I read in Ashcroft and Mermin's Solid State Physics how only those modes that have a frequency tending to 0 as k(or q) tends to 0 contribute to specific heat capacity. This would include the 3 acoustic modes and some exceptions of optical modes. Can someone please explain this to me.
Acoustic modes are like the sound waves, which for low frequency follow omega= v k
v is speed of wave, v=c for light. k=2 pi/lambda, omega =2 pi/T
It means a lot of atoms moving together in the same direction.
An optical mode means for example every second atom moving opposite the first., omega is finite.
Hi Hitanshu
In thermal equilibrium, lattice vibrations modes or phonon modes will be thermally excited. So, phonon modes with energies less than or equal to kBT (kB being the Boltzmann's constant, and T the lattice temperature in Kelvin) will primarily contribute to the specific heat at a given temperature. This is because the mean number of phonons occupying these modes will be higher, in accordance with the Bose Einstein statistics.
The three "acoustic phonons" are called so, because in the long wavelength limit (wave vector k->0), they possess a linear dispersion relation with slope (also called the group velocity) corresponding to the speed of sound propagation in the solid. Because of the linear dispersion at k->0, these modes will have energy ->0.
So, at low temperature , it's very easy to thermally excite the three acoustic phonon modes, at least the modes which correspond to long wavelength (k->0).
However, all the the optical modes need not be excited, because the optical phonons, even at zero k, possess a non-zero energy. This energy must then be less than kBT at that particular temperature for the optical modes to be excited and to contribute to the specific heat.
Hope this helps.
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