Heat or thermal energy as understood is nothing but motion of molecules of the matter. If the molecules are tightly bound (in case of solids), it is to-and-fro molecular vibrations, otherwise it is continuous random motion of molecules (in case of liquids/gases/plasma).

Sound, being waves, is also vibrations in matter. Why, then, if we heat one end of a solid rod, assuming rod is at least few feet in length, does it take ages for the heat to reach the other end, whereas sound reaches in no time ? (sound travels at 1400 m/s in solid)

Doesn't it show that heat is more an intra-atomic feature rather than an atomic or molecular motion? Observation that electrical good conductors are also good conductors for heat, can we assume that heat is chaotic motion of electrons (the "electron gas")? (Of course, all the known phenomena related to heat will be needed to be explained with this hypothesis.)


The analogy is a very good one, because heat transfer is in fact modelled by phonons, which you could also use to describe sound waves.

The crucial difference is that sound waves have a much longer wavelength (at least in the range of some millimetres) than thermal phonons (not more than a few orders of magnitude bigger than the atomic lattice scale). These small-wavelength phonons can easily scatter at any lattice impurities, while the sound waves need macroscopic pertubations (like air gaps in an insulated glazing) to do so.

  • $\begingroup$ I don't really know about phonons, but I see a fundamental problem here- why should all vibrations associate with a wavelength? All waves are vibrations, but all vibrations are not necessarily waves. $\endgroup$ – pongapundit Jul 3 '11 at 5:44
  • $\begingroup$ Yes they are. You can fourier-expand any vibration into frequency components, and each of these frequencies corresponds to some wavelength. $\endgroup$ – leftaroundabout Jul 3 '11 at 10:04
  • $\begingroup$ Very enlightening! $\endgroup$ – Richard Terrett Jul 5 '11 at 6:58

leftaroundabout gave an excellent explanation for the thermal conduction of insulators. However, in the case of metals, a significant amount of energy is carried by the excitations of electrons (the width of their Fermi-Dirac distribution). The thermal conductivity is then related to how far an excited electron can travel before being scattered, and is therefore related to the electrical conductivity. In most metals, the electrons will have a greater contribution to the thermal conductivity than the phonons.

  • $\begingroup$ A very important point! $\endgroup$ – leftaroundabout Jul 4 '11 at 12:29

Here is a numerical simulation of heat transfer in a thread which is only one atom thick, each number is an atom, 8 is an atom with 8 units of kinetic energy, and so on:

First left side has more heat energy:

8 0 0 0

After one collision between neighbor atoms:

4 4 0 0

After one more collision between neighbor atoms:

4 2 2 0

After one more collision between neighbor atoms:

3 3 1 1

As you see, there is some heat traveling at speed of sound. At the beginning of the simulation, when there was a heat difference of 8 between neighbor atoms, large fraction of heat energy traveled at speed of sound.

(Smell does not travel at speed of sound. If some smell was transferred when molecules touch, then smell would travel like heat energy)

  • $\begingroup$ contrast that to sound traveling: 8000, then 0800, then 0080, and finally 0008 (ignoring minor volume loss) $\endgroup$ – Dale Jul 5 '11 at 3:14
  • $\begingroup$ -1 as this simulation does not really address the question. You are basically solving a discretized diffusion equation. The real question is why does heat obey a diffusion equation while sound obeys a wave equation, given that both have the same physical origin, at least for the phonon contribution to thermal conductivity. $\endgroup$ – Edgar Bonet Jul 5 '11 at 8:29
  • $\begingroup$ (continued) Also, your picture is wrong at the atomic scale, as thermal phonons propagate like waves. This is because vibrations in a solid are a collective motion. They are not independent atomic vibrations, with only random energy transfers between neighbor atoms, as your simulation suggests. pongapundit understood this issue, and that's why he is posting this question. The diffusion model only becomes valid at scales larger than the phonon mean free path, and leftaroundabout explained the physics quite clearly. $\endgroup$ – Edgar Bonet Jul 5 '11 at 8:30
  • $\begingroup$ Well my view is that conduction of heat in solids resembles conduction of heat in gases, much more than conduction of heat in solids resembles propagation of sound in gases or solids. $\endgroup$ – jartza Jul 7 '11 at 7:11
  • $\begingroup$ At the macroscopic scale you are perfectly right! But at the atomic scale, solids and gases are quite different. Atoms in a gas move more or less independently from one another, whereas in a solid they have collective motions. The difference between sound and heat in solids is then more a matter of whether you look at scales shorter or larger than the phonon mean free path. So your simulation is OK provided you realize that the “simulation cell” has to be larger than this mean free path. Now, the original question was explicitly addressing the atomic scale. $\endgroup$ – Edgar Bonet Jul 7 '11 at 9:29

I feel that when a physicist speak about Heat he/she has a flow of energy in mind. Suppose that you have a rod and that the two extrema are held at different temperatures. Then the Fourier law states that there must be a flow of Heat from the hotter end to the cooler. When a physicist speak, instead, of the molecular motion he/she is thinking to the internal energy of the body.

Now when molecules and atoms are involved, it more likely that we must enter into the quantum world. By the way, we can make some few heuristic semiclassical consideration, namely we can apply Boltzmann statistics to the quantum structure of atomic and molecular spectra. A body that is immersed in a certain environment will be in thermal equilibrium state. Atoms and molecules receives energy from the thermal bath, but they also radiate energy in such a way the the total balance is "no energy exchange", therefore no energy flow, i.e. no heat flow.

We must note though that when we deal with atomic or molecular excitation levels, we are considering relatively tiny amounts of energy. Take as a reference the binding energy of the electron in the hydrogen atom, this being roughly 13.6 eV. Sounds excitations involves way more energy than this and in this case you can forget that the body has a quantum nature. You can treat it as a continuum and apply the laws of classical mechanics, i.e. the theory of elasticity and forth.

  • $\begingroup$ It's more like the other way around: sounds excitations involves way less energy. You don't get around $E=\hbar\omega$ there, which means that a $440\:\mathrm{Hz}$ has an energy of only $1.8\cdot10^{-12}\:\mathrm{eV}$! [wolframalpha.com/input/?i=440Hz*2pi+hbar+in+eV] $\endgroup$ – leftaroundabout Jul 2 '11 at 20:39
  • $\begingroup$ What I was trying to say is that you have to hit a rod of steel with a hammer to produce an audible sound, and this is way more than just a few eV. In this regime you can surely apply classical mechanics. $\endgroup$ – Phoenix87 Jul 3 '11 at 8:09
  • $\begingroup$ The hammer may have a noteworthy energy, but without sufficient frequency this cannot be used for any kind of excitation at all. Consider an asteriod far from the sun: this has plenty of kinetic energy WRT the solar system, but the frequency is almost zero, so no excitation whatsoever can take place. $\endgroup$ – leftaroundabout Jul 3 '11 at 10:09
  • $\begingroup$ I'm sorry but I don't really get your comment. What do you mean by "sufficient frequency"? Also I don't get the asteroid example. If the asteroid is travelling at constant speed w.r.t the remote stars than there must be a frame of reference where the asteroid is not moving... $\endgroup$ – Phoenix87 Jul 3 '11 at 16:16
  • $\begingroup$ Exactly. Such a frame of reference only exists for a "sufficiently constant" motion, that is, for one with a low enough frequency. $\endgroup$ – leftaroundabout Jul 3 '11 at 17:49

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