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What is the physical picture of heat conduction in a poor conductor? In particular, I'm curious about the role of phonons in conduction in poor conductors. I know that phonons (within the harmonic approximation) move without being scattered and would lead to infinite thermal conductivity. This problem is resolved by including anharmonic terms in the Hamiltonian so that there are phonon-phonon scatterings.

  1. But how do the phonon-phonon scatterings reduce the thermal conductivity? I wish to understand this both physically and mathematically. The expression for conductivity can depend upon various quantities and scatterings must be affecting one of those.

  2. How does this phonon picture explain the fact that when we heat a poor conductor the heat propagates gradually from the hotter to the cooler end? If they are delocalized collective excitations, shouldn't they heat up all parts of the substance at the same time?

I don't have a condensed matter background and therefore, a detailed but not-too-technical answer will be helpful.

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  • $\begingroup$ (1) conduction of heat by the electrons is important at low temperatures. By room temperature the phonons are doing most of the job. (2) Just because phonons are collective excitations does not mean that they instantaneously propagate. Remember that electrons in crystals obey Bloch functions (just like the phonons do!), but also don't 'exist' everywhere. $\endgroup$
    – Jon Custer
    Commented Nov 21, 2016 at 17:30
  • $\begingroup$ @JonCuster- Why do you say that the Bloch functions don't exist everywhere? $\endgroup$
    – SRS
    Commented Nov 26, 2016 at 13:46
  • $\begingroup$ The point is that electrons, which are Bloch states, do not propagate instantly, so one should not expect phonons to do so either. $\endgroup$
    – Jon Custer
    Commented Nov 26, 2016 at 16:17
  • $\begingroup$ @JonCuster In (1), you must be confusing thermal conductivity with heat capacity. In metals, thermal conductivity is totally dominated by electrons, as the Wiedemann-Franz law shows. $\endgroup$
    – user137289
    Commented Nov 29, 2016 at 20:20
  • $\begingroup$ @Pieter - "... in the intermediate temperature range (roughly ten to a few hundred degrees K), where inelastic collisions are both prevalent and capable of producing electronic energy losses of order $k_{B}T$, one expects and observes failures of the Widemann-Franz law." (Ashcroft and Mermin. cja[ter 16). Or, put differently, why is the thermal conductivity of silicon (few free electrons) and nickel essentially the same near room temperature? $\endgroup$
    – Jon Custer
    Commented Nov 29, 2016 at 21:14

5 Answers 5

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As a starting point, I suggest that you take a look at the Wikipedia page on heat transfer physics, phonon scattering and Umklapp scattering.

In general the thermal conductivity $\kappa$ associated to some carrier satisfies

$$\kappa = \frac 1 3 n c_v u \lambda \tag{1}\label{1}$$

where $n$ is the carrier number density, $c_v$ is the heat capacity per carrier, $u$ is the carrier speed and $\lambda$ is the mean free path. The mean free path is related to the scattering relaxation time of the carrier $\tau$ by

$$\lambda = u \tau \tag{2}\label{2}$$

You can interpret $\tau$ as the average time between two successive "collisions" (scattering events) of the carrier. Therefore, from \ref{1} and \ref{2}, you can immediately see that in an infinite perfect harmonic crystal, in which no scattering takes place, $\kappa$ would be infinite, since there would be no scattering and therefore $\tau = \infty$ (in a finite crystal, you would still have scattering from the boundaries).

In a realistic physical systems, phonons are scattered by other phonons, electrons, defects (impurities) and boundaries. To take these effects into account, you have to leave the harmonic approximation and consider also anharmonic terms in the Hamiltonian. It was shown by Pierls [a] that it is the anharmonicity, together with the discrete nature of the crystal lattice, that generates thermal resistance.

In particular, when talking about phonon-phonon and phonon-electron scattering, a distinction is usually made between "normal" processes, which conserve the total wave vector, and "Umklapp" processes (or U-processes), where the total wave vector is changed by a reciprocal lattice vector. To know more about this distinction and its (debatable) utility, you can take a look here.

All these events contribute to the relaxation time $\tau$, and their contributions are taken into account through Matthiessen's rule, which states that the total relaxation time $\tau$ can be calculated as

$$\frac 1 \tau = \sum_i \frac 1 {\tau_i}$$

where $\tau_i$ are the relaxation times associated to the possible different scattering events. So this is how, "mathematically", scattering events influence the thermal conductivity of a material.

How does this phonon picture explain the fact that when we heat a bad conductor the heat propagates gradually from the hotter to the cooler end? If they are delocalized collective excitations, shouldn't they heat up all parts of the substance at the same time?

I think that here the answer is simply that you need some time for the "delocalized collective excitation" you are talking about to set in. This must be true even for a perfect crystal, even if its infinite thermal conductivity would seem to suggest the opposite, otherwise we would have instant propagation of a signal (the vibration of the atoms). You think about the phonons as "delocalized collective excitation", but in reality they are much more similar to wave packets arising from a superposition of these collective excitation (the normal modes of the crystal). Maybe I am not being 100% precise in my terminology here, but I hope that I managed to convey the general meaning of what I have in mind.

References

[a] R. Peierls, “Zur kinetischen Theorie der Wärmeleitung in Kristallen” ("On the Kinetic Theory of Thermal Conduction in Crystals") Ann. Phys. 395, 1055–1101 (1929)

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  • $\begingroup$ Are you saying that phonons in an imperfect crystal are not delocalized? @valerio $\endgroup$
    – SRS
    Commented Apr 3, 2018 at 12:42
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    $\begingroup$ @SRS Yes, basically. When deriving the phonon mathematically, you rely heavily on the periodicity of the lattice, but a real crystal is never perfectly periodic. This means that the phonons you are used to can only be defined locally, and you cannot have collective vibrations at the level of the whole crystal. I should warn you, however, that my solid state physics is a bit rusty and these thoughts mainly come from the few things I remember and from my intuition. $\endgroup$
    – valerio
    Commented Apr 3, 2018 at 12:52
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Ballistic propagation can be observed but that needs special conditions. Normally, transport is diffusive. At low temperatures, scattering is dominated by defects in the lattice. Even isotopes have an effect, diamond with a reduced content of $^{13}$C has higher thermal conductivity than diamond with the natural isotope composition. This scattering determines the mean free path. While $\lambda_{free}$ may be approximately independent of temperature at low temperatures, the conductivity increases with temperature because the phonons carry more energy (proportional to $C_v$).

At high temperatures, thermal conductivity decreases because the shorter mean free path caused by phonon-phonon interaction. When the variations in atomic bond lengths get larger, anharmonic terms become important. The wave equation is not linear anymore, waves do not always pass through each other anymore, there is a probability that new waves (phonons) will be created.

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Thermal conductivity is defined as a ratio of energy flux and temperature gradient (up to a factor). If phonons move freely, arbitrary energy flux can exist without temperature gradient (so the finite phonon velocity does not make the thermal conductivity final). See details on the role of scattering, say, at http://www.physics.iisc.ernet.in/~aveek_bid/PH208/Lecture%208%20phonons-thermal%20properties.pdf

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I'm essentially going to explain what is explained in the video here though this answer does specifically refer to poor conductors.

There are two types of phonon-phonon scattering: normal scattering and Umklapp scattering. While both scatterings conserve energy the former conserves crystal momentum while the latter does not. Two acoustic phonons with equal and opposite momentum $\textbf{q}$ (or more precisely, $\hbar\textbf{q}$) and $-\textbf{q}$ in the first Brillouin zone can scatter and combine to give a zero momentum ($\textbf{q}=0$) phonon. It will be an optical phonon because energy has to be conserved and adds. On the other hand, two acoustic phonons of unequal momenta $\textbf{q}_1$ and $\textbf{q}_2$ in the first Brillouin zone scatter without conserving the crystal momentum but only energy.

Now when we heat one end of a bar of some material and cool the other, the heated end produces phonons as local excitations. As phonons approach from the hotter to the colder end, there is a significant phonon momentum $\textbf{q}$ in that direction. For normal scattering, we expect only the momenta of the phonons to be redistributed, and there will be no resistance to heat flow. On the other hand, since Umklapp scatterings don't conserve momentum, if the source end of the bar locally produces a significant number of phonons with rightward momentum, after a few Umklapp scattering a good fraction of them will move in the opposite direction impeding the heat transmission which explains why the heating of all parts is not simultaneous. Umklapp scattering is the dominant mechanism of phonon scattering at room temperature and above and the scattering rate proportional to phonon concentration. It turns out that the phonon mean free path $\lambda$ is inversely proportional to temperature $T$. Increasing the temperature will increase the phonon concentration and decrease the mean free path.

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  • $\begingroup$ Thanks for posting this. Related, I still don't quite understand how polarizations change thermal transport. Naively, it seems like the direction of the energy transferred should be different when I think about atoms move perpendicular/parallel to propagating direction (k-vector direction). Although I do understand we treat phonon as a quasi-"particle" so whatever direction k-vector has will be the direction phonon is propagating. In this sense, transverse/longitudinal is just a flavor for the particle. Do you think this is correct? Or do you know a reference that elaborates on this? Thanks! $\endgroup$
    – Histoscienology
    Commented Jul 12, 2020 at 20:44
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Regarding Phonon Scattering despite phonons being eigenmodes of infinite harmonic lattice:

It is wavepackets that scatter. Just as plane waves are infinitely extended but their superposition can give rise to finite gaussian wavepackets. The scattering of phonons refers to scattering of wavepackets. The analysis works out by analyzing plane wave phonons because the frequency/wavenumber content of a gaussian wavepacket is centered around some w/k. This is why the quantity to look at would be group velocity.

I think the way to look at it is from Boltzmann equations where you have distributions of wavepackets.

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