Update docs/2_debye_model_solutions.md

docs/2_debye_model_solutions.md

@@ -112,53 +112,56 @@ Here, we analyze the phonon energy and heat capacity of a two-dimensional Debye
 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 
     
     $$ 
-    E = \int_0^\omega_D\frac{\hbar\omega g(\omega)}{e^{\hbar\omega/k_B T}- 1}d\omega = \alpha\frac{k^3T^3}{\hbar^2}\int_)^\infty\frac{x^2}{e^x-1}dx 
+    E = \int_0^\omega_D\frac{\hbar\omega g(\omega)}{e^{\hbar\omega/k_B T}- 1}d\omega \approx \alpha\frac{k^3T^3}{\hbar^2}\int_0)^\infty\frac{x^2}{e^x-1}dx 
     $$
-    From which we find  $C_v = dE/dT =K T^2$ with 
+    
+    From which we find  $C_v = dE/dT = K T^2$, with 
     
     $$
     K = 3\alpha\frac{k^3}{\hbar^2}\int_)^\infty\frac{x^2}{e^x-1}dx 
     $$
 
 ###  Exercise 3: Longitudinal and transverse vibrations with different sounds 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:
-Neglecting the zero-point energy, the energy in the longitudinal modes is given by
+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: $E = \int_0^{\omega_D}\frac{\hbar \omega g(\omega)}{e^{\beta\hbar\omega} - 1}d\omega + E_Z$. Using $g(\omega) = g_\parallel(\omega) + g_perp(\omega)$ with
     
-    $$ 
-    E_\parallel = \int_0^{\omega_D}g_\parallel(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} )d\omega
-    $$
+    \begin{align}
+    g_\parallel & = 1_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\parallel^3}, \\
+    g_\perp = 2_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\perp^3},
+    \end{align}
 
-    with $g_\parallel = 1_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\parallel^3} $. Similarly, the energy in the transverse modes is 
+    we get 
     
     $$ 
-    E_\perp = \int_0^{\omega_D}g_\perp(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} )d\omega
+    E = \frac{L^3 k_B^4 T^4}{2\pi^2\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}   {v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx.
     $$
 
-    with $g_\perp = 2_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\perp^3} $. Therefore, the total energy is 
+
+2. As in exercise 1, in the high-T limit, we have $\beta \rightarrow 0$. Therefore, $n_B \approx = k_BT/\hbar\omega$ and the integral for $E$ becomes: 
     
     $$
-    E = \frac{L^3}{2\pi^2} \left(\frac{1_p}{v_\parallel} + \frac{2_p}{v\perp} \right) \int_0^{\omega_D}  \frac{\hbar\omega^3}{e^{\beta\hbar\omega} - 1} d\omega = \frac{L^3 k_B^4 T^4}{2\pi^2\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}   {v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx.
+    E = \int_0^\omega_D \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_0^\omega_D k_B T g(\omega) d\omega +E_Z = N_\text{modes} k_B T + E_Z
     $$
 
+    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
     
-2. As in exercise 1, in the high-T limit, we have $\beta \rightarrow 0$. Therefore, $n_B \approx = k_BT/\hbar\omega$. The integral becomes: 
     $$
-    E = \int_0^\omega_D \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_0^\omega_D k_B T g(\omega) d\omega = N_\text{modes} k_B T
+    C \approx \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3$, where we used that $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}.
     $$
-    and we are left with the Dulong-Petit law $C = 3N_\text{atoms}k_B$.
 
-3. In the low temperature limit we have $C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3$. We used that $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
 
 ### Exercise 4: Anisotropic sound velocities.
+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
     
     $$
-E = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + C,
+    E = 3_p\left(\frac{L}{2\pi}\right)^3\int \hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) d\mathbf{k} = 3_p \left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int \frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1}d\mathbf{\kappa} + E_Z,
     $$
 
     where we used the substitutions $\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z$. Finally
 
     $$
-E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + C = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + C,
+    E = \frac{3_p\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} \frac{\kappa^3}{e^{\beta\hbar\kappa} - 1}d \kappa + E_Z = \frac{3_pL^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} \frac{x^3}{e^{x} - 1}dx + E_Z,
     $$
 
-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 $1/v^3$.
+    Therefore, $C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} \frac{x^3}{e^{x} - 1}dx$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/v_xv_yv_z$ instead of $1/v^3$.