## PHY452H1S Basic Statistical Mechanics. Problem Set 6: Max entropy, fugacity, and Fermi gas

Posted by peeterjoot on April 3, 2013

# Disclaimer

This is an ungraded set of answers to the problems posed.

## Question: Maximum entropy principle

Consider the “Gibbs entropy”

where is the equilibrium probability of occurrence of a microstate in the ensemble.

For a microcanonical ensemble with configurations (each having the same energy), assigning an equal probability to each microstate leads to . Show that this result follows from maximizing the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles ), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

where is the energy of microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied. What is the resulting ?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

and a given fixed average particle number

Here represent the energy and number of particles in microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constrains be satisfied. What is the resulting ?

## Answer

Writing

our unconstrained minimization requires

Solving for we have

The probabilities for each state are constant. To fix that constant we employ our constraint

or

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

Using this we our Gibbs entropy can be summed easily

or

For the “action” like quantity that we want to minimize, let’s write

for which we seek , such that

or

Our probability constraint is

or

Taking logs we have

We could continue to solve for explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

or

To determine we must look implicitly to the energy constraint, which is

or

The constraint () is given implicitly by this energy constraint.

Again write

The unconstrained minimization requires

or

The unit probability constraint requires

or

Our probability is then

The average energy and average number of particles are given by

The values and are fixed implicitly by requiring simultaneous solutions of these equations.

## Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

where

denotes the gamma function.

Obtain the behavior of for keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of for again keeping the two leading terms.

For Bosons, we must have (why?), obtain the leading term of for .

## Answer

For we can rewrite the integrand in a form that allows for series expansion

For the th power of in this series our integral is

Putting everything back together we have for small

We’ll expand about , writing

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the ranges. Observe that in the first integral we have

Since this term is of order 1, let’s consider the difference of this from , writing

or

This gives us

Now let’s make a change of variables in the first integral and in the second. This gives

As gets large in the first integral the integrand is approximately . The exponential dominates this integrand. Since we are considering large , we can approximate the upper bound of the integral by extending it to . Also expanding in series we have

For the remaining integral, Mathematica gives

where for

This gives

or

or

Evaluating the numerical portions explicitly, with

so to two terms (), we have

In order for the Boson occupation numbers to be non-singular we require less than all . If that lowest energy level is set to zero, this is equivalent to . Given this restriction, a substitution is convenient for investigation of the case. Following the text, we'll write

For , this is integrable

so that

Taylor expanding we have

Noting that , we have for the limit

or

For values of , the denominator is

To first order this gives us

Of this integral Mathematica says it can be evaluated for , and has the value

From [1] 6.1.17 we find

with which we can write

## Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number . i.e., having total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin- (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius , where , calculate the Fermi energy and the average energy per nucleon in MeV.

## Answer

Our nucleon particle density is

With for the mass of either the proton or the neutron, and , the Fermi energy for these particles is

With , and for either the proton or the neutron, this is

This gives us

In lecture 16

we found that the total average energy for a Fermion gas of particles was

so the average energy per nucleon is approximately

## Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at moving in the gravitational field of a heavy point mass at the center. Show that the pressure obeys the equation

where is the gravitational constant, is the distance from the center, and is the density which only depends on distance from the center.

## Answer

In the grand canonical scheme the pressure for a Fermion system is given by

The kinetic energy of the particle is adjusted by the gravitational potential

Differentiating eq. 1.75 with respect to the radius, we have

Noting that is the average density of the particles, presumed radial, we have

# References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[2] Kerson Huang. *Introduction to statistical physics*. CRC Press, 2001.

[3] RK Pathria. *Statistical mechanics*. Butterworth Heinemann, Oxford, UK, 1996.

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