Question

Fill in the steps to derive equations $7.112$and$7.117$.

Short answer

The derived equations are $U=\frac{9Nk{T}^4}{{T_D}^3}{\int }_0^{x_{max}}\frac{x^3}{e^x-1}dx$and$C_V=\frac{3\pi }{2}\frac{1}{{\left(k{T}_D\right)}^3}\left(\frac{6N}{\pi }\right)k^4{T}^3{\int }_0^{x_{max}}\frac{x^4{e}^x}{{(e^x-1)}^2}dx$.

Step-by-step solution

Given Information

We need to fill in the steps to derive equations.

Simplify

The total thermal energy equals the sum of planck distribution all over the modes $n_x,n_{y}and{n}_z$multiplied by the energy ofeach mode, and sum is multiplied by factor of 3 because we have three polarization modes in crystal, that is:

role="math" localid="1650022039736" $U=3\sum _{n_x}\sum _{n_y}\sum _{n_z}\epsilon \overline{n}P1\left(\epsilon \right)=\sum _{n_x{n}_y{n}_z}\epsilon \overline{n}P1\left(\epsilon \right)$

but,

$\overline{n}P1\left(\epsilon \right)=\frac{1}{e^{\epsilon /kT}-1}$

consider we have a cubic box with volume of V and side width of L, the allowed energy for the photon is:

$\epsilon =\frac{h{c}_{s}n}{2L}$

where $c_s$is the speed of sound in the crystal substitute into the above equation to get:

$U=3\left(\frac{h{c}_s}{2L}\right)\sum _{n_{x,}{n}_{y,}{n}_z}\frac{n}{e^{h{c}_{s}n/2LkT}-1}$

now we need to change the sum to an integral in spherical coordinates, by multiplying this by the spherical integration factor $n^2$sin$\left(\theta \right)$, so we get:

$U=3\left(\frac{h{c}_s}{2L}\right){\int }_0^{\pi /2}d\varphi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{n_{max}}\frac{n^3}{e^{h{c}_{s}n/2LkT}-1}dn$

the first two integrals are easy to evalute, and they give a factor of$\pi /2$,so:

now let,$$\begin{gathered} U=\left(\frac{3\pi h{c}_s}{4L}\right){\int }_0^{n_{max}}\frac{n^3}{e^{h{c}_{s}n/2LkT}-1}dn \\ x=\frac{h{c}_{s}n}{2LkT}\to dx=\frac{h{c}_s}{2LkT}dn \end{gathered}$$

thus,

$$\begin{gathered} U=\left(\frac{3\pi h{c}_s}{4L}\right){\left(\frac{2LkT}{h{c}_s}\right)}^3{\int }_0^{n_{max}}\frac{x^3}{e^x-1}dx \\ U=\left(\frac{3\pi }{2}\right)\left(\frac{2L}{h{c}_s}\right){\left(kT\right)}^4{\int }_0^x \end{gathered}$$_maxx³e^x-1dx

simplify

We need to change the limits of the integral, the lover 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:

$x_{max}=\frac{T_D}{T}=\frac{h{c}_s{n}_{max}}{2LkT}=\frac{h{c}_s}{2LkT}{\left(\frac{6N}{\pi }\right)}^{1/3}$

$T_D=\frac{h{c}_s}{2Lk}{\left(\frac{6N}{\pi }\right)}^{1/3}$

$$\begin{gathered} {T_D}^3=\frac{1}{k^3}{\left(\frac{h{c}_s}{2L}\right)}^3\left(\frac{6N}{\pi }\right) \\ {\left(\frac{2L}{h{c}_s}\right)}^3=\frac{1}{{\left(k{T}_D\right)}^3}\left(\frac{6N}{\pi }\right) \end{gathered}$$

substitute into (2) to get:

$U=\left(\frac{3\pi }{2}\right)\frac{1}{{\left(k{T}_D\right)}^3}\left(\frac{6N}{\pi }\right){\left(kT\right)}^4{\int }_0^x$_maxx3ex-1dxU=9NkT4TD3∫0xmaxx3ex-1dx

Simplify

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

$C_V=\frac{aU}{aT}$

substitute form(1) to get$C_V=9Nk{\left(\frac{T}{T_D}\right)}^3{\int }_0^{x_{max}}\frac{x^3{e}^x}{{(e^x-1)}^2}dx$

$C_V=\left(\frac{3\pi h{c}_s}{4L}\right){\int }_0^{n_{max}}\frac{\partial U}{\partial T}\left[\frac{n^3}{e^{h{c}_{s}n/2LkT}-1}\right]dn$

$C_V=\left(\frac{3\pi h{c}_s}{4L}\right){\int }_0^{n_{max}}\frac{h{c}_{s}n}{2Lk{T}^2}\frac{n^3{e}^{h{c}_{s}n/2LkT}}{{(e^{h{c}_{s}n/2LkT}-1)}^2}dn$

$C_V=\frac{3\pi }{2}{\left(\frac{h{c}_s}{2L}\right)}^2\frac{1}{k{T}^2}{\int }_0^{n_{max}}\frac{n^4{e}^{h{c}_{s}n/2LkT}}{(e^{h{c}_{s}n/2LkT}-1)}dn$

now let,

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

thus,

$C_V=\frac{3\pi }{2}{\left(\frac{h{c}_s}{2L}\right)}^2\frac{1}{k{T}^2}{\left(\frac{2LkT}{h{c}_s}\right)}^5{\int }_0^{x_{max}}\frac{x^4{e}^x}{{(e^x-1)}^2}dx$

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

$C_V=\frac{3\pi }{2}\frac{1}{{\left(k{T}_D\right)}^3}\left(\frac{6N}{\pi }\right)k^4{T}^3{\int }_0^{x_{max}}\frac{x^4{e}^x}{{(e^x-1)}^2}dx$