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: Bose Condensation in 2D

  1. Mar 28, 2016 #1
    • Moved from a technical forum, so homework template missing

    I was given the following question:

    Show that Bose condensation does not occur in 2D. Hint: The integral you will get when you write the formula for N is doable in elementary functions. You should find that that N ∝ ln(1 − e βµ).

    I do indeed find that N ∝ ln(1/(1 − e βµ)) ∝ ln(1 − e βµ), but I not sure how this tells me that Bose condensation isn't occurring.

    I have a feeling that Bose condensation is not occurring because the equation n/n_Q = f(βµ) always has a solution (n/n_Q ∝ ln(1/(1 − e βµ)), and ln(1/(1 − e βµ)) takes any value from 0+ (i.e. infinitesimally above zero) to +infinity if you have a look at it), where n_Q is the quantum concentration, some constant, and n = N/A, where A is the area of the 2D system, N is the total number of particles in it. By contrast, when in 3D, you find that there isn't always a solution and there is some critical temperature where a transition takes place; the f(βµ) in that case looks a little like ln(1/(1 − e βµ)) except for the fact that it hits the vertical axis at a value of about 2.6 ish before cutting off for βµ > 0. Is this correct?

    Further to this, how can N vary with temperature, i.e. where are the particles going to/coming from when the temperature changes? Is this just some weird unphysical result?

    Sorry if this is in the wrong forum - I know it's sort of quantum mechanics but it's part of a statistical mechanics course.
  2. jcsd
  3. Apr 2, 2016 #2
    Hi Jeff,
    Let us pick a start pt. For me, it comes from the occupation density for bosons (which can be derived from basic counting microstates available given boson statistics restrictions on the counting, see Mehran kardar's notes).

    This is:

    $$ n_{\epsilon} \propto {1 \over e^{\beta \epsilon} - 1}$$

    Now a main condition is that the total number of particles has to be the integral of the distribution of particles across available energies. Note I write the formula for generic d-dimensional space. If this statement can't be satisfied, it means that there must be a macroscopic # of particles congregating in a particular state so that $n_{\epsilon}$ has singular behaviors. This is Bose-Einstein Condensation in a nutshell. Also remark, the whole analysis I did had nothing to do with "Changing N" the number of particles.

    $$ N \propto \int_0^\infty d \epsilon \epsilon^{{d \over 2} -1} {1 \over e^{\beta \epsilon -1} -1 } = \zeta({d \over 2}) \Gamma({d \over 2})$$

    see http://www.physik.uni-regensburg.de/forschung/fabian/pages/mainframes/teaching/teaching_files/files of mf_statistical_physics/BE_integrals.pdf

    Now note, the above statement relies on the fact there is no macroscopic number of states in the ground state. We therefore know Bose-Einstein condensation must happen when the above integral is finite, which means there exist an N large enough that the above equation needs an additional term (macroscopic # of particles in a ground state). In other words, when $$\zeta({d \over 2})$$ is infinite, bose-einstein condensation can't happen. The lowest ${d \over 2}$ where this happens is $1$, so the highest dimension where BEC can't happen is $$d = 2$$.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted