7.5: Low-temperature Heat Capacity

If

$$G(\omega) d \omega=\text { number of normal modes with frequencies from } \omega \text { to } \omega+d \omega$$

then

$$E^{\mathrm{crystal}}=\int_{0}^{\infty} G(\omega) e^{\mathrm{SHO}}(\omega) d \omega \quad \text { and } \quad C_{V}^{\mathrm{crystal}}=\int_{0}^{\infty} G(\omega) c_{V}^{\mathrm{SHO}}(\omega) d \omega$$

and so forth.

Density of modes:

$$\begin{aligned} G(\omega) d \omega &=\sum_{\text { branches }}[\text { vol. of shell in } k \text { -space }] \text { (density of modes in } k \text { -space) } \\ &=\sum_{\text { branches }}\left[4 \pi\left(k_{b}(\omega)\right)^{2} d k_{b}\right]\left(\frac{V}{8 \pi^{3}}\right) \end{aligned}$$

This formula holds for any isotropic dispersion relation k_b(ω). For small values of ω the dispersion relation for each branch is linear (with sound speed c_b) so

$$k_{b}=\frac{\omega}{c_{b}} \quad \text { and } \quad d k_{b}=\frac{d \omega}{c_{b}},$$

whence

$G(\omega) d \omega=\sum_{\text { branches }}\left[4 \pi\left(\frac{\omega}{c_{b}}\right)^{2} \frac{d \omega}{c_{b}}\right]\left(\frac{V}{8 \pi^{3}}\right)$

$$=\frac{V}{2 \pi^{2}}\left(\sum_{b=1}^{3} \frac{1}{c_{b}^{3}}\right) \omega^{2} d \omega.$$

If we define the “average sound speed” c_s through the so-called “harmonic cubed average”,

$$\frac{1}{c_{s}^{3}} \equiv \frac{1}{3} \sum_{b=1}^{3} \frac{1}{c_{b}^{3}},$$

then we have the small-ω density of modes

$$G(\omega) d \omega=\frac{3 V}{2 \pi^{2}} \frac{\omega^{2}}{c_{s}^{3}} d \omega.$$

At any temperature,

$$C_{V}^{\text { crystal }}=\int_{0}^{\infty} G(\omega) c_{V}^{\mathrm{SHO}}(\omega) d \omega.$$

Recall from equation (5.78) that

$$c_{V}^{\mathrm{SHO}}(\omega)=k_{B}\left(\frac{\hbar \omega}{k_{B} T}\right)^{2} \frac{e^{-\hbar \omega / k_{B} T}}{\left(1-e^{-h \omega / k_{B} T}\right)^{2}},$$

and using the small-ω result (7.11), we have the low-temperature result

$$C_{V}^{\mathrm{crystal}}=\frac{3 V}{2 \pi^{2}} \frac{1}{c_{s}^{3}} k_{B} \int_{0}^{\infty} \omega^{2} d \omega\left(\frac{\hbar \omega}{k_{B} T}\right)^{2} \frac{e^{-\hbar \omega / k_{B} T}}{\left(1-e^{-\hbar \omega / k_{B} T}\right)^{2}}.$$

For our first step, avoid despair — instead convert to the dimensionless variable

$$x=\frac{\hbar \omega}{k_{B} T}$$

and find

$$C_{V}^{\text { crystal }}=\frac{3 V}{2 \pi^{2}} \frac{1}{c_{s}^{3}} k_{B}\left(\frac{k_{B} T}{\hbar}\right)^{3} \int_{0}^{\infty} \frac{x^{4} e^{-x}}{\left(1-e^{-x}\right)^{2}} d x$$

The integral is rather hard to do, but we don’t need to do it — the integral is just a number. We have achieved our aim, namely to show that at low temperatures, C_V ∼ T³.

However, if you want to chase down the right numbers, after some fiddling you’ll find that

$$\int_{0}^{\infty} \frac{x^{4} e^{-x}}{\left(1-e^{-x}\right)^{2}} d x=4 \Gamma(4) \zeta(4)=\frac{4}{15} \pi^{4}.$$

Thus, the low-temperature specific-heat of a solid due to a lattice vibration is

$$C_{V}^{\mathrm{crystal}}=k_{B} V \frac{2 \pi^{2}}{5}\left(\frac{k_{B} T}{\hbar c_{s}}\right)^{3}.$$

7.3 How far do the atoms vibrate?

Consider a simplified classical Einstein model in which N atoms, each of mass m, move classically on a simple cubic lattice with nearest neighbor separation of a. Each atom is bound to its lattice site by a spring of spring constant K, and all the values of K are the same. At temperature T, what is the root mean square average distance of each atom from its equilibrium site? (Note: I am asking for an ensemble average, not a time average.)