From 3c5575a14890c705c29a587312e1d281402d58ab Mon Sep 17 00:00:00 2001
From: Pim Vree <p.h.vree@tudelft.nl>
Date: Tue, 25 Mar 2025 16:20:26 +0000
Subject: [PATCH] Edit 2_debye_model_solutions.md
Fixed an error that we took the polarization 3_p into account twice
---
docs/2_debye_model_solutions.md | 16 ++++++++--------
1 file changed, 8 insertions(+), 8 deletions(-)
diff --git a/docs/2_debye_model_solutions.md b/docs/2_debye_model_solutions.md
index 5dfe740..43fbe93 100644
--- a/docs/2_debye_model_solutions.md
+++ b/docs/2_debye_model_solutions.md
@@ -164,29 +164,29 @@ E =& 3_p\left(\frac{L}{2\pi}\right)^3 \int \frac{\hbar\omega(\mathbf{k})}{e^{\be
where we made the substitutions $\kappa_x = k_x v_x,\kappa_y = k_y v_y, \kappa_z = k_z v_z$ so that $\omega = \kappa$. Going to spherical coordinates:
\begin{align*}
-E &= 3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar\kappa^3}{e^{\beta\hbar\kappa} - 1} d\kappa + E_Z
+E &= \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar\kappa^3}{e^{\beta\hbar\kappa} - 1} d\kappa + E_Z
\end{align*}
To calculate the specific heat $C = \frac{dE}{dT}$, let's differentiate this expression directly:
\begin{align*}
-C &= \frac{dE}{dT} = 3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{d}{dT}\left(\frac{\hbar\kappa^3}{e^{\beta\hbar\kappa} - 1}\right) d\kappa \\
-&= 3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar\kappa^3 \cdot \hbar\kappa \cdot e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} \cdot \frac{d\beta}{dT} d\kappa
+C &= \frac{dE}{dT} = \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{d}{dT}\left(\frac{\hbar\kappa^3}{e^{\beta\hbar\kappa} - 1}\right) d\kappa \\
+&= \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar\kappa^3 \cdot \hbar\kappa \cdot e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} \cdot \frac{d\beta}{dT} d\kappa
\end{align*}
Since $\beta = \frac{1}{k_B T}$, we have $\frac{d\beta}{dT} = -\frac{1}{k_B T^2} = -\frac{\beta}{T}$. Substituting this:
\begin{align*}
-C &= -3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar^2\kappa^4 e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} \cdot \frac{\beta}{T} d\kappa \\
-&= -3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T}\int_{0}^{\kappa_D} \frac{\kappa^4 e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} d\kappa
+C &= - \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar^2\kappa^4 e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} \cdot \frac{\beta}{T} d\kappa \\
+&= - \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T}\int_{0}^{\kappa_D} \frac{\kappa^4 e^{\beta\hbar\kappa}}{(e^{\beta\hbar\kappa} - 1)^2} d\kappa
\end{align*}
Now, we can make the substitution $x = \beta\hbar\kappa$, which gives $d\kappa = \frac{dx}{\beta\hbar}$ and changes our limits to $\int_{0}^{\beta\hbar\kappa_D}$:
\begin{align*}
-C &= -3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T} \cdot \int_{0}^{\beta\hbar\kappa_D} \left(\frac{x}{\beta\hbar}\right)^4 \frac{e^x}{(e^x - 1)^2} \cdot \frac{dx}{\beta\hbar} \\
-&= -3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T} \cdot \frac{1}{(\beta\hbar)^5} \int_{0}^{\beta\hbar\kappa_D} \frac{x^4 e^x}{(e^x - 1)^2} dx \\
-&= -3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{1}{T\beta^4\hbar^3} \int_{0}^{\beta\hbar\kappa_D} \frac{x^4 e^x}{(e^x - 1)^2} dx
+C &= - \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T} \cdot \int_{0}^{\beta\hbar\kappa_D} \left(\frac{x}{\beta\hbar}\right)^4 \frac{e^x}{(e^x - 1)^2} \cdot \frac{dx}{\beta\hbar} \\
+&= - \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{\hbar^2\beta}{T} \cdot \frac{1}{(\beta\hbar)^5} \int_{0}^{\beta\hbar\kappa_D} \frac{x^4 e^x}{(e^x - 1)^2} dx \\
+&= - \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z} \cdot \frac{1}{T\beta^4\hbar^3} \int_{0}^{\beta\hbar\kappa_D} \frac{x^4 e^x}{(e^x - 1)^2} dx
\end{align*}
Since $\frac{1}{T\beta^4} = \frac{k_B^4 T^3}{1} = k_B^4 T^3$, we get:
--
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