The specific heat of solids II

Exercises

Preliminary provocations

1. Why are there only 3 polarizations when there are 6 degrees of freedom in three-dimensions for an oscillator?

   The polarization is related to the direction of the amplitudes of the waves with respect to the direction of the wave. In 3D, there are only 3 different amplitude directions possible.

2. Express the two-dimensional integral $\int\mathrm{d}k_x\mathrm{d}k_y$ in terms of polar coordinates. You can assume rotational symmetry.

   One can convert the integral as follows: $$\int k_x k_y \rightarrow \int_{0}^{2\pi} \mathrm{d} \theta \int_{0}^{\infty} k \mathrm{d} k = 2\pi \int_{0}^{\infty} k \mathrm{d} k$$

3. The Einstein model has a material-dependent frequency $\omega_0 = k_\mathrm{B} T_E/\hbar$ of the quantum harmonic oscillators as a free fitting parameter. What is the material-dependent parameter that plays a similar role in the Debye model?

   The Debye frequency $\omega_d$.

4. Derive an expression for the shortest possible wavelength in the Debye model it in terms of the interatomic distance $a$. Hint: assume that the number of atoms is given by $N=V/a^3$. Discuss if the answer is reasonable.

   From the definition of the Debye frequency, one can calculate that the wavelength is of the order of the interatomic spacing: $$\lambda = \left(\frac{4}{3}\pi\right)^{1/3} a.$$

Exercise 1: Debye model - concepts

Consider the probability to find an atom of a 1D solid that originally had a position $x$ at a displacement $\delta x$ shown below:

1. Describe which $k$-states are occupied. Explain your answer.

   It is clear that $n=4$, and thus $k = \frac{2 \pi n}{L} = \pm \frac{4\pi}{L}$.

2. Describe the concept of $k$-space. What momenta are allowed in a 2D system with dimensions $L\times L$?

   $k$ space is the space of possible momentum states. In 2 dimensions, a period system with dimensions $L\times L$ will have allowed states

   $$k_{x,y} = \frac{2\pi}{L} n_{x,y}$$

3. Explain the concept of density of states.

   The density of states $g(\omega)$ is the number of states per frequency

4. Calculate the density of states $g(\omega)$ for oscillations of a 3D, 2D and 1D solid with linear dispersion $\omega=v_s|\mathbf{k}|$.

   $$g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \frac{1}{v}\frac{dN}{dk}.$$

   We assume that in $d$ dimensions there are $d$ polarizations.

   For 1D we have that $N = \frac{L}{2\pi}\int_{-k}^{k} dk$, hence $g(\omega) = \frac{L}{\pi v}$.

   For 2D we have that $N = 2\left(\frac{L}{2\pi}\right)^2\int d^2k = 2\left(\frac{L}{2\pi}\right)^2\int 2\pi kdk$, hence $g(\omega) = \frac{L^2\omega}{\pi v^2}$.

   For 3D we have that $N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{L}{2\pi}\right)^3\int 4\pi k^2dk$, hence $g(\omega) = \frac{3L^3\omega^2}{2\pi^2v^3}$.

Exercise 2: Debye model in 2D

1. State the assumptions of the Debye model.

   The Debye model treats oscillations as sound waves with no fitting parameters

   • $\omega = v|k|$ similar to light (but with 3 polarisations)
   • Quantisation of waves similar to Planck quantisation of light
   • Maximum cut-off frequency which is necessary to obtain total of $3N$ degrees of freedom

2. Determine the energy of a two-dimensional solid as a function of $T$ using the Debye approximation. You do not have to solve the integral.

   $$\begin{align} E &= \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega \\ &= \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + C. \end{align}$$

3. Calculate the heat capacity in the high $T$ limit.

   High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + C$, and then $C = \frac{k_BL^2\omega^2_D}{2\pi v^2} = 2Nk_B$. We've used the value for $\omega_D$ calculated from $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.

4. At low $T$, show that $C_V=KT^{n}$. Find $n$. Express $K$ as an indefinite integral.

   In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + C = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + C$. Finally $C = \frac{6\zeta(3)k^3_BL^2}{\pi v^2\hbar^2}T^2$. We used the fact that $\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx = 2\zeta(3)$ where $\zeta$ is the Riemann zeta function.

Exercise 3: Different oscillation modes

(adapted from exercise 2.6a of The Oxford Solid State Basics)

During the lecture we derived the low-temperature heat capacity assuming that all the modes of oscillation have the same sound velocity $v$. In reality, the longitudinal and transverse modes have different sound velocities (see Wikipedia for an illustration of different sound wave types).

Assume that there are two types of excitations:

• One longitudinal mode with $\omega = v_\parallel |k|$
• Two transverse modes with $\omega = v_\bot |k|$

1. Write down the total energy of oscillations in this material.

   $$g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2$$ $$E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + C.$$

2. Verify that at high $T$ you reproduce the Dulong-Petit law.

   Note that we can get $\omega_D$ from $3N = \int_{0}^{\omega_D}g(\omega)$ so everything cancels as usual and we are left with the Dulong-Petit law $C = 3Nk_B$.

3. Compute the behaviour of heat capacity at low $T$.

   In the low temperature limit we have that $C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3$. We used that $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.

Exercise 4: Anisotropic sound velocities

(adapted from exercise 2.6b of The Oxford Solid State Basics)

Suppose now that the velocity is anisotropic ($v_x \neq v_y \neq v_z$) and $\omega = \sqrt{v_x^2 k_x^2 + v_y^2 k_y^2 + v_z^2 k_z^2}$. How does this change the Debye result for the heat capacity?

Hint

Write down the total energy as an integral over $k$, then change the integration variables so that the spherical symmetry of the integrand is restored.

$$\begin{align} E & = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) \\ & = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + C, \end{align}$$

where we used the substitutions $\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z$. Finally

$$E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + C = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + C,$$

hence $C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1}$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/v_xv_yv_z$ instead of $1/v^3$.

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Last update: October 30, 2023

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