How come electrons and phonons lead to the heat equation?

  • #1
fluidistic
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Main Question or Discussion Point

Phonons on their own lead to the common heat equation. One sees that for example in insulators or non doped semiconductors.
However in metals (or conductors), the electrons are the ones that are mostly responsible for the heat transfer, which extremely surprisingly to me, is also of the form of the common heat equation.
In heavily doped semiconductors, both electrons and phonons play a major role on the thermal transfer, and again the heat equation applies.

So... knowing that the electrons responsible for the heat transfer in a metal move at very high speeds (Fermi velocity), while phonons have a speed about 3 order of magnitude lower than electron's, and also that they are quite different from each other, how come they both produce the same heat equation?
 

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  • #3
Henryk
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The heat equation derivation requires that the heat flux is proportional to the temperature gradient. That's all. It is a mathematical formula and as long as the proportionality between heat flux and the temperature gradient is maintained, the equation is valid. It does not depend on the mechanism of the heat transfer.
Yes, good electrical conductors are good thermal conductors (Wiedemann-Frantz law) and your find out that for good heat conduction copper or aluminum is chosen (two of the best electrical conductors). But phonon thermal conductivity can be large too. Actually, at around room temperature, the heat conductivity of diamond (a very good insulator) exceeds that of copper.
 
  • #4
fluidistic
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Not sure if this will help but I found this discussion on Physics Stack Exchange on deriving the heat equations and the assumptions made:

https://physics.stackexchange.com/q...quation-arise-from-phonons-and-from-electrons

From there perhaps you can draw your own conclusions as to why they generate the same heat equation.
I am the one who asked that question on that website. I asked it here too (formulated in a different way) because PSE is not a forum and does not encourage discussion. Comments are "2nd class citizens" which get deleted without any warning and without leaving any trace. I still did not manage to get any satisfying answer.
 
  • #5
ZapperZ
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I must say that I don't quite understand the issue here.

As @Henryk has stated, the phenomenological equation of heat transfer (if this is what you meant by "heat equation") depends on the temperature gradient.

Now, if it turns out that both metals and insulators end up with the same thermal conductivity coefficient, then I would be puzzled. However, they obviously don't, and they certainly differ from one another by quite a bit. Because of this difference, I don't quite get the issue you're having here. It is, after all, a phenomenological model that describes the phenomenon after-the-fact, where all the details are lumped and hidden inside the coefficient.

Zz.
 
  • #6
fluidistic
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The heat equation derivation requires that the heat flux is proportional to the temperature gradient. That's all. It is a mathematical formula and as long as the proportionality between heat flux and the temperature gradient is maintained, the equation is valid. It does not depend on the mechanism of the heat transfer.
Yes, good electrical conductors are good thermal conductors (Wiedemann-Frantz law) and your find out that for good heat conduction copper or aluminum is chosen (two of the best electrical conductors). But phonon thermal conductivity can be large too. Actually, at around room temperature, the heat conductivity of diamond (a very good insulator) exceeds that of copper.
I don't think this answers the question. Using your terms, it could be formulated as "why do phonons and electrons lead both lead to a heat flux proportional to the temperature gradient?" Answering by "the equation is valid, it does not depend on the mechanism of the heat transfer" is just not given any real answer, and it is also wrong. It's wrong because there exist ballistic heat transfers that arise from electrons, and they do not lead to a diffusive heat equation. They lead to a wavelike heat equation. ref: https://arxiv.org/abs/1306.4972. So there certainly is something that makes electrons (and phonons) creating a diffusive heat equation in a solid. It's not a given, and I don't know what it is. Strangely, both phonons and electrons seem to yield it, unless we go to ballistic and other "weird" stuff.

I also know about the W-F law and its rough validity for some metals. Cu and Al are chosen because of rather high ##\kappa## and low price compared to better materials like diamond as you point out (or silver). There are also some oxydes (ceramics) that beat Al's thermal conductivity thanks to phonons. But I don't see how this helps answering the question.
 
  • #7
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I think you still didn’t understand henryks’s response he said the heat flux was dependent on the temperature gradient. Whatever mechanism produces the temperature gradient produces the heat flux. It doesn’t depend on the mechanism.

Try looking at the equation using Henryk’s description. Often we ascribe too much info or assume more than we need to and it makes our understanding confused.
 
  • #8
fluidistic
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I think you still didn’t understand henryks’s response he said the heat flux was dependent on the temperature gradient. Whatever mechanism produces the temperature gradient produces the heat flux. It doesn’t depend on the mechanism.

Try looking at the equation using Henryk’s description. Often we ascribe too much info or assume more than we need to and it makes our understanding confused.
I'm still not sure I understand him then. He says that if ##\nabla T \propto \vec q## then the heat eq. is valid, no matter the underneath mechanism. Ok. But why would electrons and phonons produce ##\nabla T \propto \vec q## sometimes, and other times not (for example in ballistic heat transfer)?
 
  • #9
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The two heat transfer schemes are different. It seems that ballistic heat transfer occurs in a medium with different properties of less resistivity which allows for more quantum mechanical effects whereas the normal heat equation is predicated on Newtonian effects and some resistivity and scattering.

https://en.m.wikipedia.org/wiki/Ballistic_conduction
 
  • #10
Orodruin
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I'm still not sure I understand him then. He says that if ##\nabla T \propto \vec q## then the heat eq. is valid, no matter the underneath mechanism. Ok. But why would electrons and phonons produce ##\nabla T \propto \vec q## sometimes, and other times not (for example in ballistic heat transfer)?
In order to understand this you need to look at the underlying mechanism and the mean free path of the carriers (be it phonons or electrons). If you do not have scattering, you end up with something that is more akin to radiative loss. The following is very heuristic: If you have scattering, you essentially end up with transferring heat in a random direction with a typical size depending on the temperature, which leads to a net transfer in directions with lower temperature (the likelihood of transferring heat between two points is the same, but the hotter point is going to transfer more heat per transfer).
 
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  • #11
fluidistic
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In order to understand this you need to look at the underlying mechanism and the mean free path of the carriers (be it phonons or electrons). If you do not have scattering, you end up with something that is more akin to radiative loss. The following is very heuristic: If you have scattering, you essentially end up with transferring heat in a random direction with a typical size depending on the temperature, which leads to a net transfer in directions with lower temperature (the likelihood of transferring heat between two points is the same, but the hotter point is going to transfer more heat per transfer).
Thanks a lot, that was very informative. If I understand well, the heat equation arises from both phonons and electrons because they both share many properties, in particular scattering processes.

Now I wonder why phonons do not cause electricity. Since they're so similar to electrons, leading to the same heat equation (though with a different contribution of course, i.e. a different kappa), how come they do not lead to charge transfer? There is the electron-phonon interaction, scattering, etc... and yet there is no net charge transfer due to phonons when we apply a temperature gradient? How is that possible?
I am well aware that there is a net charge transfer when we apply a temperature gradient, but that's due to electrons only, not to phonons. Why is it so?
 
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Orodruin
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Electrons have electric charge, phonons do not.
 
  • #13
fluidistic
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Electrons have electric charge, phonons do not.
I know this, but I would expect the electron-phonon interaction to cause a motion of electrons. So, if we apply a heat gradient in a material, I would expect the phonons to "push" somehow the electrons via the phonon-electron interaction, causing a voltage (and a short lived current if the material is an open circuit). Instead the electrons would move because the chemical potential is different across the material, regardless of the existence of phonons.
 
  • #14
ZapperZ
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I know this, but I would expect the electron-phonon interaction to cause a motion of electrons. So, if we apply a heat gradient in a material, I would expect the phonons to "push" somehow the electrons via the phonon-electron interaction, causing a voltage (and a short lived current if the material is an open circuit). Instead the electrons would move because the chemical potential is different across the material, regardless of the existence of phonons.
Electron-phonon interactions do affect charge transport, but in the opposite way, i.e. in the resistivity of the material. You cannot expect these random scatter to have a preferred direction, much like electron-electron or electron-impurity scattering. They all add to the resistivity.

Phonons are not a mechanism for charge transport.

Zz.
 
  • #15
DrDu
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I know this, but I would expect the electron-phonon interaction to cause a motion of electrons. So, if we apply a heat gradient in a material, I would expect the phonons to "push" somehow the electrons via the phonon-electron interaction, causing a voltage (and a short lived current if the material is an open circuit). Instead the electrons would move because the chemical potential is different across the material, regardless of the existence of phonons.
Thermoelectric effects like the Seebeck effect exist and the chemical potential will also change due to the presence of phonons. So I guess that you are right here.
 
  • #16
fluidistic
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I know this, but I would expect the electron-phonon interaction to cause a motion of electrons. So, if we apply a heat gradient in a material, I would expect the phonons to "push" somehow the electrons via the phonon-electron interaction, causing a voltage (and a short lived current if the material is an open circuit)
Well, from what I could gather during the last few days, it seems I wasn't wrong after all. Here's Wikipedia:
Wiki the Magnificient said:
If the phonon-electron interaction is predominant, the phonons will tend to push the electrons to one end of the material, losing momentum in the process. This contributes to the already present thermoelectric field. This contribution is most important in the temperature region where phonon-electron scattering is predominant.
from the phonon drag page. I've seen a similar explanation given in an old (1957 or so) book but I wasn't sure whether this was correct until I read Wikipedia.
 

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