Phonons: Where Does Their Energy Go?

[motoroller]

When two phonons interact in a material that is being heated, where does their energy go? Are the electrons in neighbouring atoms promoted? Is radiation given off?

What about at a crystal boundary? My guess is that the phonon would be reflected, so the energy would be radiated back off again.

 

[genneth]

Remember the three ways for heat to transfer: conduction, convection and radiation. Of course, things aren't really that simple, but we can try our best. Phonon transport is essentially convection. Photon transport is basically radiation. The coupling between phonons and electrons, and the latter's coupling to photons, means that yes, some of the energy will get converted to photons and radiated away. Finally, if the crystal is in contact with a gas, there will be transfer of energy, via some complicated adsorption, phonon/polariton excitation, de-adsorption events to move energy about. Either way, some gas molecules get a kick. Thermodynamics tells us to not worry too much about how that happens, and assures us that it will happen in all possible ways compatible with exact symmetries and conservations.

 

[Manchot]

When two phonons interact in a material that is being heated, where does their energy go? Are the electrons in neighbouring atoms promoted? Is radiation given off?

I'm not really sure what you mean by "interact." To a leading degree, phonons do not interact with each other, as they are bosons. Now, there are definitely higher-order nonlinear effects that can occur, and they can also interact strongly with electrons and with the EM field, but you have to be more specific about what type of interaction you're talking about.

What about at a crystal boundary? My guess is that the phonon would be reflected, so the energy would be radiated back off again.

You are correct; a phononic wave packet will reflect off the surface of a boundary, with a reflection coefficient determined by the type of boundary. For a perfectly hard or soft boundary (e.g., a boundary with air), the reflection will just give a phase factor. Though strictly speaking, if you're going to talk about individual modes, the canonical quantization procedure has already built the boundary in. For example, in the quantum-mechanical square well, even though you can describe the time-evolution of the system perfectly well by a wave packet comprised of traveling and reflecting exponentials, the stationary states are those superpositions that repeat themselves.