# Isothermal compressibility of bose-einstein condensate

1. May 7, 2012

### gbertoli

1. The problem statement, all variables and given/known data

Variables: N (number of particles), μ (chemical potential), P (pression), V (volume).
k is Boltzmann's constant. I often use β=1/kT.

The (isothermal) compressibility is given by

$\kappa_{T} = -\frac{1}{V}\left (\frac{\partial V}{\partial P}\right )_{N,T}$

The following is also true (this was the first part of the assignment but I've already done it):

$\left ( \frac{\partial P}{\partial V} \right )_{\mu ,T} = \left ( \frac{\partial P}{\partial V} \right )_{N,T} + \left ( \frac{\partial P}{\partial N} \right )_{V,T}\left ( \frac{\partial N}{\partial V} \right )_{\mu,T}$

The questions are:
a) Show that the left-hand side of this equation vanishes
b) Show that
$\kappa_{T} = \frac{1}{V}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T} \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}$
c) Show that
$\left ( \frac{\partial \mu}{\partial P} \right )_{V,T} = \left ( \frac{\partial V}{\partial N} \right )_{\mu,T} = \frac{V}{N}$
d) Show that
$\kappa_{T} = \frac{V^{2}}{N^{2}} \frac{g_{0}}{2} \int_{0}^{\infty }\frac{\mathrm{d}\varepsilon}{\sqrt{\varepsilon}(e^{\beta\varepsilon-\beta\mu}-1)}$
e) The Bose-Einstein condensation can be approached by isothermally compressing the
gas. Describe qualitatively the behavior of κ in that process. What do you consider
an essential diﬀerence with the behavior of the ideal gas at low temperatures?

f) Show that in the high temperature the ideal gas result of $\kappa_{T} = \frac{1}{P} = \frac{V}{NkT}$ is recovered.

2. Relevant equations

To keep the V dependence explicit, the density of states can be written as

$g(\varepsilon) = V g_{0} \sqrt{\varepsilon }$

Here is $g_{0} = 2\pi (2m)^{\frac{3}{2}}h^{-3}$

An expression for N is:

$N = \int_{0}^{\infty }\mathrm{d}\varepsilon g(\varepsilon) \frac{1}{e^{\beta \varepsilon - \beta \mu} -1} = V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{1}{e^{\beta \varepsilon - \beta \mu}-1}$

The pressure can be found from $\beta PV = \ln Z$, with Z the gibbs sum, leading to the equation

$P = kTg_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\log(1-e^{\beta \mu - \beta \varepsilon})$

3. The attempt at a solution

My guesses are:
a) The expression for P does not explicitely depend on V and μ and T are held constant. This gives that $\left ( \frac{\partial P}{\partial V} \right )_{\mu ,T}$ is equal to 0.

b) If what I said by a) is true, I can just use the equality
$\left ( \frac{\partial P}{\partial V} \right )_{N,T} = - \left ( \frac{\partial P}{\partial N} \right )_{V,T}\left ( \frac{\partial N}{\partial V} \right )_{\mu,T}$
and "turn it upside down" (sorry for poor english) to get
$\left ( \frac{\partial V}{\partial P} \right )_{N,T} = - \left ( \frac{\partial N}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}$
Then I think I can use (chain rule)
$\left ( \frac{\partial N}{\partial P} \right )_{V,T} = \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T}$
Combining all this should give the right answer:
$\kappa_{T} = -\frac{1}{V}\left (\frac{\partial V}{\partial P}\right )_{N,T} = -\frac{1}{V} \times - \left ( \frac{\partial N}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T} =\frac{1}{V} \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}$

c) I really don't know, maybe I could get an expression for μ but that seems unlikely.

d)Assuming I already have the results from c), the equation found in b) becomes
$\kappa_{T} = \frac{1}{V} \frac{V^{2}}{N^{2}}\left ( \frac{\partial N}{\partial \mu} \right )_{V,T}$
so I just need to take the derivative of N with respect to μ. I can use integration by parts but it gets quite messy because you integrate in ε but derive in μ. Anyway, this is what I have:
$\frac{\partial }{\partial \mu}V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} = V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{\partial }{\partial \mu}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} = \left [ V g_{0} \sqrt{\varepsilon}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} \right ]_{0}^{\infty} - V g_{0} \int_{0}^{\infty }\mathrm{d}\varepsilon \frac{1}{2\sqrt{\varepsilon }}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1}$
I put the derivative inside the integral and derive only the Bose-Einstein distribution because the square root is not μ-dependent. The get the correct answer, the evaluated term should go to zero but i still get a minus sign. Help?

e) I dont really know. Probably κ would diverge but I can only say that from the fact that if T goes to 0 then β goes to infinity. I cant quite grasp the physical insight of the question.

f) No very good ideas. For a small β the exponential goes to 1 but that doesnt really help.
This is my first post so go easy on me.

Last edited: May 7, 2012
2. May 10, 2012

### gbertoli

ok, i know where the minus sign comes in question (d). The thing is that taking the derivative in mu gives the same result as taking the derivative in epsilon except for a minus sign that comes from the exponent. so you can just substitute (d/dmu) with (-d/deps) inside the integral.

3. May 13, 2012

### TvB

It seems we are in class together, because I'm trying to finish this exact same test.
I see you still don't have a clue of what to do at c. I did not complete it yet, but I gave an expression for N and then took the derivative with respect to V. Afterwards I used the equation for P and derived it with respect to mu. (I end up with 2 functions which are the same apart from a - sign.

I don't know what you mean with the 2nd post about part d. I used: kt=1/V V^2/N^2 (dn/dmu) and (dn/dmu) gives some integral. Afterwards you can start with the expression given at problem d, if you partially integrate that there is not going to be any minus sign. In the solution you posted you forget to take the derivative to mu from that last fraction, this will give you an extra minus sign. [remember: int (uV) = UV - int (Vu)]

For part e, I think the big difference is that for an ideal gas, kt only depends on V where for a Bose gas, kt depends on V^2. Part f is still giving me some trouble. At first glace I think that the integral part of 'f' will vanish because beta will be 0 and the exp(be-bmu) will be 1. Then the problem is to find a way to equate the remaining to V/NkT.

Last edited: May 13, 2012
4. May 14, 2012

### gbertoli

Watch out. when you derive in mu you dont get any minus sign. the exponent is negative but it is at the denominator, so its actually positive. and i think this is also why you get a minus sign in part c.