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When dealing with phonons and specific heat of solids, it seems the really important quantity to obtain is the density of states $N(\omega)$. When we have it, we can find the internal energy as

$$U(T)=\int N(\omega) \dfrac{\hbar \omega}{e^{-\hbar\omega/k_BT}-1}d\omega,$$

and having it we can also find the specific heat

$$C(T)=\dfrac{\partial U}{\partial T}.$$

Now, the way to find $N(\omega)$ is usually this: if the real lattice has primitive cell with volume $V$, the volume of the primitive cell of the $k$-space is $(2\pi)^3/V$. This means that, since there's just one point of the Bravais lattice per primitive cell, there are $V/(2\pi)^3$ points of the $k$-lattice per unit volume.

This in turns leads to integrals of the form

$$N(\omega)d\omega=\dfrac{V}{(2\pi)^3}\int d\mathbf{k},$$

where the integral is taken over the region between $\omega $ and $\omega+d\omega$, or equivalently

$$N(\omega)=\dfrac{V}{(2\pi)^3}\int\dfrac{dS}{|\nabla_\mathbf{k}\omega|},$$

where the integral is taken over the surface $\omega$.

This all is fine, given one dispersion relation $\omega(\mathbf{k})$ we can find $N(\omega)$ using those integrals, and with $N(\omega)$ we can find $U(T)$ and hence $C(T)$.

On the other hand, what if the basis of the Bravais lattice has more than one atom? For instance, a 2 atom basis?

This is quite common, but I'm failing to see how this affects all of this derivation. The naive guess would be that $N(\omega)$ would be multiplied by $2$, but this is just a guess. So, how the number of atoms in the basis affects this reasoning, and hence, the thermodynamic properties of a crystal, like specific heat?

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  • $\begingroup$ I dont think it depends on the number of atoms. It depends on the number of primitive cells considered inside a K-space volume. $\endgroup$ – Prasad Mani Oct 19 '16 at 0:55
  • $\begingroup$ So if the basis has two atoms or more, that is, in each primitive cell there's more than one atom, it won't affect the reuslts? $\endgroup$ – user1620696 Oct 19 '16 at 1:04
  • $\begingroup$ The density of states is found by differentiating the number of allowed modes(of vibration) 'N' with wave vector $<K$ (in $3D$ by $N=$ $\frac{V}{2\pi^3}$.$\frac{4\pi K^3}{3}$)...this automatically accounts for everything. I may be wrong but the statement 'number of allowed modes of vibration ie the number of phonon modes does depend on the number of atoms but when we translate to the K space, the unit cell(along with the number of atoms in it) gets translated accordingly so that you dont have to factor for it separately. Again, I may be wrong. Good question though. $\endgroup$ – Prasad Mani Oct 19 '16 at 1:38
  • $\begingroup$ Just a side note here; in the quantum free electron model, if there is a contribution of more than one electron by the atom in consideration, we do multiply with the number of conduction electrons accordingly when finding density of states! $\endgroup$ – Prasad Mani Oct 19 '16 at 1:43
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I am going to use debye model to clarify as far as i can your doubt.

Debye assumed that the number of modes of vibration in a crystalline solid is limited to $3N$ , the number of translational degrees of freedom of $N$ atoms(see how it conviniently overlooks the number of atoms in the primitive lattice and all other details about it), to account for the actual atomic nature of a crystalline solid. There is a minimum wavelength in the problem set by the spacing between atoms. It is not possible for sound waves to propagate through a solid with wavelength smaller than the atomic spacing because there's nothing in the middle there to shake. The allowed modes varied in frequency then from zero to some maximum frequency. To get $\omega_D$, Debye set (i am using $g$ instead of your $N$ to denote density of states.)

$\int \limits_{0}^{\omega_D} g(\omega) d\omega$ = $3N$

Now i hope you know that $g(\omega)$ = $\frac{3V}{2\pi^2v^3}$$\omega^2$ for phonon modes (3 in the numerator accounts for two transverse and one longitudinal polartization and $v$ in the denominator is the speed of sound).

To summarize, we just look at the whole solid crystal and count the number of atoms in it (apparently) - dont care about the primitive unit cell or the number of atoms it contains - and then just do the above integral where we actually begin to account for the number of atoms in the crystal (and hence still not technically in the primitive cell; we just gloss over it)

I realize your question was whether the number of atoms in the primitive cell somehow affects the density of states and the answer is no. Because we are counting the number of PHONONS per unit frequency range $\omega$ to $\omega$+$d\omega$ and not atoms or anything else

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