The Homework of Chapter six

Author： 宗玥 2014301020004

Statistical-and-Thermal-Physics

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Problem 6.1. Mean distance between particles

(a) Consider a system of N particles confined to a line of length L. What is the definition of the particle density ρ? The mean distance between particles is L/N. How does this distance depend on ρ?

(a) $\rho=N/L$
The mean distance between two particles is $d=L/N=1/\rho$

(b) Consider a system of N particles confined to a square of linear dimension L. How does the mean distance between particles depend on ρ?

(b) $\rho=N/L^2$, the mean distance of two paricles is $d^2=L^2/N$, thus $d=1/\rho^{1/2}$

(c) Use similar considerations to determine the density dependence of the mean distance between particles in three dimensions.

(c) $d=\rho^{-1/3}$

Problem 6.2. Independence of the partition function on the shape of the box

The volume dependence of Z1 should be independent of the shape of the box. Show that the same result for Z1 is obtained if the box has linear dimensions Lx, Ly, and Lz with V = LxLyLz.

According to Eq.6.9 we have $S_x=L_x(\frac{2\pi m}{\beta h^2 })^{1/2}-1$. Because in (6.9) is order $L_x/\lambda \gg1$, we can obtain$S_x=L_x(\frac{2\pi m}{\beta h^2 })^{1/2}$
$Z_1=S_xS_yS_z=L_xL_yL_z(\frac{2\pi m}{\beta h^2 })^{3/2}=V(\frac{2\pi m}{\beta h^2 })^{3/2}$

Problem 6.3. Semiclassical limit of the single particle partition function

We obtained the semiclassical limit of the partition function Z1 for one particle in a box by writing it as a sum over single particle states and then converting the sum to an integral. Show that the semiclassical partition function Z1 for a particle in a one-dimensional box can be expressed as

$$Z_1 =\int\int \frac{dpdx}{ h} e^{−βp^2/2m}. (6.18)$$
The integral over p in (6.18) extends from −∞ to +∞.

$$\begin{eqnarray*} Z_1&=&\int \frac{dx}{h}\int_{-\infty}^{\infty}e^{-\beta p^2/2m}dp\\ &=&\int \frac{dx}{h}2 \frac12(\frac{2m\pi}{\beta})^{1/2}\\ &=&\int(\frac{2m\pi}{\beta h^2})^{1/2}dx\\ &=&L(\frac{2m\pi}{\beta h^2})^{1/2} \end{eqnarray*}$$

Thus the result is same as Eq 6.9

Problem 6.4. Equations of state of an ideal classical gas

Use the result (6.26) to find the pressure equation of state and the mean energy of an ideal gas. Do the equations of state depend on whether the particles are indistinguishable or distinguishable?

$$\begin{eqnarray*} P&=&-\frac{\partial F}{\partial V}\\ &=&kTN\frac{dln\frac VN}{dV}\\ &=&kTN\frac{1}{V}\\ PV&=&NkT \end{eqnarray*}$$
$$\begin{eqnarray*} lnZ_N&=&ln\frac{V}{N}+\frac32ln(\frac{2\pi m}{\beta h^2})+1\\ \overline E&=&-\frac{\partial}{\partial \beta}lnZ\\ &=&-\frac32\frac{1}{\beta}=-\frac32kT \end{eqnarray*}$$

The equations of state don't depend on whether the particles are indistinguishable or distinguishable

Problem 6.5. Entropy of an ideal classical gas

(a) The entropy can be found from the relations F = E −TS or S = −∂F/∂T. Show that

$$S(T,V,N) = Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πmkT}{ h^2}) +\frac 5 2] . (6.27)$$
The form of S in (6.27) is known as the Sackur-Tetrode equation (see Problem 4.20, page 197). Is this form of S applicable for low temperatures?

(a)

$$\begin{eqnarray*} S&=&-\frac{\partial F}{\partial T}\\ &=&NklnZ+NkT\frac{\partial lnZ}{\partial T}\\ &=&Nk[ln\frac{V}{N}+\frac32ln(\frac{2\pi mkT}{ h^2})+1]+NkT\frac{3}{2T}\\ &=&Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πmkT}{ h^2}) +\frac 5 2] \end{eqnarray*}$$

(b) Express kT in terms of E and show that S(E,V,N) can be expressed as

$$S(E,V,N) = Nk[ln\frac{ V }{N} + \frac3 2 ln\frac{4πmE}{ 3Nh^2})+\frac 5 2], (6.28)$$
in agreement with the result (4.63) found using the microcanonical ensemble.

$$\begin{eqnarray*} E&=&\frac32NkT\\ S&=&Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πmkT}{ h^2}) +\frac 5 2]\\ &=&Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πm}{ h^2}\frac{2E}{3N}) +\frac 5 2]\\ S(E,V,N) &=& Nk[ln\frac{ V }{N} + \frac3 2 ln\frac{4πmE}{ 3Nh^2})+\frac 5 2] \end{eqnarray*}$$

Problem 6.6. The chemical potential of an ideal classical gas

(a) Use the relation $µ = ∂F/∂N$ and the result (6.26) to show that the chemical potential of an ideal classical gas is given by

$$\mu = −kT [ln\frac{V}{ N}(\frac{2πmkT}{ h^2})^{ 3/2}]. (6.29)$$

(a)

$$\begin{eqnarray*} \mu&=&\frac{\partial F}{\partial N}\\ &=&-kT[ln\frac{V}{N}+\frac32ln(\frac{2\pi mkT}{ h^2})+1]+kTN[\frac{1}{N}]\\ &=&-kT[ln\frac{V}{N}+\frac32ln(\frac{2\pi mkT}{ h^2})]\\ &=&-kTln[\frac{V}{ N}(\frac{2πmkT}{ h^2})^{ 3/2}] \end{eqnarray*}$$

(b) We will see in Chapter 7 that if two systems are placed into contact with different initial chemical potentials, particles will go from the system with higher chemical potential to the system with lower chemical potential. (This behavior is analogous to energy going from high to low temperatures.) Does “high” chemical potential for an ideal classical gas imply “high” or “low” density?

(b) $\rho^{-1/3}=V/N$. So it should imply low density

(c) Calculate the entropy and chemical potential of one mole of helium gas at standard temperature and pressure. Take V = 2.24 × 10−2 m3, N = 6.02 × 1023, m = 6.65 × 10−27 kg, and T = 273K.

(c)

$$\begin{eqnarray*} S(T,V,N)&=&Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πmkT}{ h^2}) +\frac 5 2]\\\ &=&124.097J/K\\ \mu&=&-kTln[\frac{V}{ N}(\frac{2πmkT}{ h^2})^{ 3/2}]\\ &=&-4.686*10^{-20} \end{eqnarray*}$$

Problem 6.7. Entropy as an extensive quantity

(a) Because the entropy is an extensive quantity, we know that if we double the volume and double the number of particles (thus keeping the density constant), the entropy must double. This condition can be written formally as

$$S(T,λV,λN) = λS(T,V,N). (6.30)$$
Although this behavior of the entropy is completely general, there is no guarantee that an approximate calculation of S will satisfy this condition. Show that the Sackur-Tetrode form of the entropy of an ideal gas of identical particles, (6.27), satisfies this general condition.

(a)

$$\begin{eqnarray*} S(T,V,N) &=& Nk[ln\frac{ V}{ N} + \frac3 2 ln(\frac{2πmkT}{ h^2}) +\frac 5 2] \\ S(T,\lambda V,\lambda N)&=&\lambda Nk[ln\frac{\lambda V}{\lambda N}+\frac32ln(\frac{2\pi mkT}{h^2})+\frac52]\\ &=&\lambda S(T,V,N) \end{eqnarray*}$$

(b) Show that if the N! term were absent from (6.25) for ZN, S would be given by

$$S = Nk[lnV +\frac 3 2 ln(\frac{2πmkT}{ h^2}) +\frac 3 2]. (6.31)$$ Is this form of S extensive?

(b)

$$\begin{eqnarray*} Z_N&=&V^N(\frac{2\pi mkT}{h^2})^{3N/2}\\ F&=&-kTlnZ_N\\ &=&-kTN[lnV+\frac32ln(\frac{2\pi mkT}{h^2})]\\ S&=&-\frac{\partial F}{\partial T}\\ &=&Nk[lnV+\frac32ln(\frac{2\pi mkT}{h^2})]+Nk\frac32\\ &=&Nk[lnV +\frac 3 2 ln(\frac{2πmkT}{ h^2}) +\frac 3 2] \end{eqnarray*}$$
Yes, it is

(c) The fact that (6.31) yields an entropy that is not extensive does not indicate that identical particles must be indistinguishable. Instead the problem arises from our identification of S with lnZ as mentioned in Section 4.6, page 199. Recall that we considered a system with fixed N and made the identification that [see (4.106)]
dS/k = d(lnZ + βE). (6.32)
It is straightforward to integrate (6.32) and obtain
S = k(lnZ + βE) + g(N), (6.33)
where g(N) is an arbitrary function only of N. Although we usually set g(N) = 0, it is important to remember that g(N) is arbitrary. What must be the form of g(N) in order that the entropy of an ideal classical gas be extensive?

$$\begin{eqnarray*} S&=&Nk[lnV+\frac32ln(\frac{2\pi mkT}{h^2})+\frac32]\\ S&=&k(lnZ+\beta E)+g(N)\\ &=&k[N[ln\frac{V}{N}+\frac32ln(\frac{2\pi mkT}{h^2})+1]+\frac{E}{kT}]+g(N)\\ &=&Nk[lnV+\frac32ln\frac{2\pi mkT}{h^2}+\frac32]-NklnN-\frac12Nk+\frac32Nk+g(N)\\ g(N)&=&Nk(lnN-1) \end{eqnarray*}$$

Problem 6.8. Entropy of mixing of identical particles

(a) Use (6.37) to show that ∆S = 0 if the two gases have equal densities before separation. Write NA = ρVA and NB = ρVB.

(a)

$$\begin{eqnarray*} N_A&=&\rho V_A\\ N_B&=&\rho V_B\\ \Delta S&=&k[(N_A+N_B)ln\frac{V_A+V_B}{N_A+N_B}-N_Aln\frac{V_A}{N_A}-N_Bln\frac{V_B}{N_B}]\\ &=&k[\rho(V_A+V_B)ln\frac{V_A+V_B}{\rho(V_A+V_B)}-\rho V_Aln\frac{V_A}{\rho V_A}-\rho V_Bln\frac{V_B}{\rho V_B}]\\ &=&k[\rho(V_A+V_B)ln\frac{1}{\rho}-\rho(V_A+V_B)ln\frac{1}{\rho}]\\ &=&0 \end{eqnarray*}$$

(b) Why is the entropy of mixing nonzero if the two gases initially have different densities even though the particles are identical?

(b)
If the particles are identical,$N_A=N_B=N$

$$\begin{eqnarray*} \Delta S&=&k[(N_A+N_B)ln\frac{V_A+V_B}{N_A+N_B}-N_Aln\frac{V_A}{N_A}-N_Bln\frac{V_B}{N_B}]\\ &=&k[2Nln\frac{V_A+V_B}{2N}-Nln\frac{V_AV_B}{N^2}]\\ &=&k[Nln(\frac{\rho_A+\rho_B}{2\rho_A\rho_B})^2-Nln\frac{1}{\rho_A\rho_B}]\\ &=&NKln\frac{(\rho_A+\rho_B)^2}{4\rho_A\rho_B}\\ &\ne&0 \end{eqnarray*}$$

Problem 6.9. More on the entropy of mixing

(a) Explain the result (6.40) for nonidentical particles in simple terms.

(a) When $N_A=N_B=N$, $V_A=V_B=V$, for equation 6.39, we can get a special result.

$$\Delta S=2Nkln2$$

(b) Consider the special case NA = NB = N and VA = VB = V and show that if we use the result (6.31) instead of (6.27), the entropy of mixing for identical particles is nonzero. This incorrect result is known as Gibbs paradox. Does it imply that classical physics, which assumes that particles of the same type are distinguishable, is incorrect?

(b)

$$\begin{eqnarray*} S&=&Nk[lnV+\frac32ln(\frac{2\pi mkT}{h^2})+\frac32]\\ &=&k(N_A+N_B)[ln(V_A+V_B)+\frac32ln(\frac{2\pi (m_A+m_B)kT}{h^2})+\frac32]\\ S_A&=&N_Ak[lnV_A+\frac32ln(\frac{2\pi m_AkT}{h^2})+\frac32]\\ S_B&=&N_Bk[lnV_B+\frac32ln(\frac{2\pi m_BkT}{h^2})+\frac32]\\ \Delta S&=&k[(N_A+N_B)ln(V_A+V_B)-N_AlnV_A-N_BlnV_B] \end{eqnarray*}$$
If $N_A=N_B=N$, $V_A=V_B=V$
$$\begin{eqnarray*} \Delta S&=&k[2Nln(2V)-2NlnV]\\ &=&2Nkln2 \end{eqnarray*}$$

Problem 6.10. Is there an upper limit to the velocity?

The upper limit to the velocity of a particle is the velocity of light. Yet the Maxwell velocity distribution imposes no upper limit to the velocity. Does this contradiction lead to difficulties?

Of course not. Though the Maxwell velocity distrabution imposes no upper limit to the velocity, the higher speed, the less probability. And we know when the area of $5\sigma$, the probability is almost 99.999994%. Thus we also can think that the velocity of light is the upper limit.

Problem 6.11. Simulations of the Maxwell velocity distribution

(a) Program LJ2DFluidMD simulates a system of particles interacting via the Lennard-Jones potential (1.1) in two dimensions by solving Newton’s equations of motion numerically. The program computes the distribution of velocities in the x-direction among other quantities. Compare the form of the velocity distribution to the form of the Maxwell velocity distribution in (6.60). How does its width depend on the temperature?

(a）




Higher temperature, wider it becomes.

(b) Program IdealThermometerIdealGas implements the demon algorithm for an ideal classical gas in one dimension (see Section 4.9). All the particles have the same initial velocity. The program computes the distribution of velocities among other quantities. What is the form of the velocity distribution? Give an argument based on the central limit theorem (see Section 3.7) to explain why the distribution has the observed form. Is this form consistent with (6.60)?

(b)

The form of the velocity distribution is guassian.
Because the system has reached equilibrium, the mean velocity of all particles must be 0. The probability that one particle has velocity. This form is consistent with 6.60.

Problem 6.12. Maxwell speed distribution

(a) Compare the form of the Maxwell speed distribution (6.61) with the form of the Maxwell velocity distribution (6.59).

(a)
The Maxwell velocity distrabution is

$$f(v)dv=4\pi Av^2e^{-mv^2/2kT}dv$$
The Maxwell speed distribution is
$$f(v_x,v_y,v_z)=(\frac{m}{2\pi kT})^{3/2}e^{-m(v_x^2+v_y^2+v_z^2)/2kT}dv_xdv_ydv_z$$

(b) Use the normalization condition $\int_0^{\infty}f(v)dv = 1$ to calculate A and show that

$$f(v)dv = 4πv^2(\frac{ m}{ 2πkT})^{3/2}e^{−mv^2/2kT}dv$$ (Maxwell speed distribution). (6.62)

(b)

$$\begin{eqnarray*} 1&=&\int_0^{\infty}4πv^2Ae^{−mv^2/2kT}dv\\ &=&4πA\int_0^{\infty}v^2e^{-\frac{m}{2kT}v^2}dv\\ &=&4πA\frac14\sqrt{\frac{\pi}{(\frac{m}{2kT})^3}}\\ &=&A(\frac{2kT\pi}{m})^{3/2}\\ A&=&(\frac{m}{2\pi kT})^{3/2}\\ f(v)dv&= &4πv^2(\frac{ m}{ 2πkT})^{3/2}e^{−mv^2/2kT}dv \end{eqnarray*}$$

(c) Calculate the mean speed $\overline v$, the most probable speed $\widetilde v$, and the root-mean-square speed vrms and discuss their relative magnitudes.

(c)

$$\begin{eqnarray*} \overline v&=&\int_0^{\infty}v·4πv^2(\frac{ m}{ 2πkT})^{3/2}e^{−mv^2/2kT}dv\\ &=&4π(\frac{ m}{ 2πkT})^{3/2}\int_0^{\infty}v^3e^{-\frac{m}{2kT}v^2}dv\\ &=&4π(\frac{ m}{ 2πkT})^{3/2}\frac{1}{2(\frac{m}{2kT})^2}\\ &=&\sqrt{\frac{8kT}{\pi m}}\\ \overline {v^2}&=&4π(\frac{ m}{ 2πkT})^{3/2}\int_0^{\infty}v^4e^{-\frac{m}{2kT}v^2}dv\\ &=&4π(\frac{ m}{ 2πkT})^{3/2}·\frac38·\sqrt{\frac{\pi}{(\frac{m}{2kT})^5}}\\ &=&\frac{3 kT}{m}\\ v_{rms}&=&\sqrt{\overline{v^2}}=\sqrt{\frac{3 kT}{m}}\\ \frac{df(v)}{dv}&=&4\pi(\frac{m}{2\pi kT})^{3/2}(2ve^{-mv^2/2kT}-v^2e^{-mv^2/2kT})=0\\ \widetilde v&=&{}\sqrt{\frac{2kT}{m}} \end{eqnarray*}$$

(d) Make the change of variables $u = v/\sqrt{(2kT/m)}$ and show that

$$f(v)dv = f(u)du = (4/\sqrt π)u^2e^{−u^2}du, (6.63)$$
where we have again used the same notation for two different, but physically related probability densities. The (dimensionless) speed probability density f(u) is shown in Figure 6.2.

$$\begin{eqnarray*} u&=&\frac{v}{\sqrt{\frac{2kT}{m}}}\\ f(v)dv&=&4πv^2(\frac{ m}{ 2πkT})^{3/2}e^{−mv^2/2kT}dv \\ &=&4\pi \frac{v^2}{\frac{2kT}{m}}(\frac{1}{\pi})^{\frac32}(\frac{m}{2kT})^{1/2}dv\\ &=&\frac{4}{\sqrt{\pi}}u^2e^{-u^2}du\\ &=&f(u)du \end{eqnarray*}$$

Problem 6.13. Maxwell speed distribution in one or two dimensions

Find the Maxwell speed distribution for particles restricted to one and two dimensions.

$$\begin{eqnarray*} f(\vec v)&=&f(v_x)f(v_y)f(v_z)\\ &=&(\frac{m}{2\pi kT})^{3/2}e^{-\frac{mv^2}{2kT}}dv_xdv_ydv_z \end{eqnarray*}$$
Thus for each direction is $f(v_i)=(\frac{m}{2\pi kT})^{1/2}e^{-\frac{mv_i^2}{2kT}}dv_i$
One dimention:
$$\begin{eqnarray*} f(\vec v)&=&(\frac{m}{2\pi kT})^{1/2}e^{-\frac{mv^2}{2kT}}dv\\ f(v)&=&v(\frac{m}{w\pi kT})^{1/2}e^{-\frac{mv_i^2}{2kT}}dv \end{eqnarray*}$$

Two dimentions:

$$\begin{eqnarray*} f(\vec v)&=&\frac{m}{2\pi kT}e^{-\frac{mv^2}{2kT}}dv\\ f(v)&=\int_0^{2\pi}&\frac{m}{2\pi kT}e^{-\frac{mv^2}{2kT}}vdvd\theta\\ &=&\frac{mv}{kT}e^{-\frac{mv^2}{2kT}}dv \end{eqnarray*}$$

Problem 6.14.

Calculate n1, the mean number of fermions in the single particle microstate 1 with energy 1, for the system in Example 6.1.

0.75

Problem 6.15. Mean energy of a toy model of an ideal Bose gas

(a) Calculate the mean energy of an ideal gas of N = 2 identical bosons in equilibrium with a heat bath at temperature T, assuming that each particle can be in one of three microstates with energies 0, ∆, and 2∆.

(a)

0	$\Delta$	2$\Delta$
2	0	0
0	2	0
0	0	2
1	1	0
1	0	1
0	1	1

$$\begin{eqnarray*} Z_2&=&1+e^{-2\beta\Delta}+e^{-4\beta\Delta}+e^{-\beta\Delta}+e^{-2\beta\Delta}+e^{-3\beta\Delta}\\ \overline E&=&\frac{1}{Z_2}\times0+\frac{e^{-\beta\Delta}}{Z_2}\times\Delta+\frac{2e^{-2\beta\Delta}}{Z_2}\times2\Delta+\frac{e^{-3\beta\Delta}}{Z_2}\times3\Delta+\frac{e^{-4\beta\Delta}}{Z_2}\times4\Delta\\ &=&\frac{e^{-\beta\Delta}+4e^{-2\beta\Delta}+3e^{-3\beta\Delta}+4e^{-4\beta\Delta}}{1+e^{-2\beta\Delta}+e^{-4\beta\Delta}+e^{-\beta\Delta}+e^{-2\beta\Delta}+e^{-3\beta\Delta}}\Delta \end{eqnarray*}$$

(b) Calculate the mean energy for N = 2 distinguishable particles assuming that each particle can be in one of three possible microstates.

(b)

$$\begin{eqnarray*} \overline E&=&-\frac{\partial lnZ_2}{\partial \beta}=\frac{e^{-\beta\Delta}+2e^{-2\beta\Delta}}{1+e^{-\beta\Delta}+e^{-2\beta\Delta}}·2\Delta \end{eqnarray*}$$

(c) If ¯ E1 is the mean energy for one particle and ¯ E2 is the mean energy for the two-particle system, is ¯ E2 = 2 ¯ E1 for either bosons or distinguishable particles?

(c)
For bose system, $\bar E_1=\frac{e^{-\beta\Delta}+2e^{-2\beta\Delta}}{1+e^{-\beta\Delta}+e^{-2\beta\Delta}}·\Delta$, $\bar E_2=\frac{e^{-\beta\Delta}+4e^{-2\beta\Delta}+3e^{-3\beta\Delta}+4e^{-4\beta\Delta}}{1+e^{-2\beta\Delta}+e^{-4\beta\Delta}+e^{-\beta\Delta}+e^{-2\beta\Delta}+e^{-3\beta\Delta}}\Delta$
$\overline E_1\neq2\overline E_2$
For distinguishable particles, $\bar E_1=\frac{e^{-\beta\Delta}+2e^{-2\beta\Delta}}{1+e^{-\beta\Delta}+e^{-2\beta\Delta}}·\Delta$, $\bar E_2=\frac{e^{-\beta\Delta}+2e^{-2\beta\Delta}}{1+e^{-\beta\Delta}+e^{-2\beta\Delta}}·2\Delta$
$2\overline E_1=\overline E_2$

Problem 6.16. Single particle density of states in one and two dimensions

Find the form of the density of states in k-space for standing waves in a two-dimensional and in a one-dimensional box.

First of all, we know that in one dimention, the condition for a standing wave is that the wavelength satisfied the condition

$$\lambda=\frac{2L}{n}$$
Thus we get the wave number k as $k=\frac{2\pi}{\lambda}$
For one dimention $\vec k=n\frac{\pi}{L}$, and $\Gamma(n)=n$
$$\begin{eqnarray*} \Gamma(n)&=&n\\ &=&\frac{kL}{\pi}\\ g(k)dk&=&\frac{d\Gamma(k)}{dk}dk\\ &=&\frac{Ldk}{\pi} \end{eqnarray*}$$
For two dimention $\vec k=(n_x,n_y)\frac{\pi}{L}$, and $\Gamma(n)=\frac14\pi n^2$
$$\begin{eqnarray*} \Gamma(n)&=&\frac14\pi n^2\\ &=&\frac14 \pi \frac{k^2 L^2}{\pi^2}\\ g(k)dk&=&\frac{\Gamma(k)}{dk}dk\\ &=&\frac{kL^2}{2\pi}dk \end{eqnarray*}$$

Problem 6.17. Density of states in one and two dimensions

Calculate the density of states $g(\epsilon)$ for a nonrelativistic particle of mass m in in one and two dimensions (see Problem 6.16). Sketch $g(\epsilon)$ on one graph for d = 1, 2, and 3 and comment on the different dependence of $g(\epsilon)$ on $\epsilon$ for different spatial dimensions.

As we know that $\epsilon=\frac{\hbar^2k}{2m}$, $d\epsilon=\frac{\hbar^2k}{m}dk$
For one dimention

$$\begin{eqnarray*} g(k)dk&=&\frac{L}{\pi}dk\\ g(\epsilon)d\epsilon&=&\frac{L}{\pi}\frac{m}{\hbar^2}\sqrt{\frac{\hbar^2}{2m\epsilon}}d\epsilon\\ &=&\frac{\sqrt{m/2}}{\hbar \epsilon^{1/2}}d\epsilon \end{eqnarray*}$$
For two dimentions
$$\begin{eqnarray*} g(k)dk&=&\frac{kL}{2\pi}dk\\ g(\epsilon)d\epsilon&=&\frac{L}{2\pi}\frac{m}{\hbar^2}d\epsilon \end{eqnarray*}$$

Problem 6.18. Relativistic particles

Calculate the density of states $g(\epsilon)$ in three dimensions for a relativistic particle of rest mass m for which $\epsilon^2 = p^2c^2 + m^2c^4$. Don’t try to simplify your result.

$$\begin{eqnarray*} g(\epsilon)d\epsilon&=&n_sV\frac{\epsilon^2d\epsilon}{2\pi^2\hbar^3c^3}\\ &=&n_s\frac{V}{2\pi^2\hbar^3c^3}(p^2c^2+m^2c^4)d\epsilon \end{eqnarray*}$$

Problem 6.20. Relation between the energy and pressure equations of state for photons

Use similar considerations as in Problem 6.19 to show that for photons:

$$PV =\frac1 3E. (6.111)$$
Equation (6.111) holds at any temperature and is consistent with Maxwell’s equations. Thus, the pressure due to electromagnetic radiation is related to the energy density by P = u(T)/3.

$$\begin{eqnarray*} E&=&\int_0^{\infty}\epsilon \overline n(\epsilon)g(\epsilon)d\epsilon\\ &=&\int_0^{\infty}\frac{V\epsilon^3}{2\pi^2\hbar^3c^3}\frac{1}{e^{\beta(\epsilon-\mu)}-1}d\epsilon\\ &=&\frac{V}{2\pi^2\hbar^3c^3}\int_0^{\infty}\frac{\epsilon^3d\epsilon}{e^{\beta(\epsilon-\mu)}-1}\\ \Omega&=&kT\int_0^{\infty}g(\epsilon)ln[1-e^{-\beta(\epsilon-\mu)}]d\epsilon\\ &=&kT\int_0^{\infty}\frac{V\epsilon^2}{2\pi^2\hbar^3c^3}ln[1-e^{-\beta(\epsilon-\mu)}]d\epsilon\\ &=&\frac{kTV}{2\pi^2\hbar^3c^3}\int_0^{\infty}\epsilon^2ln[1-e^{-\beta(\epsilon-\mu}]d\epsilon\\ &=&-\frac13\frac{V}{2\pi^2\hbar^3c^3}\int_0^{\infty}\frac{\epsilon^3d\epsilon}{e^{\beta(\epsilon-\mu)}-1}\\ PV&=&-\Omega\\ &=&\frac13E \end{eqnarray*}$$

Problem 6.21. The chemical potential

(a) Estimate the chemical potential of one mole of an ideal monatomic classical gas at standard temperature and pressure and show that $\mu \ll0$.

(a) For standard temperature and pressure, $T=298.15K, P=101325 Pa=1 atm, N=6.02\times 10^{23}$

$$\begin{eqnarray*} \mu &=&kTln[\frac{\bar N}{V}(\frac{2\pi \hbar^2\beta}{m})^{3/2}]\\ &=&kTln[\frac{\bar N}{V}(\frac{2\pi \hbar^2}{mkT})^{3/2}]\\ &\approx&-5\times10^{-20}\\ &\ll&0 \end{eqnarray*}$$

(b) Show that N can be expressed as [see (6.114)]

$$N =\frac{V}{ λ^3}e^{β\mu}, (6.116)$$
and hence
$$µ(T,V ) = −kT ln\frac{1}{ ρλ^3}, (6.117)$$
where $ρ =\bar N/V$ .

(b)

$$\begin{eqnarray*} \lambda&=&(\frac{h^2}{2\pi mkT})^{1/2}=(\frac{2\pi \hbar^2}{mkT})^{1/2}\\ N&=&V(\frac{m}{2\pi \hbar^2\beta})^{3/2}e^{\beta\mu}\\ &=&V(\frac{mkT}{2\pi\hbar^2})^{3/2}e^{\beta\mu}\\ &=&\frac{V}{\lambda^3}e^{\beta\mu}\\ \mu &=&kTln[\frac{\bar N}{V}(\frac{2\pi\hbar^2\beta}{m})^{3/2}]\\ &=&kTln[\frac{\bar N\lambda^3}{V}]\\ &=&-kTln[(\rho \lambda)^{-1}]\\ &=&-kTln\frac{1}{\rho\lambda^3} \end{eqnarray*}$$

(c) In Section 6.1 we argued that the semiclassical limit $λ\ll ρ^{−1/3}$ [see (6.1)] implies that $\bar n_k\ll 1$; that is, the mean number of particles in any single particle energy state is very small. Use the expression (6.117) for $\mu$ and (6.87) for $\bar n_k$ to show that the condition $\bar n_k\ll1$ implies that$λ \llρ^{−1/3}.$

(c)

$$\begin{eqnarray*} \mu(T,V)&=&-kTln\frac{1}{\rho\lambda^3}\\ \bar n_k&=&e^{-\beta(\epsilon_k-\mu)}\\ \bar n_k&\ll&1\\ e^{-\beta(\epsilon_k-\mu)}&\ll&1\\ -\beta(\epsilon_k-\mu)&\ll&0\\ -\beta\epsilon_k&\ll&ln\frac{1}{\rho\lambda^3}\\ e^{-\beta\epsilon_k}&\ll&\frac{1}{\rho\lambda^3}\\ \lambda^3&\ll&\rho^{-1}\\ \lambda&\ll&\rho^{-1/3} \end{eqnarray*}$$

Problem 6.22. Ideal gas equations of state

Show that E = (3/2)NkT and PV = NkT from the results of this section.

$$\begin{eqnarray*} \bar N&=&\frac{V}{\lambda^3}e^{\beta \mu}\\ P&=&\frac{kT}{\lambda^3}e^{\beta\mu}\\ &=&\frac{kT\bar N}{V}\\ PV&=&NkT\\ S(T,V,N)&=&Nk[\frac53-ln\frac{N}{V}-ln(\frac{2\pi\hbar^2}{mkT})^{3/2}]\\ \frac{\partial S}{\partial T}&=&E\\ E&=&-Nk(\frac{mkT}{2\pi\hbar^2})^{3/2}\frac32(\frac{2\pi\hbar^3}{mkT})^{1/2}\frac{2\pi\hbar^3}{mk}(-1)\frac{1}{T^2}\\ &=&\frac32NkT \end{eqnarray*}$$

Problem 6.23. Wien’s displacement law

The maximum of u(ν) shifts to higher frequencies with increasing temperature. Show that the maximum of u can be found by solving the equation

$$(3−x)e^x = 3, (6.136)$$
where $x = βhν_{max}$. Solve (6.136) numerically for x and show that
$$\frac{hν_{max}}{ kT}= 2.822$$ (Wien’s displacement law). (6.137)

We can find the root by using mathematica:

$x\approx 2.82144$

Problem 6.24. Derivation of the Rayleigh-Jeans and Wien’s laws

(a) Make a change of variables in (6.134) to find the energy emitted by a blackbody at a wavelength between λ and λ + dλ.

(a)

$$\begin{eqnarray*} \nu&=&\frac{c}{\lambda}\\ d\nu&=&-\frac{c}{\lambda^2}d\lambda\\ u(\nu)d\nu&=&\frac{8\pi h\nu^3}{c^3}\frac{d\nu}{e^{\beta h\nu}-1}\\ u(\lambda)d\lambda&=&\frac{8\pi h\frac{c^3}{\lambda^3}}{c^3}\frac{1}{e^{\frac{\beta hc}{\lambda}}-1}(-\frac{c}{\lambda^2}d\lambda)\\ &=&\frac{8\pi hc}{\lambda^5}\frac{d\lambda}{1-e^{\frac{\beta hc}{ \lambda}}} \end{eqnarray*}$$

(b) Determine the limiting behavior of your result in part (a) for long wavelengths. This limit is called the Rayleigh-Jeans law and is given by

$$u(λ)dλ =\frac{8πkT}{ λ^4}dλ. (6.138)$$
Does this form involve Planck’s constant? The result in (6.138) was originally derived from purely classical considerations.

(b) No, it doesn't

(c) Classical theory predicts what is known as the ultraviolet catastrophe, namely, that an infinite amount of energy is radiated at high frequencies or short wavelengths. Explain how (6.138) would give an infinite result for the total radiated energy, and thus the classical result cannot be correct for all wavelengths.

(c) When frequency is high enough or wavelength is too short, $u(\lambda)\rightarrow +\infty$, which is impossible

(d) Determine the limiting behavior of u(λ) for short wavelengths. This behavior is known as Wien’s law, after Wilhelm Wien who found it by finding a functional form to fit the experimental data.

(d)

$$\begin{eqnarray*} \lambda&\rightarrow&0\\ u(\lambda)&=&\frac{8\pi h}{\lambda^3}\frac{1}{e^{\frac{\beta hc}{\lambda}}-1}\\ &=&\frac{8\pi h}{\lambda^3}e^{-\frac{\beta hc}{\lambda}} \end{eqnarray*}$$

Problem 6.25. Thermodynamics of blackbody radiation

Use the various thermodynamic relations to show that

$$\begin{eqnarray*} E &=& V\int_0^{\infty}u(ν)dν =\frac{ 4σ}{ c}V T^4, (6.139a)\\ Ω &=& F = −\frac{4σ}{ 3c}V T^4, (6.139b)\\ S &=&\frac{16σ}{ 3c}V T^3, (6.139c)\\ P &=&\frac{4σ}{ 3c}T^4 = \frac1 3\frac{E}{ V}, (6.139d)\\ G& = &F + PV = 0. (6.139e) \end{eqnarray*}$$
The free energy $F$ in (6.139b) can be calculated from $Z$ starting from (6.128) and using (6.100). The Stefan-Boltzmann constant σ is given by
$$σ =\frac{2π^5k^4}{ 15h^3c^2}. (6.140)$$
The integral
$$\int_0^{\infty}\frac{x^3 dx}{ e^x −1}=\frac{π^4}{ 15}. (6.141)$$
is evaluated in the Appendix.

$$\begin{eqnarray*} E(\nu)d\nu&=&\frac{8\pi hV\nu^3}{c^3}\frac{d\nu}{e^{\beta h \nu}-1}\\ E&=&\int_0^{\infty}\frac{8\pi hV\nu^3}{c^3}\frac{d\nu}{e^{\beta h \nu}-1}\\ &=&V\frac{8\pi h}{c^3}\int_0^{\infty}\frac{\nu^3d\nu}{e^{\beta h \nu}-1}\\ &=&\frac{8\pi hV}{c^3}(\frac{kT}{h})^4\int_0^{\infty}\frac{x^3}{e^x-1}dx\\ &=&\frac{8\pi hV}{c^3}(\frac{kT}{h})^4\frac{\pi^4}{15}\\ &=&\frac{8\pi^5k^4}{15h^3c^3}VT^4\\ &=&\frac{4\sigma}{c}VT^4\\ \Omega&=&-\frac13E=-\frac{4\sigma}{3c}VT^4\\ F&=&-kTlnZ\\ &=&-kTln\frac{1}{1-e^{-\beta\epsilon_k}}\\ &=&\\ S&=&-\frac{\partial F}{\partial T}\\ &=&\frac{16\sigma}{3c}VT^3\\ PV&=&-F=\frac13E\\ P&=&\frac{4\sigma}{3c}T^4=\frac{E}{3V}\\ G&=&F+PV\\ &=&0 \end{eqnarray*}$$

Problem 6.26. Mean number of photons

Show that the total mean number of photons in an enclosure of volume V is given by

$$N =\frac{V}{ π^2c^3}\int_o^{\infty}\frac{ω^2dω}{ e^{\hbar ω/kT} −1} =\frac{V (kT)^3}{ π^2c^3\hbar^3}\int_0^{\infty}\frac{x^x2dx}{ e^x −1} . (6.142)$$
The integral in (6.142) can be expressed in terms of known functions (see the Appendix). The result is
$$\int_0^{\infty}\frac{x^Z∞ 0 x2dx}{ e^x −1} = 2×1.202. (6.143)$$
Hence N depends on T as $$N = 0.244V(\frac{kT}{\hbar c})^3.$$

$$\begin{eqnarray*} N(\epsilon)d\epsilon&=&\frac{V}{\pi^2\hbar^3c^3}\frac{\epsilon^2d\epsilon}{e^{\beta\epsilon}-1}\\ N&=&\int_0^{\infty}\frac{V}{\pi^2\hbar^3c^3}\frac{\epsilon^2d\epsilon}{e^{\beta\epsilon}-1}\\ &=&\frac{V(kT)^3}{\pi^2\hbar^3c^3}\int_0^{\infty}\frac{x^2dx}{e^x-1}\\ &=&\frac{2.404}{\pi^2}V(\frac{kT}{\hbar c})^3\\ &=&0.244V(\frac{kT}{\hbar c})^3 \end{eqnarray*}$$

Problem 6.27. Order of magnitude estimates

(a) Verify that the values of F given in electon volts (eV) lead to the values of TF in Table 6.3.

(a)

(b) Compare the values of TF in Table 6.3 to room temperature. What is the value of kT in eV at room temperature?

(b)

(c) Given the data in Table 6.3 verify that the electron densities for Li and Cu are ρ = 4.7 × 1028 m−3 and ρ = 8.5×1028 m−3, respectively.

(c)

(d) What is the mean distance between the electrons for Li and Cu?

(d)

(e) Use the fact that the mass of an electron is 9.1×10−31 kg to estimate the de Broglie wavelength corresponding to an electron with energy comparable to the Fermi energy.

(e)

(f) Compare your result for the de Broglie wavelength that you found in part (e) to the mean interparticle spacing that you found in part (d).

(f)

Problem 6.28. Landau potential at zero temperature

From (6.107) the Landau potential for an ideal Fermi gas at arbitrary T can be expressed as

$$Ω = −kT\int_0^{\infty}g(\epsilon) ln[1 + e^{−β(\epsilon−µ)}]d\epsilon. (6.156)$$
To obtain the T = 0 limit of Ω, we have that $\epsilon < µ$ in (6.156), β →∞, and hence $ln[1+e^{−β(\epsilon−µ)}] → lne^{−β(\epsilon−µ)} = −β(\epsilon−µ).$ Hence, show that
$$Ω =\frac{(2m)^{3/2}V}{ 2π^2\hbar^3}\int_0^{\epsilon}\epsilon^{1/2}(\epsilon-\epsilon_F)d\epsilon (6.157)$$
Calculate Ω and determine the pressure at T = 0.

$$\begin{eqnarray*} \Omega&=&-kT\int_0^{\infty}\frac{V(2m)^{3/2}}{2\pi^2\hbar^3}\epsilon^{1/2}ln[1+e^{-\beta(\epsilon-\mu)}]d\epsilon\\ &=&-kT\frac{(2m)^{3/2}V}{2\pi^2\hbar^3}\int_0^{\infty}-\beta(\epsilon-\mu)\epsilon^{1/2}d\epsilon\\ &=&\frac{(2m)^{3/2}V}{2\pi^2\hbar^3}\int_0^{\epsilon}\epsilon^{1/2}(\epsilon-\epsilon_F)d\epsilon\\ P&=&-\frac{\partial F}{\partial V}\\ &=&-\frac{\partial \Omega}{\partial V}\\ &=&-\frac{(2m)^{3/2}}{2\pi^2\hbar^3}\int_0^{\epsilon}\epsilon^{1/2}(\epsilon-\epsilon_F)d\epsilon \end{eqnarray*}$$

Problem 6.29.

Show that the limit (6.145) for $\bar n(\epsilon)$ at T = 0 follows only if µ > 0.

$$\begin{eqnarray*} \bar n(\epsilon)&=&\frac{1}{e^{\beta(\epsilon-\mu)}+1})\\ &=&\left\{ \begin{aligned} 1　　,for　\epsilon <\mu \\ 0　　,for　\epsilon >\mu \\ \end{aligned} \right. \end{eqnarray*}$$
However, $\epsilon>0$, if $\bar n(\epsilon)=1$, then $\epsilon<\mu$. Thus $\mu>0$

Problem 6.30. Numerical evaluation of the chemical potential for an ideal Fermi gas

(a) Start with $T^* = 0.2$and find $\mu^*$ such that (6.162) is satisfied. (Recall that $\mu^*= 1$ at $T^* = 0.$) Does $\mu^*$ initially increase or decrease as T is increased from zero? What is the sign of $\mu^*$for $T^* \gg 1$?

(a) To satisfied (6.162), $\mu^*=0.9645$. So $\mu^*$ initially decrease as T is increased from zero. When $T\gg 1$, the sign of $\mu^*$ is negative

(b) At what value of $T^*$ is $\mu^*\approx 0$?

(b) $T^*\approx 0.988734$

(c) Given the value of $\mu^*(T^*)$ the program computes the numerical value of E(T). Describe its qualitative T dependence and the T dependence of $C_V$ .

(c)

Problem 6.31. Low temperature behavior

(a) Fill in the missing steps in (6.163)–(6.174).

(a)

$$\begin{eqnarray*} N&=&\int_0^{\infty}\bar n(\epsilon)g(\epsilon)d\epsilon=\int_0^{\epsilon_F}g(\epsilon)d\epsilon\\ \int_0^{\epsilon_F}\epsilon_Fg(\epsilon)d\epsilon&=&\int_0^{\infty}\epsilon_F\bar n(\epsilon)g(\epsilon)d\epsilon\\ \int_0^{\epsilon_F}\epsilon_Fg(\epsilon)d\epsilon&=&\int_0^{\epsilon_F}\epsilon_F\bar n(\epsilon)g(\epsilon)d\epsilon+\int_{\epsilon_F}^{\infty}\epsilon_F\bar n(\epsilon)g(\epsilon)d\epsilon\\ 0&=&\int_0^{\epsilon_F}\epsilon_F[\bar n(\epsilon)-1]g(\epsilon)d\epsilon+\int_{\epsilon_F}^{\infty}\epsilon_F\bar n(\epsilon)g(\epsilon)d\epsilon\\ \Delta E&=&\int_0^{\infty}\epsilon\bar n(\epsilon) g(\epsilon)d\epsilon-\int_0^{\epsilon_F}\epsilon g(\epsilon)d\epsilon\\ &=&\int_0^{\epsilon_F}\epsilon[\bar n(\epsilon)-1]g(\epsilon)d\epsilon+\int_{\epsilon_F}^{\infty}\epsilon\bar n(\epsilon)g(\epsilon)d\epsilon\\ &=&\int_0^{\epsilon_F}(\epsilon-\epsilon_F)[\bar n(\epsilon)-1]g(\epsilon)d\epsilon+\int_{\epsilon_F}^{\infty}(\epsilon-\epsilon_F)\bar n(\epsilon)g(\epsilon)d\epsilon\\ &=&\int_{\epsilon_F}^{\infty}(\epsilon-\epsilon_F)\bar n(\epsilon)g(\epsilon)d\epsilon+\int_0^{\epsilon_F}(\epsilon_F-\epsilon)[1-\bar n(\epsilon)]g(\epsilon)d\epsilon. (6.166)\\ C_V&=&\frac{\partial E}{\partial T}\\ &=&\int_{\epsilon_F}^{\infty}(\epsilon-\epsilon_F)\frac{d\bar n(\epsilon)}{dT} g(\epsilon)d\epsilon+\int_0^{\infty}(\epsilon-\epsilon_F)\frac{d\bar n(\epsilon)}{dT}g(\epsilon)d\epsilon\\ &=&\int_0^{\infty}(\epsilon-\epsilon_F)\frac{d\bar n(\epsilon)}{dT}g(\epsilon)d\epsilon\\ &=&g(\epsilon_F)\int_0^{\infty}(\epsilon-\epsilon_F)\frac{d\bar n}{dT}d\epsilon\\ &=&g(\epsilon_F)\int_0^{\infty}(\epsilon-\epsilon_F)\frac{d\bar n}{d\beta}\frac{d\beta}{dT}d\epsilon\\ &=&g(\epsilon_F)\int_0^{\infty}(\epsilon-\epsilon_F)\frac{1}{kT^2}\frac{(\epsilon-\epsilon_F)e^{\beta(\epsilon-\epsilon)}}{[e^{\beta(\epsilon-\epsilon_F}+1]^2}d\epsilon\\ &=&g(\epsilon_F)\int_0^{\infty}k^2T(\frac{\epsilon-\epsilon_F}{kT})^2\frac{e^{\beta(\epsilon-\epsilon_F)}}{[e^{\beta(\epsilon-\epsilon_F)+1}]^2}d(\frac{\epsilon-\epsilon_F}{kT})\\ &=&k^2Tg(\epsilon_F)\int_0^{\infty}\frac{x^2e^x}{(e^x+1)^2}dx\\ &=&k^2Tg(\epsilon_F)\frac{\pi^2}{3}\\ &=&k^2T\frac{\pi^2}{3}\frac{3N}{2kT_F}\\ &=&\frac{\pi^2NkT}{2T_F}=\frac{\pi^2}{2}Nk\frac{T}{T_F}.(6.174) \end{eqnarray*}$$

(b) Use (6.175) and (6.179) to show that the mean pressure for $T \ll T_F$ is given by

$$P =\frac2 5\rho \epsilon_F[1 + \frac{5π^2}{ 12}(\frac{T}{ T_F})^2+ ...]. (6.180)$$

(b)

$$\begin{eqnarray*} P&=&-\frac{\partial \Omega}{\partial T}\\ &=&\frac23\frac{2^{1/2}m^{3/2}}{\pi^2\hbar^3}[\frac25\mu^{5/2}+\frac{\pi^2}{4}(kT)^2\mu^{1/2}]\\ &=&\frac23\frac{2^{1/2}m^{3/2}}{\pi^2\hbar^3}\mu^{5/2}[\frac25+\frac{\pi^2}{4}(kT)^2\mu^{-2}]\\ \epsilon_F&=&\frac{\hbar^2}{2m}(3\pi^2\rho)^{2/3}\\ \frac{2}{3\rho}\epsilon_F^{3/2}&=&\frac{\pi^2\hbar^3}{2^{1/2}m^{3/2}}\\ P&=&\frac23\frac{3\rho}{2}\epsilon_F^{-3/2}\epsilon_F^{5/2}[1-\frac{\pi^2}{12}(\frac{T}{T_F})^2]^{5/2}[\frac25+\frac{\pi^2}{4}(\frac{T}{T_F})^2]\\ &=&\frac25\rho\epsilon_F[1+\frac{5\pi^2}{8}(\frac{T}{T_F})^2][1-\frac{\pi^2}{12}(\frac{T}{T_F})^2]^{5/2}\\ &=&\frac25\rho\epsilon_F[1+\frac{5\pi^2}{12}(\frac{T}{T_F})^2+....] \end{eqnarray*}$$

(c) Use the general relation between E and PV to show that

$$E =\frac3 5N\epsilon_F[1 + \frac{5π^2}{ 12}(\epsilon{T}{ T_F})^2 + ...]. (6.181)$$

(c)

$$\begin{eqnarray*} PV&=&NkT\\ E&=&\frac32NkT=\frac32PV\\ &=&\frac32V\frac25\rho\epsilon_F[1+\frac{5\pi^2}{12}(\frac{T}{T_F})^2+...]\\ &=&\frac35N\epsilon_F[1+\frac{5\pi^2}{12}(\frac{T}{T_F})^2+...]\\ \end{eqnarray*}$$

(d) For completeness, show that the low temperature behavior of the entropy is given by

$$S=\frac{π^2}{ 2}Nk\frac{T}{ T_F} (6.182)$$

(d)

$$\begin{eqnarray*} S&=&-\frac{\partial E}{\partial T}\\ &=&\frac35N\epsilon_F\frac{5\pi^2}{12}\frac{2T}{T_F^2}\\ &=&\frac{1}{2}NkT_F\frac{\pi^2T}{T_F^2}\\ &=&\frac{\pi^2}{2}Nk\frac{T}{T_F} \end{eqnarray*}$$

Problem 6.32. Effective electron mass

From Table 6.3 we see that TF = 8.2×104 K for copper. Use (6.174) to find the predicted value of C/NkT for copper. How does this value compare with the experimental value C/NkT = 8×10−5? It is remarkable that the theoretical prediction agrees so well with the experimental result based on the free electron model. Show that the small discrepancy can be removed by defining an effective mass m∗ of the conduction electrons equal to≈ 1.3me, where me is the mass of the electron. What factors might account for the effective mass being greater than me?

$$\frac{C}{NkT}=\frac{\pi^2}{2T_F}=6.018\times10^{-5}$$
It is almost equal to the experimental value.
If we set $m^*=1.3m_e$, $C/N^*kT=6.018\times1.3\times10^{-5}\approx8\times10^{-5}$

Problem 6.33. Temperature dependence of the chemical potential in two dimensions

Consider a system of electrons restricted to a surface of area A. Show that the mean number of electrons can be written as

$$\bar N = \frac{mA}{ π\hbar^2}\int_0^{\infty}\frac{d\epsilon}{e^{β(\epsilon−µ)} + 1} . (6.183)$$
The integral in (6.183) can be evaluated in closed form using
$$\int\frac{dx}{ 1 + ae^{bx}} = \frac1 b ln \frac{e^{bx}}{ 1 + ae^{bx} }+ constant. (6.184)$$

$$\begin{eqnarray*} g(\epsilon)d\epsilon&=&\frac{mA}{\pi\hbar^2}d\epsilon\\ \bar N&=&\int_0^{\infty}\bar n(\epsilon)g(\epsilon)d\epsilon\\ &=&\frac{mA}{ π\hbar^2}\int_0^{\infty}\frac{d\epsilon}{e^{β(\epsilon−µ)} + 1}\\ &=&\frac{mA}{\pi\hbar^2}[\frac{1}{\beta}ln\frac{e^{\beta(\epsilon-\mu)}}{1+e^{\beta(\epsilon-\mu)}}]_0^{\infty}\\ &=&\frac{mAkT}{\pi\hbar^2}ln(1+e^{\beta\mu}) \end{eqnarray*}$$

(a) Show that

$$µ(T) = kT ln[e^{ρπ\hbar^2/mkT} −1], (6.185)$$ where ρ = N/A.

(a)

$$\begin{eqnarray*} \bar N&=&\frac{mAkT}{\pi\hbar^2}ln(1+e^{\beta\mu})\\ \frac{\pi\hbar^2\rho}{mkT}&=&ln(1+e^{\beta\mu})\\ \mu&=&kTln(e^{\frac{\pi\hbar^2\rho}{mkT}}-1) \end{eqnarray*}$$

(b) What is the value of the Fermi energy $\epsilon_F = \mu(T = 0)$? What is the value of µ for $T \gg T_F$?

(b)

$$\begin{eqnarray*} \epsilon_F&=&\mu(T=0)=\frac{\pi\hbar^2\rho}{m}\\ \mu&=&kTln(e^{\frac{\pi\hbar^2\rho}{mkT}}-1)\\ &=&kTln(e^{\frac{\epsilon_F}{kT}}-1)\\ &=&kTln(e^{\frac{T_F}{T}}-1)\\ T&\gg&T_F\\ \mu&\rightarrow&-\infty \end{eqnarray*}$$

(c) Plot µ versus T and discuss its qualitative dependence on T.

(c)

Problem 6.34. Limiting behavior of the heat capacity in the Einstein model

Derive the limiting behavior of CV given in (6.194).

$$\begin{eqnarray*} C_V&=&3Nk(\frac{T_E}{T})^2\frac{e^{T_E/T}}{[e^{T_E/T}-1]^2}\\ &=&3Nk(\frac{T_E}{T})^2\frac{1}{(1-e^{-T_E/E})^2}\\ &=&3Nk(\frac{T_E}{T})^2\frac{1}{[1-(1-T_E/T)]^2}\\ &=&3Nk(\frac{T_E}{T})^2(\frac{T}{T_E})^2\\ &=&3Nk \end{eqnarray*}$$

Problem 6.35. More on the Einstein and Debye theories

(a) Determine the wavelength λD corresponding to ωD and show that this wavelength is approximately equal to a lattice spacing. This equality provides another justification for a high frequency cutoff because the atoms in a crystal cannot oscillate with a wavelength smaller than a lattice spacing.

(a)

$$\begin{eqnarray*} \lambda_D&=&2\pi\bar c/\omega_D=(\frac{4\pi}{3\rho})^{1/3}\propto\rho^{-1/3}\\ \end{eqnarray*}$$

(b) Show explicitly that the energy in (6.202) is proportional to T for high temperatures and proportional to T4 for low temperatures.

(b) When the temperature is high

$$\begin{eqnarray*} E&=&9NkT(\frac{T}{T_D})^3\int_0^{T_D/T}\frac{x^3dx}{e^x-1}\\ &=&9NkT(\frac{T}{T_D})^3(\frac{T_D}{T})^3\\ &=&9NkT\\ &\varpropto&T \end{eqnarray*}$$
In the low temperature limit
$$\begin{eqnarray*} E&=&9NkT(\frac{T}{T_D})^3\int_0^{T_D/T}\frac{x^3dx}{e^x-1}\\ &=&9NkT(\frac{T}{T_D})^3\frac{\pi^4}{15}\\ &\varpropto&T^4 \end{eqnarray*}$$

(c) Plot the temperature dependence of the mean energy as given by the Einstein and Debye theories on the same graph and compare their predictions.

(c)

(d) Derive an expression for the mean energy analogous to (6.202) for one- and two-dimensional crystals. Then find explicit expressions for the high and low temperature dependence of the specific heat on the temperature.

(d)
One dimention

$$\begin{eqnarray*} g(\omega)d\omega&=&\frac{V\omega^2d\omega}{2\pi^2c^3}\\ N&=&\int_0^{\omega_D}g(\omega)d\omega\\ &=&\int_0^{\omega_D}\frac{V\omega^2 d\omega}{2\pi^2c^3}\\ &=&\frac13\frac{V\omega_D^3}{2\pi^2c^3}\\ 3N&=&\frac{V}{2\pi^2c^3}(\frac{kT_D}{\hbar})^3\\ 3N(\frac{\hbar}{kT_D})^3&=&\frac{V}{2\pi^2c^3}\\ E&=&\int\hbar\omega\bar n(\omega)g(\omega)d\omega\\ &=&\int\frac{V\hbar}{2\pi^2c^3}\frac{\omega^3d\omega}{e^{\beta\hbar\omega}-1}\\ &=&\frac{V\hbar}{2\pi^2c^3}\int_0^{kT_D/T}\frac{\omega^3d\omega}{e^{\beta\hbar\omega}-1}\\ &=&\frac{V\hbar}{2\pi^2c^3}(\frac{kT}{\hbar})^4\int_0^{T_D/T}\frac{x^cdx}{e^x-1}\\ &=&3N\hbar(\frac{\hbar}{kT_D})^3(\frac{kT}{\hbar})^4\int_0^{T_D/T}\frac{x^cdx}{e^x-1}\\ &=&3NkT(\frac{T}{T_D})^3\int_0^{T_D/T}\frac{x^cdx}{e^x-1}\\ \end{eqnarray*}$$

For two dimention

$$\begin{eqnarray*} g(\omega)d\omega&=&\frac{2V\omega^2d\omega}{2\pi^2c^3}\\ 2N&=&\int_0^{\omega_D}g(\omega)d\omega\\ &=&\frac13\frac{2V\omega_D^3}{2\pi^2c^3}\\ 3N&=&\frac{V}{2\pi^2c^3}(\frac{kT_D}{\hbar})^3\\ E&=&\int\hbar\omega\bar n(\omega)g(\omega)d\omega\\ &=&\frac{2V\hbar}{2\pi^2c^3}(\frac{kT}{\hbar})^3\int_0^{T_D/T}\frac{x^3dx}{e^x-1}\\ &=&6NkT(\frac{T}{T_D})^3\int_0^{T_D/T}\frac{x^3dx}{e^x-1}\\ \end{eqnarray*}$$

Problem 6.36. Relation of Tc to the zero-point energy

Express (6.210) in terms of the zero-point energy associated with localizing a particle of mass m in a volume a3, where a = ρ−1/3 is the mean interparticle spacing.

$$\begin{eqnarray*} kT_c&=&\frac{1}{2.612^{2/3}}\frac{2\pi\hbar^2}{m}\rho^{2/3} \end{eqnarray*}$$

Problem 6.37. Numerical evaluation of µ for an ideal Bose gas

(a) Fill in the missing steps and derive (6.211).

(a)

$$\begin{eqnarray*} \rho&=&\frac{\bar N}{V}=\frac{(2m)^{3/2}}{4\pi^2\hbar^3}\int_0^{\infty}\frac{\epsilon^{1/2}d\epsilon}{e^{\beta(\epsilon-\mu)}-1}\\ &=&\frac{(2mkT_c)^{3/2}}{4\pi^2\hbar^3}\int_0^{\infty}\frac{x^{1/2}dx}{e^x-1}\\ &=&\frac{(2m)^{3/2}}{4\pi^2\hbar^3}\int_0^{\infty}\frac{(kT_cy)^{1/2}d(kT_cy)}{e^{\frac{kT_c(y-\mu^*)}{kT_cT^*}}-1}\\ &=&\frac{(2mkT_c)^{3/2}}{4\pi^2\hbar^3}\int_0^{\infty}\frac{y^{1/2}dy}{e^{(y-\mu^*)/T^*}-1}\\ \int_0^{\infty}\frac{x^{1/2}dx}{e^x-1}&=&\int_0^{\infty}\frac{y^{1/2}dy}{e^{(y-\mu^*)/T^*}-1}\\ 2.612\frac{\pi^{1/2}}{2}&=&\int_0^{\infty}\frac{y^{1/2}dy}{e^{(y-\mu^*)/T^*}-1}\\ 1&=&\frac{2}{2.612\sqrt{\pi}}\int_0^{\infty}\frac{y^{1/2}dy}{e^{(y-\mu^*)/T^*}-1}\\ \end{eqnarray*}$$

(b) The program evaluates the left-hand side of (6.211b). The idea is to find µ∗ for a given value of T∗ such the left-hand side of (6.211b) equals 1. Begin with T∗ = 10. First choose µ∗ = −10 and find the value of the integral. Do you have to increase or decrease the value of µ∗ to make the numerical value of the left-hand side of (6.211b) closer to 1? Change µ∗ by trial and error until you find the desired result. You should find that µ∗ ≈−25.2.

(b) $T^*=10,\mu^*=10$, the integral is 5.185

Then we find that when $\mu^*=-25.23$, the left-hand side equals to 1

(c) Next choose T∗ = 5 and find the value of µ∗ so that the left-hand side of (6.211b) equals 1. Does µ∗ increase or decrease in magnitude? You can generate a plot of µ∗ versus T∗ by clicking on the Plot button each time you find an approximately correct value of µ.

(c)

When $\mu^*=-7.683$, the left-hand side equals to 1. Clearly, $\mu^*$ increase in magnitude

(d) Discuss the qualitative behavior of µ as a function of T for fixed density.

(d) As $T^*$ decreasing, $\mu^*$ is increasing.

Problem 6.38.

Show that the density ρc corresponding to µ = 0 for a given temperature is given by

$$\rho_c = \frac{2.612}{ λ^3} , (6.212)$$ where λ is given by (6.2). Is it possible for the density to exceed ρc for a given temperature?

$$\begin{eqnarray*} \rho_c&=&\frac{(2mkT_c)^{3/2}}{4\pi^2\hbar^3}\int_0^{\infty}\frac{x^{1/2}dx}{e^x-1}\\ &=&\frac{(2mkT_c)^{3/2}}{4\pi^2\hbar^3}2.612\frac{\pi^{1/2}}{2}\\ \lambda&=&(\frac{2\pi\hbar^2}{mkT})^{1/2}\\ \rho_c&=&\frac{2.612}{\lambda^3} \end{eqnarray*}$$

Problem 6.39.

Show that the thermal de Broglie wavelength is comparable to the interparticle spacing at T = Tc. What is the implication of this result?

$$\begin{eqnarray*} \rho_c&=&\frac{2.612}{\lambda^3}=\frac{\bar N}{V}\\ \lambda&\propto&V^{1/3}=L \end{eqnarray*}$$

Problem 6.40. Temperature dependence of the pressure

(a) Start from the classical pressure equation of state, PV = NkT, replace N by N for an ideal Bose gas, and give a qualitative argument why $P \propto T^{5/2}$ at low temperatures.

(a)

$$\begin{eqnarray*} PV&=&NkT\\ N_{eff}&=&V\frac{(2mkT)^{3/2}}{4\pi^2\hbar^3}\frac{2.612\pi^{1/2}}{2}\\ P&=&\frac{(2m)^{3/2}}{4\pi^2\hbar^3}\frac{2.612\pi^{1/2}}{2}(kT)^{5/2}\\ P&\propto&T^{5/2} \end{eqnarray*}$$

b) Show that the ground state contribution to the pressure is given by

$$P_0 =\frac{kT}{V}ln(N_0 + 1). (6.222)$$
Explain why P0 can be regarded as zero and why the pressure of an Bose gas for T < Tc is independent of the volume.

(b)

Problem 6.41. Estimate of the Bose condensation temperature

(a) What is the approximate value of Tc for an ideal Bose gas at a density of ρ ≈ 125kg/m3, the density of liquid 4He? Take m = 6.65×10−27 kg.

(a)

$$\begin{eqnarray*} kT_c&=&\frac{1}{2.612^{2/3}}\frac{2\pi\hbar^2}{m}\rho^{2/3}\\ T_c&=&\frac{1}{2.612^{2/3}k}\frac{2\pi\hbar^2}{m}\rho^{2/3}\\ &=&1.0049\times10^{-17} \end{eqnarray*}$$

(b) The value of Tc for a collection of 87Rb (rubidium) atoms is about 280nK (2.8×10−7 K). What is the mean separation between the atoms?

(b)

$$\begin{eqnarray*} kT_c&=&\frac{1}{2.612^{2/3}}\frac{2\pi\hbar^2}{m}\rho^{2/3}\\ \rho^{2/3}&=&\frac{2.612^{2/3}mkT_c}{2\pi\hbar^2}\\ \rho&=&\frac{2.612(mkT_c)^{3/2}}{(2\pi)^{3/2}\hbar^3}\\ &=&5.75\times10^{19} \end{eqnarray*}$$

Problem 6.42. Number fluctuations

Use the Gibbs distribution Ps in (6.73) to show that $\bar N$ can be written as

$$\overline N=\frac{\sum_s N_s e^{−β(E_s−\mu N_s)} }{\sum e^{−β(E_s−\mu N_s)}} . (6.227)$$
Then use (6.227) to show that 
$$(\frac{\partial \overline N}{\partial µ})_{T,V} = \frac{1}{ kT} [\overline {N^2} −\overline N^2], (6.228)$$
or
$$[\overline {N^2} −\overline{N}^2] = kT(\frac{\partial \overline N}{\partial \mu})_{T,V} , (6.229)$$
where
$$\overline{N^2} =\frac{\sum_s N_s^2 e^{−β(E_s-\mu N_s)}}{ \sum_s e^{−β(E_s−\mu N_s)}} .$$

$$\begin{eqnarray*} P_s&=&\frac{1}{Z_G}e^{-\beta(E_s-\mu N_s)}\\ &=&\frac{e^{-\beta(E_s-\mu N_s)}}{\sum_s e^{-\beta(E_-\mu N_s)}}\\ \overline N&=&\sum_s N_s P_s\\ &=&\frac{\sum_sN_se^{-\beta(E_s-\mu N_s)}}{\sum_s e^{-\beta(E_s-\mu N_s)}}\\ (\frac{\partial \overline N}{\partial \mu})_{T,V}&=&\frac{\sum_sN_se^{-\beta(E_s-\mu N_s)}\beta N_s}{\sum_s e^{-\beta(E_s-\mu N_s)}}-\frac{\sum_sN_se^{-\beta(E_s-\mu N_s)}}{(\sum_s e^{-\beta(E_s-\mu N_s)})^2}\sum_s e^{-\beta(E_s-\mu N_s)}\beta N_s\\ &=&\beta[\frac{\sum_sN_s^2e^{-\beta(E_s-\mu N_s)} }{\sum_s e^{-\beta(E_s-\mu N_s)}}-(\frac{\sum_sN_se^{-\beta(E_s-\mu N_s)}}{\sum_se^{-\beta(E_s-\mu N_s)}})^2]\\ &=&\frac{1}{kT}[\overline {N^2}-\overline N^2] \end{eqnarray*}$$

Problem 6.43. Another Maxwell relation

(a) Show that

$$\mu= f + ρ(\frac{∂f}{ ∂ρ})_T , (6.232)$$
$$(\frac{∂\mu}{ ∂ρ})_T = 2(\frac{∂f}{ ∂ρ})_T + ρ(\frac{∂^2f}{ ∂ρ^2})_T , (6.233)$$
and
$$P = ρ^2(\frac{∂f}{ ∂ρ})_T , (6.234)$$
 $$(\frac{∂P}{ ∂ρ})_T = 2ρ(\frac{∂f}{ ∂ρ})_T + ρ^2\frac{∂^2f} {∂ρ^2})_T = ρ(\frac{∂µ}{ ∂ρ})_T . (6.235)$$
Note that (6.235) is an example of a Maxwell relation (see Section 2.22).

(a)

$$\begin{eqnarray*} \mu&=&\frac{\partial F}{\partial N}\\ &=&f+N(\frac{\partial f}{\partial N})_T\\ &=&f+\rho(\frac{\partial f}{\partial \rho})_T\\ (\frac{\partial \mu}{\partial\rho})_T&=&(\frac{\partial f}{\partial \rho})_T+(\frac{\partial f}{\partial \rho})_T+\rho(\frac{\partial^2f}{\partial \rho^2})_T\\ &=&2(\frac{\partial f}{\partial\rho})_T+\rho(\frac{\partial^2f}{\partial \rho^2})_T\\ P&=&-\frac{\partial F}{\partial V}=-\frac{\partial (Nf)}{\partial V}\\ &=&-N\frac{\partial f}{\partial\frac{N}{\rho}}\\ &=&\rho^2(\frac{\partial f}{\partial\rho})_{T}\\ (\frac{\partial P}{\partial\rho})_T&=&2\rho(\frac{\partial f}{\partial \rho})_T+\rho^2(\frac{\partial^2f}{\partial\rho^2})_T\\ &=&\rho(\frac{\partial \mu}{\partial\rho})_T \end{eqnarray*}$$

(b) Show that

$$(\frac{∂P}{ ∂ρ})_T = −\frac{V^2}{ N}(\frac{∂P}{ ∂V})_{T,N} , (6.236a) $$
$$(\frac{∂\mu}{ ∂ρ})_T = V(\frac{∂\mu}{ ∂N})_{T,V} . (6.236b)$$

(b)

$$\begin{eqnarray*} (\frac{\partial P}{\partial V})_{T,N}&=&\frac{\partial(\rho^2\frac{\partial f}{\partial\rho})}{\partial V}\\ &=&\frac{\partial \rho^2}{\partial V}\frac{\partial f}{\partial\rho}+\rho^2\frac{\partial\frac{f}{\partial \rho}}{\partial V}\\ &=&-\frac{N}{V^2}2\rho(\frac{\partial f}{\partial\rho})_{T,N}-\rho^2\frac{N}{V^2}(\frac{\partial^2 f}{\partial\rho^2})_{T,N}\\ &=&-\frac{N}{V^2}[2\rho(\frac{\partial f}{\partial\rho})_{T,N}+\frac{N}{V^2}(\frac{\partial^2 f}{\partial\rho^2})_{T,N}]\\ (\frac{\partial P}{\partial V})_{T,N}&=&-\frac{N}{V^2}(\frac{\partial P}{\partial\rho})_T\\ (\frac{\partial\mu}{N})_{T,V}&=&\frac{\partial f}{\partial N}+\frac{\partial\rho}{\partial N}(\frac{\partial f}{\partial \rho})_T+\rho(\frac{(\frac{\partial f}{\partial \rho})_T}{\partial N})\\ &=&\frac{2}{V}(\frac{\partial f}{\partial\rho})_T+\frac{\rho}{V}(\frac{\partial^2 f}{\partial \rho^2})_T\\ V(\frac{\partial\mu}{N})_{T,V}&=&(\frac{\partial\mu}{\partial\rho})_T \end{eqnarray*}$$

(c) Use (6.235) and (6.236) to show that

$$N(\frac{∂µ}{ ∂N})_{T,V} = \frac{1}{ ρκ} . (6.237)$$
Hence it follows from (6.228) that
$$κ =\frac{1}{ ρkT}\frac{(\overline{N^2} −\overline N^2)}{\overline N} . (6.238)$$

(c)

$$\begin{eqnarray*} (\frac{\partial\mu}{\partial\rho})_T&=&V(\frac{\partial \mu}{\partial N})_{T,V}\\ \rho(\frac{\partial\mu}{\partial\rho})_T&=&N(\frac{\partial \mu}{\partial N})_{T,V}\\ (\frac{\partial P}{\partial\rho})_T&=&N(\frac{\partial \mu}{\partial N})_{T,V}\\ N(\frac{\partial \mu}{\partial N})_{T,V}&=&-\frac{V^2}{N}(\frac{\partial P}{\partial V})\\ &=&\frac{V^2}{N}·\frac{1}{\kappa V}\\ &=&\frac{1}{\rho\kappa} \end{eqnarray*}$$

Problem 6.44. Number fluctuations in a noninteracting classical gas

(a) Show that the grand partition function of a noninteracting classical gas can be expressed as

$$Z_G = \sum_{N=0}^{\infty}\frac{(zZ1)^N}{ N!}= e^{zZ_1}, (6.241)$$
where the activity $z = e^{\beta\mu}.$

(a)

$$\begin{eqnarray*} Z_G&=&\sum_{N=1}^{\infty}e^{\beta\mu N}\frac{1}{N!}Z_1^N\\ &=&\sum_{N=0}^{\infty}\frac{(e^{\beta \mu}Z_1)^N}{N!}-1\\ &\approx&\sum_{N=0}^{\infty}\frac{(e^{\beta \mu}Z_1)^N}{N!}\\ &=&\sum_{N=0}^{\infty}\frac{(zZ_1)^N}{N!}=e^{zZ_1} \end{eqnarray*}$$

(b) Show that the mean value of N is given by

$$\overline N = zZ_1, (6.242)$$
and the probability that there are N particles in the system is given by a Poisson distribution:
$$P_N =\frac{z^NZ_N}{ Z_G}=\frac{(zZ_1)^N}{ N!Z_G}=\frac{\overline N^N}{N!}e^{−\overline N}. (6.243)$$

(b)

$$\begin{eqnarray*} P_N&=&\frac{z^NZ_N}{Z_G}\\ \overline N&=&\sum_{N=0}^{\infty}NP_N=\sum_{N=0}^{\infty}\frac{Nz^NZ_N}{Z_G}\\ &=&\sum_{N=0}^{\infty}\frac{N(zZ_1)^N}{N!e^{zZ_1}}=zZ_1 \end{eqnarray*}$$

(c) What is the N dependence of the variance, $\overline {(N −\overline N)^2}$?

$$\begin{eqnarray*} \overline {(N −\overline N)^2}&=&\overline{N^2}-\overline N^2\\ \overline{N^2}&=&\sum_{N=0}^{\infty}N^2P_N\\ &=&\sum_{N=0}^{\infty}\frac{N^2z^NZ_N}{Z_G}\\ &=&\sum_{N=0}^{\infty}\frac{N^2(zZ_1)^N}{N!e^{zZ_1}}\\ &=&zZ_1+(zZ_1)^2\\ \overline {(N −\overline N)^2}&=&zZ_1+(zZ_1)^2-(zZ_1)^2\\ &=&zZ_1 \end{eqnarray*}$$