Merge branch 'master' into local

src/1_einstein_model.md

@@ -269,17 +269,17 @@ ax.text(1.05, 0.95, r'$k_{\rm B}T=\hbar \omega_0 =$', ha='left', color='r');
 We now calculate the heat capacity per atom $C$ explicitly. To do so, we need to differentiate $\langle E \rangle$ with respect to $T$.
 $$
 \begin{multline}
-C = \frac{\partial\bar{\varepsilon}}{\partial T}
-= -\frac{\hbar\omega}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2}\frac{\partial}{\partial T}\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)\\
-= \frac{\hbar^2\omega^2}{k_{\rm B}T^2}\frac{ {\rm e}^{\hbar\omega/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2}
-=k_{\rm B}\left(\frac{\hbar\omega}{k_{\rm B}T}\right)^2\frac{ {\rm e}^{\hbar\omega/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2}
+C = \frac{\partial\langle E \rangle{\partial T}
+= -\frac{\hbar\omega_0}{\left({\rm e}^{\hbar\omega_0/k_{\rm B}T}-1\right)^2}\frac{\partial}{\partial T}\left({\rm e}^{\hbar\omega_0/k_{\rm B}T}-1\right)\\
+= \frac{\hbar^2\omega_0^2}{k_{\rm B}T^2}\frac{ {\rm e}^{\hbar\omega_0/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega_0/k_{\rm B}T}-1\right)^2}
+=k_{\rm B}\left(\frac{\hbar\omega_0}{k_{\rm B}T}\right)^2\frac{ {\rm e}^{\hbar\omega_0/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega_0/k_{\rm B}T}-1\right)^2}
 \end{multline}
 $$
 There is still one step to do, rewriting this equation more elegantly:
 $$
 C = k_b \left(\frac{T_E}{T}\right)^2\frac{e^{T_E/T}}{(e^{T_E/T} - 1)^2)},
 $$
-where we introduce the *Einstein temperature* $T_E \equiv \hbar \omega / k_B$.
+where we introduce the *Einstein temperature* $T_E \equiv \hbar \omega_0 / k_B$.
 This is a characteristic temperature when the thermal excitations "freeze out" in the harmonic oscillator.
 Consequently, it is also the temperature scale when the heat capacity of an Einstein solid starts significantly decreasing.
 
@@ -306,9 +306,9 @@ draw_classic_axes(ax)
 ```
 
 The dashed line is the classical value, $k_{\rm B}$.
-Shaded area $=\frac{1}{2}\hbar\omega$, the zero point energy that cannot be removed through cooling.
+Shaded area $=\frac{1}{2}\hbar\omega_0$, the zero point energy that cannot be removed through cooling.
 
-This is for just one atom. In order to obtain the heat capacity of a full material, we would have to multiply $C$ (or $\bar{\varepsilon}$) by $3N$, _i.e._ the number of harmonic oscillators according to Einstein model.
+This is for just one atom. In order to obtain the heat capacity of a full material, we would have to multiply $C$ (or $\langle E \rangle$) by $3N$, _i.e._ the number of harmonic oscillators according to Einstein model.
 
 ```python
 fit = curve_fit(c_einstein, T, c, 1000)