Update 2_debye_model.md - wrote sound velocity as v_s everywhere

src/2_debye_model.md

@@ -98,11 +98,11 @@ All wave vectors are points in _reciprocal space_ or _k-space_.
 
 These waves don't all have the same frequency $\omega_0$ as the atoms did have in the Einstein model, but rather a _dispersion relation_
 $$
-\omega = v|\mathbf{k}|.
+\omega = v_{\rm s}|\mathbf{k}|.
 $$
-Here $v$ is the *sound velocity*.
+Here $v_{\rm s}$ is the *sound velocity*.
 
-Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v|\mathbf{k}|$. The total energy is given by the sum over the energies of all the oscillators:
+Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v_{\rm s}|\mathbf{k}|$. The total energy is given by the sum over the energies of all the oscillators:
 
 $$
 E=\sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right)
@@ -158,7 +158,7 @@ Continuing with our calculation of the total energy, we get:
 $$
 \begin{align}
 E &= \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z × 3×\left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right),\\
-\omega(\mathbf{k}) &= v\sqrt{k_x^2 + k_y^2 + k_z^2}.
+\omega(\mathbf{k}) &= v_{\rm s}\sqrt{k_x^2 + k_y^2 + k_z^2}.
 \end{align}
 $$
 The factor $3$ accounts for three possible directions of displacement (wave polarizations).
@@ -185,7 +185,7 @@ E = ∫\limits_0^∞\left(\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{ {\rm e}^{\h
 $$
 with
 $$
-g(ω) = \frac{L^3}{(2\pi)^3}×4 π × 3 × v^{-3} × \omega^2.
+g(ω) = \frac{L^3}{(2\pi)^3}×4 π × 3 × v_{\rm s}^{-3} × \omega^2.
 $$
 
 We can trace back all the factors in the density of states to their origin:
@@ -199,7 +199,7 @@ We can trace back all the factors in the density of states to their origin:
 
 ## Low $T$
 
-Using $g(\omega)=V\omega^2/2\pi^2v^3$, the total energy becomes:
+Using $g(\omega)=V\omega^2/2\pi^2v_{\rm s}^3$, the total energy becomes:
 
 $$ E=E_{\rm Z}+\frac{3V}{2\pi^2 v_{\rm s}^3}\int\limits_0^\infty\left(\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}\right)\omega^2{\rm d}\omega$$
 
@@ -217,9 +217,9 @@ Therefore we conclude that $C=\frac{ {\rm d}E}{ {\rm d}T}\propto T^3$.
 Can we understand this without calculating any terms? Turns out we can!
 
 1. At temperature $T$ only modes with $\hbar \omega \lesssim k_B T$ get thermally excited.
-2. These modes have wave vectors $|k| \lesssim k_B T /\hbar v$. Therefore their total number is proportional to the volume of a sphere with radius $|k|$ multiplied by the density of modes in $k$-space. This gives us $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
+2. These modes have wave vectors $|k| \lesssim k_B T /\hbar v{\rm s}$. Therefore their total number is proportional to the volume of a sphere with radius $|k|$ multiplied by the density of modes in $k$-space. This gives us $N_\textrm{modes} \sim (k_B T L/\hbar v{\rm s})^3$.
 3. As these modes are thermally excited, they behave like classical harmonic oscillators and contribute $\sim k_B$ to the heat capacity each (similar to the Einstein model).
-4. Multiplying $N_\textrm{modes}$ by the contribution of each mode, we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
+4. Multiplying $N_\textrm{modes}$ by the contribution of each mode, we obtain $C\propto k_B (k_B T L/\hbar v{\rm s})^3$.
 
 
 ## Debye's interpolation for medium $T$