bounding the system in order to ask a question. They are not the point of the question. The question is whether photon-photon interactions will thermalize a photon gas that is out of equilibrium.

Assume a photon gas of practically infinite extent so that boundary conditions are practically negligible - will photon-photon interactions thermalize that photon gas?

Assume a closed universe composed only of photons, so that there are no boundary conditions - will photon-photon interactions thermalize that photon gas?

Do you see what I am asking?

This abstract http://meetings.aps.org/Meeting/DPP11/Event/152079 would suggest that the answer is yes, which I don't think is unexpected.
 
  • #13

1. Can photon-photon interactions alone produce black body radiation?

No, photon-photon interactions alone cannot produce black body radiation. Black body radiation is produced by the thermal motion of charged particles within a material, and photons do not have a charge and therefore do not contribute to this process.

2. What is the role of photon-photon interactions in black body radiation?

Photon-photon interactions play a minor role in black body radiation. They can contribute to the scattering and absorption of photons within a material, but the majority of black body radiation is produced by the thermal motion of charged particles.

3. Can photon-photon interactions be observed in black body radiation?

Yes, photon-photon interactions can be observed in black body radiation. When photons interact with each other, they can either scatter or combine to form new photons with different energies. These interactions can be detected through spectroscopic analysis of black body radiation.

4. How do photon-photon interactions affect the spectrum of black body radiation?

Photon-photon interactions do not significantly affect the spectrum of black body radiation. The spectrum of black body radiation is determined primarily by the temperature and composition of the material, rather than by interactions between photons.

5. Are there any practical applications of studying photon-photon interactions in black body radiation?

Studying photon-photon interactions in black body radiation can provide insights into the behavior of photons in different materials and environments. This knowledge can be useful in developing technologies such as solar cells, optical sensors, and photonics devices.

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