Phonons propagate without problems in a lattice, until they scatter on something, like a defect, an electron, or another phonon. But in a typical solid at room temperature, how much (or how long) is the mean free path of a phonon? I know that depending on the temperature one type of scattering can be more probable than the other, but do mean free paths of these different types of scattering tend to be very different from each other? (With different orders of magnitude?)
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Phonons are lattice vibrations. The distance between two consecutive phonons is of the order of $1/N$ where $N$ is the number of atoms in the lattice. At room temperature a phonon travels approx 10 to 100 lattice constants before scattering.
In this article they say that a phonon travels $< 1 \mu m$ before scattering, that is approx 100 lattice constants (the lattice constant is approx 1 nm)
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$\begingroup$ "The distance between two consecutive phonons is of the order of 1/N". What is the unit of this estimate? Meters? $\endgroup$– LautronCommented Jan 4, 2020 at 6:43
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$\begingroup$ 1/m^3 for 3D (1/m for 1D) N is the number of the atoms in the lattice The distance between two phonons is approx the distance between 2 atoms $\endgroup$ Commented Jan 4, 2020 at 17:06
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That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths.
At lower temperatures, phonon mean free paths can be very long. Some researchers synthesized isotopically pure C-12 diamond with high thermal conductivities. The mean free path can be millimeters.
This is for high-frequency phonons that conduct most thermal energy. Phonons with longer wavelength have longer scattering lengths.