Statistical mechanics solution sheet 5

1. The curves of constant energy are

p2

2m+

mω2q2

2= E

which are ellipses with semi-axes of length (a, b) = (√

2mE,√

2E/mω2). The areaof an ellipse is πab so the phase volume between E and E + δ is

Ω =2π

hω(E + δ − E) =

2πδ

hω

where we have normalised by Planck’s constant as required. Thus the entropy is

S = k ln Ω = k ln2πδ

hω

Temperature is infinite as the entropy is in this unusual case independent of energy.

1/T =∂S

∂E= 0

For macroscopic systems Ω ≈ 1034, so the entropy is about 80k, a typical entropyof 80 microscopic particles.

2. (a) The potential energy is unbounded below. This means that if two stars getclose enough together there will be sufficient energy released for other starsto escape the system. This means that phase space has infinite volume andthere is no maximum entropy. The final state is theoretically two stars ina very close orbit with all the others drifting away to infinity; in practiceobservations are consistent with evaporation of some of the stars and in somecases the formation of a central black hole.

(b) The virial theorem relates the average kinetic energies of stars (observable)to a quantity involving forces and positions (the latter also observable). Theaverage acceleration (force/stellar mass) then permits the total mass of thecluster to be estimated. The masses of individual stars can be estimated fromtheir brightness and surface temperature, but we don’t actually need these asthey appear both in the kinetic energy and the force and so can be dividedout of the virial theorem. Note that evaporation (discussed above) violatesthe conditions of the virial theorem (which requires that the displacements bebounded), but is usually small enough to be neglected in practice.

3. A typical trajectory of an ergodic system will spend most of its time in regions oflarge volume, ie high Boltzmann entropy. So if it starts in a low entropy state, theentropy will probably increase. However if it starts in a high entropy state, theentropy will rarely decrease significantly. Of course we can reverse this argument,and talk about the system finishing in a low entropy state; the second law is reallythe empirical fact that initial (rather than final) conditions are the least probable.

4. In four dimensions we have

V (r) = V4r4 =

2π2r4

4Γ(2)=

π2r4

2

S(r) = V ′(r) = 2π2r3

Here we used Γ(n + 1) = n! for n an integer. In five dimensions we have

V (r) = V5r5 =

2π5/2r5

5Γ5/2)=

8π2r5

15

S(r) = V ′(r) =8π2r4

3

Here we used Γ(1/2) =√

π and Γ(x + 1) = xΓ(x).

5. Let’s do the second part first. The volume in the shell is given by

V (1) − V (1 − 10−19) = V1020(1 − (1 − 10−19)1020

)

which should be compared with V1020 . Thus the proportion is

1 − exp[1020 ln(1 − 10−19)] = 1 − exp[−10] = 99.996%

by expanding the logarithm in its Taylor series. High dimensional spheres aremostly concentrated at the surface.

Now, the first part. We have

VN =2πN/2

NΓ(N/2)

for some large N which we need to estimate using Stirling’s formula. The expo-nential terms are

VN ≈(

2πe

N

)N/2

≈ 10−1021

A more detailed calculation (involving more of Stirling’s formula, logarithms andhigh precision arithmetic) yields

VN ≈ 3.616 × 10−938376282486931656150

High dimensional spheres have much smaller volumes than cubes.

6. The phase space volume Ω increases by a factor (2j + 1)N , thus the entropy S =k ln Ω increases by an additive constant Nk ln(2j + 1). This will not contribute toenergy or volume derivatives (ie temperature or pressure), but it will change thechemical potential

µ = −T

(

∂S

∂N

)

V E

by an amount −kT ln(2j + 1). However, in practice neither the entropy nor thechemical potential are directly measurable classically.

7. The maximum work extractable is the entropy of mixing times temperature, ie

kT [N1 ln(V/V1) + N2 ln(V/V2)]

Assuming we have perfect semi-permeable membranes we can do this by allowingeach gas to expand to the full volume isothermally (ie absorbing heat from the

environment), thus extracting the work, and then putting them in a containerseparated by the membranes. The container can (theoretically) be compressedwithout doing any work if the total volume available to each component remainsconstant. If the two gases are the same no such semi-permeable membrane exists,and no work can be extracted. This example illustrates that the second law isphysically associated with differences in chemical potential, not just temperature.

8. We have from lecturesdS

dt= 〈k∇ · v〉

and from the solution to 4.1

∇ · v = (3N − 1)p · ∇qV

p · p

thus if the latter has a negative time average, S will decrease towards −∞. Thisappears nonsensical as entropy is supposed to increase. However note that theargument we used for Hamiltonian systems - that our knowledge of the system istheoretically constant, is valid. Also, the heat removed by the thermostat also takesentropy from the system, and the Gibbs formula says nothing about the entropyof the environment. So, this result (sort of) makes sense.

9. I am mostly interested in the amount of relevant content (as opposed to structureand style) in the answer to this type of question. Some relevant points are:

• Microscopic dynamics itself cannot justify irreversible processes in thermodyam-ics, since the equations are reversible. However irreversibility can (and does)come about through initial conditions (rather than final conditions).

• Poincare recurrence implies that most low entropy states will return to thatstate after sufficient time. In practice the times are astronomically large foreven moderately large systems. Small fluctuations do occur on reasonabletime scales, however.

• Ergodic theory provides some basis for statistical mechanics ensembles forequilibrium systems, in particular mixing (if it can be proved) provides forapproach to equilibrium. However, time scales must be reasonable, also somenon-mixing behaviour in realistic systems might occur in regions of phasespace too small to observe, that are otherwise well described by statisticalmechanics.

• The Boltzmann entropy gives an explanation for the usual increase of entropy,albeit with rare significant decreases. The Gibbs entropy of a mixing systemdoes also, if observed at finite resolution. Both give correct expressions for egthe properties of the ideal gas as discussed in thermodyamics, but both aresomewhat ambiguously defined.

• In summary, microscopic dynamics is consistent with thermodynamics, sub-ject to a number of difficult mathematical and philosophical issues.