Physics Friday 93

Continuing from last Friday, we now find the behavior of an ideal Bose gas at a temperature below the condensation temperature , where our previous analysis fails due to the integral approximation neglecting the ground state.
First, let us examine the spacing between the energy levels. For our quantum particle in a box, we have energy levels , where nx, ny, and nz are positive integers. Using , we rewrite as . The ground state is . The first excited state is three-fold degenerate: , and the difference between these states is
. For a volume of one liter (=10-3 cubic meter), and particles the mass of 4He (6.6×10-27 kg), this energy difference is about 2.5×10-39 J, which divides by k to get an equivalent temperature in the vicinity of 2×10-16 K, and so the spacing is close enough that for any reasonable temperature, replacing the sum by an integral makes sense. Thus, our problem is purely the neglect of the ground state.
Let us examine what happens as the chemical potential μ approaches the ground state  from below. We name the number of particles in the ground state as n0; our Bose-Einstein distribution tells us that . If our occupation number of the ground state is large, , then we see that ; we thus expand the exponential in first order:
, so that
. If the total number of particles is Ñ, then we see that the population of the orbital ground state is comparable to this when , and the chemical potential cannot get closer to the ground state than ; the higher states are shielded from Gibbs potential by the ground state.
When the ground state population is significant, which occurs below Tc, we can see that the occupation number of any other individual state is relatively small; compare  to . Thus, the integral approximation remains valid for representing all states except the ground state. Thus, we simply add an additional separate term explicitly listing the ground state in the sum of states.
The total number of particles is , where  is here the number of particles in the excited states, and thus the reason for the subscript we added before. Writing n0 in terms of fugacity, and using ε=0 for the ground state, we have
.
Since particles in the ground state have zero energy, the energy of the gas remains , and the reinterpreting of Ñe is the key correction.
Correcting the grand canonical potential , we see that our explicit ground state adds a term , and so our fundamental equation is
.

Now, we can use the above to analyze the properties of the Bose gas below the condensation temperature. First, for T<Tc, the maximum number of particles in the excited states is
. As TTc, we have ÑeÑ, and so we have , where λc is the value of the thermal de Broglie wavelength at the condensation temperature. Dividing these,
; this tells us that the fraction of the gas particles in the ground state as a function of temperature is
.

Now, examine the energy. For T>Tc, we have as before, . For T<Tc, we have
.

From this, we find the heat capacity at constant volume by
.
Below the condensation temperature, we take the derivative of the above, to get
, giving at  a value of , significantly higher than the classical value 1.5Ñk, which we approach in the high-temperature classical regime. For temperatures above the condensation temperature, we have to take the derivative with respect to temperature of  at constant Ñ, and eliminating  using , and consider the temperature dependence of the thermal de Broglie wavelength. Performing that calculus, the net result is that
. This is greater than the classical value, and approaches it as T becomes large.
Both the T<Tc and T>Tc equations give  at T=Tc; however, the heat capacity has a cusp at this point; CV is increasing for T<Tc and decreasing for T>Tc. This unique cusp in heat capacity is one of the signatures of Bose condensation.