Q. 7.63 - An Introduction to Thermal Physics

Question

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." 

Step-by-Step Solution

Answer

The expression for the heat capacity is $C_V=2Nk{\left(\frac{T}{T_D}\right)}^2{\int }_0^{x_{\max }}\frac{x^3{e}^x}{{\left(e^x-1\right)}^2}dx$ and the plot of the heat capacity as a function of temperature is 

Step 1:Given Information

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.  

Step 2: Simplify

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 is$\sqrt{N}$ , $N$for a large we change the summation to an integral, that is:

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

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 $n_{\max }$

as shown in the following figure:

Step 3: Simplify

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:

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

Step 4: Simplify

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, $T\ll T_D$ 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:

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

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

Step 5: Simplify

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.

Step 6: Diagram