What Temperature Initiates Bose-Einstein Condensation in a Gravitational Field?

AI Thread Summary
Bose-Einstein condensation (BEC) for an ideal Bose gas in a gravitational field occurs at a temperature influenced by the potential energy term in the Hamiltonian, specifically V=mgz. To determine the condensation temperature, one must calculate the density of states by integrating over phase space, accounting for the gravitational potential. The mean occupation number in BEC statistics is expressed in terms of momentum and height, complicating the integral needed to find the number of particles. Approximations are necessary to solve the integral, especially in relating the upper limit of the z-integral to momentum. The discussion emphasizes the challenge of deriving the density of states under gravitational influence, which is crucial for determining the condensation temperature.
qbslug
Messages
23
Reaction score
0

Homework Statement


For an ideal Bose gas in a uniform gravitational field, at what temperature does Bose-Einstein condensation set in. Gas is in a container of height L.

Homework Equations


Normal BEC temperature of an ideal Bose gas not under the influence of gravity is
T = \frac{h^2}{2 \pi m k}\left(\frac{N}{V\xi(3/2)}\right)^{2/3}

The Attempt at a Solution


I think one must first calculate the density of states for such a system by calculating the volume of phase space and dividing by h^(3N). But I don't know how to calculate the volume of phase space for this situation. I could be wrong of course in this attempt.
\omega = \int^\prime\cdot\cdot\cdot\int^\prime (d^{3N}q \,d^{3N}p) = ?
 
Physics news on Phys.org
I have an exam tomorrow in statistical mechanics so I should be able to solve this! :smile:

Normally you calculate the condensation temperature assuming that the particles are free, i.e. E = p^2 / 2m. Now you have a potential energy term in your hamiltonian, V=mgz, where z is the height of the particle in the container. This will influence your density of states and you will get something like:

<br /> N = \int^{\infty}_{p=0} \int^h_{z=0} &lt;n(p,z)&gt; f(p,z) dz dp<br />

for the number of particles, where <n> is the mean occupation number in BE statistics (expressed in terms of p and z), and f(p,z) the density of states, also expressed in terms of p and z.

Do the appropriate approximations in that integral and solve it. Then you can find the condensation temperature.

(Note: I haven't done the calculations but I'm guessing this would be one way to solve the problem, alert me if something seems to be wrong)
 
yeah the hard part is deriving the density of states for a bose gas under gravity. It doesn't seem to be so trivial.
 
No, I sat some time trying to solve it and had some serious trouble. How about the density of states as:
<br /> f(p,z)=\frac{4 \pi A zdz p^2dp}{h^3}<br />
?

(The 4pi comes from integrating out angular dependence in p (spherical coordinates) and the A from the spatial part, except z (x and y, or \rho and \varphi in cylindrical coordinates).)

Chief concern for me is then how to solve the integral. Using that \mu \rightarrow 0 at condensation we have
<br /> &lt;n&gt;= \frac{1}{e^{p^2/2m + mgz} -1}<br />
, right?

So how is the upper limit in the integral for z, L, translated into something dependent on p? We have E=p^2/2m + mgz. I think that can be used to find the upper limit L in the z-integral in terms of p.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top