Equipartition of energy in the modes of a resonant cavity

In summary: The modes are so excited that they are all at equilibrium with each other and the spectrum of the light coming out of the cavity is a continuous approximation of the input light.Summary: The apparent conflict between the dominance of the fundamental in strings and things and the black body continuous spectrum is a practical one.A string or an air / microwave cavity will not sustain the energy from an impulse for many cycles and the modes which are examined are pretty low order. That also implies that, for a continuous exciting waveform, the observed vibrations will be dominated by selected frequencies of the input waveform (any others will be at a much lower level).In the case of a practical black body cavity, it has been excited by a
  • #1
Guilherme Franco
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TL;DR Summary
Why was energy expected to be equally distributed between the modes of a resonant cavity? Why couldn't they be independent like in a string?
I think the answer for this may be straightforward, but I don't see anywhere that explains this from the scratch:

A large resonant cavity with a small hole is used to approximate an ideal black body.

I understand the conditions for the modes inside the cavity. But there are two points that aren't clear to me:
  1. I don't understand why it was considered that the energy should be equipartitioned between those modes. Because I don't see a reason why they couldn't be independent. At least not if it was ideally reflecting body inside. In that case, just like in ideal vibrating stings, there could be no exchange of energy between the modes and the spectrum of the light inside it would be just like the spectrum of the light entering it. I think the story has to do with the body not being perfectly reflecting and being in equilibrium with the modes inside the cavity. But even then: Why couldn't it just stay in equilibrium with the modes that has already being formed by the light that entered the cavity? Is the equipartition being mediated by the material portion of the cavity?
  2. Why exactly does particular solution serves as a model for entirely solid radiators? Does this EM field modes exist inside opaque materials?
 
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  • #2
Guilherme Franco said:
Summary: Why was energy expected to be equally distributed between the modes of a resonant cavity? Why couldn't they be independent like in a string?

I don't understand why it was considered that the energy should be equipartitioned between those modes. Because I don't see a reason why they couldn't be independent.
I imagine that it's assumed that there is a lot of coupling between modes, for practical reasons due to interaction with the surface. At lower frequencies (microwave cavities, for instance) I think the energy in the mode that was excited in an unloaded (ideal highly conductive) cavity would not couple to other modes very quickly.
 
  • #3
The equipartition of energy is seen in a system at thermal equilibrium. This is the necessary condition and is assumed for a black body cavity.
 
  • #4
Guilherme Franco said:
Summary: Why was energy expected to be equally distributed between the modes of a resonant cavity? Why couldn't they be independent like in a string?

Because I don't see a reason why they couldn't be independent
The reason is that no modes are perfectly isolated. There is always some coupling between the nodes. It is also true for a string. A violin always sounds like a violin because the ration of vibrations between different modes is at equilibrium.
 
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  • #5
Henryk said:
The reason is that no modes are perfectly isolated. There is always some coupling between the nodes. It is also true for a string. A violin always sounds like a violin because the ration of vibrations between different modes is at equilibrium.
The apparent conflict between the dominance of the fundamental in strings and things and the black body continuous spectrum is a practical one.
A string or an air / microwave cavity will not sustain the energy from an impulse for many cycles and the modes which are examined are pretty low order. That also implies that, for a continuous exciting waveform, the observed vibrations will be dominated by selected frequencies of the input waveform (any others will be at a much lower level).
In the case of a practical black body cavity, it has been excited by a continuous noise like source - such as white light or a hot wall. That will excite a continuum of modes (very high order, when we're talking about IR spectrum of the fields inside).
 

1. What is the equipartition of energy in the modes of a resonant cavity?

The equipartition of energy in the modes of a resonant cavity refers to the equal distribution of energy among all the different modes or oscillations of the cavity. This means that each mode has the same amount of energy, resulting in a balanced and stable system.

2. Why is the equipartition of energy important in a resonant cavity?

The equipartition of energy is important in a resonant cavity because it ensures that the energy is evenly distributed, preventing any one mode from becoming dominant and destabilizing the system. This allows the cavity to maintain a constant and predictable resonance frequency.

3. How is the equipartition of energy achieved in a resonant cavity?

The equipartition of energy is achieved in a resonant cavity through the process of thermalization. This means that as the cavity absorbs energy from an external source, it will distribute the energy equally among all the different modes, resulting in equipartition.

4. Can the equipartition of energy be disrupted in a resonant cavity?

Yes, the equipartition of energy can be disrupted in a resonant cavity if there is an external force or disturbance that affects the distribution of energy among the modes. This can lead to a change in the resonance frequency and potentially destabilize the system.

5. How does the equipartition of energy relate to the quality factor of a resonant cavity?

The quality factor (Q-factor) of a resonant cavity is a measure of how well the cavity can store and release energy. A higher Q-factor indicates a more efficient cavity with less energy loss. The equipartition of energy is important in maintaining a high Q-factor as it ensures that the energy is evenly distributed among the modes, minimizing energy loss and maximizing the efficiency of the cavity.

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