91Ó°ÊÓ

We need to find an expression for the heat capacity, and use a computer or a calculator to plot the heat capacity as a function of temperature.

Consider we have two-dimensional material which is simply a square chunk with area of $A=L^2$the thermal energy equals the sum of the energies multiplied by the Planck distribution ${n^{-}}_{P}L$that is:

$U=\underset{n_x}{∑}\underset{n_y}{∑}\mathrm{Ï\mu }{\stackrel{-}{n}}_{\text{Pl}}$

note that we multiplied with a factor of $1$, since we have only one mode of polarization, the Planck distribution is given by:

${\stackrel{-}{n}}_{\text{Pl}}=\frac{1}{e^{\mathrm{Ï\mu }/kT}−1}$

thus,

$U=\underset{n_x}{∑}\underset{n_y}{∑}\frac{\mathrm{Ï\mu }}{e^{\mathrm{Ï\mu }/kT}−1}$

the allowed wavelengths is:

$\mathrm{λ}=\frac{2L}{n}$

therefore the allowed energies are:

$\mathrm{Ï\mu }=\frac{hc}{\mathrm{λ}}=\frac{hcn}{2L}$

therefore:

$U=\frac{hc}{2L}\underset{n_x}{∑}\underset{n_y}{∑}\frac{n}{e^{hen/kT}−1}$

assume we have $N$atoms is the shape of the square, therefore the width of this square islocalid="1650284294248" $\sqrt{N}$, $N$for a large we change the summation to an integral, that is:

localid="1650284284032" $U=\frac{hc}{2L}{∫}_0^{\sqrt{N}}d{n}_x{∫}_0^{\sqrt{N}}d{n}_y\frac{n}{e^{hcn/kT}−1}$

now we need to change this into a polar coordinates. First change the square to a quarter circle that has the same area of the square, with radius of localid="1650284300903" $n_{\max }$

as shown in the following figure:

To find $n_{\max }$we set the area of the quarter circle $\frac{1}{4}\mathrm{Ï€}{n}_{\max }^2$to the area of the square which is $N$so we get:

$$\begin{gathered} N=\frac{1}{4}\mathrm{Ï€}{n}_{\max }^2 \\ {n}_{\max }=\sqrt{\frac{4N}{\mathrm{Ï€}}} \end{gathered}$$

$\begin{array}{rr}U& =\frac{h{c}_s}{2L}{∫}_0^{\mathrm{π}/2}d\mathrm{θ}{∫}_0^{n_{\max }}\frac{n^2}{e^{h{c}_{s}n/kT}−1}dn\\ U& =\frac{\mathrm{π}}{2}\left(\frac{h{c}_s}{2L}\right){∫}_0^{n_{\max }}\frac{n^2}{e^{h{c}_{s}n/kT}−1}dn\end{array}$

now let,

$x=\frac{h{c}_{s}n}{2LkT}→dx=\frac{h{c}_s}{2LkT}dn$

thus,

$\begin{array}{c}U=\frac{\mathrm{π}}{2}\left(\frac{h{c}_s}{2L}\right){\left(\frac{2LkT}{h{c}_s}\right)}^3{∫}_0^{n_{\max }}\frac{x^2}{e^x−1}dx\\ U=\frac{\mathrm{π}}{2}{\left(\frac{2L}{h{c}_s}\right)}^2{\left(kT\right)}^3{∫}_0^{n_{\max }}\frac{x^2}{e^x−1}dx\end{array}$

We need to change the limits of the integral, the lower limit will stay the same while the upper limit is:

$x_{\max }=\frac{h{c}_s{n}_{\max }}{2LkT}$

also we can write $x_{\max }$as:

$\begin{array}{c}{x}_{\max }=\frac{T_D}{T}=\frac{h{c}_s{n}_{\max }}{2LkT}=\frac{h{c}_s}{2LkT}\sqrt{\frac{4N}{\mathrm{Ï€}}}\\ T_D=\frac{h{c}_s}{2Lk}\sqrt{\frac{4N}{\mathrm{Ï€}}}\\ T_D^2=\frac{1}{k^2}{\left(\frac{h{c}_s}{2L}\right)}^2\left(\frac{4N}{\mathrm{Ï€}}\right)\\ {\left(\frac{2L}{h{c}_s}\right)}^2=\frac{1}{{\left(k{T}_D\right)}^2}\left(\frac{4N}{\mathrm{Ï€}}\right)\end{array}$

substitute into ($2$) to get:

$U=\frac{\mathrm{π}}{2}\frac{1}{{\left(k{T}_D\right)}^2}\left(\frac{4N}{\mathrm{π}}\right){\left(kT\right)}^3{∫}_0^{x_{\max }}\frac{x^2}{e^x−1}dx$

$U=\frac{2Nk{T}^3}{T_D^2}{∫}_0^{x_{\max }}\frac{x^2}{e^x−1}dx$

at a very low temperature, role="math" localid="1650283166300" $T≪T_D$then $x_{\max }→\mathrm{∞}$since $x_{\max }=T_D/T$so we get:

$U=\frac{2Nk{T}^3}{T_D^2}{∫}_0^{\mathrm{∞}}\frac{x^2}{e^x−1}dx$

this integral can be evaluated numerically and its value is $2.404$then:

$U_{low}=\left(2.404\right)\frac{2Nk{T}^3}{T_D^2}$

the heat capacity at low temperature is therefore:

$\begin{array}{c}{C}_V=\frac{∂U}{∂T}=\left(2.404\right)\frac{6Nk{T}^2}{T_D^2}\\ C_V=\left(2.404\right)\frac{6Nk{T}^2}{T_D^2}\end{array}$

at high temperature limit is very small, so we can expand the exponential using the power series that is:

$U=\frac{2Nk{T}^3}{T_D^2}{∫}_0^{x_{\max }}\frac{x^2}{1+x−1}dx$

$$\begin{gathered} \frac{2Nk{T}^3}{T_D^2}{∫}_0^{x_{\max }}\frac{x^2}{1+x−1}dxU=\frac{2Nk{T}^3}{T_D^2}{∫}_0^{x_{\max }}xdx \\ U=\frac{Nk{T}^3}{T_D^2}{x}_{\max }^2 \\ U=\frac{Nk{T}^3}{T_D^2}{\left(\frac{T_D}{T}\right)}^2 \\ {U}_{high}=NkT \\ {C}_V=Nk \end{gathered}$$

To find the heat capacity at the intermediate temperature we take the derivative of the equation($1$) with respect to the temperature, so we get:

$\begin{array}{rr}{C}_V& =\frac{\mathrm{π}}{2}\left(\frac{h{c}_s}{2L}\right){∫}_0^{n_{\max }}\frac{∂}{∂T}\left[\frac{n^2}{e^{h{c}_{s}n/2LkT}−1}\right]dn\\ C_V& =\frac{\mathrm{π}}{2}\left(\frac{h{c}_s}{2L}\right){∫}_0^{n_{\max }}\frac{h{c}_{s}n}{2Lk{T}^2}\frac{n^2{e}^{h{c}_{s}n/2LkT}}{{\left(e^{h{c}_{s}n/2LkT}−1\right)}^2}dn\\ C_V& =\frac{\mathrm{π}}{2}{\left(\frac{h{c}_s}{2LT}\right)}^2\frac{1}{k}{∫}_0^{n_{\max }}\frac{n^3{e}^{h{c}_{s}n/2LkT}}{{\left(e^{h{c}_{s}n/2LkT}−1\right)}^2}dn\end{array}$

now let,

$x=\frac{h{c}_{s}n}{2LkT}→dx=\frac{h{c}_s}{2LkT}dn$

thus,

$\begin{array}{c}{C}_V=\frac{\mathrm{π}}{2}{\left(\frac{h{c}_s}{2LT}\right)}^2\frac{1}{k}{\left(\frac{2LkT}{h{c}_s}\right)}^4{∫}_0^{x_{\max }}\frac{x^3{e}^x}{{\left(e^x−1\right)}^2}dx\\ C_V=\frac{\mathrm{π}}{2}{\left(\frac{2L}{h{c}_s}\right)}^2{k}^3{T}^2{∫}_0^{x_{\max }}\frac{x^3{e}^x}{{\left(e^x−1\right)}^2}dx\end{array}$

substitute with,

${\left(\frac{2L}{hc}\right)}^2=\frac{1}{{\left(k{T}_n\right)}^2}\left(\frac{4N}{\mathrm{Ï€}}\right)$

to get:

$\begin{array}{c}{C}_V=\frac{\mathrm{π}}{2}\frac{1}{{\left(k{T}_D\right)}^2}\left(\frac{4N}{\mathrm{π}}\right)k^3{T}^2{∫}_0^{x_{\max }}\frac{x^3{e}^x}{{\left(e^x−1\right)}^2}dx\\ C_V=2Nk{\left(\frac{T}{T_D}\right)}^2{∫}_0^{x_{\max }}\frac{x^3{e}^x}{{\left(e^x−1\right)}^2}dx\end{array}$

To plot this function i used the code is shown in the picture.