1
$\begingroup$

Let's say I determine a material's dispersion relation from experiments. Would it then be possible to use the information contained within the dispersion relation to calculate the material's static lattice energy?

$\endgroup$
1
$\begingroup$

No, it is impossible. The reason is very simple if one recalls the meaning of phonon dispersion curves. They give the set of possible frequencies, $\omega_s({\bf k})$, for the eigenmodes of the $s-$th dispersion branch.

The energy density of a harmonic crystal, at a temperature $T$ is: $$ u = u_{static} + \frac{1}{V}\sum_{{\bf k},s} \frac{1}{2} \hbar \omega_s({\bf k}) + \frac{1}{V}\sum_{{\bf k},s} \frac{ \hbar \omega_s({\bf k})}{e^{ \frac{ \hbar \omega_s({\bf k})}{k_B T}}-1}, $$ where the second term is independent on $T$ and corresponds to the so-called zero-point contribution. $u_{static}$ is the energy of a (static) perfect lattice.

It is clear that the whole information on phonons is contained in the second and third terms. The first term is the (static) potential energy of a set of particles at the equilibrium positions in the lattice. Even though both $u_{static}$ and $ \omega_s({\bf k}) $ are obtained from the same potential energy as a function of the atomic positions, they contain different information and one cannot obtain one of them from knowledge of the other.

$\endgroup$
3
  • $\begingroup$ Very clear explanation, thank you. It makes sense. Could you make the following approximation though? From the dispersion relation, you know the highest frequency and its associated wave vector. From the highest frequency + the mass of the atom, we can back calculate a spring constant, and from the wave vector we know the interatomic spacing. Let's say we assume cubic (or some isotropic lattice), then would the static energy not just be the sum over 1/2 * k * spacing^2 of nearest neighbors? $\endgroup$ Jan 10 '20 at 18:58
  • 1
    $\begingroup$ @DrewLilley The are too many problems with your proposed approximation. I'll list them in increasing order of importance. The first problem is that the highest frequency could be at k=0 (the usual case if there are optical branches). More important, your proposed formula has only the right physical dimension. It is true that you can get a rough estimate of elastic force constants. But there are two major problems: 1) the average displacement is not of the order of the inter-atomic spacing. In a solid close to its melting point, ... $\endgroup$
    – GiorgioP
    Jan 10 '20 at 21:16
  • 1
    $\begingroup$ ....this quantity is of the order of one tenth of nearest neighbors distance. 2) (the most important objection) using elastic constant and a displacement means that the resulting energy is not the energy of the static lattice with atoms at their equilibrium positions (which is the required quantity). $\endgroup$
    – GiorgioP
    Jan 10 '20 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.