1. For low T, $1/T \rightarrow \infty$. The heat capacity is then given as:
$$
C \overset{\mathrm{low \: T}}{\approx} 9Nk_{\mathrm{B}}\left(\frac{T}{T_{D}}\right)^3\int_0^{\infty}\frac{x^4{\mathrm{e}}^x}{({\mathrm{e}}^x-1)^2}{\mathrm{d}}x.
$$
2. See plot below (shown for $T_{D,1} < T_{D,2}$)
3. The polarization describes the direction of the motion of the atoms in the wave with respect to the direction in which the wave travels. In 3D, there are only 3 different polarizations possible.
4. The integral can be expressed as
4. The integral can be expressed as
$$
\int dk_x dk_y = \int k dk d\phi
$$
...
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@@ -68,18 +68,19 @@ ax.set_yticklabels(['$0$'])
ax.legend();
```
### Exercise 1: Deriving the density of states for the linear dispersion relation of the Debye model.
### Exercise 1: Deriving the density of states for the linear dispersion relation of the Debye model
1. $\omega = v_s|\mathbf{k}|$
2. The distance between nearest-neighbour points in $\mathbf{k}$-space is $2\pi/L$. The density of $\mathbf{k}$-points in 1, 2, and 3 dimensions is $L/(2\pi)$, $L^2/(2\pi)^2$, and $L^3/(2\pi)^3$ respectively.
3. Express the number of states between frequencies $0<\omega<\omega_0$ as an integral over k-space. Do so for 1D, 2D and 3D. Do not forget the possible polarizations. We assume that in $d$ dimensions there are $d_p$ polarizations.
2. The high-$T$ limit implies $\beta \rightarrow 0$. Therefore, $n_B \approx k_B T/\hbar\omega$, and the integral becomes particularly illuminating:
$$
E = \int_{0}^{\omega_D} \hbar\omega n_B(\omega) g(\omega) d\omega \approx \int_{0}^{\omega_D} k_B T g(\omega) d\omega = N_\text{modes} k_B T
$$
where we neglected the zero-point energy. In 2D, we have $N_\text{modes} = 2_p N_\text{atoms}$, so that we recover the 2D law of Dulong–Petit $C_v = dE/dT = 2 k_B$ per atom.
3. In the low temperature limit, the high-energy modes are not excited so we can safely let the upper boundary of the integral go to infinity. For convenience, we write $g(\omega) = \alpha \omega$, with $\alpha = \frac{L^2}{\pi v_s^2}$. We get
$$
E = \int_0^{\omega_D}\frac{\hbar\omega g(\omega)}{e^{\hbar\omega/k_B T}- 1}d\omega \approx \alpha\frac{k^3 T^3}{\hbar^2}\int_0^\infty\frac{x^2}{e^x-1}dx
3. In the low temperature limit, the high-energy modes are not excited so we can safely let the upper boundary of the integral go to infinity. For convenience, we write $g(\omega) = \alpha \omega$, with $\alpha = \frac{L^2}{\pi v_s^2}$. We get
$$
From which we find $C_v = dE/dT = K T^2$, with
E = \int_0^{\omega_D}\frac{\hbar\omega g(\omega)}{e^{\hbar\omega/k_B T}- 1}d\omega \approx \alpha\frac{k^3 T^3}{\hbar^2}\int_0^\infty\frac{x^2}{e^x-1}dx
$$
K = 3\alpha\frac{k^3}{\hbar^2}\int_0^\infty\frac{x^2}{e^x-1}dx
From which we find $C_v = dE/dT = K T^2$, with
$$
K = 3\alpha\frac{k^3}{\hbar^2}\int_0^\infty\frac{x^2}{e^x-1}dx
$$
### Exercise 3: Longitudinal and transverse vibrations with different sound velocities
1. The key idea is that the total energy in the individual harmonic oscillators (the vibrational modes) is the sum of the energies in the individual oscillators: $E = \int_0^{\omega_D}\frac{\hbar \omega g(\omega)}{e^{\beta\hbar\omega} - 1}d\omega + E_Z$, where $g(\omega) = g_\parallel(\omega) + g_\perp(\omega)$. Using
### Exercise 3: Longitudinal and transverse vibrations with different sound velocities
1. The key idea is that the total energy in the individual harmonic oscillators (the vibrational modes) is the sum of the energies in the individual oscillators: $E = \int_0^{\omega_D}\frac{\hbar \omega g(\omega)}{e^{\beta\hbar\omega} - 1}d\omega + E_Z$, where $g(\omega) = g_\parallel(\omega) + g_\perp(\omega)$. Using
E = \frac{L^3 k_B^4 T^4}{2\pi^2\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1} {v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx.
$$
2. As in exercise 1, in the high-T limit, we have $\beta \rightarrow 0. Therefore, $n_B \approx k_B T/\hbar\omega$ and the integral for $E$ becomes:
2. As in exercise 1, in the high-T limit, we have $\beta \rightarrow 0$. Therefore, $n_B \approx k_B T/\hbar\omega$ and the integral for $E$ becomes:
$$
E = \int_{0}^{\omega_D} \hbar\omega n_B(\omega) g(\omega) d\omega \approx \int_{0}^{\omega_D} k_B T g(\omega) d\omega +E_Z = N_\text{modes} k_B T + E_Z
$$
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...
@@ -144,18 +145,17 @@ ax.legend();
and we are left with the Dulong-Petit law $C = 3N_\text{atoms} k_B$.
3. In the low temperature limit, we can let the upper integral boundary go to infinity as in exercise 1. This yields
$$
C \approx \frac{2\pi^2 k_B^4 L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3
$$
where we used $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
### Exercise 4: Anisotropic sound velocities
### Exercise 4: Anisotropic sound velocities.
In this case, the velocity depends on the direction. Note however that, in contrast with the previous exercise, the polarization does not affect the dispersion of the waves. We get
Now, we can make the substitution $x = \beta\hbar\kappa$, which gives $d\kappa = \frac{dx}{\beta\hbar}$ and changes our limits to $\int_{0}^{\beta\hbar\kappa_D}$:
Therefore, $C = \frac{dE}{dT} = \frac{6k_B^4 L^3 T^3}{\pi^2\hbar^3}\frac{1}{v_x v_y v_z}\int_0^{\beta\hbar\kappa_D} \frac{x^3}{e^x - 1}dx$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/(v_x v_y v_z)$ instead of $1/v^3$.
This is similar to the result with isotropic linear dispersion, with the difference being the factor $1/(v_x v_y v_z)$ instead of $1/v^3$.
