typos

docs/2_debye_model.md

@@ -441,7 +441,7 @@ Here, we analyze the vibrational energy and heat capacity of a _two-dimensional_
 In the lecture, we derived the low-temperature Debye heat capacity assuming that all the vibrational modes have the same sound velocity $v_s$.
 In reality, longitudinal and transverse modes have different sound velocities (see [Wikipedia](https://en.wikipedia.org/wiki/Sound#Longitudinal_and_transverse_waves) for an illustration of different sound wave types).
 
-Assume that there are two types of excitations: One longitudinal mode with $\omega = v_\parallel |k|$, and two transverse modes with $\omega = v_\bot |k|$
+Assume that there are two types of excitations: One longitudinal mode with $\omega = v_\parallel |\mathbf{k}|$, and two transverse modes with $\omega = v_\bot |\mathbf{k}|$
 
 1. Write down the total energy of the thermally excited phonons in this material *(hint: use the same reasoning as in the [Lithium exercise](1_einstein_model.md#exercise-3-total-heat-capacity-of-a-diatomic-material))*.
 2. Verify that at high $T$ you reproduce the Dulong-Petit law.

docs/2_debye_model_solutions.md

@@ -82,7 +82,7 @@ ax.legend();
 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, 1D} & = 1_p \frac{L}{2\pi} \int_0^{\omega_0} 2 \frac{1}{v_s} d\omega  := \int_0^{\omega_0} 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} 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}