# Entropy Production in an Optical Cavity?

• A
Gold Member

## Main Question or Discussion Point

Howdy,

Say you've got two highly reflective mirrors forming a cavity. Some broadband light goes in, but only narrowband light comes out. Entropy is definitely decreased as far as the photons are concerned. Where does it go?

This has been bugging me. I have a partial solution I was hoping you all might sanity check. When light at some frequency is incident on the mirror (which I think of as a very good but not perfect conductor), it excites charge oscillations that produce electric fields to cancel most of the transmitted light. One could imagine that in a lossless cavity at steady state the electrons in the mirrors are oscillating at those off-resonant frequencies, but the optical fields they produce interfere destructively. So you're just shuffling entropy from photon modes to electron motional modes, or something along those lines. What still bugs me is that it seems like I've double-booked the energy of the incident beam. On the one hand, the spectral density of the circulating light skyrockets at a longitudinal mode as compared to the spectral density of the incident light, and you have the energy from the non-resonant modes being transferred to the motional states of the electrons in the mirrors. Now it isn't clear to me that energy is conserved in my picture. Anyone got some better insights?

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marcusl
Gold Member
What is the spacing between mirrors? Why do you think that white light would be converted to monochromatic?

Cryo
Gold Member
Could it be a good idea to start with the thermodynamic definition of change in entropy?

$dS=\delta Q_{rev}/T$

Change in entropy is the reversible change in heat divided by temperature. So, I guess, the cavity will heat up.

Andy Resnick
Howdy,

Say you've got two highly reflective mirrors forming a cavity. Some broadband light goes in, but only narrowband light comes out. Entropy is definitely decreased as far as the photons are concerned. Where does it go?
Interesting question. Laser light is not in thermal equilibrium, but it is possible to define an order parameter and create an analog of thermodynamic entropy. When this is done, laser threshold is described as a second-order (continuous) phase transition. Multiple-mode lasers can exhibit discontinuous, first-order phase transitions.

My reference for this is Mandel and Wolf, "Optical Coherence and Quantum Optics", and another reference is:

Cooperative phenomena in systems far from thermal equilibrium and in nonphysical systems
H. Haken
Rev. Mod. Phys. 47, 67 (1975) – Published 1 January 1975

tech99
Gold Member
Howdy,

Say you've got two highly reflective mirrors forming a cavity. Some broadband light goes in, but only narrowband light comes out. Entropy is definitely decreased as far as the photons are concerned. Where does it go?
Although this is a transmission line resonator, is it not analogous to a LCR resonant circuit being driven by noise?
When the noise is first applied, the components near the resonant frequency cause oscillation to slowly build up, and as it grows to maximum, increasing energy is supplied to the resistance. The noise generator then sees the LCR circuit as a resistance at that frequency and continues to deliver energy to it.
At switch-off, the energy stored as oscillation in the LCR circuit is then slowly transferred to the resistance.
The components at frequencies different to the resonant frequency see the LCR circuit as a reactance, which reflects the energy back to the generator.

Gold Member
Sorry for the late and brief response. Grad school hit me right where it hurts: the schedule :)

What is the spacing between mirrors? Why do you think that white light would be converted to monochromatic?
The transfer function of the cavity will narrow the bandwidth of the light. If numbers are necessary assume a FSR of 10GHz and a finesse of 10.

Could it be a good idea to start with the thermodynamic definition of change in entropy?

$dS=\delta Q_{rev}/T$

Change in entropy is the reversible change in heat divided by temperature. So, I guess, the cavity will heat up.
I think you are right that the cavity will heat up due to the electrons jiggling around to produce the reflected wave. But it’s not clear to me that this is necessarily true. I think it does not change the fundamentals of you assume the mirrors to be superconducting, and thus not dissipate any heat upon reflection. What’s bugging me is that it’s not clear to me that the electrons really account for the missing entropy from an an initio standpoint.

Interesting question. Laser light is not in thermal equilibrium, but it is possible to define an order parameter and create an analog of thermodynamic entropy. When this is done, laser threshold is described as a second-order (continuous) phase transition. Multiple-mode lasers can exhibit discontinuous, first-order phase transitions.

My reference for this is Mandel and Wolf, "Optical Coherence and Quantum Optics", and another reference is:

Cooperative phenomena in systems far from thermal equilibrium and in nonphysical systems
H. Haken
Rev. Mod. Phys. 47, 67 (1975) – Published 1 January 1975
This is really interesting, even to an experimentalist chump like me haha. I’ll have to read this paper thoroughly and get back to you shortly.

Although this is a transmission line resonator, is it not analogous to a LCR resonant circuit being driven by noise?
When the noise is first applied, the components near the resonant frequency cause oscillation to slowly build up, and as it grows to maximum, increasing energy is supplied to the resistance. The noise generator then sees the LCR circuit as a resistance at that frequency and continues to deliver energy to it.
At switch-off, the energy stored as oscillation in the LCR circuit is then slowly transferred to the resistance.
The components at frequencies different to the resonant frequency see the LCR circuit as a reactance, which reflects the energy back to the generator.
I think you’re right about the LRC circuit for sure, but I think the analogy breaks down. In the cavity, there doesn’t have to be any resistance, as I mentioned above if the mirrors are superconducting then there is no dissipation. In this extreme idealized case, the energy can only be stored, reflected, or transmitted.