Bose-Einstein condensation

The total number of particles in the Bose-Einstein gas is obtained by summing over the occupation numbers $n_r$ of each state:

$$N = \sum_r \bar{n}_r = \sum_r \frac{1}{e^{\beta (\epsilon_r-\mu)}-1}.$$ (5.43)

Of these, the ground state is particularly interesting. Since there is no limit to the number of particles that can fall into any state, at zero temperature all particles will be in the ground state. Here the particles have zero energy and zero momentum, and so they contribute zero to the pressure and to the viscosity (which depends on momentum). They are in a superfluid state, known as Bose-Einstein condensate. If the temperature is very close to zero, but not exactly zero, some particles will be in excited states, and the fraction of the number of particles in the ground state decreases as the temperature increases up to some critical temperature $T_c$, above which this fraction becomes negligible. For $0 \le T \le T_c$ the gas is formed by a mixture of two fluids, a superfluid condensate and a standard gas of bosons. The two fluids are not spatially separated, as the particles have no identity.

Figure: The quantity $b(\epsilon; \beta,\mu)$ plotted as function of $\epsilon$ for various values of $\beta$ and $\mu$. The integral of $b(\epsilon; \beta,\mu)$ is proportional to the number of particles in the system. Taking $b(\epsilon; 5, -0.1)$ as target (black solid curve), we see that increasing $\beta$ to 6 decreases the value of the area under the curve for fixed $\mu=-0.1$ (red dashed curve). However, the area can be increased by changing $\mu = -0.02$ (green dotted curve) and therefore for such a value of $\beta$ it is possible to find a value of $\mu$ to match the area under the curve $b(\epsilon; 5, -0.1)$. However, if $\beta = 15$ the area under the curve is too small, even for highest possible value $\mu=0$ (blue dot dashed curve).

To see how this limiting temperature comes about we now proceed with the following argument. We have discussed earlier the density of states $f(\epsilon)$, which gives the number of available states per unit energy. This is a useful quantity because, under the assumption that the number of states in the temperature range $k_{\rm B}T$ is large, we can turn the sum  into an integral using the density of states :

$$N = (2s + 1)\frac{2\pi V}{h^3} (2m)^{3/2}\int_0^\infty \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1} d\epsilon.$$ (5.44)

The chemical potential is defined implicitly by , subject to the constraint $\mu < 0$. To see how this constraint comes into play, let us consider the integrand

$$b(\epsilon; \beta,\mu) = \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1}$$ (5.45)

in more details. This is plotted in Fig.  for three values of $\beta$. We see that, for fixed $\mu$, the area under the curve $I (\beta,\mu) = \int_0^\infty b(\epsilon;\beta,\mu) d\epsilon$ decreases for increasing $\beta$ (decreasing temperature). We also see that for the highest two temperatures this decrease in $I$ can be reversed by decreasing $\vert\mu\vert$, and in this case it is possible to adjust $\vert\mu\vert$ in order to satisfy the constraint that the total number of particle, which is proportional to $I$, is actually equal to $N$. However, at the lowest temperature (highest $\beta$) even setting $\vert\mu\vert = 0$ is not enough to restore $I$ to the value required to satisfy the constraint. This would suggest that either there is a limit to how low the temperature can go, or some particles start to disappear. Both hypothesis are obviously non physical. In fact, what is actually happening is that below a critical temperature $T_c$, defined by  with $\mu=0$: ^5.3

$$N = (2s + 1)\frac{2\pi V}{h^3} (2m)^{3/2}\int_0^\infty \frac{\epsilon^{1/2}}{e^{\epsilon/k_{\rm B}T_c}-1} d\epsilon,$$ (5.46)

the particles will start to populate the ground state with $\epsilon=0$. This state is not present in  because, as the states are weighted with $\epsilon^{1/2}$, the ground state is given zero weight and contributes nothing to the integral. This shortcoming is a consequence of our approximation of the sum  with the integral . We have come across this difficulty already, when we have derived expressions for the equation of state of the perfect gas, and found that also in that case problems appeared because of the replacement of a sum over states with an integral to compute the partition function. We must therefore go back to the sum and include the ground state explicitly, so Eq.  needs to be replaced by:

$$N = (2s+1)\left [\frac{1}{e^{-\beta\mu}-1} + \frac{2\pi V}{h^3} (... ...t_0^\infty \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1} d\epsilon \right ],$$ (5.47)

which we can rewrite as

$$N = N_1 + N_{\epsilon> 0},$$ (5.48)

with

$$N_1 = \frac{2s+1}{e^{-\beta\mu}-1},$$ (5.49)

the number of particles in the ground state and

$$N_{\epsilon> 0} = (2s+1)\frac{2\pi V}{h^3} (2m)^{3/2}\int_0^\infty \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1} d\epsilon,$$ (5.50)

the remaining particles. The number of particles in the ground state $N_1$ cannot be determined using Eq. , because the chemical potential is not known. All we know is that it must be very close to zero, but we cannot set it exactly equal to zero. However, we can obtain $N_1$ as difference, by combining Eqs.  and  with $\mu=0$. This may appear as an inconsistency at first sight, but in fact it isn't, because the integral in Eq.  does not really change much if $\mu=0$ or if $\mu < 0$ but very small. On the contrary, the expression for $N_1$ in  is singular for $\mu=0$, and so it is very sensitive to the actual value of $\mu$. The fraction of particles in the ground state is therefore given by:

$$\frac{N_1}{N} = \frac{N - N_{\epsilon> 0}}{N} = \frac{\int_0^\inf... .../2}}{e^{\beta_c\epsilon}-1} } = \frac{I(\beta_c,0) - I(\beta,0)}{I(\beta_c,0)},$$ (5.51)

with $\beta_c = 1/k_{\rm B}T_c$. The integral $I(\beta,0)$ is proportional to $\beta^{-3/2}$, as it can be easily verified with a change of variable $z=\beta \epsilon$ , and so:

$$\frac{N_1}{N} = 1 - \left(\frac{T}{T_c}\right )^{3/2},$$ (5.52)

for $T<T_c$. For $T>T_c$ the number of particles in the ground state is a negligible fraction of the total. This fraction is shown in Fig. .

Figure: The fraction of particles in the ground state, $N_1/N$, as function of temperature.