Physics 35200 R. Wald Statistical Mechanics Spring, 2002

Problem Set II Solutions

1) (a) Consider a classical gas composed of N non-interacting point particles, each of mass m, which is confined to a volume V. Calculate the total volume V[Γ_{E}] of the “energy shell” Γ_{E}. You may use the fact (shown in Appendix E of Mazenko) that the “area” of a unit sphere in a d-dimensional space is 2π^{d/2}/Γ(d/2), where Γ denotes the gamma-function (so Γ(n)=(n-1)! when n is an integer; Stirling’s formula provides an excellent approximation for Γ(x) for large x).

Solution: The volume in phase space is defined as ∫ dq^{N}dp^{n}δ (H − E). Since the Hamiltonian is independent of postion (the particles are free), the position integral can be performed immediately and yields V^{N}. The delta-

function enforces energy conservation , i.e., E = \frac{1}{2m} (∑_{i=1}^{N} ∑_{j=x}^{z} (v_{j}^{(i)})^{2}).

This defines a 3N − 1 sphere whose volume is 2\frac{π^{α}}{Γ(α)} (2mE)^{α− \frac{1}{2}} , with α = \frac{3N}{2} . However, there is also a Jacobian coming from evaluating the delta-function,

so we must include a factor of |\frac{∂H}{∂q_{i}} |^{−1}”=”p/m. Since the gradient is normal to a level-set, which in this case is a sphere, the gradient is simply the vector from the origin to the point where it is being evaluted. Its norm is equal to the radius of the sphere, and the net effect of the Jacobian is to reduce the volume by one power of \frac{\sqrt{2mE}}{m}. Combining these, we find V = V^{N} \frac{(2πm)^{α}}{Γ(α)} E^{α−1}. ■

(b) Two boxes each have volume V and each contain a classical gas of N particles with energy E, as in part (a). The boxes are now placed in “thermal contact” with each other, so that they can exchange energy with each other, but not change their volume or number of particles. On account of this energy exchange, the energy, E_{1}, within the first box now may vary between 0 and 2E. Obtain the probability distribution, p(E_{1})dE_{1}, that the energy contained within the first box will be within dE_{1} of E_{1} if one examines the system at a “random time”.

Solution: In part (a) we calculated the volume of phase on the energy shell of E. Now we consider the phase space for the product system S_{1} × S_{2}. If the first box has energy E_{1}, then second one has energy E_{2} = 2E − E_{1}. Thus, total volume of phase space will be

V(E_{1}) = V(Γ_{E_{1}}) × V(Γ_{2E−E_{1}}) = V^{2N} \frac{(2mπ)^{2α}}{Γ(α)^{2}} E_{1}^{α−1} (2E − E_{1})^{α−1} .

Since E_{1} specfies the energies of all parts of the system, the total phase space available to the combined system is simply ∫_{0}^{E} dE_{1}V(E_{1}). This integral can be done in two ways: 1) The hard way, which is to evaluate it in terms of the confluent hypergeometric function and then express that in terms of Γ functions, and 2) the easy way, which is to think about it physically (thanks to Kosta Ladavac, from whom I’m copying this solution). The combined system consists of 2N particles, each of which can be anywhere in some volume V. The only restriction we have on the system is that the total energy is 2E. Since the particles are free, this is just like having 2N particles in a box with energy 2E, so we get the same answer as in part (a) only with N → 2N, E → 2E. Thus V_{Total} = V^{2N} \frac{(2πm)^{3N}}{Γ(3N)} (2E)^{3N−1}.The probability density is simply the ratio of the phase spaces so it is

P (E_{1}) = \frac{Γ(3N)}{Γ (\frac{3N}{2} )^{2}} \frac{E_{1}^{\frac{3N}{2} −1} (2E − E_{1})^{\frac{3N}{2} −1}}{(2E)^{3N−1}} . ■

2) (Counts as 2 problems) (a) Consider a non-relativistic particle of mass m in one dimension in a box of size L, with infinite potential walls (i.e. V = 0 for 0 < x < L; V = ∞ otherwise). Find its energy eigenstates and eigenvalues.

Solution: This is a standard problem from quantum mechanics. We must find the eigenfunctions of the Hamiltonian operator Ĥ = −\frac{ℏ^{2}}{2m^{2}} \frac{d^{2}}{dx^{2}} which vanish at the boundary. With our choice of coordinates, the eigenfunctions are e_{n} = \sqrt{\frac{2}{L}} sin (\frac{nπx}{L}) with eigenvalues \frac{n^{2}π^{2}ℏ^{2}}{2mL^{2}} , n ∈ ℤ^{+}. ■

(b) Now generalize the results of part (a) to the case of N, non-interacting, distinguishable particles in 3-dimensions in a cubic box of size L with infinite potential walls.

Solution: Since the particles are distinguishable, the Hilbert space of states is simply the 3N-fold tensor product of the 1d particle in box. Thus, the eigenstates are e_{n_{x}^{1},n_{y}^{1},n_{z}^{1},...,n_{x}^{N} ,n_{y}^{N} ,n_{z}^{N}} = Π_{i=1}^{N}Π_{j∈\{x,y,z\}} sin (\frac{n^{(i)}i_{j}πx}{L}) with energies

E_{n_{x}^{1},n_{y}^{1},n_{z}^{1},...,n_{x}^{N} ,n_{y}^{N} ,n_{z}^{N}} = ∑_{i=1}^{N} ∑_{j∈\{x,y,z\}} ^{\frac{(n_{j}^{(i)})^{2}π^{2}ℏ^{2}}{2mL^{2}}} and each n_{j}^{(i)} ∈ ℤ^{+}. ■

(c) For the situation of part (b), derive a formula for the number of quantum states with total energy between E and E + ΔE, where ΔE/E << 1, but ΔE is large enough to include many states. (Hint: Find the average density of states in 3N-dimensional “k-space” from part (b) and multiply by the appropriate volume.)

Solution: The energy of the system only depends on the “length-squared” of the “wave-vector” n_{j}^{(i)}. Thus, the number of states with energy less than or equal to E will be proportional the volume of a 3N-1-ball of radius

r = (\frac{2mEL^{2}}{π^{2}ℏ^{2}})^{\frac{1}{2}} . However, since we are the states are labeled by positive

integers, we must insert a factor of \frac{1}{2}^{2α} to take into account that we only want the volume in the first “quadrant.” Hence, the number of states is S = 2^{1−2α} \frac{π^{α}}{2αΓ(α)} (\frac{2mL^{2}E}{π^{2}ℏ^{2}})^{α}, α = \frac{3N}{2} is the number of states with energy ≤ E.

Note that this simplifes to \frac{π^{α}}{αΓ(α)} (\frac{V}{h^{3}} )^{N} (2mE)^{α}, so we do get a factor of V^{N} as expected classically. Differentiating the expression, we find the density of states ρ = \frac{dS}{dE} at energy E is equal to = \frac{π^{α}}{Γ(α)} (\frac{V}{h^{3}} )^{N} \frac{(2mE)^{α}}{E}. We conclude that the number of states in the desired range is

ρΔE = \frac{π^{\frac{3N}{2}}}{Γ(\frac{3N}{2} )} (\frac{V}{h^{3}} )^{N} (2mE)^{\frac{3N}{2}} \frac{ΔE}{E} . ■

(d) For a 1m^{3} box of (monatomic) gas at ordinary pressure and room temperature conditions we have N ∼ 10^{25} and the total energy is E = \frac{3}{2}NkT ∼ 10^{12} ergs. Assume that the mass of each particle is m ∼ 10^{−22}gm. Use the result of (c) to estimate the number of quantum states with total energy within 1 erg of E = 10^{12} ergs, assuming that the particles in the gas are distinguishable.

Solution: To our desired precision α = 1.5 × 10^{25}, \frac{ΔE}{E} = 10^{−12}, and (\frac{V}{h^{3}} )^{1} (2mE)^{\frac{3}{2}} = 9.7 × 10^{69}. The hard part is computing the volume of the unit sphere, for which we need Sterling’s approximation n! ≈ e^{−n}n^{n}\sqrt{2πn}. For N = 10^{25}, N ln N −N ≈ 5.65×10^{26}. Thus we get (converting all numbers raised to ridiculous powers to e^something)

\frac{e^{ln(π)1.5×10^{25}}}{e^{5.65×10^{26}}} (e^{160})^{1×10^{25}} (10^{−12}) ≈ e^{10^{27}}. ■

(e) For the system of part (d), estimate the probability that a given “single particle energy level” will be occupied by more than one particle, i.e., compare the number of single particle levels in the relevant energy range to the number of particles. (If this probability is small, a boson or fermion gas will behave in the same manner as a gas of distinguishable particles.)

Solution: We want to find the number of states available to a single particle in the energy range 0 ≤ E ≤ 10^{12}. Since we’re looking at a single particle, what we want is the volume of the 3-ball of radius r, where r is given

by the length-squared of the “wave-vector” as before, so r = (\frac{2mL^{2}E}{π^{2}ℏ^{2}})^{\frac{1}{2}} and the number of states is

\frac{4π}{3} (\frac{2mL^{2}E}{π^{2}ℏ^{2}} )^{\frac{3}{2}} \frac{1}{8} = 4 × 10^{70}.

Thus the average occupance of a single particle state is about 2.5 × 10^{−46}, which is very small indeed; it is irrelevant if our gas is made up of bosons or fermions. ■