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

@@ -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: