Quantum Statistics

For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of "background radiation," in addition to the photons and neutrinos (see Figure 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (ea.ch with the same mass), and each has two independent polarization states. During the time period of interest, the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in Figure 7.21. When the temperature was greater than the electron mass times $\raisebox{1ex}{$c^2$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$, the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutrinos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles. the previous problem. Recall from special relativity that the energy of a massive particle is $ϵ=\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}$.

(a) Show that the energy density of electrons and positrons at temperature $T$ is given by

$u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e^{\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}+1}dx; where u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e^{\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}+1}dx$

(b) Show that $u\left(T\right)$ goes to zero when $kT\ll m{c}^2$ , and explain why this is a

reasonable result.

( c) Evaluate $u\left(T\right)$ in the limit $kT\gg m{c}^2$, and compare to the result of the

the previous problem for the neutrino radiation.

(d) Use a computer to calculate and plot $u\left(T\right)$ at intermediate temperatures.

(e) Use the method of Problem 7.46, part (d), to show that the free energy

density of the electron-positron radiation is

$\frac{F}{V}=-\frac{16\pi (kT{)}^4}{(hc{)}^3}f\left(T\right); where f\left(T\right)={\int }_0^{\infty }{x}^2\ln \left(1+e^{-\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}\right)dx$

Evaluate $f\left(T\right)$ in both limits, and use a computer to calculate and plot $f\left(T\right)$ at intermediate

temperatures.

(f) Write the entropy of the electron-positron radiation in terms of the functions

$u\left(T\right)$ and $f\left(T\right)$. Evaluate the entropy explicitly in the high-T limit.

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c² was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have ben conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

${\left(\frac{T}{T_{\nu }}\right)}^3\left[\frac{2{\pi }^4}{45}+u\left(T\right)+f\left(T\right)\right]=\text{ constant }$

as the universe expands and cools. Evaluate the constant by assuming that T=T_v when the temperatures are very high.

(c) Calculate the ratio T/T_v, in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/T_v, as a function of T, for kT/mc²ranging from 0 to 3.*

The tungsten filament of an incandescent light bulb has a temperature of approximately $3000 K$. The emissivity of tungsten is approximately $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, and you may assume that it is independent of wavelength.

(a) If the bulb gives off a total of $100$ watts, what is the surface area of its filament in square millimetres?

(b) At what value of the photon energy does the peak in the bulb's spectrum occur? What is the wavelength corresponding to this photon energy?

(c) Sketch (or use a computer to plot) the spectrum of light given off by the filament. Indicate the region on the graph that corresponds to visible wavelengths, between$400 and 700 nm$.

(d) Calculate the fraction of the bulb's energy that comes out as visible light. (Do the integral numerically on a calculator or computer.) Check your result qualitatively from the graph of part (c).

( e) To increase the efficiency of an incandescent bulb, would you want to raise or lower the temperature? (Some incandescent bulbs do attain slightly higher efficiency by using a different temperature.)

(f) Estimate the maximum possible efficiency (i.e., fraction of energy in the visible spectrum) of an incandescent bulb, and the corresponding filament temperature. Neglect the fact that tungsten melts at $3695 K$.

(a) Estimate (roughly) the total power radiated by your body, neglecting any energy that is returned by your clothes and environment. (Whatever the color of your skin, its emissivity at infrared wavelengths is quite close to $1$ ; almost any nonmetal is a near-perfect blackbody at these wavelengths.)

(b) Compare the total energy radiated by your body in one day (expressed in kilocalories) to the energy in the food you cat. Why is there such a large discrepancy?

(c) The sun has a mass of $2\times {10}^{30} \mathrm{kg}$ and radiates energy at a rate of  $3.9\times {10}^{26}\text{ watts }$. Which puts out more power per units mass-the sun or your body?

A black hole is a blackbody if ever there was one, so it should emit blackbody radiation, called Hawking radiation. A black hole of mass M has a total energy of $M{c}^2$ , a surface area of $16\pi G^2{M}^2/c^4$, and a temperature of$h{c}^3/16{\pi }^{2}kGM$ (as shown in Problem 3.7).

(a) Estimate the typical wavelength of the Hawking radiation emitted by a one-solar-mass (2 x 1030 kg) black hole. Compare your answer to the size of the black hole. 

(b) Calculate the total power radiated by a one-solar-mass black hole.

(c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing. As it loses energy, its mass must decrease; one could say it "evaporates." Derive a differential equation for the mass as a function of time, and solve this equation to obtain an expression for the lifetime of a black hole in terms of its initial mass.

(d) Calculate the lifetime of a one-solar-mass black hole, and compare to the estimated age of the known universe (1010 years).

(e) Suppose that a black hole that was created early in the history of the universe finishes evaporating today. What was its initial mass? In what part of the electromagnetic spectrum would most of its radiation have been emitted? 

The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan's law.

(a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of$2.4\mathrm{eV}$, while Sirius A is approximately $24$ times as luminous as the sun. How does the radius of Sirius A compare to the sun's radius?

(b) Sirius B, the companion of Sirius A (see Figure 7.12), is only $3\%$ as luminous as the sun. Its spectrum, plotted as a function of energy, peaks at about$7\mathrm{eV}$. How does its radius compare to that of the sun?

(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of $0.8\mathrm{eV}$, while Betelgeuse is approximately$10,000$times as luminous as the sun. How does the radius of Betelgeuse compare to the sun's radius? Why is Betelgeuse called a "red supergiant"?

The planet Venus is different from the earth in several respects. First, it is only $70\%$ as far from the sun. Second, its thick clouds reflect $77\%$of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light.

(a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.

(b) Estimate the surface temperature again, taking the reflectivity of the clouds into account.

(c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly $70$ times that of earth's atmosphere. You can therefore model the atmosphere of Venus as $70$ successive "blankets" of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of $2^{1/4}$. The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)

Suppose that the concentration of infrared-absorbing gases in earth's atmosphere were to double, effectively creating a second "blanket" to warm the surface. Estimate the equilibrium surface temperature of the earth that would result from this catastrophe. (Hint: First show that the lower atmospheric blanket is warmer than the upper one by a factor of $2^{1/4}$. The surface is warmer than the lower blanket by a smaller factor.)

The heat capacity of liquid  ${}^{4}He$ below $0.6 \mathrm{K}$is proportional to $T^3$, with the measured value$C_V/Nk=(T/4.67 \mathrm{K}{)}^3$. This behavior suggests that the dominant excitations at low temperature are long-wavelength photons. The only important difference between photons in a liquid and photons in a solid is that a liquid cannot transmit transversely polarized waves-sound waves must be longitudinal. The speed of sound in liquid ${}^4\mathrm{He}$ is $238 \mathrm{m}/\mathrm{s}$, and the density is $0.145 \mathrm{g}/{\mathrm{cm}}^3$. From these numbers, calculate the photon contribution to the heat capacity of${}^4\mathrm{He}$ in the low-temperature limit, and compare to the measured value.

Sketch the heat capacity of copper as a function of temperature from $0$ to $5 \mathrm{K}$, showing the contributions of lattice vibrations and conduction electrons separately. At what temperature are these two contributions equal?

Explain in some detail why the three graphs in Figure $7.28$ all intercept the vertical axis in about the same place, whereas their slopes differ considerably.

The speed of sound in copper is $3560 m/s$. Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure 7.28, and compare.

Evaluate the integrand in equation $7.112$ as a power series in x, keeping terms through $x^4$• Then carry out the integral to find a more accurate expression for the energy in the high-temperature limit. Differentiate this expression to obtain the heat capacity, and use the result to estimate the percent deviation of $C_v$from$3Nk at T=T_D and T=2T_{D.}$

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

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." 

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At $t=0$ the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are $N$ atoms, the total magnetization is typically$~2\mu eN$, where µa is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure $7.30$. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of $\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$21 r$}\right.$and therefore reduces the magnetization by $~2\mu e$. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$.. (in the limit of long wavelengths). Therefore, since$\in =hf$ and $p=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$.. for any "particle," the energy of a magnon is proportional

In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small. 

to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write $\in =\raisebox{1ex}{$p^2$}\!\left/ \!\raisebox{-1ex}{$2pm$}\right.$*, where$m$* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal $1.24\times {10}^{29}kg$, about$14$ times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.

(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by 

$\frac{Nm}{V}=2\pi {\left(\frac{2m\times kT}{h^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx.$

Evaluate the integral numerically.

(b) Use the result of part ($a$) to find an expression for the fractional reduction in magnetization, $\left(M\right(O) - M(T\left)\right)/M\left(O\right).$ Write your answer in the form ${(T /To)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, and estimate the constant$T_0$ for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find $Cv / N k ={ (T /Ti)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ , where $Ti$ differs from To only by a numerical constant. Estimate$Ti$ for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is $470k$.)

(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section $8.2$ we will consider a different two-dimensional model in which magnetization does occur.) 

Evaluate the integral in equation $N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$ numerically, to confirm the value quoted in the text.

Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen,  a gas of approximately  $2\times {10}^{10}$ atoms was trapped and cooled until its peak density was $1.8\times {10}^{14} atoms/c{m}^3$. Calculate the condensation temperature for this system, and compare to the measured value of $50\mu K$.

$Consider a collection of 10,000 atoms of rubidium- 87 , confined inside a box of volume {\left({10}^{-5}m\right)}^3.$ $\left(a\right) Calculate {\epsilon }_0, the energy of the ground state. (Express your answer in both joules and electron-volts.)$$\text{ (b) Calculate the condensation temperature, and compare }k{T}_c\text{ to }{ϵ}_0\text{. }$      $\left(c\right) Suppose that T=0.9 T_c. How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states?$$\left(d\right) Repeat parts \left(b\right) and \left(c\right) for the case of {10}^6 atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state.$

Consider a collection of 10,000 atoms of rubidium- 87 , confined inside a box of volume ${\left({10}^{-5}m\right)}^3$.

(a) Calculate ${\epsilon }_0$, the energy of the ground state. (Express your answer in both joules and electron-volts.)

(b) Calculate the condensation temperature, and compare $k{T}_c to {\epsilon }_0$.

(c) Suppose that $T= 0.9T_c$ How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states?

(d) Repeat parts (b) and (c) for the case of ${10}^6$ atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state.

Calculate the condensate temperature for liquid helium-4, pretending that liquid is a gas of noninteracting atoms. Compare to the observed  temperature of the superfluid transition, $2.17K$. ( the density of liquid helium-4 is $0.145 g/c{m}^3$)

If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate $\mu$ for $T>T_c$.

(a)  As usual when solving a problem on a computer, it's best to start by putting everything in terms dimensionless                      variables. So define $t=T/T_c, c=\mu /k{T}_c,\text{ and }x=ϵ/k{T}_c$. Express the integral that defines $\mu$, equation                                $N={\int }_0^{\infty }g\left(ϵ\right)\frac{1}{e^{(ϵ-\mu )/kT}-1}dϵ$, in terms of these variables, you should obtain the equation

                                                            $2.315={\int }_0^{\infty }\frac{\sqrt{x}dx}{e^{(x-c)/t}-1}$

(b) According to given figure , the correct value of $c$ when $T=2T_c$ , is approximately $-0.8$. Plug in these values and        check that the equation above is approximately satisfied.

(c) Now vary $\mu$, holding $T$ fixed, to find the precise value of $\mu$ for $T=2T_c$ . Repeat for values of$T/T_c$ ranging from $1.2$ up       to $3.0$, in increments of $0.2$. Plot a graph of $\mu$ as a function of temperature.

Problem 7.69. If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate $\mu \text{ for }T>T_c$.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define $t= T/T_c$ ,$c=\mu /k{T}_c,\text{ and }x=ϵ/k{T}_c$. Express the integral that defines , equation 7.22, in terms of these variables. You should obtain the equation

$2.315={\int }_0^{\infty }\frac{\sqrt{x}dx}{e^{(x-c)/t}-1}$

(b) According to Figure

the correct value of $c$ when $T=2T_c$ is approximately $-0.8$. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary $\mu$, holding $T$ fixed, to find the precise value of $\mu$for $T=2T_c$. Repeat for values of $T/T_c$ ranging from $1.2$ up to $3.0$, in increments of $0.2$. Plot a graph of $\mu$ as a function of temperature.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define $t=T/T_c$$c=\mu /k{T}_c,\text{ and }x=ϵ/k{T}_c$. Express the integral that defines $\mu$, equation 7.122, in terms of these variables. You should obtain the equation

(b) According to Figure 7.33, the correct value of $c$when $T=2T_c$ is approximately $-0.8$. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary $\mu$, holding $T$ fixed, to find the precise value of $\mu$ for . Repeat for values of $T/T_c$ ranging from $1.2$ up to $3.0$, in increments of $0.2$. Plot a graph of $\mu$ as a function of temperature.

Starting from the formula for $C_V$ derived in Problem 7.70(b), calculate the entropy, Helmholtz free energy, and pressure of a Bose gas for $T<T_c$. Notice that the pressure is independent of volume; how can this be the case?

Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.

(a) Write down an expression for the total energy of a gas of $N$ bosons confined to a volume $V$, in terms of an integral (analogous to equation 7.122).

(b) For $T<T_c$ you can set $\mu =0$. Evaluate the integral numerically in this case, then differentiate the result with respect to $T$ to obtain the heat capacity. Compare to Figure 7.37.

(c) Explain why the heat capacity must approach $\frac{3}{2}Nk$ in the high- $T$ limit.

(d) For $T>T_c$ you can evaluate the integral using the values of $\mu$ calculated in Problem 7.69. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with Figure 7.37.

Figure 7.37. Heat capacity of an ideal Bose gas in a three-dimensional box.

For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of $\epsilon$ (see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains significantly less than zero as long as $T$ is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must $\epsilon$ have in order for there to be an abrupt Bose-Einstein condensation?

Consider a gas of $n$ identical spin-0 bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are $\epsilon =nhf$, where $n$ is any nonnegative integer and $f$ is the classical oscillation frequency. The degeneracy of level $n\text{ is }(n+1)(n+2)/2$.

(a) Find a formula for the density of states, $g\left(\epsilon \right)$, for an atom confined by this potential. (You may assume $n>>1$.)

(b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency $f$.

(c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order $kT$ ) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and $\mathrm{\pi }$ and so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.

Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation 7.121 numerically, without approximating it as an integral.*

(a) Write equation 7.121 for this system as a sum over energy levels, taking degeneracy into account. Replace $T and \mu$ with the dimensionless variables $t=kT/hf\text{ and }c=\mu /hf$.

(b) Program a computer to calculate this sum for any given values of $t and c$. Show that, for $N=2000$, equation 7.121 is satisfied at $t=15$ provided that $c=-10.534$. (Hint: You'll need to include approximately the first 200 energy levels in the sum.)

(c) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.

(d) Now reduce $t$ to 14 , and adjust the value of $c$ until the sum again equals 2000. Plot the number of particles as a function of energy.

(e) Repeat part (d) for $t=13,12,11,\text{ and }10$. You should find that the required value of $c$ increases toward zero but never quite reaches it. Discuss the results in some detail.

Consider a gas of noninteracting spin-0 bosons at high temperatures, when $\mathit{T}\mathbf{\gg }{\mathit{T}}_{\mathbf{c}}$. (Note that “high” in this sense can still mean below 1 K.)

1. Show that, in this limit, the Bose-Einstein function can be written approximately as
   ${\overline{\mathbf{n}}}_{\mathbf{B}\mathbf{E}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\left(\in -\mu \right)\mathbf{/}\mathbf{k}\mathbf{T}}\left[1+e^{-\left(\in -\mu \right)/kT}+\cdots \right]$.
   
2. Keeping only the terms shown above, plug this result into equation 7.122 to derive the first quantum correction to the chemical potential for gas of bosons.
   
3. Use the properties of the grand free energy (Problems 5.23 and 7.7) to show that the pressure of any system is given by  In $\mathit{P}\mathbf{=}\left(kT/V\right)$, where $\mathit{Z}$ is the grand partition function. Argue that, for gas of noninteracting particles, In $\mathit{Z}$ can be computed as the sum over all modes (or single-particle states) of In ${\mathit{Z}}_{\mathbf{i}}$, where ${\mathit{Z}}_{\mathbf{i}}$; is the grand partition function for the $\mathit{i}\mathbf{th}$ mode.
   
4. Continuing with the result of part (c), write the sum over modes as an integral over energy, using the density of states. Evaluate this integral explicitly for gas of noninteracting bosons in the high-temperature limit, using the result of part (b) for the chemical potential and expanding the logarithm as appropriate. When the smoke clears, you should find
   $\mathit{p}\mathbf{=}\frac{\mathbf{N}\mathbf{k}\mathbf{T}}{\mathbf{V}}\left(1-\frac{N{v}_Q}{4\sqrt{2}V}\right)$,
   again neglecting higher-order terms. Thus, quantum statistics results in a lowering of the pressure of a boson gas, as one might expect.
5. Write the result of part (d) in the form of the virial expansion introduced in Problem 1.17, and read off the second virial coefficient, $\mathit{B}\left(T\right)$. Plot the predicted $\mathit{B}\left(T\right)$ for a hypothetical gas of noninteracting helium-4 atoms.
6. Repeat this entire problem for gas of spin-1/2 fermions. (Very few modifications are necessary.) Discuss the results, and plot the predicted virial coefficient for a hypothetical gas of noninteracting helium-3 atoms.