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A Fugacity and Bose gas

  1. Jan 8, 2016 #1
    We know that the average occupation number cannot be negative for all systems and chemical potential must be negative in Ideal Bose Gas. This fact leads us to arrive a conclusion for fugacity which is related by chemical potential, as I quoted below:

    The restriction of the fugacity to the interval 0<=z<=1 is very important in the following. For a large volume, the sum over all one-particle states can be rewritten in terms of integral. (Greiner, Neise, Stocker, "Thermodynamics and Statistical Mechanics")


    So my question is, how can negative chemical potential helps us to convert the sum to the integral for a large volume? As what can I think fugacity in an İdeal Bose Gas?
     
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  3. Jan 8, 2016 #2

    vanhees71

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    It's not clear to me, what's meant by this claim. The issue with the ideal Bose gas is as follows. To make it clear, you have to start with a finite volume. The most conventient one is to take a cube of length ##l## with periodic boundary conditions. Then the one-particle basis can be chosen as the momentum-spin basis. From your description I also assume that you deal with non-relativistic bosons.

    The single-particle basis is ##|\vec{p},\sigma \rangle## with ##\vec{p}\in \frac{2 \pi}{l} \mathbb{Z}^3## and ##\sigma \in \{-s,\ldots,s \}##. The single-particle dispersion relation reads ##E=\frac{\vec{p}^2}/(2m)##.

    Now we calculate the grand-canonical potential
    $$Z=\mathrm{Tr} \exp \left [-\frac{\hat{H}-\mu \hat{N}}{T} \right ].$$
    The trace is over all boson-Fock states, i.e., after some algebra you get
    $$\ln Z=-\sum_{\vec{p}} \sum_{\sigma} \ln[1-\exp(-(\vec{p}^2/(2m)-\mu)/T)].$$
    Now as long as you keep the volume finite, it's clear that ##T > 0## and ##\mu \leq 0##, because otherwise ##\ln Z## would get complex.

    Now the temperature and chemical potential are used to adjust the mean total energy and particle number. To get these quantities it's more convenient to introduce the new independent variables ##\beta=1/T## and ##\alpha=\beta \mu=\mu/T##. Then you have
    $$\ln Z=-\sum_{\vec{p}} \sum_{\sigma} \ln[1-\exp(-\beta \vec{p}^2/(2m)+\alpha)]$$
    and
    $$\langle E \rangle=-\partial_{\beta} \ln Z=(2s+1) \sum_{\vec{p}} \frac{\vec{p}^2}{2m} f_{\text{B}}(E-\mu), \quad \langle N \rangle = \partial_{\alpha} \ln Z=\sum_{\vec{p}} f_{\text{B}}(E-\mu),$$
    where
    $$f_{\text{B}}(E-\mu)=\frac{1}{\exp[(E-\mu)/T]-1]}, \quad E=\frac{\vec{p}^2}{2m}.$$
    Now you can get any value ##\langle{E} \rangle \geq 0## and any ##\langle N \rangle \geq 0## if you have ##T>0## and ##\mu<0##. For ##T \rightarrow 0##, what happens is that all particles must occupy the ground state with ##E=0##, and indeed the smaller you make ##T## the larger you must make ##\mu## to keep the average particle number at the fixed given value.

    Now make the box size ##l## very large. Then you can put the sum to an integral by noting that in any momentum-volume element ##\mathrm{d}^3 \vec{p}## there are ##l^3 \mathrm{\mathrm{d}^3 \vec{p}}{(2 \pi)^3}## states and thus the mean particle number
    $$\langle N \rangle=V \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} f_{\text{B}}(E-\mu).$$
    Now, because ##\mathrm{d}^3 \vec{p}=4 \pi p^2 \mathrm{d} p## the integral stays finite for ##\mu \rightarrow 0## and becomes also arbitrarily small when taking ##T \rightarrow 0##, i.e., for a given fixed particle number at one point you cannot choose ##\mu \leq 0## anymore such that you get the correct average particle number, but from our finite-box discussion, it's clear what happens: there will be a finite number of particles occupying the ground state, and in the limit of infinite volume, this means you must take ##\mu=0## and write the distribution function as
    $$f(\vec{p})=(2 \pi)^3 N_{\text{BEC}} \delta^3(\vec{p}) + f_{\text{B}}(E) \quad \text{if} \quad T< T_c.$$
    On the other hand, if you can fix the particle number at the given temperature by choosing a chemical potential ##\mu \geq 0## you have to set
    $$f(\vec{p})=f_{\text{B}}(E-\mu) \quad \text{if} \quad T>T_c.$$
    The former case is known as Bose-Einstein condensation and was predicted to happen by Einstein in the early 1920ies. The experimental discovery was in 1995 (Nobel prize 2001).
     
  4. Feb 21, 2016 #3
    @vanhees71 thank you for your explanation. I really need to put it some effort to understand what's really going on it. I'm a new researcher in this field, and I can say that there's to much things that confusing me.
    Do you know any resources that can help me to understand ideal bose systems in 1D and 2D? I have find the mean particle number and grand potential for 1D and 2D but I'm not ready to answer this kind of questions yet.

    Consider a non-interacting, ideal Bose gas in one and two dimensions.


    5.1 Show that the integral for the number of particles is logarithmically
    diverging in 2D and algebraically in 1D for μ = 0 and T → 0.

    5.2 Show that this observation implies that Bose condensation does not
    occur in 1 or 2D at any finite temperature.
     
    Last edited: Feb 21, 2016
  5. Feb 22, 2016 #4

    vanhees71

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    That's simple. Just expand the integrand around ##\vec{p}=0## and check how the integral over ##\mathrm{d}^{d} \vec{p}## behaves around ##\vec{p}=0##. If the integral diverges for ##\mu=0## your can for each ##T## (including ##T \rightarrow 0##) accomodate any particle density without a Bose-Einstein condensate, which then answers your question 5.2.

    NOTE: For homework questions like this, please use the Homework Section of this forum, even if it's not literally homework in your case.
     
  6. Mar 3, 2016 #5
    @vanhees71 Thank you for your help. I'll be more carefull in the next time but just for this time that me continue to ask what do you mean by expanding the integral? Can you clarify this please? I've found N in all three dimension but I can't show how can integrals divergence logaritmically or algebraically. All I know that E = 0 gives to the density of states g(E) = 0. As it seems it's not in the case 2D since g(E) is propotinal with E^-1/2 in 1D.
     
    Last edited: Mar 3, 2016
  7. Mar 4, 2016 #6
    I have finally found that integral is diverging as ln(1/εβ) in 2D as z→1.
     
  8. Mar 4, 2016 #7

    vanhees71

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    So you see that in 2D you can get any density at arbitrary low temperatures in taking the naive infinite-volume limit. So no Bose-Einstein condensation occurs in two dimensions. In 3D in taking the naie infinite-volume limit, however, that's no longer the case, i.e., if you forget to treat the ground state adequately in taking the infinite-volume limit you cannot adjust the density if the temperature becomes lower than a critical value ##T_c##. With the correct treatment of the ground state you get Bose-Einstein condensation, i.e., in the infinite-volume limit at temperature ##T<T_c## you have to set ##\mu=0## and introduce a finite density of particles sitting in the single-particle ground state. That's the Bose-Einstein condensate.
     
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