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

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Homework Help Overview

The discussion revolves around determining the temperature at which Bose-Einstein condensation occurs for an ideal Bose gas in a uniform gravitational field, specifically within a container of height L. Participants are exploring the implications of gravitational potential energy on the condensation temperature and the density of states.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the density of states for a Bose gas influenced by gravity, with one suggesting the need to derive the volume of phase space. Others propose integrating over momentum and height to account for the gravitational potential in the Hamiltonian.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and expressing uncertainty about the calculations involved. Some have offered potential methods for solving the problem, while others are questioning the assumptions and definitions related to the density of states and integration limits.

Contextual Notes

There is a noted complexity in deriving the density of states under gravitational influence, and participants are grappling with how to appropriately set up integrals given the gravitational potential energy term. The original poster's attempt and subsequent responses indicate a lack of consensus on the correct approach.

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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) = ?
 
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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.
 

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