Consider a two-dimensional solid, such as a stretched drumhead or a layer of mica or graphite. Find an expression (in terms of an integral) for the thermal energy of a square chunk of this material of area , and evaluate the result approximately for very low and very high temperatures. Also, find an expression for the heat capacity, and use a computer or a calculator to plot the heat capacity as a function of temperature. Assume that the material can only vibrate perpendicular to its own plane, i.e., that there is only one "polarization."

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:

$U=\sum _{n_x}\sum _{n_y}ϵ{\stackrel{-}{n}}_{\text{Pl}}$

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

${\stackrel{-}{n}}_{\text{Pl}}=\frac{1}{e^{ϵ/kT}-1}$

thus,

$U=\sum _{n_x}\sum _{n_y}\frac{ϵ}{e^{ϵ/kT}-1}$

the allowed wavelengths is:

$\lambda =\frac{2L}{n}$

therefore the allowed energies are:

$ϵ=\frac{hc}{\lambda }=\frac{hcn}{2L}$

therefore:

$U=\frac{hc}{2L}\sum _{n_x}\sum _{n_y}\frac{n}{e^{hen/kT}-1}$

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

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 $\frac{1}{4}\pi n_{\max }^2$to the area of the square which is $N$so we get:

$$\begin{gathered} N=\frac{1}{4}\pi n_{\max }^2 \\ {n}_{\max }=\sqrt{\frac{4N}{\pi }} \end{gathered}$$

$\begin{array}{rr}U& =\frac{h{c}_s}{2L}{\int }_0^{\pi /2}d\theta {\int }_0^{n_{\max }}\frac{n^2}{e^{h{c}_{s}n/kT}-1}dn\\ U& =\frac{\pi }{2}\left(\frac{h{c}_s}{2L}\right){\int }_0^{n_{\max }}\frac{n^2}{e^{h{c}_{s}n/kT}-1}dn\end{array}$

now let,

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

thus,

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

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:

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

substitute into ($2$) to get:

$U=\frac{\pi }{2}\frac{1}{{\left(k{T}_D\right)}^2}\left(\frac{4N}{\pi }\right){\left(kT\right)}^3{\int }_0^{x_{\max }}\frac{x^2}{e^x-1}dx$

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

at a very low temperature, role="math" localid="1650283166300" $T\ll T_D$then $x_{\max }\to \infty$since $x_{\max }=T_D/T$so we get:

$U=\frac{2Nk{T}^3}{T_D^2}{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

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:

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

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

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

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

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:

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

now let,

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

thus,

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

substitute with,

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

to get:

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

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