The high density, limit

The limit is the limit of maximum chemical potential, which is expected at high density. However, since , maximum chemical potential will be the limit . In this limit, the full problem, including the divergent terms, must be solved:

We will need to refer to these two sums often in this section, so let us define them to be

Thus, the problem becomes one of solving

We examine, first the density equation. The second term will diverge at . It is instructive to ask what is the behavior of the first term at . In fact is nothing but a Riemann zeta-function:

In general, a Riemann zeta-function R(n) is given by

and the values of this function are given in many standard math tables. The particles value of R(3/2) is approximately 2.612... Moreover, from the form of , it is clear that, since , is the maximum value of . A plot of is given below:

  
Figure 1:

The figure also indicates that the derivative diverges at despite the fact that the value of the function is finite. Note that, since

It is possible to solve the density equation for by noting that unless is very close to 1, the divergent term will still vanish in the thermodynamic limit as a result of its prefactor. How close to 1 must it be for this term to dominate? It can only be different from 1 by an amount on the order of 1/V. Thus, let us take to be of the form

where a is a positive constant. Substituting this ansatz into the equation for the density gives

Since does not change its value much if is displaced just a little from 1, we can replace the first term by R(3/2). Then,

can be solved for a to yield

where we have neglected a term , which vanishes in the thermodynamic limit. Since a must be positive, this solution is only valid for . For , will be different from 1 by more than an amount 1/V so in this regime, the term can be neglected, leaving the problem of solving . Therefore, the solution for can be expressed as

which, in the thermodynamic limit, becomes

A plot of vs. is shown below:

  
Figure 2:

Clearly, point R(3/2) is special, as undergoes a transition there to a constant value of 1.

Recall that the occupation of the ground state is

Thus, for , this becomes

for . At the occupation of the ground state becomes 0. To what temperature does this correspond? We can find this out by solving

so that for temperatures less than the occupation of the ground state becomes

Thus, at T=0

which is equivalent to

If we sum both sides over m, this gives

where indicates that the spin degeneracy has been summed over. For , and is not within 1/V of 1. This means that is finite and

as . Therefore, we have, for the occupation of the ground state:

which is shown in the figure below:

  
Figure 3:

The occupation of the ground state undergoes a transition from a finite value to 0 at and for all higher temperatures, remains 0. Now, represents the probability that a particle will be found in the ground state. It also represents the fraction of the total number of particles that will be found in the ground state. For , this number is very close to 1, and at T=0, it becomes exactly 1, implying that at T=0 all particles will be found in the ground state. This is a phenomenon known as Bose-Einstein condensation. The occupation number of the ground state as a function of temperature is shown in the plot below:

Note that there is also a critical density corresponding to this temperature. This will be given by the solution of

which can be solved to yield

and the occupation number, expressed in terms of the density is

The term in the pressure equation

becomes, for very close to 1

which clearly vanishes in the thermodynamic limit, since . This allows to deduce the equation of state as

where in the above equation comes from the actual solution of . What is particularly interesting to note about the equation of state is that the pressure is independent of the density for . Isotherms of the ideal Bose gas are shown below:

  
Figure 4:

Here, corresponds to the critical density . As a function of temperature, we see that , which is quite different from the classical ideal gas. This is also in contrast to the fermion ideal gas, where as the pressure remains finite. For the Boson gas, as the pressure vanishes, in keeping with the notion of an ``effective'' attraction between the particles.

Other thermodynamic quantities can be determined in a similar manner. The energy can be obtained from E=3PV/2 straightforwardly:

and the heat capacity at constant volume from

which gives

A plot of the heat capacity exhibits a cusp at :

  
Figure 5:

Experiments carried out on liquid He , which has been observed to undergo Bose-Einstein condensation at around T=2.18 K, have measured an actual discontinuity in the heat capacity at the transition temperature, suggesting that Bose-Einstein condensation is a phase transition known as the transition. The experimental heat capacity is shown roughly below:

  
Figure 6:

By contrast, the ideal Bose gas undergoes a first order phase transition. However, using the mass and density of liquid He in the expression for given above, one would predict that is about 3.14 K, which is not far off the experimental transition temperature of 2.18 K for real liquid helium.

For completeness, other thermodynamic properties of the ideal Bose gas are given as follows: The entropy is

The Gibbs free energy is given by

It is clear from the analysis of this and the fermion ideal gas that quantum statistics give rise to an enormously rich behavior, even when there are no particle interactions!