## Physics Friday 92

Last week, we began modeling the ideal Bose gas, including showing that it possesses the same classical limit as the Fermi gas, and that the mean occupation number for a given state in the Bose gas (Bose-Einstein distribution), , differs from that in the Fermi gas (Fermi-Dirac distribution), , by a change in sign in the denominator. After using the particle-in-a-box “density of states” to approximate sums over states as integrals over energy, we found several parameters in terms of polylogarithms; with fugacity , we had
,
,

(the subscript placed here on the particle number will become important for our notation later).
We also found previously that due to the divergence of  at , the Gibbs potential must always be negative. Since β is always positive, then the fugacity must be less than unity: 0<ξ<1.

In the following work, I will use various properties of the polylogarithm functions. A separate post exploring the mathematics involved can be found here.

Using the thermal de Broglie wavelength , the above expressions for energy and particle number simplify to:
.
and
.
[Note that when the fugacity is small, these are approximated to first order in fugacity by  and , which we found in the classical approximation here].
Now, dividing these to cancel the  terms, we get
,
which differs from the classical  by a factor of , which equals 1 at zero fugacity, and decreases with increasing fugacity (see the addendum). At the ξ→1 limit (μ→0 from below), we have

and
.

Now, we have a way to analyze a given Bose gas. Suppose we know Ñe, V, and T. Then
 tells us that , and this can be solved numerically for the fugacity, which in turn allows us to evaluate the various thermodynamic functions. Note that the dependence of the fugacity upon volume and particle number can be expressed as dependence on particle density : .

However, we should note that  diverges for ξ>1, so the maximum value of the polylogarithm in question is . However, one can be given values of Ñe, V, and T with the temperature low enough or density high enough that . In this situation, our proceeding analysis gives no solution for the fugacity; our analysis breaks down in this strong quantum limit. Where is the error?

The error is in how we considered the ground state. Recall that at μ=0, the occupation number n0 becomes infinite; as the Gibbs potential rises toward zero, an increasing number of particles are found in the ground state. However, recall that in approximating our sum over states as an integral, we used a “density of states” , which for zero energy gives zero state density; our approximation neglects the ground state. This wasn’t a problem for our Fermi gas, as the ground state there can only hold a small number (2s+1) of particles.

Examining this “breakdown” point, we set , and seek the temperature:
,
where Tc is called the Bose condensation temperature. Above this temperature, our integral approximation still holds; below it, a phenomenon called Bose condensation happens, where the population of the ground state becomes significant. For example, 4He has atomic mass m=4.0026 amu=6.6465×10-27 kg, and liquid 4He has a density of approximately 125 kg/m3, which gives number density n≈1.88×1028 m-3. Using these, we get Tc≈2.8 K, which is fairly close to the 2.17 K temperature where liquid helium-4 becomes superfluid.

Next week, I will explore how we correct the above analysis for temperatures below the condensation temperature.