3
$\begingroup$

We all know heat transfer occurs via energy carriers (electrons, photons, bulk flow of molecules, phonons, etc.).

In many materials, phonons are the major contributors to heat transfer.

Classic texts describe phonons as plane waves of atomic displacements in crystals, which can travel at some group velocity $v$.

Kinetic theory then describes these phonons as gas particles, which may collide with each other due to anharmonicity, resulting in some finite relaxation time $\tau$. The thermal conductivity of the material is then given by a kinetic theory. While this theory agrees with experiments for some simple crystals, the physical picture of plane waves "scattering" with each other seems weird to me.

  1. Is the classical physical picture of phonons some sort of crude approximation to what actually happens?
  2. Are there actually plane waves flowing through materials that transport heat? It would make more sense if it was a wave packet, since plane waves imply that the vibration spans throughout the entire material in the plane of the wave. EDIT: If plane waves aren't it, then what exactly is the form of the transferred heat?
  3. How does heat transfer occur in noncrystalline solids, where most modes are not plane waves?
$\endgroup$
3
  • $\begingroup$ "Seems weird to me" is not a good reason for doubting a theory! Do you really expect the plane wave model to be an accurate description of what actually happens? Aren't questions 1 & 2 the same? $\endgroup$ Jan 27, 2020 at 20:52
  • $\begingroup$ @sammygerbil You're right. No I don't expect the plane wave model to be what really happens, but I'm curious about what really happens if anyone has any info. Question 2 is meant to be more specific, i.e. instead of plane waves transferring heat, maybe its wave packets? That would make more sense, but I can't find any info on it. $\endgroup$ Jan 27, 2020 at 22:03
  • $\begingroup$ Check that we usually use plane waves because of its simplicity, but that's an approximation. Plane waves are infinite, so we se tthe solid as... infinite, or even periodic. That is obviously not true. What will you find? Border effects, of boundary effects. In the boudnaries, the approximations cease to be valid $\endgroup$
    – FGSUZ
    Jan 27, 2020 at 23:22

2 Answers 2

2
$\begingroup$

As the highly influential 20th century physicist, Rudolph Peirls said in his article "Quantum Theory of Solids" in 1951:

"It seems there is no problem in modern physics for which there are on record as many false starts, and as many theories which overlook some essential feature, as in the problem of the thermal conductivity of nonconducting crystals."

So, the question you are asking is a very deep one, which has caused no end of trouble for a long line of physicists! As @Thomas basically says in their final paragraph, this is very complicated, and not as well understood as you might expect. For example, it might surprise you to find out that:

  1. Given the properties of particles (masses, interparticle potentials, etc.) in a system, we do not know how to predict the thermal conductivity of the system in general. We know how in certain simple cases like ideal gases, perfect conductors, etc. But in general, and particularly for insulating solids, we do not know how. More generally, the prediction of transport coefficients (thermal conductivity, viscosity, etc.) is a great unsolved problem of non-equilibrium statistical mechanics.
  2. All of this assumes that Fourier's Law of heat conduction applies. Usually it does. But there are cases where it doesn't, usually in "low dimensional" systems. We do not know the criteria for Fourier's Law to apply, and we do not know how to derive Fourier's Law from more fundamental mechanics. A nice paper on this unsolved problem is by Bonetto, Lebowitz, and Rey-Bellet, and can be found on archiv here: https://arxiv.org/abs/math-ph/0002052

Basically, you are asking a question which is an area of active research, containing some unsolved "old problems". Happy reading!

$\endgroup$
0
$\begingroup$

Typically, whenever there is a simple physical picture of heat conduction it is based on energy transport by quasi-particles. A quasi-particle is an excitation of the system with dispersion relation $\omega(p)$, where $p$ is momentum (or quasi-momentum, in a crystal), and the width $\Gamma(p)\ll\omega(p)$. By the basic principles of quantum mechanics (or just the properties of the Fourier transform, in classical system) these excitations are delocalized in space (because they have sharply defined momenta). This applies to any quasi-particle, electron-like quasi-particles, phonons, polarons, etc.

If a quasi-particle picture applies, then the conductivity can be determined using kinetic theory. Kinetic theory is based on distribution function $f(x,p,t)$. There is a formal procedure for defining distribution functions in QM or QFT (based on the Wigner-transform), but a reasonable physical picture is indeed that we constructed wave packets. These distributions functions satisfy a Boltzmann equation (even in the quantum case), so that the propagation of phonons looks like motion of classical particles, with scattering cross sections computed either classically (from the non-linear elasticity of the lattice), or quantum mechanically. Note that for a good crystal at sufficiently low temperature this is not some kind of model, but a rigorous, first principles calculation without free parameters.

Thermal conductivity of amorphous solids is a completely separate story. At least in some cases it is understood -- people can perform molecular dynamics simulations and quantitatively account for heat flow. However, we should not necessarily expect a simple physical picture based on quasi-particles. Amorphous solids have vibrational modes, but they cannot necessarily be described by phonons with a sharp dispersion relation and a small width. There may be regimes where an amorphous solid is a crystaline solid with a macroscopic number of dislocations (these are also quasi-particles), but I don't know how relevant this is in practice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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