From b654f398ed7afd78b254698fa1409134168bf64c Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Tue, 11 Feb 2020 20:22:22 +0000
Subject: [PATCH] Update 2_debye_model.md - polish

---
 src/2_debye_model.md | 46 +++++++++++++++++++++-----------------------
 1 file changed, 22 insertions(+), 24 deletions(-)

diff --git a/src/2_debye_model.md b/src/2_debye_model.md
index 56d03dd4..fcea0851 100644
--- a/src/2_debye_model.md
+++ b/src/2_debye_model.md
@@ -31,10 +31,10 @@ _(based on chapter 2.2 of the book)_
 
     Before the start of this lecture, you should be able to:
 
-    - Describe how atoms are modeled in the Einstein model 
-    - Derive the low-temperature behaviour of the Einstein heat capacity per atom 
+    - Recall how atoms are modeled in the Einstein model
+    - Describe a procedure for calculating the heat capacity of a solid
     - Describe how the frequency of a sound wave depends on the wavenumber $k=2\pi/\lambda$, where $\lambda$ is the wavelength.
-    - Formulate a volume integral in spherical coordinates
+    - Formulate volume integrals using spherical coordinates
     
 !!! summary "Learning goals"
 
@@ -71,7 +71,7 @@ T_E = fit[0][0]
 fig, ax = pyplot.subplots()
 ax.scatter(T, c)
 ax.plot(temp, c_einstein(temp, T_E), label=f'Einstein model, $T_E={T_E:.5}K$')
-ax.text(T_E, 2, r'$T=T_E$', ha='left', color='r');
+ax.text(T_E+3, 1.5, r'$T=T_E$', ha='left', color='r');
 ax.plot([T_E, T_E], [0, 3], 'r--')
 ax.set_ylim(bottom=0, top=3)
 ax.set_xlim(0, 215)
@@ -79,7 +79,7 @@ ax.set_xlabel('$T(K)$')
 ax.set_ylabel(r'$C/k_B$');
 ```
 
-The einstein model works reasonably well, but it predicts a too small heat capacity at low $T$
+The Einstein model works reasonably well, but it predicts a too small heat capacity at low $T$.
 
 ??? question "How does $C$ predicted by the Einstein model behave at low $T$?"
 
@@ -104,9 +104,7 @@ $$
 $$
 Here $v$ is the *sound velocity*.
 
-Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v|\mathbf{k}|$.
-
-This makes the total energy equal to 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 $\omega(k) = v|\mathbf{k}|$. The total energy is given by the sum over the energies of all the oscillators:
 
 $$
 E=\sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right)
@@ -140,8 +138,8 @@ $$
 (of course similar periodicity applies to $y$ and $z$ coordinates)
 
 Periodicity means that not all the points in $k$-space are allowed.
-Instead only waves with each component $k_x, k_y, k_z$ of the $\mathbf{k}$-vector belonging to a set
-$$k=…, \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, …$$
+Instead only waves for which each component $k_x, k_y, k_z$ of the $\mathbf{k}$-vector belongs to the set
+$$k_{x,y,z}=…, \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, …$$
 satisfy the periodic boundary conditions.
 
 In 3D the allowed $k$-vectors form a regular grid:
@@ -152,13 +150,13 @@ There is therefore exactly one allowed ${\bf k}$ per volume $\left(\frac{2\pi}{L
 
 When we consider larger and larger box sizes $L→∞$, the volume per allowed mode becomes smaller and smaller, and eventually we obtain an integral:
 $$
-\sum_\mathbf{k} f(\mathbf{k}) ≈ \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z f(\mathbf{k})
+\sum_\mathbf{k}  \rightarrow \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z )
 $$
 
 
 ## Density of states
 
-Turning back to our problem of computing heat capacity, we get:
+Continuing with our calculation of the total energy, we get:
 $$
 \begin{align}
 E &= \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z × 3×\left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right),\\
@@ -179,9 +177,9 @@ $$
 
 In the last expression everything inside the brackets is about Bose-Einstein statistics, while all the prefactors together are specific to the problem we are studying.
 
-We can emphasize this further by introducing a new concept, _density of states_, $g(\omega)$.
+We can emphasize this further by introducing a new concept, _density of states_, $g(\omega)$, which is a central concept in this course.
 
-> Density of states $g(ω)$ is the number of available normal modes per infinitesimal interval $δω$.
+> The density of states $g(ω)$ is the number of available normal modes per infinitesimal interval $δω$.
 
 With this definition, our integral becomes
 $$
@@ -194,20 +192,20 @@ $$
 
 We can trace back all the factors in the density of states to their origin:
 
-* $(L/2\pi)^3$ is the number of possible waves per unit volume in the reciprocal space
-* $4π$ is the area of a unit sphere, the result of integration over $d \varphi$ and $dθ$
-* $ω^2$ is due to the area of this sphere being proportional to its squared radius
-* $3$ is the number of possible polarizations in 3D
+* $(L/2\pi)^3$ is the number of possible waves per unit volume in the reciprocal space.
+* $4π$ is the area of a unit sphere, the result of integration over $d \varphi$ and $dθ$.
+* $ω^2$ is due to the area of this sphere being proportional to its squared radius.
+* $3$ is the number of possible polarizations in 3D  (two transversal, one longitudinal).
 * $v^{-3}$ is due to $ω = v|k|$.
 
 
 ## Low $T$
 
-In general, ${\rm d}k/{\rm d}\omega$ can be difficult to calculate; we will see more of this later. But going back to the Debye model for now, and using $g(\omega)=V\omega^2/2\pi^2v^3$. The total energy then becomes:
+Using $g(\omega)=V\omega^2/2\pi^2v^3$, the total energy becomes:
 
 $$ E=E_{\rm Z}+\frac{3V}{2\pi^2 v_{\rm s}^3}\int\limits_0^\infty\left(\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}\right)\omega^2{\rm d}\omega$$
 
-Here, the factor 3 comes from the fact that every wave has three polarizations (two transversal, one longitudinal). The term $E_{\rm Z}$ goes to infinity through integration. This is no problem, as it doesn't count towards the heat capacity.
+The term $E_{\rm Z}$ goes to infinity through integration. This is no problem, as it doesn't count towards the heat capacity.
 
 Substitute $x\equiv\frac{\hbar\omega}{k_{\rm B}T}$:
 $$
@@ -218,12 +216,12 @@ $$
 
 Therefore we conclude that $C=\frac{ {\rm d}E}{ {\rm d}T}\propto T^3$.
 
-Can we understand this without any calculation terms? Turns out we can!
+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 phonons have wave vectors $|k| \lesssim k_B T /\hbar v$, and therefore their total number is proportional to the volume of a sphere with the same radius times the density of modes in $k$-space. This gives us $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
-3. Each of these modes is a harmonic oscillator at a reasonably high temperature. Therefore, similarly to the Einstein model, it contributes $\sim k_B$ to the heat capacity.
-4. Multiplying the number of modes by the contribution of each mode we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
+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 $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
+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 $C\propto k_B (k_B T L/\hbar v)^3$.
 
 
 ## Debye's interpolation for medium $T$
-- 
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