I have phonon dispersion curves for a crystal with two atoms per unit cell. The following figure is phonon dispersion curves for the same crystal as above. $q_{BZ}$ denotes the wavevector at the BZ boundary. The direction of $q$ is antisymmetric, so the LA, LO branches are doubly degenerate. LA/TA, LO/TO denote longitudinal or transverse and acoustic and optical branches.

How would the temperature dependence of the lattice contribution to the heat capacity of this crystal be, in the low temperature limit and high temperature limit? The final page of this document shows the longitudinal and tranverse, acoustic and optical branches.

If we know that $$C_v = 3\frac{V}{(2\pi)^3} k_B 4\pi \int_{0}^{k_D} \frac{(\hbar \omega (k)/k_B T)^2e^{\hbar\omega (k)/k_B T}}{(e^{\hbar \omega (k)/k_B T}-1)^2}$$

What is the simplification for $\omega(k)$ that I use to solve for $C_v$?

enter image description here

  • $\begingroup$ You know that phonons obey Bose-statisttics and from the plots you can find the energy of phonon, $\omega=\omega(q)$. Then, from thermodynamics, you also know that $C\sim \partial^2U/\partial T^2$, where $U$ is the total internal energy. To calculate $U$, you should perform integration of $\omega(q)$ over $q$ with Bose-statistics. High-T and low-T limits can be found as the corresponding expansions of the integrand $\endgroup$ Sep 29, 2020 at 5:35
  • $\begingroup$ @ArtemAlexandrov, thanks for your reply. How do I find $\omega (q)$? $\endgroup$
    – megamence
    Sep 29, 2020 at 13:00
  • $\begingroup$ On your plot I can see these curves, $\omega=\omega(q)$ $\endgroup$ Sep 29, 2020 at 13:36
  • $\begingroup$ But I don't have any quantitative information to do anything mathematical with it, right? $\endgroup$
    – megamence
    Sep 29, 2020 at 16:08
  • $\begingroup$ At first glance, you can perform fitting of curves and then numerical differentiation (if the curves were obtained numerically/experimentally). From the other hand, it seems that the problem of 1D crystall with two different atoms per cell can be solved analytically $\endgroup$ Sep 29, 2020 at 18:37

2 Answers 2


Finally, as I understand the question is: how can one calculate the dispersion relation for diatomic 1D chain and then derive the heat capacity?

The quite detailed derivation of the dispersion relation can be find here (p. 8-9).

Having obtained the analytical expressions for the dispersion relation, we can compute the internal energy of our crystal,

$$U\propto\int\frac{d\omega\,\omega}{e^{\beta\omega}-1},$$ where, for simplicity, we can consider the limit of small momenta and expand $\omega(q)$ in terms of $q$. Then, perform change of variables, it seems possible to evaluate appearing integral. Finally, we use $$C\propto\frac{\partial^2U}{\partial T^2}$$ to obtain the desired answer.

  • $\begingroup$ I'm still confused. what is $\omega$ here? Do we consider both the acoustic and optical modes? $\endgroup$
    – megamence
    Sep 30, 2020 at 17:12
  • $\begingroup$ @megamence acoustic mode is the mode that goes to zero at $q=0$, optical mode goes to non-zero at $q=0$. $\endgroup$ Sep 30, 2020 at 17:56
  • $\begingroup$ yes. So do I write two separate integrals with those modes and sum them up? I appreciate all your advice! $\endgroup$
    – megamence
    Sep 30, 2020 at 18:49
  • $\begingroup$ @megamence in my opinion yes $\endgroup$ Sep 30, 2020 at 19:41

The simplest approximations might be all that is required:

  • the Einstein model for the optical branches
  • the Debye model for the acoustical branches

So estimate $\omega_E$ and $\omega_D$ from the dispersion curves and add their contributions to $c_v$.

  • $\begingroup$ How do you estimate $\omega _E$ and $\omega _D$ from these curves? $\endgroup$
    – megamence
    Sep 30, 2020 at 14:11
  • $\begingroup$ @megamence The Debye frequency is the cutoff of the acoustical branch at the Brillouin zone boundary. The Einstein frequency an average over the optical branches. If this is homework, you really need to study more in your book. $\endgroup$
    – user137289
    Sep 30, 2020 at 17:31

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