Physics 35200 R. Wald Statistical Mechanics Spring, 2002

Problem Set VII Solutions v1.0

Dear Class: Due to the length of the assignment and shortness of the remainder of the quarter, these solutions will necessarily be rather brief. I hope that nonetheless they will be useful to you. Please feel free to email me or call me if you have any questions.

1) a) Calculate the chemical potential, μ, of a classical, monatomic ideal gas of particles of mass, m, as a function of m, its temperature, T, number of particles, N, and volume V. How is the formula for μ modified if the single particle energy is “offset” from zero by ϵ_{0}, i.e. if

H = ∑_{i}^{N} \{\frac{|p⃗_{i}|^{2}}{2m} + ϵ_{0}\}

Solution: The first law of thermodynamics tells us that

dE = T dS − pdV + μdN ⇒ μ = − (\frac{∂S}{∂N} )_{V,E}

Since the Sackur-Tetrode equation for the entropy of a classical ideal gas is in terms of E, V, and N, we can just differentiate it to yield

μ_{ideal gas} = −kT [ln \frac{V}{N} + \frac{3}{2} ln(2πmkT)] .

If we shift the Hamiltonian by ϵ_{0}, then the phase space volume (and hence the entropy) in the new system at energy E will be the same as the entropy in the classical ideal gas at energy E − Nϵ_{0} since all we’ve done is relabel the points in phase space. Hence

μ(N, T, V) = μ(N, E, V) = T \frac{∂S(N, E, V)}{∂N} = T \frac{∂S_{ideal gas}(N, E − Nϵ_{0}, V)}{∂N} =

μ_{ideal gas}(N, E−Nϵ_{0}, V)+T \frac{∂S_{ideal gas}}{∂(E − Nϵ_{0})} \frac{∂(E − Nϵ_{0})}{∂N} = μ_{ideal gas}(N, T, V)+ϵ_{0}. ■

b) Suppose a dynamical system contains 3 species of particles, labeled A,B,C. Suppose that the chemical reaction A + B ⇌ C can occur. Show that a necessary condition for thermal equilibrium of the system is μ_{A} + μ_{B} = μ_{C}.

Solution: In thermal equilibrium, all three species are at the same temperature and the entropy is extremized. Hence

0 = dS = \frac{∂S}{∂N_{A}} dN_{A} + \frac{∂S}{∂N_{B}} dN_{B} + \frac{∂S}{∂N_{C}} dN_{C}.

Imposing the equality of temperature and the number conservation relation dN_{A} = dN_{B} = −dN_{C} yields

0 = T dN_{A}(μ_{A} + μ_{B} − μ_{C}) ⇒ μ_{A} + μ_{B} = μ_{C}. ■

c) Suppose each of the particle species A, B, C of part (b) can be treated as a monatomic, ideal gas. (For example, A could represent free electrons, B could represent free protons, and C could represent neutral hydrogen atoms.) Suppose further, that the total energy is E = E_{A} + E_{B} + E_{C} − ϵN_{C}. Here, E_{A}, E_{B}, and E_{C} are the usual kinetic energies of the three gases, whereas ϵ is the binding energy of the C-particle (so the energy of the C-particle is “offset” from zero). Assume, for simplicity, that N_{A} = N_{B}. The system is in thermal equilibrium at temperature T. Calculate the ratio N_{A}^{2}/N_{C} as a function of T, ϵ, and the masses of the particles.

Solution: Unfortunately, this problem is not well-posed in classical mechanics. Since the particles are interacting, we need to write down the interaction Hamiltonian, and the answer will depend on the details of the Hamiltonian. This shows up as a dimensionful answer to this dimensionless problem. These dimensions come from the having dropped the cell size in the expression for entropy (incidentitally, this shows that cell size also appears in the chemical potential, since if it is included in the entropy there is an addtional factor of kT N ln v which survives the differentiation with respect to N. However, since this offset is the same for all species it can be ignored in classical mechanics). If we know from either experiment or our quantum mechanics classes that the cell size is h, then we could simply set μ_{A} + μ_{B} = μ_{C} and solve for the ratio, yielding

\frac{N_{A}^{2}}{N_{C}} = V (\frac{m_{A}m_{B}kT}{m_{C}h^{2}} )^{\frac{3}{2}} e^{− \frac{ϵ}{kT}} . ■

2) Suppose we have a dynamical system in contact with a “heat bath” but now, in addition to being able to exchange energy with the heat bath, suppose the system can “exchange volume” with the bath, i.e. the wall separating the systems is mobile. (Physically, this corresponds to “holding our system at fixed pressure”.) Show that the probability that our system is in a given macroscopic state (with respect to a coarse-grained observable O) is proportional to exp(−G/kT), with G ≡ E − T_{0}S_{O} + P_{0}V, where T_{0} and P_{0} are the temperature and pressure of the heat bath, E and V are the energy and volume of our system, and S_{O} is its observable entropy with respect to O for the given value of E and V.

Solution: The entropy of the total system is S_{T} = S_{B} + S_{O}. Then

S_{B}(E_{T} −E_{0}, V_{T} −V_{0}, N) ≈ S(E_{T} , V_{T} , N)−\frac{∂S_{B}}{∂E_{B}} E_{0}−\frac{∂S_{B}}{∂V_{T}} V_{0} = S_{B}(E_{T} , V_{T} , N)−\frac{E}{T_{0}} −\frac{P_{0}V}{T_{0}} ,

where we have assumed E_{T} >> E_{0}, V_{T} >> V_{0}. Since the probability is proportional to the phase space, i.e., the exponential of entropy, we get

p ∝ e^{S_{T}} = e^{S_{B}(E_{T} ,V_{T} ,N)− \frac{E}{T_{0}} − \frac{P_{0}V}{T_{0}} +S_{O}} ∝ e^{−βG}

as S_{B}(E_{T} , V_{T} , N) is constant. ■

3) A cubic box of side L containing a classical gas of N particles of mass m is placed in a uniform gravitational field of strength (i.e., acceleration) g. (The box is oriented so that the gravitational field is normal to a pair of faces.) The gas is in thermal equilibrium at temperature T.

a) Calculate the number density distribution, n(x⃗), of the gas.

Solution: The number density will clearly only be a function of z, so

n(x, y, z) = \frac{N}{L^{2}} f(z).

If we think of each z=constant plane as an allowed state, the relative occupation (in the microcanonical ensemble) will equal to e^{−βE} = e^{−βmgz}. Now all we need to is normalize, so we get

f(z) = e^{−βmgz} [∫_{0}^{L} e^{−βmgz}dz]^{−1} = \frac{βmge^{−βmgz}}{1 − e^{−βmgL}} .

Thus

n(x, y, z) = \frac{N}{L^{2}} \frac{βmge^{−βmgz}}{1 − e^{−βmgL}} . ■

b) Calculate the heat capacity (at constant volume) of the gas.

Solution: Again, we approach this problem via the microcanonical ensemble, using each z=constant plane as a state. At each level, we have an ideal gas whose zero energy has been shifted by mgz. Thus, the energy contained in that plane is

n(z)E_{single particle}(z) = \frac{N}{L^{2}} \frac{βmge^{−βmgz}}{1 − e^{−βmgL}} [\frac{3}{2β} + mgz] L^{2}dz.

Integrating yields

E = \frac{5N}{2β} − \frac{mgLN}{e^{βmgL} − 1}.

Hence,

C_{V} = \frac{∂E}{∂T} = kN [\frac{5}{2} − (\frac{βmgL}{2 sinh (\frac{βmgL}{2})})] . ■

Note: Most or all, of the class did this problem using the partition function. That is perfectly acceptable; it is simply not the quickest way to the answer.

4) One wishes to measure the mass, m, of a classical body by weighing it. The body is hung on a vertical spring of spring constant K in a uniform gravitational field, g, and the mass of the body is then determined from the formula m = −Kz/g, where z denotes the displacement of the spring from its equilibrium position. Suppose K and g are known exactly and that z can be measured with perfect accuracy, so that the only source of error is the buffeting of the mass by air molecules at temperature T. Suppose that only one reading of z is taken. What uncertainty, Δm, should be assigned to the measurement of m?

Solution: The key to this problem is that any macroscopic parameter has well-defined expectation value and variance. Hence, the we want to compute ⟨z^{2}⟩ − ⟨z⟩^{2}, which is the standard deviation of actual value of z, i.e., z changes in time due to buffeting and is measured perfectly at a single instant, and we need to know the distribution of those measurements. Since the only degree of freedom we’re interested in is z, we may consider the momentum of the particle and any other degrees of freedom as part of the “thermal bath.” Choosing choosing the zero of gravitational potential and the coordinate z so that z = 0 is zero energy, we find that the Hamiltonian for our “system” is

H(z) = mgz + \frac{1}{2}Kz^{2}.

In the microcanonical ensemble, the probability is then

P (z) = e^{−β( \frac{1}{2} kz^{2}+mgz)} [∫_{∞}^{∞} e^{−β( \frac{1}{2} kz^{2}+mgz)}dz]^{−1} = \sqrt{\frac{βK}{2π}} e^{− \frac{βg^{2}m^{2}}{2K}} e^{−β( \frac{1}{2} Kz^{2}+mgz)}

This leads to the expectation values

⟨z⟩ = −\frac{mg}{K} ⟨z^{2}⟩ = \frac{kT}{K} + (\frac{mg}{K} )^{2} Δz = \sqrt{⟨z^{2}⟩ − ⟨z⟩^{2}} = \sqrt{\frac{kT}{K}}

so that the uncertainty is

ΔM = \frac{K}{g} Δz = \frac{\sqrt{kKT}}{g} . ■

5) a) Consider a classical spinning particle with magnetic moment μ⃗ = μ_{0}S⃗. A magnetic field B⃗ is applied, resulting in an interaction energy E = −μ⃗ · B⃗. The particle is now placed in contact with a heat bath at temperature T. Find the probability distribution p(θ)dθ for finding the angle between S⃗ and B⃗ to be within dθ of θ. If N such particles are present, what is the expected magnetization?

Solution: The relative probability of being within a certain solid angle dΩ is e^{βμ_{0}SB cos θ}dΩ. Normalizing and integrate out the trivial ϕ dependence yields

p(θ)dθ = \frac{βμ_{0}SB}{2} \frac{e^{βμSB cos θ} sin θdθ}{sinh(βμ_{0}SB)}.

The magnetization will N \left⟨{ \atop μ}\right⟩. Clearly μ_{x} = μ_{y} = 0. Since μ_{z} = μ_{0}S cos θ, we have

M = μ_{0}SN [coth (βμ_{0}SB) − \frac{1}{βμ_{0}SB} ] ẑ ■

b) Now consider a quantum mechanical particle of spin s, with μ⃗_{0} = μ_{0}S⃗. Then S_{z} can take any of the (2s + 1) values, −s, −s + 1, …, s − 1, s. Again a magnetic field B⃗ (in the z-direction) is applied and the particle is placed in contact with a heat bath at temperature T. Calculate the probability p(S_{z}) of each of the allowed values of S_{z}. If N such (distinguishable) particles are present, what is the expected magnetization? Compare your answers to these questions to the classical case in the limit of large s.

Solution: The density matrix is

ρ̂ = \frac{e^{−βĤ}}{Tre^{−βĤ}} = \frac{∑_{S_{z}=−s}^{s} e^{−βμ_{0}BℏS_{z}} |S_{z}⟩ ⟨S_{z}|}{∑_{S_{z}=−s}^{S_{z}=s} e^{−βμ_{0}BℏS_{z}}}.

So the probability of being in a particular eigenstate is

p(S_{z}) = \frac{sinh (\frac{βμ_{0}Bℏ}{2}) e^{βμ_{0}BℏS_{z}}}{sinh (βμ_{0}Bℏ (s + \frac{1}{2}))}.

The magnetization is then

M_{z} = NTrμ̂_{z}ρ̂ = NTr [(μ_{0}Ŝ_{z})ρ̂] = \frac{∑_{S_{z}=−s}^{s} e^{−βμ_{0}BℏS_{z}}μ_{0}ℏS_{z}}{∑_{S_{z}=−s}^{s} e^{−βμ_{0}BℏS_{z}}}

= Nμ_{0}ℏ [(s + \frac{1}{2}) coth (βμ_{0}Bℏ (s + \frac{1}{2})) − \frac{1}{2} coth (\frac{βμ_{0}Bℏ}{2})] .

In the s → ∞ limit the second term in the brackets becomes negligible and (s + \frac{1}{2})ℏ → S so the quantum mechanical and classical calculations agree (since the 2nd term in the classical result also becomes negligible as S gets large). To compare the probabilities, we use the fact that S_{z}ℏ → S cos θ for large s, so ΔS_{z} ≈ \frac{S}{ℏ} sin θdθ. Thus,

p_{QM}(S_{z})ΔS_{z} → p_{QM}(θ)dθ = \frac{S}{ℏ} sinh (\frac{βμ_{0}Bℏ}{2}) \frac{e^{βμSB cos θ} sin θdθ}{sinh(βμ_{0}SB)},

where we have again made the substitution (s + \frac{1}{2})ℏ → S. This has almost the same function form as the classical probability, but differs have an additional sinh β instead of being linear in β. In order to make the two identical, we must further use the smallness of ℏ, which allows us to expand the sinh(βμ_{0}Bℏ/2) ≈ βμ_{0}ℏB/2, and so the two are identical. ■