91Ó°ÊÓ

We need to find calculate the integrand in the equation$7.112$as a power series in$x$keeping terms through $x^4$.

Equation is given as:

$U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}\frac{x^3}{e^x−1}dx$

where,

$x=\frac{h{c}_{s}n}{2LkT}{x}_{\max }=\frac{T_D}{T}$

at low-temperature limit $Tâ\copyright ½T_D$, the value of is very small so we can expand the exponential in the denominator using:

$e^x≈1+x+x^2/2+x^3/6$

therefore:

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}\frac{x^3}{1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3−1} \\ U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}\frac{x^3}{x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3} \\ U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}{x}^2{\left[1+\frac{1}{2}x+\frac{1}{6}{x}^2\right]}^{−1} \end{gathered}$$

now we can use ${(1+y)}^{−1}≈1+y+y^2$, where $y=−\frac{1}{2}x−\frac{1}{6}{x}^2$we get :

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}{x}^2\left[1−\frac{1}{2}x−\frac{1}{6}{x}^2+{\left(−\frac{1}{2}x−\frac{1}{6}{x}^2\right)}^2\right] \\ U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}{x}^2\left[1−\frac{1}{2}x−\frac{1}{6}{x}^2+\frac{1}{4}{x}^2+\frac{1}{36}{x}^4+\frac{1}{6}{x}^3\right] \end{gathered}$$

neglect the $x$with power of $3and4$inside the bracket because $x$ is very small ,so:

$U=\frac{9Nk{T}^4}{T_D^3}{∫}_0^{x_{\max }}\left[x^2−\frac{1}{2}{x}^3+\frac{1}{12}{x}^4\right]$

now the function inside the integration is easy to be integrated, so integrate from $0$to $x_{max}$to get:

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{\left[\frac{1}{3}{x}^3−\frac{1}{8}{x}^4+\frac{1}{60}{x}^5\right]}_0^{x_{\max }} \\ U=\frac{9Nk{T}^4}{T_D^3}\left[\frac{1}{3}{x}_{\max }^3−\frac{1}{8}{x}_{\max }^4+\frac{1}{60}{x}_{\max }^5\right] \end{gathered}$$

but $x_{max}=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$T_D$}\right.$, so,

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}\left[\frac{1}{3}{\left(\frac{T_D}{T}\right)}^3−\frac{1}{8}{\left(\frac{T_D}{T}\right)}^4+\frac{1}{60}{\left(\frac{T_D}{T}\right)}^5\right] \\ U=9Nk{T}_D\left[\frac{1}{3}\left(\frac{T}{T_D}\right)−\frac{1}{8}+\frac{1}{60}\left(\frac{T_D}{T}\right)\right] \end{gathered}$$

the heat capacity at constant volume is the just the partial derivative of the energy with respect to the temperature, that is:

$C_v=\frac{∂U}{∂T}$

thus,

localid="1650887002266" $$\begin{gathered} C_v=9Nk{T}_D\frac{∂}{∂T}\left[\frac{1}{3}\left(\frac{T}{T_D}\right)−\frac{1}{8}+\frac{1}{60}\left(\frac{T_D}{T}\right)\right] \\ {C}_v=9Nk{T}_D\left[\frac{1}{3}\left(\frac{1}{T_D}\right)−\frac{1}{60}\left(\frac{T_D}{T^2}\right)\right] \\ {C}_v=3Nk\left[1−\frac{1}{20}{\left(\frac{T_D}{T}\right)}^2\right] \end{gathered}$$

Now calculating the deviation from the asymptotic value $3NkT,atT=T_D$,

substitute into the above equation we get:

$\begin{array}{c}{\left.C_V\right|}_{T=T_D}=3Nk\left[1−\frac{1}{20}{\left(\frac{T_D}{T_D}\right)}^2\right]\\ {\left.C_V\right|}_{T=T_D}=\left(0.95\right)3Nk\end{array}$

so it deviates by $5\%AtT=2T_D$.

$\begin{array}{c}{\left.C_V\right|}_{T=2T_D}=3Nk\left[1−\frac{1}{20}{\left(\frac{T_D}{2T_D}\right)}^2\right]\\ {\left.C_V\right|}_{T=2T_D}=\left(0.9875\right)3Nk\end{array}$

so it deviates by$1.25\%$