Introduction to the Statistical Theory of Matter

Abstract

In this Chapter we provide an introduction to the basic concepts of the statistical theory of matter, which played a prominent role in the development of the Quantum Theory.

Problems

3.1

We have to place four distinct objects into 3 boxes. How many possible different distributions can we choose? What is the multiplicity $\mathcal{M}$ of each distribution? And its probability p?

Answer: We can make 3 different choices for each object, therefore the total number of possible choices is $3^4 = 81$. The total number of possible distributions is instead given by all the possible choices of non-negative integers $n_1, n_2, n_3$ with $n_1 + n_2 + n_3 = 4$, i.e. $(4 +1)(4 + 2)/2 = 15$. There are in particular 3 distributions like (4, 0, 0), each with $p = \mathcal{M}/81 = 1/81$; 6 like (3, 1, 0), with $p = \mathcal{M}/81 = 4/81$; 3 like (2, 2, 0), with $p = \mathcal{M}/81 = 6/81$; 3 like (2, 1, 1), with $p = \mathcal{M}/81 = 12/81$.

3.2

The integer number k can take values in the range between 0 and 8 according to the binomial distribution:

$$ P(k)={1\over 2^8} \left( {\begin{array}{c}8\\ k\end{array}}\right) . $$

Compute the mean value of k and its mean quadratic deviation.

Answer: $\bar{k} = 4; \;\;\; \langle (k - \bar{k})^2 \rangle = 2$.

3.3

Let us consider a system which can be found in 4 possible states, enumerated by the index $k = 0,1,2,3$ and with energy $E_k = \epsilon k$, where $\epsilon = 10^{-2}\ \mathrm{eV}$. The system is at thermal equilibrium at room temperature $T \simeq 300\,^{\circ }\mathrm{K}$. What is the probability for the system being in the highest energy state?

Answer: $Z = \sum _{k = 0}^3 \mathrm{e}^{- \beta \epsilon k}$; $\quad U = (1/Z) \sum _{k = 0}^3 \epsilon k \mathrm{e}^{- \beta \epsilon k} \simeq 1.035\ 10^{-2}\ \mathrm{eV}$; $\quad P_{k = 3} = (1/Z) \mathrm{e}^{- 3 \beta \epsilon } \simeq 0.127$.

3.4

A diatomic molecule is made up of two particles of equal mass $M= 10^{-27}\ \mathrm{kg}$ which are kept at a fixed distance $L= 4\,\times \,10^{-10}\ \mathrm{m}$. A set of $N = 10^9$ such systems, which are not interacting among themselves, is in thermal equilibrium at a temperature $T = 1\,^{\circ }\mathrm{K}$. Estimate the number of systems which have a non-vanishing angular momentum (computed with respect to their center of mass), i.e. the number of rotating molecules, making use of the fact that the number of states with angular momentum $n \hbar$ is equal to $2n + 1$.

Answer: If we quantize rotational energy according to Bohr, then the possible energy levels are $E_n = n^2 \hbar ^2/(M L^2) \simeq 4.3\,\times \,10^{-4} n^2\ \mathrm{eV}$, each corresponding to $2 n + 1$ different states. These states are occupied according to the Canonical Distribution. The partition function is $Z = \sum _{n=0}^\infty (2n +1) \mathrm{e}^{-E_n/kT}$. If $T = 1\,^{\circ }\mathrm{K}$, then $kT \simeq 0.862\,\times \,10^{-4}\ \mathrm{eV}$, hence $\mathrm{e}^{-E_n/kT} \simeq \mathrm{e}^{- 5.04 n^2} \simeq (6.47\,\times \,10^{-3})^{n^2}$. Therefore the two terms with $n = 0,1$ give a very good approximation of the partition function, $Z \simeq 1 + 1.94\,\times \,10^{-2}$. The probability that a molecule has $n = 0$ is 1 / Z, hence the number of rotating molecules is $N_R = N (1 - 1/Z) \simeq 1.9\,\times \,10^7$. If we instead make use of Sommerfeld’s perfected theory, implying $n^2 \rightarrow n(n+1)$ in the expression for $E_n$, we obtain $Z \simeq 1 + 1.25\,\times \,10^{-4}$ and $N_R = N (1 - 1/Z) \simeq 1.25\,\times \,10^5$. In the present situation, being quantum effects quite relevant, the use of Sommerfeld’s correct formula for angular momentum quantization in place of simple Bohr’s rule makes a great difference .

3.5

Consider again Problem 3.4 in case the molecules are in equilibrium at room temperature, $T \simeq 300\,^{\circ }\mathrm{K}$. Compute also the average energy of each molecule.

Answer: In this case the partition function is, according to Sommerfeld’s theory:

$$ Z = \sum _{n=0}^\infty (2n +1) \mathrm{e}^{-\alpha n (n + 1)}, $$

with $\alpha \simeq 0.0168$. Since $\alpha \ll 1$, $(2n +1) \mathrm{e}^{-\alpha n^2}$ is the product of a linear term times a slowly varying function of n, hence the sum can be replaced by an integral

$$ Z \simeq \int _{0}^\infty dn (2n +1) \mathrm{e}^{-\alpha n (n + 1)} = {1 \over \alpha } \simeq 60 $$

hence the number of non-rotating molecules is $N_{NR} = N/Z \simeq 1.67\,\times \,10^7$. From $Z \simeq 1/\alpha = k T M L^2 / \hbar ^2$, we get $U = -\partial /\partial \beta \ln Z = k T$, in agreement with equipartition of energy.

3.6

A system in thermal equilibrium admits 4 possible states: the ground state, having zero energy, plus three degenerate excited states of energy $\epsilon$. Discuss the dependence of its mean energy on the temperature T.

Answer:

$$ U = \frac{3 \epsilon \ \mathrm{e}^{-\epsilon /kT}}{1 + 3 \mathrm{e}^{-\epsilon /kT}} \,\, ; \quad \quad \lim _{T\rightarrow 0} U(T) = 0 \,\, ; \quad \quad \lim _{T\rightarrow \infty } U(T) = {3 \over 4} \epsilon \ . $$

3.7

A simple pendulum of length $l=10\ \mathrm{cm}\$ and mass $m=10\ \mathrm{g}\$ is placed on Earth’s surface in thermal equilibrium at room temperature, $T=300\,^{\circ }\mathrm{K}$. What is the mean quadratic displacement of the pendulum from its equilibrium point?

Answer: The potential energy of the pendulum is, for small displacements $s<<l$, ${mgs^2/2l}$. From the energy equipartition theorem we infer ${mgs^2/2l} \simeq {kT/2}$, hence $\sqrt{\langle s^2 \rangle } \simeq \sqrt{kTl/mg}=2\,\times \,10^{-10}\ \mathrm{m}$.

3.8

What is the length of a pendulum for which quantum effects are important at room temperature?

Answer: $\hbar \omega \sim k T$, so that $l = g/\omega ^2 \sim \hbar ^2 g/(kT)^2 \sim 6.5\,\times \,10^{-28}$ m.

3.9

A massless particle is constrained to move along a segment of length L; therefore its wave function vanishes at the ends of the segment. The system is in equilibrium at a temperature T. Compute its mean energy as well as the specific heat at fixed L. What is the force exerted by the particle on the ends of the segment?

Answer: Energy levels are given by $E_n=nc\pi \hbar /L$, hence $Z=\sum _{n=1}^\infty \mathrm{e}^{-\beta E_n}=1/(\mathrm{e}^{\beta c\pi \hbar /L}-1)$ from which the mean energy follows

$$ U=-{\partial \ln Z\over \partial \beta }={c\pi \hbar \over L}{1 \over (1-\mathrm{e}^{- c\pi \hbar / LkT})}\ ,$$

and the specific heat

$$ C_L=k(c\pi \hbar /LkT)^2{\mathrm{e}^{- c\pi \hbar /LkT}\over (1-\mathrm{e}^{- c\pi \hbar /LkT})^2}\ , $$

which vanishes at low temperatures and approaches k at high temperatures; notice that the equipartition principle does not hold in its usual form in this example, since the energy is not quadratic in the momentum, hence we have k instead of k / 2. The equation of state can be obtained making use of (3.29) and (3.35), giving for the force $F=(1/\beta ){(\partial \ln Z/\partial L) }=U/ L.$ Hence at high temperatures we have $FL=kT.$

3.10

Consider a system made up of N distinguishable and non-interacting particles which can be found each in two possible states of energy 0 and $\epsilon$. The system is in thermal equilibrium at a temperature T . Compute the mean energy and the specific heat of the system.

Answer: The partition function for a single particle is $Z_1=1+\mathrm{e}^{-{\epsilon /kT}}$. That for N independent particles is $Z_N=Z_1^N$. Therefore the average energy is

$$ U={N\epsilon \over 1+\mathrm{e}^{\epsilon /kT}} $$

and the specific heat is

$$ C=Nk\left( {\epsilon \over kT}\right) ^2{ \mathrm{e}^{\epsilon /kT}\over (\mathrm{e}^{\epsilon / kT}+1)^2}. $$

3.11

A system consists of a particle of mass m moving in a one-dimensional potential which is harmonic for $x > 0$ ($V={k}x^2/2$) and infinite for $x<0$. If the system is at thermal equilibrium at a temperature T, compute its average energy and its specific heat.

Answer: The wave function must vanish in the origin, hence the possible energy levels are those of the harmonic oscillator having an odd wave function. In particular, setting $\omega = \sqrt{k/m}$, we have $E_n=(2n+{3/2})\hbar \omega$, with $n=0, 1, \ldots$. The partition function is

$$ Z={\mathrm{e}^{-{3\beta \hbar \omega /2}}\over 1-\mathrm{e}^{- 2\beta \hbar \omega }}, $$

so that the average energy is

$$ U={3\hbar \omega \over 2} +{2\hbar \omega \over \mathrm{e}^{2\beta \hbar \omega }-1} $$

and the specific heat is

$$ C=k (2\beta \hbar \omega )^2{\mathrm{e}^{2\beta \hbar \omega }\over ( \mathrm{e}^{2\beta \hbar \omega }-1)^2}\ . $$

3.12

Compute the average energy of a classical three-dimensional isotropic harmonic oscillator of mass m and oscillation frequency $\nu = 2 \pi \omega$ in equilibrium at temperature T.

Answer: The state of the classical system is assigned in terms of the momentum p and the coordinate x of the oscillator, it is therefore represented by a point in phase space corresponding to an energy $E(p,x) = p^2/2m + m \omega ^2 x^2/2$. The canonical partition function can therefore be written as an integral over phase space

$$ Z = \int \frac{d^3 p\ d^3 x}{\varDelta } \mathrm{e}^{-\beta p^2/2m} \mathrm{e}^{-\beta m \omega ^2 x^2 /2} $$

where $\varDelta$ is an arbitrary effective volume in phase space needed to fix how we count states (that is actually not arbitrary according to the quantum theory, which requires $\varDelta \sim h^3$). A simple computation of Gaussian integrals gives $Z = \varDelta ^{-1} (2 \pi / \omega \beta )^3$, hence $U = - (\partial /\partial \beta ) \ln Z = 3 k T$, in agreement with equipartition of energy.

3.13

A particle of mass m moves in the $x-y$ plane under the influence of an anisotropic harmonic potential $V(x,y) = m ( \omega _x^2 x^2/2 + \omega _y^2 y^2/2)$, with $\omega _y \ll \omega _x$. Therefore the energy levels coincide with those of a system made up of two distinct particles moving in two different one-dimensional harmonic potentials corresponding respectively to $\omega _x$ and $\omega _y$. The system is in thermal equilibrium at a temperature T. Compute the specific heat and discuss its behaviour as a function of T.

Answer: The partition function is the product of the partition functions of the two distinct harmonic oscillators, hence the average energy and the specific heat will be the sum of the respective quantities. In particular

$$ C = {(\hbar \omega _x)^2 \over k T^2} {\mathrm{e}^{\beta \hbar \omega _x} \over {(\mathrm{e}^{\beta \hbar \omega _x - 1})^2}} + {(\hbar \omega _y)^2 \over k T^2} {\mathrm{e}^{\beta \hbar \omega _y} \over {(\mathrm{e}^{\beta \hbar \omega _y - 1})^2}}. $$

We have three different regimes: $C \sim 0$ if $k T \ll \hbar \omega _y$, $C \sim k$ if $\hbar \omega _y \ll k T \ll \hbar \omega _x$ and finally $C \sim 2 k$ if $k T \gg \hbar \omega _x$.

3.14

Compute the mean quadratic velocity for a rarefied and ideal gas of particles of mass $M=10^{-20}\ \mathrm{kg}$ in equilibrium at room temperature.

Answer: According to Maxwell distribution, $\langle {v^2} \rangle =3kT/ M \simeq 1.2\ \mathrm{m}^2/\mathrm{s}^2$.

3.15

Taking into account that a generic molecule has two rotational degrees of freedom, compute, using the theorem of equipartition of energy, the average square angular momentum $\bar{J^2}$ of a molecule whose moment of inertia about the center of mass is $I= 10^{-39}\ \mathrm{kg}\ \mathrm{m}^2$, independently of the rotation axis, if the temperature is $T= 300\ {}^{\circ }\mathrm{K}$. Discuss the validity of the theorem of equipartition of energy in the given conditions.

Answer: On account of the equipartition of energy, the average rotational kinetic energy of the molecule is $2 k T=8.29\,\times \,10^{-21}\ \mathrm{J} = \bar{J^2}/2I$, thus $\bar{J^2}=1.66\,\times \,10^{-59}\ \mathrm{J}\ \mathrm{s}$. The theorem of equipartition of energy is based on the assumption that the energy, and hence the square angular momentum, be a continuous variable, while, as a matter of fact (see e.g. Problem 2.1), it is quantized according to the formula $J^2 =n(n+1)\hbar ^2$. Therefore the validity of the energy equipartition requires that the difference between two neighboring values of $J^2$ be much smaller than $\bar{J^2}$, i.e. $(2n+1)/(n(n+1))\simeq 2/n\ll 1$. In the given conditions $n\simeq J/\hbar \simeq \sqrt{1.49\,\times \,10^{9}}$, therefore previous inequality is satisfied.

3.16

Consider a diatomic gas, whose molecules can be described schematically as a pair of pointlike particles of mass $M = 10^{-27}\ \mathrm{kg}$, which are kept at an equilibrium distance $d = 2\,\times \,10^{-10}\ \mathrm{m}$ by an elastic force of constant $K = 11.25\ \mathrm{N/m}$. A quantity equal to 1.66 g atoms of such gas is contained in volume $V = 1\ \mathrm{m^3}$. Discuss the qualitative behaviour of the specific heat of the system as a function of temperature. Consider the molecules as non-interacting and as if each were contained in a cubic box with reflecting walls of size $L^3 = V/N$, where N is the total number of molecules.

Answer: Three different energy scales must be considered. The effective volume available for each molecule sets an energy scale $E_1 = \hbar ^2 \pi ^2/(4 M L^2) \simeq 1.7\,\times \,10^{-6}\ \mathrm{eV}$, which is equal to the ground state energy for a particle of mass 2 M in a cubic box, corresponding to a temperature $T_1 = E_1/k \simeq 0.02\,^{\circ }\mathrm{K}$. The minimum rotational energy is instead, according to Sommerfeld, $E_2 = \hbar ^2/(M d^2) \simeq 3.5\,\times \,10^{-3}\ \mathrm{eV}$, corresponding to a temperature $T_2 = E_2/k \simeq 40\,^{\circ }\mathrm{K}$. Finally, the fundamental oscillation energy is $E_3 = \hbar \sqrt{2 K/M} \simeq 0.098\ \mathrm{eV}$, corresponding to a temperature $T_3 = E_2/k \simeq 1140\,^{\circ }\mathrm{K}$. For $T_1 \ll T \ll T_2$ the system can be described as a classical perfect gas of pointlike particles, since rotational and vibrational modes are not yet excited, hence the specific heat per molecule is $C \sim 3 k /2$. For $T_2 \ll T \ll T_3$ the system can be described as a classical perfect gas of rigid rotators, hence $C \sim 5 k /2$. Finally, for $T \gg T_3$ also the (one-dimensional) vibrational mode is excited and $C \sim 7 k /2$. This roughly reproduces, from a qualitative point of view and with an appropriate rescaling of parameters, the observed behavior of real diatomic gases.

3.17

We have a total mass $M=10^{-6}\ \mathrm{kg}$ of a dust of particles of mass $m= 10^{-17}\ \mathrm{kg}$. The dust particles can move in the vertical $x-z$ semi-plane defined by $x > 0$, and above a line forming an angle $\alpha$, with $\tan \alpha =10^{-3}$, with the positive x axis. The dust is in thermal equilibrium in air at room temperature ($T= 300\ {}^{\circ }\mathrm{K}$) and hence the particles, which do not interact among themselves, have planar Brownian motion above the mentioned line. What is the distribution of particles along the positive x axis and which their average distance $\bar{x}$ from the vertical z axis?

Answer: We start assuming that, in the mentioned conditions, quantum effects are negligible and hence the particle distribution in the velocity and position plane is given by the Maxwell-Boltzmann law: $d^4n/ ((d^2v) dxdz)=N\exp (-E/kT)=N \exp (-(mv^2/2+mgz)/(kT))$ where g is the gravitational acceleration and N is a normalization factor. Then the x-distribution of particles is given by:

$$\begin{aligned} \nonumber \frac{dn}{dx}= & {} \int _{-\infty }^\infty dv_x \int _{-\infty }^\infty dv_y \int _{x \tan \alpha }^\infty dz \frac{d^4n}{d^2v dxdz} \\ \nonumber= & {} \frac{2\pi kT N}{m} \int _{x \tan \alpha }^\infty dz \mathrm{e}^{-mgz/(kT)} = \frac{2\pi (kT)^2 N}{m^2g} \mathrm{e}^{-mg x \tan \alpha /(kT)}. \end{aligned}$$

We can compute N using $\int _0^\infty (dn/dx)dx=2\pi (kT)^3 N/(\tan \alpha \ m^3g^2)=M/m=10^{11}$ and $\bar{x}=kT/(mg\tan \alpha )\simeq 4.22\,\times \,10^{-2}\ \mathrm{m}$. Now we discuss the validity of the classical approximation. The average inter-particle distance is of the order of magnitude of $m\bar{x}/M\sim 4\,\times \,10^{-13}\ \mathrm{m}$, which should be much larger than the average de Broglie wave length of the particles, which is of the order of magnitude of $h/\sqrt{2mkT} \sim 2\,\times \,10^{-15}\ \mathrm{m}$. We conclude that the classical approximation is valid.

3.18

The possible stationary states of a system are distributed in energy as follows:

$$ \frac{d\ n}{dE} = \alpha E^3 \mathrm{e}^{E/E_0} $$

where $E_0$ is some given energy scale. Compute the average energy and the specific heat of the system for temperatures $T < E_0/k$, then discuss the possibility of reaching thermal equilibrium at $T = E_0/k$.

Answer: Let us set $\beta _0 = 1/E_0$. The density of states diverges exponentially with energy and the partition function of the system is finite only if the temperature is low enough in order for the Boltzmann factor, which instead decreases exponentially with energy, to be dominant at high energies. For $T < E_0/k$ ($\beta > \beta _0$) we have:

$$ Z = \int _0^\infty dE \alpha E^3 \mathrm{e}^{-(\beta - \beta _0)E} = \frac{6 \alpha }{(\beta - \beta _0)^4} $$

from which the internal energy U and the specific heat $C = dU/dT$ easily follow:

$$ U = \frac{4}{\beta - \beta _0} = \frac{4 k T E_0}{E_0 - k T} \,\,\, ; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, C = \frac{4 k E_0^2}{(E_0 - k T)^2}. $$

The specific heat diverges at $T = E_0/k$: in general that may happen in the presence of a phase transition, but in this specific case also the internal energy diverges as $T \rightarrow E_0/k$, meaning that an infinite amount of energy must be spent in order to bring the system at equilibrium at that temperature, i.e. it is not possible to reach thermal equilibrium at that temperature.

3.19

Consider a system made up of two identical fermionic particles which can occupy 4 different states. Enumerate all the possible choices for the occupation numbers of the single particle states. Assuming that the 4 states have the following energies: $E_1= E_2 =0$ and $E_3=E_4 =\epsilon$ and that the system is in thermal equilibrium at a temperature T, compute the mean occupation number of one of the first two states as a function of temperature.

Answer: There are six different possible states for the whole system characterized by the following occupation numbers $(n_1,n_2,n_3, n_4)$ for the single particle states: (1, 1, 0, 0) , (1, 0, 1, 0) , (1, 0, 0, 1) , (0, 1, 1, 0) , (0, 1, 0, 1) , (0, 0, 1, 1) . The corresponding energies are $0, \epsilon , \epsilon , \epsilon , \epsilon , 2\epsilon$. The mean occupation number of the first single particle state (i.e. $\langle n_1 \rangle$) is then given by averaging the value of $n_1$ over the 6 possible states weighted using the Canonical Distribution, i.e.

$$ \langle n_1 \rangle ={(1+2 \mathrm{e}^{-\beta \epsilon })\over (1+4\mathrm{e}^{-\beta \epsilon }+\mathrm{e}^{-2\beta \epsilon })}\ . $$

3.20

A system, characterized by 3 different single particle states, is filled with 4 identical bosons. Enumerate the possible states of the system specifying the corresponding occupation numbers. Discuss also the case of 4 identical fermions.

Answer: The possible states can be enumerated by indicating all possible choices for the occupation numbers $n_1,n_2,n_3$ satisfying $n_1 + n_2 + n_3 = 4$. That leads to 15 different states. In the case of fermions, since $n_i = 0,1$, the constraint on the total number of particles cannot be satisfied and there is actually no possible state for this system.

3.21

A system, characterized by two single particle states of energy $E_1=0$ and $E_2=\epsilon$, is filled with 4 identical bosons. Enumerate all possible choices for the occupation numbers. Assuming that the system is in thermal contact with a reservoir at temperature T and that $\mathrm{e}^{-\beta \epsilon }={1\over 2}$, compute the probability of all particles being in the ground state. Compare the answer with that for distinguishable particles.

Answer: Since the occupation numbers must satisfy $N_1 + N_2 =4$, the possible states are identified by the value of, for instance, $N_2$, in the case of bosons ($N_2 = 0,1,2,3,4$), and have energy $\epsilon N_2$. In the case of distinguishable particles there are instead ${4!}/({N_2! (4 - N_2)!})$ different states for each value of $N_2$. The probability of all the particles being in the ground state is 16 / 31 in the first case and $({2/3})^4$ in the second case: notice that this probability is highly enhanced in the case of bosons.

3.22

Consider a gas of electrons at zero temperature. What is the density at which relativistic effects show up? Specify the answer by finding the density for which electrons occupy states corresponding to velocities $v={\sqrt{3}\, c /2}$.

Answer: At $T = 0$ electrons occupy all levels below the Fermi energy $E_F$, or equivalently below the corresponding Fermi momentum $p_F$. To answer the question we must impose that

$$ p_F = {m_e v \over \sqrt{1 - v^2/c^2}} = \sqrt{3}\ m_e c. $$

On the other hand, the number of states below the Fermi momentum, assuming the gas is contained in a cubic box of size L, is

$$ N = {p_F^3 L^3 \over 3 \hbar ^3 \pi ^2}, $$

hence $\rho = p_F^3 / (3 \pi ^2 \hbar ^3) \simeq 3.04\,\times \,10^{36}\ \mathrm{particles}/\mathrm{m}^3$.

3.23

The density of states as a function of energy in the case of free electrons is given in (3.64). However in a conduction band the distribution may have a different dependence on energy. Let us consider for instance the simple case in which the density is constant, ${dn_E / d E}=\gamma \ V$, where $\gamma =8\,\times \,10^{47}\ \mathrm{m}^{-3}\ \mathrm{J}^{-1}$, the energy varies from zero to $E_0=1\ \mathrm{eV}$ and the electronic density is $\rho \equiv {\bar{n} / V}= 6\,\times \,10^{28}\ \mathrm{m}^{-3}$. For T not much greater than room temperature it is possible to assume that the bands above the conduction one are completely free, while those below are completely occupied, hence the thermal properties can be studied solely on the basis of its conduction band. Under these assumptions, compute how the chemical potential $\mu$ depends on temperature.

Answer: The average total number of particles comes out to be $\bar{N} = \int _0^{E_0} n(\epsilon ) g(\epsilon ) d\epsilon \;$. The density of levels is $g(\epsilon ) = {dn_E / d E}=\gamma \ V$ and the average occupation number is $n(\epsilon ) = 1/(\mathrm{e}^{\beta (\epsilon - \mu )} + 1)$. After computing the integral and solving for $\mu$ we obtain

$$ \mu = k T \ln \left( \frac{\mathrm{e}^{\rho /(\gamma k T)} - 1}{1 - \mathrm{e}^{\rho /(\gamma k T)} \mathrm{e}^{-E_0/kT}}\right) \;. $$

It can be verified that, since by assumption $\rho /\gamma < E_0$, in the limit $T \rightarrow 0$ $\mu$ is equal to the Fermi energy $E_F = \rho /\gamma$. Instead, in the opposite large temperature limit, $\mu \rightarrow - kT \ln ( 1 - \gamma E_0/\rho )$, hence the distribution of electrons in energy would be constant over the band and simply given by $n(\epsilon ) g(\epsilon ) = \rho V / E_0$, but of course in this limit we cannot neglect the presence of other bands. Notice also that in this case, due to the different distribution of levels in energy, we have $\mu \rightarrow + \infty$, instead of $\mu \rightarrow - \infty$, as $T \rightarrow \infty$.

3.24

The modern theory of cosmogenesis suggests that cosmic space contains about $10^8$ neutrinos per cubic meter and for each species of these particles. Neutrinos can be considered, in a first approximation, as massless fermions having a single spin state instead of two, as for electrons; they belong to 6 different species. Assuming that each species be independent of the others, compute the corresponding Fermi energy.

Answer: Considering a gas of neutrinos placed in a cubic box of size L, the number of single particle states with energy below the Fermi energy $E_F$ is given, for massless particles, by $N_{E_F} = (\pi /6) E_F^3 L^3 / (\pi \hbar c)^3$. Putting that equal to the average number of particles in the box, $\bar{N} = \rho L^3$, we have $E_F = \hbar c\ (6 \pi ^2 \rho )^{1/3} \simeq 3.38\,\times \,10^{-4}\; { \mathrm{eV}}$.

3.25

Suppose now that neutrinos must be described as particles of mass $m_\nu \ne 0$. Consider again Problem 3.24 and give the exact relativistic formula expressing the Fermi energy in terms of the gas density $\rho$.

Answer: The formula expressing the total number of particles $N = L^3 \rho$ in terms of the Fermi momentum $p_F$ is:

$$ N_{E_F} = (\pi /6) p_F^3 L^3 / (\pi \hbar )^3, $$

hence

$$ E_F = \sqrt{m_\nu ^2 c^4 + p_F^2 c^2} = \sqrt{m_\nu ^2 c^4 + (6 \pi ^2 \rho )^{2/3} (\hbar c)^2}. $$

3.26

Compute the internal energy and the pressure at zero temperature for the system described in Problem 3.24, i.e. for a gas of massless fermions with a single spin state and a density $\rho \equiv {\bar{n} V}= 10^{8}\ \mathrm{m}^{-3}$.

Answer: The density of internal energy is

$$ U/V\ =(81\pi ^2\rho ^4/32)^{1\over 3}\hbar c\simeq 4.29\,\times \,10^{-15}\ \mathrm{J}/\mathrm{m}^3, $$

and the pressure

$$ P=U/3V \simeq 1.43\,\times \,10^{-15}\ \mathrm{Pa}. $$

Notice that last result is different from what obtained for electrons, Eq. (3.79): the factor 1/3 in place of 2/3 is a direct consequence of the linear dependence of energy on momentum taking place for massless or ultrarelativistic particles, in contrast with the quadratic behavior which is valid for (massive) non-relativistic particles.

3.27

$10^3$ bosons move in a harmonic potential corresponding to a frequency $\nu$ such that $h\nu =1 \ \mathrm{eV} .$ Considering that the mean occupation number of the m-th level of the oscillator is given by the Bose–Einstein distribution: $n_m= (\mathrm{e}^{\beta (h m\nu -\mu )}-1)^{-1}$, compute the chemical potential assuming $T=300\,^{\circ }\mathrm{K}\cdot$

Answer: The total number of particles, $N = 1000$, can be written as

$$ N = \sum _m n_m = {1 \over \mathrm{e}^{- \beta \mu } - 1} + {1 \over K \mathrm{e}^{- \beta \mu } - 1} + {1 \over K^2 \mathrm{e}^{- \beta \mu } - 1} + \cdots $$

where $K = \exp (h \nu / k T) \simeq \mathrm{e}^{40}$. Since K is very large and $\exp (- \beta \mu ) > 1$ ($\mu \le 0$ for bosons), it is clear that only the first term is appreciably different from zero. Hence

$$ \exp (- \beta \mu ) = 1 + {1 \over N} $$

and finally $\mu \simeq - 2.5\,\times \,10^{-5}\ \mathrm{eV}.$

3.28

Consider a system of $\bar{n}\gg 1$ spin-less non-interacting bosons. Each boson has two stationary states, the first state with null energy, the second one with energy $\epsilon$. If $\mu$ is the chemical potential of the system, $\exp (\beta \mu )=f$ is its fugacity and one has $f\le 1$. Compute $z=f^{-1}$ as a function of the temperature $T=1/(\beta k)$ (in fact z is a function of $\beta \epsilon$). In particular identify the range of values of z when T varies from 0 to $\infty$.

Answer: It is convenient to study $\zeta \equiv z\exp (\beta \epsilon /2)$ instead of z. $\zeta$ must diverge in the $T\rightarrow 0$ limit since $z\ge 1$. The condition that the sum of state occupation numbers be equal to $\bar{n}$ gives the equation:

$$ \zeta ^2 -2\cosh (\beta \epsilon /2)(1+1/\bar{n})\zeta +2/\bar{n} +1=0\ , $$

whose solutions are

$$ \zeta _\pm =\cosh (\beta \epsilon /2)(1+1/\bar{n}) \pm \sqrt{\cosh ^2(\beta \epsilon /2)(1+1/\bar{n})^2-2/\bar{n}-1}. $$

One must choose $\zeta _+$ since $\zeta _-$ vanishes in the $T\rightarrow 0$ limit. Then one has in the $T\rightarrow 0$ limit $\zeta \rightarrow 2\cosh (\beta \epsilon /2)(1+1/\bar{n})\rightarrow \exp (\beta \epsilon /2)(1+1/\bar{n})$ and hence the average occupation number of the zero energy state $\bar{n}_0=1/(\zeta \exp (-\beta \epsilon /2)-1)\rightarrow \bar{n}$. In the $T\rightarrow \infty$ limit one has $\zeta \rightarrow 1+2/\bar{n}$ and hence $\bar{n}_0\rightarrow 2/\bar{n}$. Therefore z ranges from $1+1/\bar{n}$ to $1+2/\bar{n}$ and the chemical potential ranges approximately from $-kT/\bar{n}$, for T small, to $-2kT/\bar{n}$ for T large. There is no Bose condensation.

3.29

A system is made up of N identical bosonic particles of mass m moving in a one-dimensional harmonic potential $V(x) = {m \omega ^2 x^2}/{2}$. What is the distribution of occupation numbers corresponding to the ground state of the system? And that corresponding to the first excited state? Determine the energy of both states.

If $\hbar \omega = 0.1\ \mathrm{eV}$ and if the system is in thermal equilibrium at room temperature, $T = 300\,^{\circ }\mathrm{K}$, what is the ratio R of the probability of the system being in the first excited state to that of the system being in the ground state? How the last answer changes in the case of distinguishable particles?

Answer: In the ground state all particles occupy the single particle state of lowest energy ${\hbar \omega /2}$, hence $E = N {\hbar \omega /2}$, while in the first excited state one of the N particles has energy ${3\hbar \omega /2}$, hence $E = (N +\, 2 ){\hbar \omega /2}$. The ground state has degeneracy 1 both for identical and distinguishable particles. The first excited state has denegeracy 1 in the case of bosons while the degeneracy is N in the other case, since it makes sense to ask which of the N particles has energy ${3\hbar \omega /2}$. Therefore $R=\mathrm{e}^{-{\hbar \omega / kT}} \simeq 0.021$ for bosons and $R \simeq N\ 0.021$ in the second case. For large N the probability of the system being excited is much suppressed in the case of bosons with respect to the case of distinguishable particles.

3.30

Consider again Problem 3.29 in the case of fermions having a single spin state and for $\hbar \omega = 1 \; \mathrm{eV}$ and $T \simeq 1000\,^{\circ }\mathrm{K}$.

Answer: In the ground state of the system the first N levels of the harmonic oscillator are occupied, hence its energy is $E_0=\sum _{i=0}^{N-1}(n+{1/2})\hbar \omega ={(N^2/2)}\hbar \omega$. The minimum possible excitation of this state corresponds to moving the fermion of highest energy up to the next free level, hence the energy of the first excited state is $E_1=E_0+\hbar \omega$. The ratio R is equal to $\mathrm{e}^{-{\hbar \omega / kT}}=9.12\,\times \,10^{-6} .$

3.31

A system is made up of $N = 10^8$ electrons which are free to move along a conducting cable of length $L =1\ \mathrm{cm}$, which can be roughly described as a one-dimensional segment with reflecting endpoints. Compute the Fermi energy of the system, taking also into account the spin degree of freedom.

Answer: $E_F={\hbar ^2 N^2\pi ^2/(8mL^2)}= 1.5\,\times \,10^{-18}\ \mathrm{J}$.

3.32

Let us consider a system made up of two non-interacting particles at thermal equilibrium at temperature T. Both particles can be found in a set of single particle energy levels $\epsilon _n$, where n is a non-negative integer. Compute the partition function of the system, expressing it in terms of the partition function $Z_1(T)$ of a single particle occupying the same energy levels, for the following three cases: distinguishable particles, identical bosonic particles, identical fermionic particles.

Answer: If the particles are distinguishable, all states are enumerated by specifying the energy level occupied by each particle, hence we can sum over the energy levels of the two particles independently:

$$ Z(T) = \sum _{n} \sum _{m} \mathrm{e}^{- \beta (\epsilon _n + \epsilon _m)} = \left( \sum _n \mathrm{e}^{-\beta \epsilon _n} \right) ^2 = (Z_1(T))^2. $$

Instead, in case of two bosons, states corresponding to a particle exchange must be counted only once, hence we must treat separately states where the particles are in the same level or not

$$\begin{aligned} Z(T)= & {} \frac{1}{2} \sum _{n \ne m} \mathrm{e}^{- \beta (\epsilon _n + \epsilon _m)} + \sum _n \mathrm{e}^{- \beta (\epsilon _n + \epsilon _n)} \nonumber \\= & {} \frac{1}{2} \sum _{n,m} \mathrm{e}^{- \beta (\epsilon _n + \epsilon _m)} - \frac{1}{2}\sum _n \mathrm{e}^{- 2 \beta \epsilon _n } + \sum _n \mathrm{e}^{- 2 \beta \epsilon _n } = \frac{1}{2} \left( (Z_1(T))^2 + Z_1(T/2) \right) \nonumber . \end{aligned}$$

If the particles are fermions, we must count only states where the particles are in different energy levels

$$\begin{aligned} Z(T)= & {} \frac{1}{2} \sum _{n \ne m} \mathrm{e}^{- \beta (\epsilon _n + \epsilon _m)} \nonumber \\= & {} \frac{1}{2} \sum _{n,m} \mathrm{e}^{- \beta (\epsilon _n + \epsilon _m)} - \frac{1}{2}\sum _n \mathrm{e}^{- 2 \beta \epsilon _n } = \frac{1}{2} \left( (Z_1(T))^2 - Z_1(T/2) \right) \nonumber . \end{aligned}$$

3.33

A particle of mass $m = 9\,\times \,10^{-31}$ is placed at thermal equilibrium, at temperature $T = 10^3\ {}^{\circ }\mathrm{K}$, in a potential which can be described as a distribution of spherical wells. Each spherical well has a negligible radius, a single bound state of energy $E_0 = - 1$ eV, and the density of spherical wells is $\rho = 10^{24}\ \mathrm{m}^{-1}$. Assuming that the spectrum of unbound states is unchanged with respect to the free particle case, determine the probability of finding the particle in a “ionized” (i.e., unbound) state.

Answer: Let us discuss at first the case of a single spherical well at the center of a cubic box of volume $V = L^3$. The bound state of the well is not influenced by the walls of the box if $L \gg \hbar /\sqrt{2 m |E_0|} \simeq 1.55\,\times \,10^{-10}$ m. At the given temperature the particle is non-relativistic. The density of free energy levels in the cubic box is $d n_E / d E = V \sqrt{E} (2 m)^{3/2}/(4 \pi ^2 \hbar ^3)$. Taking into account also the bound state, the partition function is

$$ Z = \mathrm{e}^{- \beta E_0} + V \frac{(2 m)^{3/2}}{4 \pi ^2 \hbar ^3} \int _0^\infty dE\ \sqrt{E} \mathrm{e}^{-\beta E} = \mathrm{e}^{- \beta E_0} + V \frac{(2 m k T)^{3/2}}{4 \pi ^2 \hbar ^3} \frac{\sqrt{\pi }}{2} = \mathrm{e}^{- \beta E_0} + \frac{V}{\lambda _T^3} $$

where $\lambda _T \equiv \sqrt{2 \pi \hbar ^2/m k T}$ is the de Broglie thermal wavelength of the particle, i.e., its typical wavelength at thermal equilibrium. The probability for the particle being in the bound state is $P_b = \mathrm{e}^{-\beta E_0}/ ( \mathrm{e}^{-\beta E_0} + V/\lambda _T^3)$ and it is apparent that, for any given T, $\lim _{V \rightarrow \infty } P_b = 0$, i.e., the particle stays mostly in a “ionized” state if the box is large enough. At very low temperatures that may seem strange, since it may be extremely unlikely to provide enough energy to unbind the particle simply by thermal fluctuations; however, once the particle is free, it escapes with an even smaller probability of getting back to the well, if the box is large enough.

In the present case, however, there is a finite density of spherical wells: that is equivalent to considering a single well in a box of volume $V = 1/\rho$. Therefore $P_b = \left( 1 + \mathrm{e}^{-\beta |E_0|}/(\rho \lambda _T^3)\right) ^{-1}$, while the probability for being in an unbound state is $P_{free} = ( 1 + \mathrm{e}^{\beta |E_0|} \rho \lambda _T^3)^{-1} \simeq \mathrm{e}^{-\beta |E_0|} (\rho \lambda _T)^{-3} = 2.75\,\times \,10^{-2}$.

3.34

A gas of monoatomic hydrogen is placed at thermal equilibrium at temperature $T = 2\,\times \,10^3\ {}^{\circ }\mathrm{K}$. Assuming that the density is low enough to neglect atom-atom interactions, estimate the rate of atoms which can be found in the first excited energy level (principal quantum number $n = 2$).

Answer: According to Gibbs distribution and to the degeneracy of the energy levels of the hydrogen atom, the ratio of the probabilities for a single atom to be in the level with $n = n_2$ or in that with $n = n_1$ is

$$ P_{n_2,n_1} = \frac{n_2^2}{n_1^2} \exp \left( -\frac{E_0}{k T} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \right) $$

where $E_0$ is the energy of the ground state, $E_0 = - {m e^4 / (32 \pi ^2 \epsilon _0^2\hbar ^2)} \simeq - 13.6\ \mathrm{eV}$. Since $k T \simeq 0.1724$ eV, we get $P_{2,1} \simeq 0.8\,\times \,10^{-24}$. Since in those conditions most atoms will be found in the ground state with $n = 1$, this number can be taken as a good estimate for the rate of atoms in the level with $n = 2$.

The above answer is correct in practice, but needs some further considerations. Indeed, if we try to compute the exact hydrogen atom partition function we find a divergent behaviour even when summing only over bound states: energy levels $E_n$ become denser and denser towards zero energy, where they have an equal Boltzmann weight and become infinite in number: one would conclude that the probability of finding an atom in the ground state is zero at every temperature. However, as n increases, also the atom radius $r_n$ increases: in this situation the atom, which is called a Rydberg atom, interacts strongly with the surrounding black body radiation. Therefore the infinite number of highly excited states should not be taken into account since these strongly interacting states have very short lifetimes. One should however take into account ionized states, in which the electron is free to move far away from the binding proton. Concerning the statistical weight of these states, we can make reference to Problem (3.33): following a similar argument we realize that ionized states become statistically relevant, with respect to the ground state, when the density of atoms is of the order of $\mathrm{e}^{-\beta |E_0|}/\lambda _T^3 = \mathrm{e}^{-\beta |E_0|} \left( m k T/(2 \pi \hbar ^2)\right) ^{3/2} \sim \mathrm{e}^{-\beta |E_0|} 10^{28}\ \mathrm{m}^{-3} \sim 10^{-6}\ \mathrm{m}^{-3}$.

3.35

Consider a homogeneous gas of non-interacting, non-relativistic bosons, which are constrained to move freely on a plane surface. Compute the relation linking the density $\rho$, the temperature T and the chemical potential $\mu$ of the system. Does Bose-Einstein condensation occur? Does the answer to the last question change if an isotropic harmonic potential acts in the directions parallel to the surface?

Answer: An easy computation shows that the density of states for free particles in two dimensions is independent of the energy and given by

$$ \frac{d n_E}{d E} = \frac{m A}{2 \pi \hbar ^2} $$

where m is the particle mass and A is the total area of the surface. According to Bose-Einstein distribution, the total particle density is then given by:

$$ \rho = \frac{\overline{N}}{A} = \frac{m}{2\pi \hbar ^2} \int _0^\infty dE\ \frac{1}{\mathrm{e}^{(E -\mu )/(k T)} - 1} = - \frac{k T m}{2 \pi \hbar ^2} \ln \left( 1 - \mathrm{e}^{\mu /(k T)} \right) . $$

The chemical potential turns out to be negative, as expected. It is interesting to consider the limit of low densities, in which the typical inter-particle distance is much greater than the typical thermal de Broglie wavelength ($\rho ^{-1/2} \gg h/\sqrt{m k T}$) and

$$ \mu \simeq k T \ln \left( \frac{2 \pi \hbar ^2 \rho }{m k T}\right) , $$

while in the opposite case of high densities one obtains

$$ \mu = -k T \exp \left( - \frac{2 \pi \hbar ^2 \rho }{m k T} \right) . $$

At variance with the three-dimensional case, $\rho$ diverges as $\mu \rightarrow 0$, meaning that the system can account for an arbitrarily large number of particles, with no need for a macroscopic number of particles in the ground state, therefore Bose-Einstein condensation does not occur in two dimensions, due to the different density of states.

If the system is placed in a two-dimensional isotropic harmonic potential of angular frequency $\omega$, the energy levels are $E_n = \hbar \omega (n + 1)$, with degeneracy $n+1$, so that the total number of energy levels found below a given energy E is $\sim E^2/(2 \hbar ^2 \omega ^2)$ and the density of states becomes $d n_E / d E \simeq E/(\hbar ^2 \omega ^2)$. The average number of particles (the system is not homogeneous and we cannot define a density) is

$$ \overline{N} = \frac{1}{\hbar ^2 \omega ^2} \int _0^\infty dE \frac{E}{\mathrm{e}^{(E -\mu )/(k T)} - 1} = \frac{k^2T^2}{\hbar ^2\omega ^2} \int _{-\mu /(kT)}^\infty dx\ \frac{x + \mu /(kT)}{\mathrm{e}^x - 1}. $$

In this case the integral in the last member stays finite even in the limit $\mu \rightarrow 0$, meaning that condensation in the ground state is necessary to allow for an arbitrary average number of particles $\overline{N}$. The condensation temperature is $T_c \sim \sqrt{\overline{N}} \hbar \omega /k$ and goes to zero, at fixed $\overline{N}$, as $\omega \rightarrow 0$ (i.e., going back to the free case).

3.36

A system of massless particles at thermal equilibrium is characterized by the known equation of state $U/V - 3 P = 0$, linking the pressure P to the density of internal energy U / V. For a homogeneous system of spinless, non-interacting and distinguishable particles of density $\rho$, placed at thermal equilibrium at temperature T, compute the lowest order violation to the above relation due to a non-zero particle mass m.

Answer: For one particle in a cubic box of volume $V = L^3$, energy levels are written as $E = \sqrt{p^2 c^2 + m^2 c^4} = \sqrt{m^2 c^4 + (\pi ^2 \hbar ^2 c^2 /L^2) (n_x^2 + n_y^2 + n_z^2)}$, with $n_x$, $n_y$ and $n_z$ positive integers. The number of energy levels below a given threshold $\bar{E} = \sqrt{\bar{p}^2 c^2 + m^2 c^4}$ is given by $\bar{p}^3 V / (3 \pi ^2 \hbar ^3)$, from which one obtains the density of states $d n_E/dE = V p E / (c^2 \hbar ^2 \hbar ^3)$. The derivative of one energy level with respect to the volume is $\partial E / \partial V = p^2 c^2 / (3 V E)$.

The internal energy and the pressure are given, for a single particle, by

$$ U = \frac{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E} E}{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E}} \,\, ; \,\,\,\,\,\,\,\,\,\, P = \frac{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E} (\partial E/\partial V)}{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E}} $$

so that

$$ U - 3 P V = \frac{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E} (E - p^2 c^2/E)}{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E}} = \frac{m^2 c^4 \int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E}/E }{\int _{m c^2}^\infty dE ({d n_E}/{d E}) \mathrm{e}^{-\beta E}}. $$

To keep the lowest order in m it is sufficient to evaluate the integrals in the last expression in the limit $m = 0$. Finally, multiplying for the total number of particles in the box, $N = \rho V$, we get

$$ \frac{U}{V} - 3 P = \frac{\rho \ m^2 c^4}{2 k T}. $$

This is the first term of an expansion in terms of the parameter $m c^2/k T$.

3.37

Consider a rarefied gas of particles of mass m in equilibrium at temperature T. The probability distribution of particle velocities is given by the Gaussian Maxwell-Boltzmann (MB) formula

$$ p({\varvec{v}})\ d^3 v =\left( \frac{m}{2\pi kT}\right) ^{3/2} \mathrm{e}^{-mv^2/(2 kT)} d^3 v . $$

Considering a pair of particles in the gas, labelled by 1 and 2, compute the distribution of the relative velocities ${\varvec{v}}_R\equiv {\varvec{v}}_1-{\varvec{v}}_2$, and that of the velocity of their center of mass ${\varvec{v}}_B\equiv ({\varvec{v}}_1+{\varvec{v}}_2)/2$.

Answer: The particles in the chosen pair behave as independent systems, hence the probability density in the six-dimensional $({\varvec{v}}_1,{\varvec{v}}_2)$ space is just the product of the two densities:

$$ p ({\varvec{v}}_1, {\varvec{v}}_2) d^3 v_1 d^3 v_2 = \left( \frac{m}{2\pi kT}\right) ^{3} \mathrm{e}^{-m(v_1^2+v_2^2)/(2 kT)} d^3 v_1 d^3 v_2\ . $$

If we change variables passing from $({\varvec{v}}_1,{\varvec{v}}_2)$ to $({\varvec{v}}_R,{\varvec{v}}_B)$, the new probability density is given by

$$ \tilde{p} ({\varvec{v}}_R, {\varvec{v}}_B) d^3 v_R d^3 v_B = p ({\varvec{v}}_1, {\varvec{v}}_2) d^3 v_1 d^3 v_2 $$

hence $\tilde{p} ({\varvec{v}}_R, {\varvec{v}}_B) = J({\varvec{v}}_R,{\varvec{v}}_B) p ({\varvec{v}}_1, {\varvec{v}}_2)$ where J is the Jacobian matrix, that is, the absolute value of the determinant of the six-dimensional matrix whose elements are $\partial v^i_{a}/\partial v^j_{S}$ where a is either 1 or 2 and S is either R or B and i, j label the Cartesian components. Furthermore $p ({\varvec{v}}_1, {\varvec{v}}_2)$ must be considered a function of $({\varvec{v}}_R,{\varvec{v}}_B)$. It is soon verified that $J=1$ and that $v_1^2+v_2^2=(1/2)v_R^2+2v_B^2$. Therefore we have

$$ \tilde{p} ({\varvec{v}}_R, {\varvec{v}}_B) d^3 v_R d^3 v_B = \left( \frac{\mu }{2\pi kT}\right) ^{3/2} \mathrm{e}^{- \mu v_R^2/(2 kT)}\ \left( \frac{M}{2\pi kT} \right) ^{3/2} \mathrm{e}^{-Mv_B^2/(2 kT)} $$

where $\mu =m/2$ is the reduced mass of the pair of particles and $M=2m$ is their total mass.

Then we see that our result corresponds to the product of the MB distribution of a single particle of mass $\mu$ and velocity ${\varvec{v}}_R$ and that of a single particle of mass M and velocity ${\varvec{v}}_B$. If we search for the ${\varvec{v}}_R$ distribution of the pair, we must integrate over ${\varvec{v}}_B$ and we get the first MB distribution, corresponding to the mass $\mu$. If instead we search for the ${\varvec{v}}_B$ distribution of the pair, we must integrate over ${\varvec{v}}_R$ and we get the second MB distribution, corresponding to the mass M. It can be easily verified that the result extends unchanged to the case in which the two particles have different masses $m_1$ and $m_2$, defining as usual $M = m_1 + m_2$, $\mu = m_1 m_2/M$, ${\varvec{v}}_B = (m_1 {\varvec{v}}_1 + m_2 {\varvec{v}}_2)/M$, ${\varvec{v}}_R = {\varvec{v}}_1 - {\varvec{v}}_2$.

3.38

We have a rarefied gas of electrons for which $mc^2=0.511\ \mathrm{MeV}$, in thermal equilibrium at $kT = 5\,\times \,10^3\ \mathrm{eV}$. Since $kT/mc^2 \ll 1$, the rarefied gas in non-relativistic. Therefore the particle momentum ${\varvec{p}}$ distribution is given by the Maxwell-Boltzmann formula

$$ \frac{d\mathcal{P}}{d{\varvec{p}}}=\left( \frac{1}{2\pi mkT} \right) ^{3/2} \mathrm{e}^{- p^2/(2mkT)}\ . $$

The electron gas is mixed with a rarefied positron gas in thermal equilibrium at the same temperature. Positrons are anti-particles of electrons, therefore a positron-electron collision can produce an annihilation into two photons and hence the two gas mixture is a photon source. We would like to compute the photon energy (frequency) distribution.

Answer: If a non-relativistic positron annihilates with a non-relativistic electron and their relative momentum is ${\varvec{p}}$ while the center of mass one is ${\varvec{P}}$, forgetting relativistic corrections, we find that the energy of a produced photon is $E=(mc^2 +p^2/(2m)) (1+P_z/Mc)$ where we have chosen the z-axis parallel to the momentum of the photon and the second factor is the transformation factor from the center of mass to the laboratory frame, that is, it accounts for the non-relativistic Doppler effect. As shown in the preceding problem, considering a generic electron-positron pair, one has the center of mass ${\varvec{P}}$ and relative momentum ${\varvec{p}}$ distribution $d^6 \mathcal{P}/ (d {\varvec{P}} d {\varvec{p}})=1/( 2\pi m kT)^3 \exp (- p^2/(m kT)) \exp (-P^2/(4m kT))$. Changing the $P_z$ variable into E and integrating over $P_x$ and $P_y$ we get (the only non-trivial element of the Jacobian matrix is $\partial P_z/\partial E$):

$$ \frac{d^4 \mathcal{P}}{d {\varvec{p}} d E} = \frac{2m^2c}{(2m^2c^2+p^2)( \pi m kT)^2 } \mathrm{e}^{- p^2/(m kT)} \exp \left( \frac{-4m^3c^2(E-mc^2-p^2/(2m))^2}{(2m^2c^2+p^2)^2kT} \right) $$

which should be integrated over ${\varvec{p}}$. The Gaussian factor in ${\varvec{p}}$ corresponds to a mean square value $\bar{p}^2=3mkT/2$ and the standard deviation $\varDelta ^2_{ p^2}=3(mkT)^2/2$. Since in the above distribution, the Gaussian factor put apart, $p^2$ appears only in the linear combination $mc^2+p^2/(2m)$ and since $mc^2=0.511\ \mathrm{MeV} \gg 3kT/4$ we can replace the linear combination with $mc^2+3kT/4$ getting the final photon energy distribution:

$$ \frac{d \mathcal{P}}{dE} = \frac{1}{\sqrt{\pi m c^2 kT}(1+3kT/(4m c^2))} \exp \left[ -\frac{(E-mc^2-3kT/4)^2}{m c^2 kT(1+3kT/(4m c^2))^2}\right] \ . $$

Notice that, however exponentially decaying, the distribution does not vanish for negative E values, this is due to the fact that for these values the non relativistic Doppler formula does not apply. Thus negative E values should not be considered. Notice also that, while the most probable E value is shifted by a factor $3kT/(4mc^2)\sim 10^{-2}$, the width of the distribution is proportional to $\sqrt{m c^2 kT}$, allowing for fluctuations in E of the order of $\sqrt{kT/m c^2} \sim$ 10 %, i.e., much larger than the shift. The physical origin of the shift is mostly in the fact that the annihilating pair has an average energy, in the center of mass, which is larger than $2 m c^2$, due to thermal motion, by an amount $\sim p^2/m \sim k T$. The broad distribution of energy is instead mostly due to Doppler effect, due to the transformation from the center of mass to the laboratory frame (see also Problem 1.31), and the broadening is proportional to the typical center of mass velocity, which is of the order of $c \sqrt{kT/mc^2}$.

3.39

Compute the magnetic susceptibility of a salt whose magnetic ions (e.g., $F_e^{3+}$) have molecular number density $\rho$, magnetic moment $2\mu _B{\varvec{J}}/\hbar$ where $J=\hbar /2$, $\mu _B=e\hbar \mu _0/(2m)$ is Bohr’s magneton and $\mu _0$ is the magnetic permeability of the vacuum.

Answer: The magnetic interaction energy of a single ion is $V=-{\varvec{\mu }}\cdot {\varvec{H}}$. If we choose the z axis parallel to the magnetic field we get two interaction energy levels for each ion, the corresponding energies being $\pm \mu _B H$. Therefore, if $\mu _BH/( kT)\ll 1$, the average total magnetic moment per unit volume of the salt is parallel to the magnetic field, the salt being paramagnetic, and its absolute value is, according to Eq. (3.15)

$$ M=\rho \mu _B{e^{\mu _BH/ (kT)}- e^{-\mu _BH/ (kT)}\over e^{\mu _BH/ (kT)}+ e^{-\mu _BH/ (kT)}}=\rho \mu _B\tanh (\mu _BH/ kT)\simeq \rho \mu _B^2H/ (kT)\ . $$

Therefore the magnetic susceptibility is $\chi (T,\rho )=\rho \mu _B^2/ (kT)$.

3.40

Compute the thermodynamic potentials of a perfect gas whose molecules are diatomic and rigid, that is, they are made of two spinless atoms with masses $m_1$ and $m_2$ whose relative distance is fixed and equal to d, when the temperature T satisfies the condition $kT\gg \hbar ^2(m_1+m_2)/(2d^2m_1m_2)\equiv \hbar ^2/(2m_r d^2)\equiv \hbar ^2/(2I)$.

Answer: From its very definition it turns out that the partition function of an ideal gas whose molecules have a given spectrum of excited states is given by

$$ Z_{id}= \left( {(2\pi m kT)^{3\over 2}V\over h^3 N}\sum _{n=0}^\infty \nu _ne^{-{E_n\over kT}}\right) ^N\equiv \left( {V\over N}\varPhi (T)\right) ^N $$

where $\nu _n$ is the degeneracy of the level with energy $E_n$ and $m=m_1+m_2$. Indeed the energy of a molecule with momentum ${\varvec{p}}$ in the n-th excited state is $p^2/(2m)+E_n$. In the case of the rigid diatomic molecule one has $\nu _n=2n+1$ and $E_n=-B+\hbar ^2 n(n+1)(m_1+m_2)/(2d^2m_1m_2)\equiv -B+ \hbar ^2 n(n+1)/(2I)$, where $B>0$ is the binding energy of the two atoms. It follows that the free energy is given by

$$ F(T,V)=-NkT\left[ \ln \varPhi _d(T)+\ln {V\over N}\right] \ . $$

We must compute $\varPhi _d$, which, with the above condition for the temperature, is given by

$$\begin{aligned} \varPhi _d(T)&=e^{B\over kT}{(2\pi m kT)^{3\over 2}\over h^3 }\sum _{n=0}^\infty (2n+1)e^{-{ \hbar ^2 n(n+1)\over 2IkT}}\simeq e^{B\over kT} {(2\pi m kT)^{3\over 2}\over h^3 }\int _0^\infty dx\ e^{-{ \hbar ^2 x\over 2IkT}}\\&=e^{B\over kT} {(2\pi )^{7\over 2}m^{3\over 2}2I( kT)^{5\over 2}\over h^5 } \end{aligned}$$

Therefore the internal energy of the rigid diatomic gas is

$$ U(T)= k T^2N{d\ln \varPhi _d(T)\over dT}={5\over 2}NkT+NB, $$

it only depends on the temperature, as it happens for any perfect gas, and the specific heat is $c_v={5\over 2}Nk\ .$ We also have for the free enthalpy

$$ G(T,P)=-NkT\left[ \ln \varPhi _d(T)+\ln {kT\over P}-1\right] . $$

A final remark is here in order. In the case of two identical atoms, like, e.g., the molecules $H_2$, $N_2$ and $O_2$, one has that only even/odd values of n are allowed, depending on the atomic statistics. As a consequence, the contribution $\varPhi _d(T)$ from the angular momentum sum must be divided by two.

3.41

Compute the entropy of the van der Waals gas.

Answer: Using Eqs. (3.114) and (3.127), that we write as $S=k(\ln Z+T\partial _T\ln Z)$, we have

$$\begin{aligned} S= & {} kN\left( \ln {(2\pi mkT)^{3\over 2}(V-b)\over h^3 N}+{CN\over 2VkT}+{3\over 2}-{CN\over 2VkT}\right) \nonumber \\= & {} kN\left( \ln {(2\pi mkT)^{3\over 2}(V-b)\over h^3 N}+{3\over 2}\right) \ . \nonumber \end{aligned}$$

3.42

Compute the equation of a reversible adiabatic transformation for the van der Waals gas.

Answer: Using the entropy computed in the former exercise and noticing that the entropy is constant along a reversible adiabatic transformation we find that its equation is $T^{3\over 2}(V-b)=constant$.

3.43

Two rigid vessels of volumes $V_1$ and $V_2$ contain respectively $N_1$ and $N_2$ molecules of two different ideal gases at the same temperature T. The vessels are connected by a tiny tube and a tap whose volumes can be disregarded. Opening the tap, both gases freely diffuse through the tube, until equilibrium is reached, without any temperature change, when the partial pressure of each gas in the two vessels equalize. There has been no heat exchange with the outside, because the gases are ideal and the molecules have kept their kinetic energy during the diffusion. The described transformation is irreversible. The total entropy of the system must have increased of a certain amount called mixture entropy. Compute it.

Answer: The initial entropy is given by the sum of the entropies of the two gases. According to Eq. (3.142) it is given by

$$ S_I=kN_1\left( {d(T\ln \varPhi _1(T))\over dT}+\ln {V_1\over N_1}\right) +kN_2\left( {d(T\ln \varPhi _2(T))\over dT}+\ln {V_2\over N_2}\right) .$$

The final state is a mixture of the two ideal gases at temperature T and volume $V_1+V_2$, then, according to (3.144), we have

$$ S_F=kN_1\left( {d(T\ln \varPhi _1(T))\over dT}+\ln {V_1+V_2\over N_1}\right) +kN_2\left( {d(T\ln \varPhi _2(T))\over dT}+\ln {V_1+V_2\over N_2}\right) \, $$

and hence the difference is

$$ S_F-S_I=kN_1\ln {V_1+V_2\over V_1}+kN_2\ln {V_1+V_2\over V_2}\ , $$

which, of course, is positive.

3.44

Gas and liquid of the same substance are in equilibrium at temperature T and pressure P. The equilibrium condition is given by the identity between the specific free enthalpies G (the molar free enthalpies) of the two phases, that is the identity between the chemical potentials. Indeed, the free enthalpy of the whole system, which is given by $(N-\varDelta )\mu _1(T,P)+\varDelta \ \mu _2(T,P)$, must be stationary, that is $\varDelta$ independent. This implies $\mu _1(T,P)=\mu _2(T,P)$. Translate this condition into a differential equation for the equilibrium curve in the $T-P$ plane (Clapeyron equation).

Answer: Equation (3.134), translated in terms of the specific thermodynamic functions, becomes $\mu _{g/l}(T,P)=h_{g/l}(T,P)-T\ s_{g/l}(T,P)$ where the indices g / l distinguish the two phases. The equilibrium equation is

$$ h_{g}(T,P)-h_{l}(T,P)=T(s_{g}(T,P)-s_{l}(T,P))\ . $$

In differential form, from Eqs. (3.135)–(3.138), we have

$$ d\mu _g(T,P)=-s_g(T,P)dT+v_g(T,P) dP=d\mu _l(T,P)=-s_l(T,P)dT+v_l(T,P) dP $$

therefore

$$ {dP\over dT}={s_{g}(T,P)-s_{l}(T,P)\over v_{g}(T,P)-v_{l}(T,P)}= {h_{g}(T,P)-h_{l}(T,P)\over T( v_{g}(T,P)-v_{l}(T,P))}. $$

Now the difference of specific enthalpies coincides, by definition, with the specific latent heat ($\lambda \equiv h_{g}(T,P)-h_{l}(T,P)$) thus we have the Clapeyron equation

$$ {dP\over dT}={\lambda (T,P)\over T( v_{g}(T,P)-v_{l}(T,P))}. $$

Whenever the specific volume of the gas phase is much larger than that of the liquid and it can be approximated by $v_g\simeq RT/P$ we have

$$ {dP\over dT}=P{\lambda (T,P)\over T^2R}. $$

3.45

Discuss the equilibrium condition for the chemical dissociation of a rigid diatomic ideal gas into the monoatomic component gases. Consider a case analogous to the chemical reaction $O_2\rightarrow 2O$, where the atoms correspond to the nuclear isotope with mass 16. Note that the result given in Problem 3.40 must be changed, since both the constituent atoms and the $O_2$ molecule have electronic spin one, this implies a factor $2S\,+\,1=3$ multiplying $\varPhi _{O_2}(T)$ and $\varPhi _{O}(T)$. Furthermore $I=md^2/2$, where m is the atomic mass, and since the constituent atoms are identical, $\varPhi _{O_2}(T)$ must be divided by two. The temperature T and the pressure P are assumed to be high and, respectively, low enough to justify the use of the ideal gas formulae.

Answer: Consider the Eqs. (3.146) and (3.147). Let the initial conditions correspond to N diatomic molecules, if $\varDelta$ such molecules dissociate into pair of constituent atoms we have a partial pressure $P_d(\varDelta )=P(N-\varDelta )/(N+\varDelta )$ for the diatomic component of the mixture, and $P_a(\varDelta )=2P \varDelta /(N+\varDelta )$ for the monoatomic ones, therefore the Gibbs potential of the gas mixture resulting from the dissociation is

$$ G_{mix}(T,P, \varDelta )=(N-\varDelta )\mu _d(T,P_d)+2\varDelta \mu _a(T,P_a). $$

The equilibrium condition is $\partial G_{mix}(T,P, \varDelta )/\partial \varDelta =0$. From Eqs. (3.135)–(3.138) and (3.140) we have for a mixture of gases $\partial \mu _i/\partial P_i=V/N_i$, therefore we have the equilibrium condition

$$ 2 \mu _a(T,P_a)-\mu _d(T,P_d)+V\left( {\partial P_d\over \partial \varDelta }+{\partial P_a\over \partial \varDelta }\right) =2\mu _a(T,P_a)-\mu _d(T,P_d)=0.$$

If we introduce the stoichiometric coefficients $\nu _i=d N_i/d\varDelta$ we have $\sum _i \nu _i\mu _i(T,P_i)=0$ and hence (Eq. (3.148)) $\sum _i\nu _i(\ln P_i-\ln (kT\varPhi _i(T))+1)$ from which we have

$$ \prod _iP_i^{\nu _i}=\left( {kT\over e}\right) ^{\sum _i\nu _i}\prod _i\varPhi _i^{\nu _i}(T). $$

Taking into account the results of Problem 3.40 and Eq. (3.143) we have

$$ \varPhi _a(T)=3 \left( {2\pi m kT\over h^2}\right) ^{3\over 2}, \quad \varPhi _d(T)=e^{B\over kT}{3\pi d^2\over 2} \left( {2\pi m kT\over h^2}\right) ^{5\over 2}, $$

and hence we finally get

$$ {P_a^2\over P_d}=e^{-1}kT{\varPhi _a^2(T)\over \varPhi _d(T)}= (kT)^{3\over 2}\left( {2\pi m \over h^2}\right) ^{1\over 2}{6\over \pi d^2}e^{-({B\over kT}+1)}\equiv K_p^{-1}(T), $$

where $K_p$ is called the (pressure) equilibrium constant of the dissociation reaction.

3.46

We have a sample of the paramagnetic salt considered in Problem 3.39 in the magnetic field H at temperature $T_i$. We switch off the magnetic field with a reversible adiabatic transformation. Compute the final temperature $T_f$ of the sample assuming its entropy density for $H=0$ equal to $s(T, \rho , 0)=3k\rho \ln (T/T_0)$.

Answer: Using Eq. (3.159) and the magnetic susceptibility computed in Problem 3.39, we compute the initial entropy per unit volume of the salt in the magnetic field H

$$ s(T_i, \rho , {\varvec{H}})= s(T_i, \rho , 0)+{\partial \chi (T_i, \rho )\over \partial T}{| {\varvec{H}}|^2\over 2}= 3k\rho \ln (T_i/T_0)-{\rho \mu _B^2\over kT_i^2}{| {\varvec{H}}|^2\over 2}. $$

At the end of the reversible adiabatic field switch off we have $s(T_i, \rho , {\varvec{H}})= s(T_f, \rho , 0)$, therefore we have

$$ T_f=T_i \exp \left( -{\mu _B^2| {\varvec{H}}|^2\over 6( kT_i)^2}\right) \simeq T_i -{\mu _B^2| {\varvec{H}}|^2\over 6 k^2T_i}. $$

It is apparent that we find a reduction of the temperature of the sample.