Bose-Einstein Condensation & de Broglie wavelength

[jncarter]

Homework Statement

Bose-Einstein condensation of a fluid occurs when the de Broglie wavelength of a "typical" particle becomes greater than the average nearest-neighbor distance. One can interpret the momentum in the de Broglie equation as
$p=\sqrt{<p^{2}>}$
where $<p^{2}>$ means the thermal average for a single particle

(a) Write down the de Broglie equation and use the above to obtain an approximate expression for the Bose-Einstein condensation temperature $T_{B}$ in terms of the particle density (N/V), the particle mass m, Boltzmann's constant k and Planck's constant h

(b)For temperatures T less than the transition temperature $T_{B}$, the fraction, f, of particles in the ground state (f=N(0)/N) is a number between zero and one. Find the chemical potential, $\mu$ to order 1/N as a function of T when $T <<T_{B}$. Assume the ground state energy is zero.

Homework Equations

de Broglie momentum: $p= \frac{h}{\lambda}$
Boltzmann distribution function: $<n_{s}> = \frac{1}{e^{(\mu - \epsilon)\beta} - 1}$
where $\beta = \frac{1}{kT}$
Energy: $E = \frac{p^{2}}{2m}$
Expectation value: $<A> = \Sigma_{s} A <n_{s}>$

The Attempt at a Solution

I interpreted the average nearest-neighbor distance to be the ratio of the number of particles, N, and the volume, V, to the 1/3 power, $\lambda > \frac{N}{V^{1/3}}$.
Put this into our equation for momentum:
$p < \frac{h V^{1/3}}{N}$
But we have been told $p=\sqrt{<p^{2}>}$, thus $<p^{2}> <(\frac{h V^{1/3}}{N})^{2}$.
Now we can use the energy-momentum relationship to find $<E> < \frac{1}{2m}(\frac{h V^{1/3}}{N})^{2}$.
So the question comes down to how to approximate the expected energy for a single particle in state with possible states s. Normally, I would use the expected value function above and go through the approximation using density of states and find a function for the chemical potential and then finally solve for temperature. I just can't kick the feeling that there is an easier way. This was taken from a previous comprehensive exam, so I should be able to do it without notes in about an hour (or at least most of the problem). Is there a trick I'm missing?

 

[mark726]

Thank you for your question. The Bose-Einstein condensation temperature T_{B} can be approximated using the de Broglie equation and the average nearest-neighbor distance as follows:

p=\sqrt{<p^{2}>}=\frac{h}{\lambda}=\frac{h}{(\frac{N}{V^{1/3}})^{1/2}}=\frac{h V^{1/6}}{N^{1/2}}

Using the energy-momentum relationship, we can write the average energy as:

<E>=\frac{<p^{2}>}{2m}=\frac{\frac{h^{2} V^{1/3}}{N}}{2m}=\frac{h^{2} V^{1/3}}{2mN}

Now, we can use the Boltzmann distribution function to find the fraction of particles in the ground state:

f=\frac{<n_{0}>}{N}=\frac{1}{e^{\frac{\mu}{kT}}-1}

Since we are considering temperatures T<T_{B}, we can assume that the ground state energy is zero. Therefore, the chemical potential can be written as:

\mu=kT\ln{\frac{N}{f}}

To approximate the chemical potential to order 1/N, we can use the Taylor series expansion of ln(N/f) around N=0:

\ln{\frac{N}{f}}=\ln{\frac{N}{N(0)}}=\ln(1+\frac{N(0)}{N})

Using the first two terms of the Taylor series, we get:

\ln{\frac{N}{f}}\approx\frac{N(0)}{N}-\frac{1}{2}(\frac{N(0)}{N})^{2}

Substituting this into our expression for the chemical potential, we get:

\mu=kT(\frac{N(0)}{N}-\frac{1}{2}(\frac{N(0)}{N})^{2})

Now, we can solve for the Bose-Einstein condensation temperature T_{B} by setting the fraction f equal to 1:

1=\frac{1}{e^{\frac{\mu}{kT_{B}}}-1}

Substituting our expression for the chemical potential and solving for T_{B}, we get:

T_{B