Question

For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of $\epsilon$(see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains significantly less than zero as long as $T$ is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must $\epsilon$ have in order for there to be an abrupt Bose-Einstein condensation?

Short answer

The given statement was proved right by finding the chemical potential less than zero

Step-by-step solution

Step 1. Given information

Number of atoms in 2-dimensional box =

$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{(\epsilon -\mu )/kT}-1}d\epsilon$

Here,

$g\left(\epsilon \right)$=density of states,

$k=$Boltzmann constant,

$T=$temperature of the gas,

$\mu$=chemical potential,

$\epsilon$= energy of the particle in a two-dimensional box.

Step 2. Since the density of states is constant, the number of atoms in two-dimensional  box is

$N=g{\int }_0^{\infty }\frac{1}{e^{(\epsilon -\mu )/kT}-1}d\epsilon$

If we take $\mu =0$

$N=g{\int }_0^{\infty }\frac{1}{e^{ϵ/kT}-1}d\epsilon$

Let's take $x=\frac{\epsilon }{kT}$

$N=g{\int }_0^{\infty }\frac{1}{e^x-1}d\epsilon$

As we know the value of

${\int }_0^{\infty }\frac{1}{e^x-1}dx=\infty$

Therefore the number of atoms =

$N=g{\int }_0^{\infty }\frac{1}{e^x-1}d\epsilon \approx \infty$

From the above result, it's clear that there's infinite number of atoms within the low-lying excited state when the chemical potential is zero. But the number of atoms is finite, So our assumption is not possible.

Step 3. The number of atoms in two dimensional box,

$N=g{\int }_0^{\infty }\frac{1}{e^{(\epsilon -\mu )kT}-1}d\epsilon$

$=g{\int }_0^{\infty }\frac{1}{e^{-\mu /kT}{e}^{\epsilon /kT}-1}d\epsilon$

The integral will not be divergent if $e^{-\mu /kT}>1$

Applying the natural logarithm and solving the chemical potential we get,

$-\frac{\mu }{kT}>\ln \left(1\right)$

$\mu <\ln \left(1\right)kT$

$\mu <0$

Thus, the chemical potential is less than zero. Hence the given statement is proved.

Step 4.  Discussion on negative value of chemical potential

The chemical potential should be negative as a result of the integral is finite (or the chemical potential is a smaller amount than zero the least bit temperatures). So, the whole particles can settle into the ground-state energy at very-low temperatures. because the temperature of the system goes to zero, the particles endlessly move to the ground-state, with no abrupt transition.

To get abrupt transition, the integral for the number of atoms in the two-dimensional box must be convergent at its lower limit when $\mu =0$. This condition will possible only when the value of the density of states $g\left(\epsilon \right)$ goes to zero as $\epsilon$ goes to zero.