1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ideal Bose condensation gas in gravitational field, Statistical Physic

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    We have an ideal Bose gas in gravitational field. Show that the critical temperature( the temperature in which the condensation is starting is:
    [itex] T_c = T^0_c( 1 + \frac{8}{9} \frac{1}{ζ(3/2)} (\frac{\pi mgh}{kT^0_c})^{1/2} ) [/itex]

    Lege artis; Firstly I want to estimate the number [itex] N [/itex] of particles in that [itex] T_c [/itex] temperature. As we know at that state the chemical potential is zero so [itex] \mu = 0 [/itex], so the average number of particles will be given by the Bose - Einstein distribution
    [itex] n_{mean} = \frac{1}{e^{ε/kT_c}-1} [/itex]
    I want to find all that particles in such state, so I form an integral over the volume in which the gas is( we assume it is a container with axial symmetry, with area of the circle [itex] S [/itex] and hight [itex] L [/itex] ) and over all possible energy states. I will skip the mathematical formulation and give the integral I need to evaluate:
    [itex] N = 4 \pi g_0 S h^{-3} \int_0^{∞} dp \int_0^L dz \frac{p^2}{e^{ \frac{p^2}{2 m k T_c} - \frac{mgz}{kT_c} } - 1 } [/itex]
    where [itex] S [/itex] is area of the circle for the cylinder, [itex] g_0 = 2s + 1 [/itex] is the number of states and [itex] h [/itex] is the Planck constant
    After changing the variables I get [itex] \frac{ 4 \pi g_0 S (2mkT_c)^{3/2}kT_c}{h^3mg} \int^{∞}_0 dp' \int^{\frac{mgL}{kT_c}}_0 \frac{1}{e^{p'^2}e^{-z'}-1} [/itex]
    I have no idea how to handle this integral:
    I just get the [itex] e^{z'} [/itex] out and get [itex] \int \frac{e^{z'}}{e^{p'^2}-e^{z'}} [/itex] which leads to something like [itex] \int p'^2 ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } dp' [/itex] on which my best attempt was [itex] \int ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } \frac{dp'^3}{3} = \frac{p'^3}{3}.ln(...) - ....[/itex] and the first is in range of [itex] 0 [/itex] which is a problematic dot and [itex] ∞ [/itex] which doesn't seem pretty too.
    I hope for my post to be accurate enough. Thanks in advance.

    PS: I found the same problem but it can't help me to solve the integral
    Last edited: Feb 20, 2013
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted