Exercises

1. Let
   $$I_n = \int_0^\infty {\rm e}^{-x} x^{ n} dx,$$
   where $n$ is a non-negative integer. Integrating by parts, demonstrate that
   $$I_n = n I_{n-1}.$$
   Furthermore, show that
   $$I_0 = 1.$$
   Hence, deduce that
   $$I_n = n!.$$

2. Let
   $$J_n = \int_0^\infty \frac{x^{ n-1} dx}{{\rm e}^{ x}-1},$$
   where $n$ is an integer greater than unity. Demonstrate that
   $$J_n= (n-1)! \zeta(n),$$
   where
   $$\zeta(n)=\sum_{k=1,\infty} \frac{1}{k^{ n}}$$
   is a Riemann zeta function. [Hint: $(1-r)^{-1} = \sum_{k=0,\infty} r^{ k}$ , provided that $\vert r\vert<1$ .] In general, the zeta function must be evaluated numerically. However, it is possible to prove that $\zeta(2)=\pi^{ 2}/6$ and $\zeta(4)=\pi^{ 4}/90$ .

3. Show that
   $$\int_{-\infty}^\infty \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}} x^{ 2} dx$$
   can be written in the form
   $$4\int_0^\infty \frac{x}{{\rm e}^{ x}+1} dx.$$
   By expanding the integrand in powers of ${\rm e}^{-x}$ , demonstrate that
   $$\int_0^\infty \frac{x}{{\rm e}^{ x}+1} dx=\int_0^\infty\left(x\... ...{-x}-x {\rm e}^{-2 x}+x {\rm e}^{-3 x}-x {\rm e}^{-4 x}+\cdots\right)dx,$$
   and, hence, that
   $$\int_0^\infty \frac{x}{{\rm e}^{ x}+1} dx=1-\frac{1}{2^{ 2}}+\frac{1}{3^{ 2}}-\frac{1}{4^{ 2}}+\cdots.$$
   Rearrange the right-hand side of the previous expression to give

   $$\int_0^\infty \frac{x}{{\rm e}^{ x}+1} dx =\left(1+\frac{1}{2^{ 2}}+\frac{1}{3^{ 2}}+\frac{1}{4^{ 2}}+\c... ...ft(\frac{1}{2^{ 2}}+\frac{1}{4^{ 2}}+\frac{1}{6^{ 2}}+\cdots\right)\nonumber$$

   $$= \left(1+\frac{1}{2^{ 2}}+\frac{1}{3^{ 2}}+\frac{1}{4^{ 2}} +... ...(1+\frac{1}{2^{ 2}}+\frac{1}{3^{ 2}}+\frac{1}{4^{ 2}}+\cdots\right)\nonumber$$

   $$= \zeta(2) - \frac{1}{2} \zeta(2) = \frac{1}{2} \zeta(2) = \frac{\pi^{ 2}}{12}.$$

   
   
   (See Exercise 2.) Thus, deduce that
   $$\int_{-\infty}^\infty \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}} x^{ 2} dx=\frac{\pi^{ 2}}{3}.$$

4. Consider the integral
   $$\int_{-\infty}^\infty \frac{x^{ s-1}}{{\rm e}^{ x}-1} dx,$$
   where $s>1$ . By expanding in powers of ${\rm e}^{-x}$ , show that
   $$\int_{-\infty}^\infty \frac{x^{ s-1}}{e^{ x}-1} dx={\mit\Gamma}(s) \zeta(s),$$
   where ${\mit\Gamma}(s)=\int_0^\infty x^{ s-1} {\rm e}^{-x} dx$ is a Gamma function, and $\zeta(s)=\sum_{n=1,\infty}n^{-s}$ a Riemann zeta function. Likewise, demonstrate that
   $$\int_{-\infty}^\infty \frac{x^{ s-1}}{e^{ x-\alpha}-1} dx={\mit\Gamma}(s)\sum_{n=1,\infty}\frac{{\rm e}^{ n \alpha}}{n^{ s}},$$
   where $\alpha<0$ .

5. Consider a gas consisting of identical non-interacting particles. The quantum states of a single particle are labeled by the index $r$ . Let the energy of a particle in state $r$ be $\epsilon_r$ . Let $n_r$ be the number of particles in quantum state $r$ . The partition function of the gas is thus
   $$Z=\sum_R \exp\left(-\beta\sum_r n_r \epsilon_r\right),$$
   where the first sum is over all allowable values of the $n_r$ , and the second is over all single-particle quantum states. Here, $\beta = 1/(k T)$ , where $T$ is the absolute temperature.
   1. Demonstrate that
      $$\bar{n}_r=-\frac{1}{\beta} \frac{\partial \ln Z}{\partial\epsilon_r}.$$

   2. Show that
      $$\overline{n_r^{ 2}}=\frac{1}{\beta^{ 2} Z} \frac{\partial^{ ... ...artial\ln Z}{\partial\epsilon_r}\right)+\beta^{ 2} (\bar{n}_r)^{ 2}\right].$$

   3. Hence, deduce that
      $$\overline{({\mit\Delta}n_r)^{ 2}} = \frac{1}{\beta^{ 2}} \frac... ...ilon_r^{ 2}}=-\frac{1}{\beta} \frac{\partial\bar{n}_r}{\partial \epsilon_r}.$$

6. Use the results of the previous exercise to show that:
   1. For identical, non-interacting, particles distributed according to the Maxwell-Boltzmann distribution,
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\frac{... ...r}\left(1-\frac{\bar{n}_r}{\sum_s \bar{n}_s}\right)\simeq \frac{1}{\bar{n}_r}.$$

   2. For photons,
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\frac{1}{\bar{n}_r}+1.$$

   3. For identical, non-interacting, massive particles distributed according to the Bose-Einstein distribution,
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\left(... ...ght)\left(1+\frac{1}{\beta} \frac{\partial\alpha}{\partial\epsilon_r}\right).$$
      Show, also, that
      $$\frac{1}{\beta} \frac{\partial\alpha}{\partial\epsilon_r}=-\frac{\bar{n}_r (1+\bar{n}_r)}{\sum_s \bar{n}_s (1+\bar{n}_s)}.$$
      Hence, deduce that
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\left(... ...bar{n}_r)}{\sum_s \bar{n}_s (1+\bar{n}_s)}\right]\simeq\frac{1}{\bar{n}_r}+1.$$

   4. For identical, non-interacting, massive particles distributed according to the Fermi-Dirac distribution,
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\left(... ...ght)\left(1+\frac{1}{\beta} \frac{\partial\alpha}{\partial\epsilon_r}\right).$$
      Show, also, that
      $$\frac{1}{\beta} \frac{\partial\alpha}{\partial\epsilon_r}=-\frac{\bar{n}_r (1-\bar{n}_r)}{\sum_s \bar{n}_s (1-\bar{n}_s)}.$$
      Hence, deduce that
      $$\frac{\overline{({\mit\Delta}n_r)^{ 2}}}{\bar{n}_r^{ 2}}=\left(... ...bar{n}_r)}{\sum_s \bar{n}_s (1-\bar{n}_s)}\right]\simeq\frac{1}{\bar{n}_r}-1.$$

   Note that, in the case of the Bose-Einstein distribution, the relative dispersion in $\bar{n}_r$ is larger than in the Maxwell-Boltzmann case, whereas in the case of the Fermi-Dirac distribution the relative dispersion is smaller.

7. Consider a non-relativistic free particle of mass $m$ in a cubical container of edge-length $L$ , and volume $V=L^{ 3}$ .
   1. Show that the energy of a general quantum state, $r$ , can be written
      $$\epsilon_r= \frac{h^{ 2}}{V^{ 2/3}} (n_x^{ 2}+n_y^{ 2}+n_z^{ 2}),$$
      where $n_x$ , $n_y$ , and $n_z$ are positive integers. Hence, deduce that the contribution to the gas pressure of a particle in this state is
      $$p_r=-\frac{\partial\epsilon_r}{\partial V} = \frac{2}{3} \frac{\epsilon_r}{V}.$$

   2. Demonstrate that the mean pressure of a gas of weakly-interacting non-relativistic particles is
      $$\bar{p}= \frac{2}{3} \frac{\overline{E}}{V},$$
      where $\overline{E}$ is its mean total kinetic energy, irrespective of whether the particles obey classical, Bose-Einstein, or Fermi-Dirac statistics.

8. As an electron moves in a molecule, there exists, at any instance in time, a separation of positive and negative charges within the molecule.The molecule therefore possesses a time-varying electric dipole moment, ${\bf p}$ . Assuming that the molecule is located at the origin, its instantaneous dipole moment generates an electric field
   $${\bf E}({\bf r}) = \frac{3 ({\bf p}\cdot{\bf r}) {\bf r}-r^{ 2} {\bf p}}{4\pi \epsilon_0 r^{ 5}},$$
   where ${\bf r}$ is the vector displacement from the origin. A second molecule, located at position ${\bf r}$ , relative to the first, develops an induced dipole moment, ${\bf p}'=\alpha {\bf E}({\bf r})$ , in response to the electric field of the first. Here, $\alpha$ is the molecular polarizability. Moreover, the electric field generates a force ${\bf f}'=\nabla[{\bf p}'\cdot {\bf E}({\bf r})]$ acting on the second molecule. (Of course, an equal and opposite force acts on the first molecule.) Use these facts to deduce that the mean (i.e., averaged over time, and all possible orientations of ${\bf p}$ ) inter-molecular potential between a pair of neutral molecules is attractive, and takes the form
   $$u(r) = -\frac{\alpha \langle p^{ 2}\rangle}{16\pi^{ 2} \epsilon_0^{ 2} r^{ 6}},$$
   where $\langle\cdots\rangle$ denotes a time average.

9. Consider a van der Waals gas whose equation of state is
   $$\bar{p} =\frac{R T}{v-b}-\frac{a}{v^{ 2}}.$$
   Here, $a>0$ , $>0$ are constants, and $v$ is the molar volume. Show that the various thermodynamic potentials of the gas are:

   $$F =-\nu R T\left\{\ln\left[\frac{n_Q}{N_A} (v-b)\right]+1\right\}-\frac{\nu a}{v},$$

   $$S =\nu R\left\{\ln\left[\frac{n_Q}{N_A} (v-b)\right]+\frac{5}{2}\right\},$$

   $$\overline{E} = \frac{3}{2} \nu R T-\frac{\nu a}{v},$$

   $$H =\nu R T\left(\frac{v}{v-b}+\frac{3}{2}\right)+\frac{2 \nu a}{v},$$

   $$G =\nu R T\left\{-\ln\left[\frac{n_Q}{N_A} (v-b)\right]+\frac{v}{v-b}-1\right\}-\frac{2 \nu a}{v}.$$

   
   
   Here, $n_Q=(2\pi m k T/h^{ 2})^{3/2}$ , where $m$ is the molecular mass, is known as the quantum concentration, and is the particle concentration at which the de Broglie wavelength is equal to the mean inter-particle spacing. Obviously, the previous expression are only valid when $N_A/v\ll n_Q$ .

10. Show that the energy density of radiation inside a radiation-filled cavity whose walls are held at absolute temperature $T$ is
    $$\bar{u}= \frac{\pi^{ 2}}{15} \frac{(k T)^{ 4}}{(c \hbar)^{ 3}}.$$
    Demonstrate that the mean pressure that the radiation exerts on the walls of the cavity is
    $$\bar{p} = \frac{1}{3} \bar{u}.$$
    [Hint: Modify and reuse the analysis of Exercise 7.]

11. Apply the thermodynamic relation $T dS=d\overline{E}+\bar{p} dV$ to a photon gas. Here, we can write $\overline{E}=V \bar{u}$ , where $\bar{u}(T)$ is the mean energy density of the radiation (which is independent of the volume). The radiation pressure is $\bar{p}=\bar{u}/3$ .
    1. Considering $S$ as a function of $T$ and $V$ , show that

       $$\left(\frac{\partial S}{\partial V}\right)_T =\frac{4}{3} \frac{\bar{u}}{T},$$

       $$\left(\frac{\partial S}{\partial T}\right)_V = \frac{V}{T} \frac{d\bar{u}}{d T}.$$

       
       

    2. Demonstrate that the mathematical identity $(\partial^{ 2} S/\partial V \partial T)=(\partial^{ 2} S/\partial T \partial V)$ leads to a differential equation for $\bar{u}(T)$ that can be integrated to give
       $$\bar{u} = a T^{ 4},$$
       where $a$ is a constant.
    3. Hence, deduce that
       $$S = \frac{4}{3} \frac{\overline{E}}{T}.$$

12. Show that the total number of photons in a radiation-filled cavity of volume $V$ , whose walls are held at temperature $T$ , is
    $$N = 16\pi \zeta(3) V\left(\frac{k T}{h c}\right)^3,$$
    where
    $$\zeta(n)= \frac{1}{(n-1)!}\int_0^\infty \frac{x^{ n-1} dx}{{\rm e}^{ x}-1}$$
    is a Riemann zeta function. (See Exercise 2.) Note that $\zeta(3)\simeq 1.202$ . Demonstrate that the mean energy per photon and the entropy per photon are

    $$\frac{\overline{E}}{N} = \frac{\pi^{ 4}}{30 \zeta(3)} k T = 2.70 k T,$$

    $$\frac{S}{N} = \frac{2 \pi^{ 4}}{45 \zeta(3)} k = 3.60 k,$$

    
    
    respectively. Note that the entropy per photon is a constant, independent of the temperature.

13. Electromagnetic radiation at temperature $T$ fills a cavity of volume $V$ . If thecavity is thermally insulated, and expands quasi-statically, show that
    $$T\propto \frac{1}{V^{ 1/3}}.$$
    (Neglect the heat capacity of the cavity walls.)

14. 1. The partition function for a photon gas is
       $$\ln Z = -\sum_r \ln\left(1-{\rm e}^{-\beta \epsilon_r}\right),$$
       where $\epsilon_r = \hbar \omega_r$ , and $\omega_r$ is the frequency of an individual photon state. The spacing between the allowed frequencies can be made arbitrarily small by increasing the volume, $V$ , of the container. Hence, the sum can be approximated as an integral. Now, the number of photon states per unit volume whose frequencies lie between $\omega$ and $\omega+d \omega$ is $\sigma(\omega) d\omega$ , where
       $$\sigma(\omega)= 2 \frac{\omega^{ 2}}{2\pi^{ 2} c^{ 3}}.$$
       (See Section 8.12.) Thus, we can write
       $$\ln Z = -V\int_0^\infty\sigma(\omega) \ln\left(1-{\rm e}^{-\beta \hbar \omega}\right)d\omega.$$
       Show that
       $$\ln Z = -8\pi V\left(\frac{k T}{h c}\right)^3\int_0^\infty y^{ 2}\ln\left(1-{\rm e}^{-y}\right)dy.$$
       Integrating by parts, demonstrate that
       $$\ln Z = \frac{8\pi}{3} V \left(\frac{k T}{h c}\right)^3\int_0^\infty \frac{y^{ 3}}{{\rm e}^{ y}-1} dy.$$
       Hence, deduce that
       $$\ln Z = \frac{8\pi^{ 5}}{45}\left(\frac{k T}{h c}\right)^3V.$$
       [Hint: See Exercise 2.]

    2. Using the standard results

       $$\overline{E} = -\frac{\partial \ln Z}{\partial \beta},$$

       $$\bar{p} = \frac{1}{\beta} \frac{\partial\ln Z}{\partial V},$$

       $$S =k (\ln Z + \beta \overline{E}),$$

       
       
       show that

       $$\overline{E} = \frac{8\pi^{ 5}}{15} \frac{(k T)^4}{(h c)^3} V,$$

       $$\bar{p} = \frac{\overline{E}}{3 V},$$

       $$S = \frac{4}{3} \frac{\overline{E}}{T}.$$

       
       

15. Show that the power per unit area radiated by a black-body of temperature $T$ peaks atangular frequency $\omega_c$ , where $\hbar \omega_c = y k T$ , and $y$ is the solution of the transcendental equation
    $$y = 3 (1-{\rm e}^{-y}).$$
    Solve this equation by iteration [i.e., $y_{n+1}= 3 (1-{\rm e}^{-y_{n}})$ , where $y_n$ is the $n$ th guess], and, thereby, show that
    $$\hbar \omega_c = 2.82 k T.$$

16. A black (non-reflective) plane at temperature $T_u$ is parallel to a black plane at temperature $T_l<T_u$ . The net energy flux density in vacuum between the two planes is $J_U = \sigma (T_u^{ 4} - T_l^{ 4})$ , where $\sigma$ is the Stefan-Boltzmann constant. A third black plane is inserted between the other two, and is allowed to come to a steady-state temperature $T_m$ . Find $T_m$ in terms of $T_u$ and $T_l$ , and show that the net energy flux is cut in half because of the presence of this plane. This is the principle of the heat shield, and is widely used to reduce radiant heat transfer.

17. Consider the conduction electrons in silver, which is monovalent (one free electron per atom), has a mass density $10.5\times 10^{ 3} {\rm kg} {\rm m}^{-3}$ , and an atomic weight of 107. Show that the Fermi energy (at $T=0$ ) is
    $$\mu_F = 5.6 {\rm eV},$$
    and the equivalent temperature--the so-called Fermi temperature--is
    $$T_F = \frac{\mu_F}{k} = 65,000 {\rm K}.$$
    Thus, if $T\ll T_F$ (as is definitely the case at room temperature) then the conduction electrons are highly degenerate. Furthermore, show that at room temperature ($T=300$ K), the contribution of the conduction electrons to the molar specific heat is only
    $$\frac{c_V^{ (e)}}{R} = 2.2\times 10^{-2}.$$

18. Demonstrate that the mean pressure of the conduction electrons in a metal can be written
    $$\bar{p} = \frac{2}{5} \frac{N k T_F}{V}\left[1+\frac{5\pi^{ 2}}{12}\left(\frac{T}{T_F}\right)^{ 2} + \cdots\right],$$
    where $T_F$ is the Fermi temperature. (Hint: See Exercise 17.) Show that for silver (see previous exercise),
    $$\bar{p} = 2.1\times 10^{ 10} {\rm Pa}.$$

19. Show that the contribution of the conduction electrons in a metal to the isothermal compressibility is
    $$\kappa_T = \frac{3}{5 \bar{p}},$$
    where $\bar{p}$ is the electron's mean pressure. Estimate $\kappa_T$ for silver. (See the previous exercise.) Compare your estimate to the experimental value $0.99\times 10^{-11} {\rm Pa}^{-1}$ .

20. Justify Equation (8.197).

21. A system of $N$ bosons of mass $m$ and zero spin is in a container of volume $V$ , at an absolute temperature $T$ . The number ofparticles is
    $$N = \frac{V}{4\pi^{ 2}}\left(\frac{2 m}{\hbar^{ 2}}\right)\int_0^\infty \frac{\epsilon^{ 1/2}}{{\rm e}^{ (x-\mu)/k T}-1} d\epsilon.$$
    [See Equation (8.227).] In the limit in which the gas is dilute, $\exp(-\mu/k T)\gg 1$ , and the Bose-Einstein distribution becomes the Maxwell-Boltzmann distribution. Evaluate the integral in this approximation, and show that
    $$\exp\left(-\frac{\mu}{k T}\right)=\left(\frac{d}{\lambda}\right)^{3},$$
    where $\lambda = h/(2\pi m k T)^{1/2}$ is the particles' de Broglie wavelength, and $d=(V/N)^{ 1/3}$ the mean inter-particle distance. Hence, deduce that, in the classical limit, the average distance between particles is much larger than the de Broglie wavelength.