phonons → phonon modes where appropriate

src/2_debye_model.md

@@ -119,7 +119,7 @@ $$
 $$
 where $v_s$ is the _sound velocity_ of a material.
 
-To summarize, instead of having $3N$ oscillators with the same frequency $\omega_0$, we now have $N$ possible phonons with frequencies depending on $\textbf{k}$ through the dispersion relation $\omega(\mathbf{k}) = v_s|\mathbf{k}|$. 
+To summarize, instead of having $3N$ oscillators with the same frequency $\omega_0$, we now have $3N$ possible phonon modes with frequencies depending on $\textbf{k}$ through the dispersion relation $\omega(\mathbf{k}) = v_s|\mathbf{k}|$. 
 The expected value of the total energy (which we, for simplicity, from now on will denote as the total energy) is given by the sum over the energy of all possible phonon modes characterized by a wavevector $\mathbf{k}$:
 
 $$
@@ -239,7 +239,7 @@ We interpret the integral above as follows: we multiply the number of modes $g(\
 
 Let us separate $g(\omega)$ into a product of individual factors:
 
-* $3$ comes from the number of possible phonon polarizations in 3D (two transversal, one longitudinal). 
+* $3$ comes from the number of possible polarizations in 3D (two transversal, one longitudinal). 
 * $(\frac{L}{2\pi})^3$ is the density of $\textbf{k}$ points in $k$-space.
 * $4\pi$ is the area of a unit sphere.
 * $\omega^2$ is due to the area of a sphere in $k$-space being proportional to its squared radius $k^2$ and by having a linear dispersion relation $\omega = v_sk$.
@@ -255,7 +255,7 @@ $$
 \langle E \rangle = E_{\textrm Z} + \frac{3L^3}{2\pi^2 v_s^3}\int\limits_0^\infty \left(\frac{\hbar\omega}{ e^{\hbar\omega/k_BT}-1}\right)\omega^2 {\textrm{d}}\omega.
 $$
 
-The term $E_{\textrm Z}$ is the temperature-independent zero-point energy of the phonons. 
+The term $E_{\textrm Z}$ is the temperature-independent zero-point energy of the phonon modes. 
 Despite $E_{\textrm Z}$ diverging towards infinity, does not contribute to $C$.
 
 The integral depends on the temperature through the $e^{\hbar\omega/k_BT}$ term.
@@ -277,25 +277,25 @@ We recover the empirical $T^3$ dependence of $C$ at low temperatures!
 
 Can we understand this without going through the integration? Turns out we can!
 
-1. At temperature $T$, only phonons with an energy below the thermal energy $E_{\textrm{T}} = k_B T$ become thermally excited. 
-These phonons have $\hbar \omega(\mathbf{k}) \lesssim k_B T$.
-2. By substituting the dispersion relation into the above inequality, we conclude that these phonons have wave vectors $|\mathbf{k}| \lesssim k_B T /\hbar v_s$. 
-Therefore number of excited phonons is proportional to the volume of a sphere $V_{\textbf{k}} = \frac{4 \pi}{3} |\mathbf{k}|^3$, multiplied by the density of phonons in $k$-space, $\left(\frac{L}{2 \pi}\right)^3$. 
-Thus the total number of excited phonons is 
+1. At temperature $T$, only phonon modes with an energy below the thermal energy $E_{\textrm{T}} = k_B T$ become thermally excited. 
+These modes have $\hbar \omega(\mathbf{k}) \lesssim k_B T$.
+2. By substituting the dispersion relation into the above inequality, we conclude that these modes have wave vectors $|\mathbf{k}| \lesssim k_B T /\hbar v_s$. 
+Therefore the number of excited modes is proportional to the volume of a sphere $V_{\textbf{k}} = \frac{4 \pi}{3} |\mathbf{k}|^3$, multiplied by the density of modes in $k$-space, $\left(\frac{L}{2 \pi}\right)^3$. 
+Thus the total number of excited modes is 
 $$
 \begin{align}
-N_\textrm{phonons} &= V_{\textbf{k}} \left(\frac{L}{2\pi}\right)^3\\
+N_\textrm{modes} &= V_{\textbf{k}} \left(\frac{L}{2\pi}\right)^3\\
 &\sim \left( |\mathbf{k}| L \right)^3\\
 &\sim (k_B T L/\hbar v_s)^3
 \end{align}
 $$
 where we have substituted $|\mathbf{k}| \simeq k_B T /\hbar v_s$ and left out all numerical factors.
-3. When thermally excited, the motion of these phonons resembles that of classical harmonic oscillators. 
-Therefore, each phonon contributes $k_B$ to the heat capacity (Equipartition theorem). 
+3. When thermally excited, the motion of these modes resembles that of classical harmonic oscillators. 
+Therefore, each mode contributes $k_B$ to the heat capacity (Equipartition theorem). 
 As a result, the heat capacity is
 $$
 \begin{align}
-C &= N_{\textrm{phonon}} k_B \\
+C &= N_{\textrm{modes}} k_B \\
 &\propto k_B (k_B T L/\hbar v_s)^3.
 \end{align}
 $$
@@ -305,16 +305,15 @@ $$
 
 We observed that the above approximation yields a correct scaling of the heat capacity at low temperatures.
 We also know that for a 3D material with $N$ atoms, $C$ should converge to $3Nk_B$ at high temperatures (the law of Dulong–Petit).
-However at high temperatures, phonons with all values of $\omega$ become thermally excited, and the number of these phonons tends towards infinity:
+However at high temperatures, phonon modes with all values of $\omega$ become thermally excited, and the number of these modes tends towards infinity:
 $$
-N_{\textrm{phonon}} = \int_0^{\infty} \textrm{d} \omega g(\omega) \to \infty
+N_{\textrm{modes}} = \int_0^{\infty} \textrm{d} \omega g(\omega) \to \infty
 $$
-As a result, we now incorrectly predict that the heat capacity also goes to infinity $C \propto N_{\textrm{phonon}} k_B \to \infty$.
+As a result, we now incorrectly predict that the heat capacity also goes to infinity $C \propto N_{\textrm{modes}} k_B \to \infty$.
 Hence, the model breaks down for high temperatures.
 
-To fix this problem Debye realised that there should be as many phonons in the system as there are degrees of freedom. 
-In a 3D material with $N$ atoms, there are a total of $3N$ degrees of freedom.
-Therefore, the material can only contain $3N$ possible phonon modes, and not more.
+To fix this problem Debye realised that there should be as many phonon modes in the system as there are degrees of freedom. 
+In a 3D material with $N$ atoms, there are a total of $3N$ normal modes, and not more.
 
 In view of this fact, Debye proposed a fix to the problem: assume that there is a maximal frequency $\omega_D$ *(Debye frequency)*, beyond which there are no phonons.
 We have no good justification for this assumption yet, but it is reasonable because the atoms certainly cannot move with infinite frequency.
@@ -384,12 +383,12 @@ Debye model clearly wins!
 
 ## Conclusions
 
-1. The Debye model assumes that atoms in materials move in a collective fashion, described by quantized normal modes / phonons with a dispersion relation $\omega = v_s|\mathbf{k}|$.
-2. The phonons have a constant density of $(L/2\pi)^3$ in the reciprocal / $k$-space.
+1. The Debye model assumes that atoms in materials move in a collective fashion, described by quantized normal modes with a dispersion relation $\omega = v_s|\mathbf{k}|$.
+2. The phonon modes have a constant density of $(L/2\pi)^3$ in the reciprocal / $k$-space.
 3. The total energy and heat capacity are obtained by integrating the contribution of the individual modes over $k$-space.
 4. The density of states $g(\omega)$ is the number of states per frequency. With a dispersion relation $ω = v_s|\mathbf{k}|$, $g(\omega)$ is proportional to $\omega^2$ for a 3D bosonic system.
 5. At low temperatures the phonon heat capacity is proportional to $T^3$.
-6. Phonons are only acoustic up until the Debye frequency $\omega_D$, after which there are no phonons in the system.
+6. Phonon modes only exist up until the Debye frequency $\omega_D$, after which there are no modes in the system.
 
 
 ## Exercises