Update docs/2_debye_model_solutions.md

docs/2_debye_model_solutions.md

@@ -32,6 +32,7 @@ configure_plotting()
 ### Warm-up exercises
 
 1. For low T, $1/T \rightarrow \infty$. The heat capacity is then given as:
+    
     $$
     C \overset{\mathrm{low \: T}}{\approx} 9Nk_{\mathrm{B}}\left(\frac{T}{T_{D}}\right)^3\int_0^{\infty}\frac{x^4{\mathrm{e}}^x}{({\mathrm{e}}^x-1)^2}{\mathrm{d}}x.
     $$
@@ -79,12 +80,15 @@ ax.legend();
     \end{align}
 
 4.   We use $k=\mathbf{k}| = \omega/v_s$ and $dk = d\omega/v_s$ to get 
+
     \begin{align}
     N_\text{states, 1D} & = 1_p \frac{L}{2\pi} \int_0^{\omega_0} 2 \frac{1}{v_s} d\omega  := \int_0^{\omega_0} 2 g_{1D}(\omega) d\omega \\
-    N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{omega_0} 2\pi \frac{\omega}{v_s^2} d\omega := \int_0^{\omega_0} 2 g_{2D}(\omega) d\omega\\
-    N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{omega_0} \frac{\omega^2}{v_s^3} k^2 d\omega := \int_0^{\omega_0} 2 g_{3D}(\omega) d\omega
+    N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{\omega_0} 2\pi \frac{\omega}{v_s^2} d\omega := \int_0^{\omega_0} g_{2D}(\omega) d\omega \\
+    N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{\omega_0} 4\pi \frac{\omega^2}{v_s^3} d\omega := \int_0^{\omega_0} g_{3D}(\omega) d\omega
     \end{align}
+
     The integral boundaries set the frequency region in which you calculate the density of states.
+
 5. The density of states is the number of states per unit frequency. It has units of 1 over frequency
 
 ###  Exercise 2: Debye model in 2D.
@@ -103,7 +107,7 @@ Here, we analyze the phonon energy and heat capacity of a two-dimensional Debye
     $$
     E = \int_0^\omega_D \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_0^\omega_D k_B T g(\omega) d\omega = N_\text{modes} k_B T
     $$
-    where we neglected the zero-point energy. In 3D, we have $N_\text{modes} = 3_p N_\text{atoms}$
+    where we neglected the zero-point energy. In 3D, we have $N_\text{modes} = 3_p N_\text{atoms}$ (Dulong Petit)
 
 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 
     
@@ -116,18 +120,34 @@ Here, we analyze the phonon energy and heat capacity of a two-dimensional Debye
     K = 3\alpha\frac{k^3}{\hbar^2}\int_)^\infty\frac{x^2}{e^x-1}dx 
     $$
 
-###  Exercise 3: Different phonon modes.
-#### Question 1.
+###  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
+
+    $$ 
+    E_\parallel = \int_0^{\omega_D}g_\parallel(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} )d\omega
+    $$
+
+    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 
 
-\begin{gather}
-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.
-\end{gather}
+    $$ 
+    E_\perp = \int_0^{\omega_D}g_\perp(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} )d\omega
+    $$
+
+    with $g_\perp = 2_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\perp^3} $. Therefore, the total energy is 
+    
+    $$ 
+    E = \frac{L^3}{2\pi^2} \left(\frac{1_p}{v_\parallel} + \frac{2_p}{v\perp} \right) \int_0^{\omega_D}  \frac{\hbar\omega^3}{e^{\beta\hbar\omega} - 1} d\omega = \frac{L^3 k_B^4 T^4}{2\pi^2\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}   {v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx.
+    $$
+
+
+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: 
+    $$
+    E = \int_0^\omega_D \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_0^\omega_D k_B T g(\omega) d\omega = N_\text{modes} k_B T
+    $$
+    and we are left with the Dulong-Petit law $C = 3N_\text{atoms}k_B$.
 
-#### Question 2.
-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 $C = 3Nk_B$.
-#### Question 3.
-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 $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
+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.