Isothermal compressibility of bose-Einstein condensate

[gbertoli]

Homework Statement

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.

Homework 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

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

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 don't 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 can't quite grasp the physical insight of the question.

f) No very good ideas. For a small β the exponential goes to 1 but that doesn't really help.
This is my first post so go easy on me.

 

[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.

 

[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.

 

[gbertoli]

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)]

Watch out. when you derive in mu you don't 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.

 

[paranormal]

As a scientist, my response to this content would be as follows:

It is clear that the topic being discussed here is the isothermal compressibility of a Bose-Einstein condensate, with the variables being the number of particles (N), chemical potential (μ), pressure (P), and volume (V). The isothermal compressibility is defined as the ratio of the relative change in volume to the relative change in pressure, with temperature being held constant. This can be expressed as -1/V times the partial derivative of volume with respect to pressure, at constant N and T.

In the first part of the assignment, it is shown that the left-hand side of the equation given in the second part vanishes, as the pressure does not explicitly depend on volume and both μ and T are held constant. This is an important result that can simplify further calculations.

Next, it is shown that the isothermal compressibility can be expressed in terms of the chemical potential, number of particles, and volume, using the chain rule. This result is then used to derive an expression for the isothermal compressibility in terms of the density of states and the Bose-Einstein distribution. This expression involves an integral that can be evaluated numerically to obtain the isothermal compressibility.

In part c), it is asked to show that the partial derivative of the chemical potential with respect to pressure is equal to the partial derivative of volume with respect to number of particles, both at constant temperature and volume. This can be shown by using the definitions of pressure and chemical potential in terms of the density of states and the Bose-Einstein distribution.

Part d) involves using the results from part c) to derive an expression for the isothermal compressibility in terms of the density of states and the Bose-Einstein distribution. This expression involves an integral that can be evaluated numerically to obtain the isothermal compressibility.

Part e) asks for a qualitative description of the behavior of the isothermal compressibility in the process of approaching Bose-Einstein condensation through isothermal compression. This process would lead to a diverging isothermal compressibility, as the temperature approaches absolute zero and the Bose-Einstein distribution becomes highly peaked at zero energy. This behavior is in contrast to that of an ideal gas at low temperatures, where the isothermal compressibility would approach a constant value.

Finally, in part f), it is shown that in the high temperature limit, the isothermal compressibility of an ideal gas is recovered. This result