5
$\begingroup$

I have a couple of conceptual questions regarding the thermodynamics of scattering. Any partial answer or argument will be appreciated.

For the sake of discussion, consider the scattering of electromagnetic waves. I understand that the physics would be different for other particles or fields, feel free to base your discussion on those to provide your perspective.

The first questions are on the definition of entropy and its governing laws in a scattering process. More specifically:

(1) How do we define entropy for electromagnetic waves? I guess, one can always start from the general definition of entropy and perhaps calculate the density matrix of electromagnetic waves (though I've not seen such calculations). While such a first-principles approach might be helpful, I'm more of looking for a phenomenological definition that would capture the physics in relatively simple terms and mathematics. There seems to be arguments saying that electromagnetic waves do not have entropy, but given the fundamental nature of entropy, I think everything has entropy but it may not be well defined (please correct me if I'm wrong).

(2) How does entropy change in a scattering event? The second law would be a correct answer to this question, but is there a tighter bound regarding its change? Or, assuming that we now know how to define entropy for the electromagnetic waves, how are the entropies of incoming and outgoing waves related to the properties of the scatterer? (To better illustrate my question, in the case of energy, the difference between energies of incoming and outgoing waves is the dissipation in the scatterer.)

My final question is what I'm really after, though it might sound stupid: are the 1st and 2nd laws of thermodynamics independent in such scattering theories? (Can we derive the 2nd law from the 1st law plus something else in the framework of scattering theory?) This is not a random question but has basis in my research. I've been unable to reach clarity in this thought, but feel free to ask me to elaborate on any confusing statement.

$\endgroup$
  • $\begingroup$ It just occurred to me (after writing my answer), that you may simply not be aware of the physics of thermal radiation? $\endgroup$ – CuriousOne Sep 5 '14 at 5:55
  • $\begingroup$ This is an excellent question (esp. your #2) since it has much relevance to the CERN experiments, which, at the moment of collision, is a highly irreversible process, yet Einstein's theory, upon which relativistic scattering theory is based, is only a reversible theory. $\endgroup$ – Geremia Sep 6 '14 at 21:14
1
$\begingroup$

We define entropy in the electromagnetic case the same way we define it everywhere else. Entropy is the cause that (in absence of other physical changes) heat flows from hot to cold. Consider two temperature baths of different temperature facing each other. The surfaces of both temperature baths are being treated as black body radiators. Using the Stefan-Boltzmann law for black bodies, we can predict, that the hotter surface will emit more radiation than the colder surface, which causes a heat flow from the hot to the cold bath. Now that you have temperatures and a heat flow, the calculation of the entropy in this process is the same as for any other heat flow mechanism (be that convection or direct thermal contact). Thermodynamics doesn't care that electromagnetic waves are involved and all its laws are still valid.

$\endgroup$
  • $\begingroup$ OP asked specifically about scattering events. Could you perhaps expand your answer to address this? $\endgroup$ – Danu Sep 5 '14 at 7:52
  • $\begingroup$ @Danu: "Scattering events" for the thermodynamic case are absorption followed by emission. If the OP can formulate his actual problem with a suitable thermodynamic system (what's the temperature of the scattering particles?), then we may have something. It seems to me, though, that there is some confusion about the difference between scattering on particles, which is dominated by either classical electromagnetism or QM and thermodynamic effects, which require ensembles that are at least near equilibrium. $\endgroup$ – CuriousOne Sep 5 '14 at 8:21
  • $\begingroup$ I think those last few lines are a good analysis. This, too, would look good as part of your answer (IMO, at least). $\endgroup$ – Danu Sep 5 '14 at 8:22
  • $\begingroup$ @Danu... actually, I was thinking about deleting my answer, because I am less and less sure, what the OP is trying to ask and my answer might be completely orthogonal to what the question is about. What do you think? $\endgroup$ – CuriousOne Sep 5 '14 at 8:26
  • $\begingroup$ I think what you said about confusing thermodynamics with QM/EM scattering is certainly relevant, making it possible to really answer OP's question. $\endgroup$ – Danu Sep 5 '14 at 8:45
0
$\begingroup$

This question has bothered me for a long time, asked some real EM/optics experts but never got a satisfying answer for it, so do not expect one from me. At best I heard that it was not a stupid but an interesting question. Despite what I was ever taught I for one believe that diffraction is a manifestation of irreversibility of EM. If you start with a hole in the wall and illuminate it with a plane wave from one side I cannot imagine any instrument that could reconstruct that plane wave after it has passed to the other side. It would really take a demon at every point of the hole working with and an infinite large mirror to reverse the diffracted and scattered wave back in to a plane wave; there is no such demon. If I take the reference plane wave to have zero entropy then the shape of the diffracting hole must some way contribute to the entropy increase that is manifested by the difficulty with which one can reverse the wave or the level of minimum error with which a causal and finite size instrument (mirror) can get close to a back propagating plane wave. This is as far as I got, speculative but that is all.

$\endgroup$
  • $\begingroup$ The instrument you are asking for is called "a lens". For monochromatic light, even a simple convex lens with the hole placed at its focal point will restore the plane wave almost perfectly. The remaining imperfections will be dominated not by the hole, but by the size of the lens, which forms a second scattering aperture. It is theoretically possible to design perfect lenses, which do not have any imaging errors, in practice it's just too expensive to approximate those solutions, and it is almost never necessary. There are no thermodynamic ensembles in this case, at all. $\endgroup$ – CuriousOne Sep 5 '14 at 18:35
  • $\begingroup$ Your crucial qualifier of perfectly here is almost. So it is almost reversible not reversible. All finite size objects induce some amount of diffraction, just like simple and very slow quasistatic processes are almost reversible, not completely but almost. This almost has nothing to do with price, it has to do with size. $\endgroup$ – hyportnex Sep 5 '14 at 20:42
  • $\begingroup$ The process you are describing is perfectly reversible, we just don't care to reverse it perfectly with technical means. A hole in a screen is a simple linear transformation and the laws of optics guarantee, that it has an inverse, as long as you don't add any noise to it. Your wireless devices, by the way, have plenty of digital signal processing built in, which does exactly that: they try to remove the linear transformation by the environment from the signal between the transmit and receive antennas... they are simply doing it at 2.5GHz, rather than THz frequencies. $\endgroup$ – CuriousOne Sep 5 '14 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.