# Ideal Bose condensation gas in gravitational field, Statistical Physic

1. Feb 20, 2013

### dpopchev

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:
$T_c = T^0_c( 1 + \frac{8}{9} \frac{1}{ζ(3/2)} (\frac{\pi mgh}{kT^0_c})^{1/2} )$

Attempt
Lege artis; Firstly I want to estimate the number $N$ of particles in that $T_c$ temperature. As we know at that state the chemical potential is zero so $\mu = 0$, so the average number of particles will be given by the Bose - Einstein distribution
$n_{mean} = \frac{1}{e^{ε/kT_c}-1}$
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 $S$ and hight $L$ ) and over all possible energy states. I will skip the mathematical formulation and give the integral I need to evaluate:
$N = 4 \pi g_0 S h^{-3} \int_0^{∞} dp \int_0^L dz \frac{p^2}{e^{ \frac{p^2}{2 m k T_c} - \frac{mgz}{kT_c} } - 1 }$
where $S$ is area of the circle for the cylinder, $g_0 = 2s + 1$ is the number of states and $h$ is the Planck constant
After changing the variables I get $\frac{ 4 \pi g_0 S (2mkT_c)^{3/2}kT_c}{h^3mg} \int^{∞}_0 dp' \int^{\frac{mgL}{kT_c}}_0 \frac{1}{e^{p'^2}e^{-z'}-1}$
I have no idea how to handle this integral:
I just get the $e^{z'}$ out and get $\int \frac{e^{z'}}{e^{p'^2}-e^{z'}}$ which leads to something like $\int p'^2 ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } dp'$ on which my best attempt was $\int ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } \frac{dp'^3}{3} = \frac{p'^3}{3}.ln(...) - ....$ and the first is in range of $0$ which is a problematic dot and $∞$ 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