4
$\begingroup$

A perfect harmonic crystal (PHC) has infinite thermal conductivity (see here for example, or also (1)).

Does this mean that in a PHC there is instantaneous heat transfer?

Fourier's law would seem to suggest that this is the case, since when the thermal conductivity is infinite the time derivative of heat blows up. However, I don't know how faithfully this equation describes heat propagation, even in an idealized system like a PHC.

If infinite thermal conductivity really implies instantaneous heat propagation, this bothers me a little, even if I know that the PHC is an idealization. What bothers me is that in my mind even in a PHC you need some time for the vibration to propagate, and therefore it seems absurd that heat propagation is infinite.

To be clearer, what I am imagining is this: I take a still PHC (no vibrations...we are considering the classical model here), and I impart momentum to a finite group of atoms, making them vibrate. Will this vibration propagate in the whole system instantaneously?

Notice that I avoided talking about temperature, since a temperature cannot be defined in a PHC, because the system can never thermalize (1).

(1)

Since the work of Born and Debye, it has been known that the idealised model, with harmonic forces between atoms, is useless for an investigation of thermal conductivity, since it would lead to an infinitely large conductivity.

For in the model with harmonic forces [...] one can think of the motions of the atoms as built up from mutually independent sound waves. Once an arbitrary energy distribution of these vibrations exists, it will exist forever. A thermal equilibrium will then not be established, and in general one can therefore not speak of a temperature at all.

-R. Peierls, “Zur kinetischen Theorie der Wärmeleitung in Kristallen” Ann. Phys. 395, 1055–1101 (1929)

Improve this question
$\endgroup$
5
  • $\begingroup$ Building on your quote about thermalization, in a PHC the elementary excitations are Fourier modes, completely delocalized across the entire crystal, so how are you going to talk about a "local energy distribution" and if you can't talk about energy in a local way are you surprised that you have an infinite thermal conductivity? $\endgroup$ – By Symmetry Feb 2 '18 at 13:11
  • $\begingroup$ @BySymmetry I am just imagining this: I take a perfect crystal which is "still" (no vibration), and I impart momentum to a certain number of atoms. Will the resulting vibration propagate instantaneously in the whole system? I will update the question to clarify. $\endgroup$ – valerio Feb 2 '18 at 13:16
  • $\begingroup$ @valerio92- Even in the perfectly harmonic crystal the initially localized excitation (which is definitely possible, as you can see from a Fourier expansion!) will not propagate instantaneously through the whole system. Think only of the acoustic branches of the lattice vibrations. They propagate at most with sound velocity through the crystal. $\endgroup$ – freecharly Feb 2 '18 at 16:53
  • $\begingroup$ An infinite heat conductivity does not lead to instantaneous propagation. By analogy, an infinite electrical conductivity, as is realized (for all practical purposes) in superconductors, does not mean current propagates instantly. $\endgroup$ – Rococo Feb 2 '18 at 16:55
  • $\begingroup$ @Rococo So the next question would be: how does Fourier's law (or Ohm's law in a electrical conductor) exactly fail to describe heat conduction in a PHC? Because in Fourier's/Ohm's law, an infinite conductivity seems to imply instantaneous propagation of heat/current. $\endgroup$ – valerio Feb 2 '18 at 17:07
1
$\begingroup$

Even in a perfectly harmonic crystal, an initially localized excitation of the atoms from their equilibrium position will not propagate through the system instantaneously! If you look at the acoustic branches of the crystal oscillation dispersion relationship, you will see that the fastest velocity of propagation of any disturbance will be the sound velocity (longitudinal or transverse).

Improve this answer
$\endgroup$
5
  • $\begingroup$ This makes perfect sense to me. So how does exactly Fourier's law fail to describe heat conduction in a PHC? There, an infinite thermal conductivity seems to imply instantaneous heat propagation. $\endgroup$ – valerio Feb 2 '18 at 17:04
  • $\begingroup$ @valerio92 - You already mentioned the reason why Fourier's law has to fail in a perfectly harmonic crystal: You cannot define a temperature and thus also no gradient of temperature! Furthermore, Fourier's law also requires that the material, where the heat conduction takes place, can be described to have a "local thermal equilibrium" and thus also a local temperature. This is definitely impossible in a perfectly harmonic crystal. $\endgroup$ – freecharly Feb 2 '18 at 17:22
  • $\begingroup$ @freecharly, you can definitely define a temperature for a perfectly harmonic crystal, in the same way you can define a temperature for the harmonic oscillator! However, gradients of temperature are an entirely different matter. Without damping of phonons (through 3rd order effects for example), a thermal difference will always propagate throughout the entire lattice without any problem, and you get a single temperature everywhere in the equilibrium limit. Of course, you will get transients for time and length scales small compared to the max phonon velocity. $\endgroup$ – KF Gauss Feb 2 '18 at 17:40
  • $\begingroup$ Of course...quite stupid of me not to realize this. $\endgroup$ – valerio Feb 2 '18 at 18:29
  • $\begingroup$ @valerio92 - Of course, you can impose a Bose-Einstein equilibrium distribution of the phonons with a given temperature as an initial condition. As far as I understand, in a perfectly harmonic crystal, you have no interaction (scattering) between and damping of the phonons. Therefore, any initial non-equilibrium distribution will not "thermalize" to an equilibrium distribution with a temperature. That's also what is said in the above citation of Peierls. $\endgroup$ – freecharly Feb 2 '18 at 18:38
1
$\begingroup$

Let's assume you have a 1D lattice of songs with spacing $a$ and one atom per unit cell. Let's further assume the harmonic approximation.

If you simulate heat by displacing a single atom, the resulting wave from this displacement will excite phonons of all wavevectors $k$. This is because you are starting with essentially a delta function in space, or equivalently something that is uniform in wavevector space.

The dispersion relation for phonons in a 1D chain is $\omega(k)=2\omega_0|\mathrm{sin}(ka/2)|$ which implies $\frac{d \omega}{d k}= a \omega_0 \mathrm{cos}(ka/2)$. This means the largest phonon speed is $a\omega_0$, not infinity.

Improve this answer
$\endgroup$
0
$\begingroup$

My take on this is that I do not really see a way to assign a speed to the heat propagation because temperature is not even defined, as stated by Peierls. So this would be my answer, and I post it as a complement to the already two magnificient answers posted by @freecharly and @user157879.

Nevertheless, here are some comments:

  • I think you may have a misunderstanding with the thermal conductivity $\kappa$ and the speed at which heat is propagated. In the regular conductive heat equation $\frac{\partial T}{\partial t} = \nabla \cdot \kappa\nabla T$, the speed at which heat is propagated is infinite. There are several ways to see this. One way is to solve the equation for a particular setup (i.e. boundary conditions and initial condition). You'll realize that the solution has a non retarded time dependence. In other words, the temperature will vary instantly due to a source, regardless of the distance between the source and the point considered. This is valid regardless of the value of $\kappa$, so clearly $\kappa$ is not the responsible for the speed of heat propagation.

  • The greater the thermal conductivity, the greater the amount of heat is propagated. Not the fastest, where speed is thought of the distance at which the initial perturbation is propagated, divided by time. Now sure, one can define a diffusivity constant and imagine the diffusion of heat through time and a greater kappa would make a material seem to thermalize faster than a material with a lesser kappa. So if one define a speed of heat propagation based on the time a material thermalize (where the criteria to define thermalization is subjective), indeed, an infinite $\kappa$ would be related to an infinite speed due to the instantaneous thermalization.

  • For some reason, you seem focused on phonons as mode of heat transfer. In metals, the electronic conduction plays a major role. So it seems you are implicitely dealing with a non metallic crystal with your question as stated.

  • Non relativistic QM, which is used in describing the harmonic crystal, is not Lorentz invariant. Many times there is no reason to expect things to have finite speed of propagation in non Lorentz invariant theories. Take the free electron, intially localized in some region. One instant later (or say dt), no matter how far from the original region one is, one will find a non vanishing wavefunction. This means the electron can be found at any arbitrary large distance soon after being measured to be at any particular position.
    Taking the analogy of heat conduction with electricity conduction, one finds a finite speed (and lesser than c) at which current and EM waves travel in a material, regardless of the value of the conductivity $\sigma$ (which is analogous to $\kappa$ for heat conduction). Note that Maxwell equations describing EM waves propagation is Lorentz invariant, unlike the Schrödinger equation used to describe the Hamiltonian of a harmonic crystal, and the conduction heat equation where Fourier law applies.

Improve this answer
$\endgroup$

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.