Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partition function expansion

  1. Nov 28, 2008 #1

    I've got a problem. If you scroll down to page 118 on the following link


    there you will find an expansion (eq. 10.27) of the partition function [tex]Z_{\pi}[/tex], but I don't know how this expansion is explicity being done. Could anyone help me, please?

    As described on page 116 I think the sum was replaced by an integral ([tex] \Omega [/tex] denotes the volume). But I still don't know how to get rid of the logarithm and the exponential function.
  2. jcsd
  3. Nov 30, 2008 #2


    User Avatar
    Homework Helper

    Ho-kay... I see how this is done. It's going to involve layers upon layers of small terms expansions, so I'm not going to derive the whole series shown because I don't want to keep track of that many terms or do that many integrals. So, here goes:

    The first thing to do is, as you guess, convert the sum into an integral:

    [tex]\ln Z_\pi = -\nu \sum_{\mathbf{k}} \ln\left(1 - e^{-\beta \varepsilon_k}\right) \rightarrow -\frac{\nu \Omega}{(2\pi)^3}\int_0^{2\pi} d\phi \int_{-1}^{1}d (\cos\theta) \int_0^{\infty} dk~k^2 \ln\left(1 - e^{-\beta \varepsilon(k)}\right)[/tex]

    Now, do the angular integrals since the energy depends only on the magnitude, and integrate the magnitude integral by parts:

    [tex]-\frac{4\pi\nu\Omega}{(2\pi)^3}\left\{ \left.\frac{k^3}{3}\ln\left(1-e^{-\beta\sqrt{k^2+m_\pi^2}}\right)\right|^{\infty}_0 + \frac{\beta}{3}\int_0^{\infty}dk k^4 \frac{e^{-\beta \sqrt{k^2+m_\pi^2}}/\sqrt{k^2 + m_\pi^2}}{1-e^{-\beta \sqrt{k^2-m_\pi^2}}}\right\}[/tex]

    The integrated out piece from the integral vanishes, leaving you with a nasty, nasty integral that would probably cause Mathematica to punch you in the face if you tried to enter it into it. So, we do the next best thing, because it's the only thing we know how to do: assume m_\pi is small compared to k and expand as a series. To start, I'll rewrite the integral a little bit:

    [tex]\ln Z_\pi \simeq \frac{\nu \Omega \beta}{6\pi^2}\int_0^{\infty}dk~k^3\frac{\left(1+ \left(\frac{m_\pi}{k}\right)^2\right)^{-1/2}}{e^{\beta k\left(1 + \left(\frac{m_\pi}{k}\right)^2\right)^{1/2}} - 1}[/tex]

    So, we're gonna be expanding the square root terms. I assume you're familar with this so I'm not gonna do those steps to save myself some latexing:

    [tex]\frac{\nu \Omega \beta}{6\pi^2}\int_0^{\infty}dk~k^3\frac{\left(1 - \frac{1}{2}\left(\frac{m_\pi}{k}\right)^2 + \cdots\right)}{e^{\beta k}e^{\beta k\left( \frac{1}{2}\left(\frac{m_\pi}{k}\right)^2 + \cdots \right)} - 1}[/tex]

    Now we're gonna want to expand the second expontential in the denominator there, as its argument is small (at least if we add on the additional assumption that \beta m_\pi is small.

    [tex]\frac{\nu \Omega \beta}{6\pi^2}\int_0^{\infty}dk~k^3\frac{\left(1 - \frac{1}{2}\left(\frac{m_\pi}{k}\right)^2 + \cdots\right)}{e^{\beta k}\left[1 + \left(\beta k\left( \frac{1}{2}\left(\frac{m_\pi}{k}\right)^2 + \cdots \right)\right) + \frac{1}{2}\left(\beta k\left( \frac{1}{2}\left(\frac{m_\pi}{k}\right)^2 + \cdots \right)\right)^2 + \cdots \right] - 1}[/tex]

    So, as you can see, this thing is getting uuuuugggglyyy. So I'm just gonna go up to O(m_\pi^2). Splitting the numerator into two terms, we see that in the second term I can neglect the expansion in the numerator as that'll give me terms of order higher than I really want. In the first term I need to expand the denominator. Factor out an exp(\beta k) - 1 from the denominator and expand.

    The zeroth order term is easy to evaluate:

    [tex]\frac{\nu \Omega \beta}{6\pi^2}\int_0^{\infty}dk~\frac{k^3}{e^{\beta k} - 1} = \frac{\nu \Omega \beta}{6\pi^2}\beta^{-4} 3! \zeta(4) = (\nu \Omega \beta)T^4 \frac{\pi^2}{90}[/tex]

    The integral is evaluated noting that it is proportional to the polylogarithm [itex]\mbox{Li}_{4}[/itex] evaulated at 1, which is equivalent to the Reimann zeta function evaluated at 4. the proportionality factor is 3! in this case. (The factor of beta^(-4) came from defining x = beta*k to de-dimensionalize the integral). Evidently Boltzmann's constant is set to 1 in this book.

    Now for the second order terms:

    [tex]-\frac{m_\pi}{2}\int_0^{\infty}dk~k\left[\frac{e^{\beta k}\beta k}{\left(e^{\beta k} -1\right)^2} + \frac{1}{e^{\beta k} - 1} \right][/tex]

    The first term came from expanding the denominator of the ugly part above. The second was from the second order term of the numerator. Cleaning up the integral a bit gives the total second order contribution as

    [tex]-\frac{\nu \Omega \beta T^4}{12 \pi^2}\frac{m_\pi^2}{T^2}\int_0^{\infty}dx~x\left[\frac{xe^x + e^x - 1}{\left(e^x-1\right)^2}\right][/tex]

    Now, there's probably some clever way to evaluate the integral there in terms of polylogarithms. I could probably do it with some thinking, but I think for now I won't. Wolfram seems to be able to do it, but I can't evaluate the definite integral online, so instead I just evaluated it numerical, from x = 0.000000000001 (or so) to 49, which gives me 4.9348... . Dividing this by 12 \pi^2 gives a number pretty close to 1/24.

    Hence, there's the second term of the series expansion. You can keep doing this game to generate higher order terms, though each one is probably going to give you more and more annoying-to-evaluate integrals.

    So, to second order,

    [tex]\ln Z_\pi \simeq (\nu \Omega \beta)T^4 \left( \frac{\pi^2}{90} - \frac{m_\pi^2}{24T^2} + \cdots~\right)[/tex]
    Last edited: Nov 30, 2008
  4. Dec 1, 2008 #3
    Thank you very much, that was a great help !!! :smile:

    But I still have one question: how did you obtain the result for second order terms:

    [tex] -\frac{m_\pi}{2}\int_0^{\infty}dk~k\left[\frac{e^{\beta k}\beta k}{\left(e^{\beta k} -1\right)^2} + \frac{1}{e^{\beta k} - 1} \right] [/tex] ?

    Splitting the numerator (as you mentioned it) into two terms, I obtained (in [tex] \mathcal{O}(m_{\pi}^{2}) [/tex]):

    \dfrac{1}{e^{\beta k} \left( 1+ \beta k \dfrac{1}{2} \left(\dfrac{m_{\pi}}{k}\right)^2} \right) - 1} - \dfrac{\dfrac{1}{2} \left(\dfrac{m_{\pi}}{k}\right)^{2}}{e^{\beta k} \left( 1+ \beta k \dfrac{1}{2} \left(\dfrac{m_{\pi}}{k}\right)^2} \right) - 1}

    But now I'm stuck at this point. How exactly did you expand this? I used a calculator and I didn't obtain the same result (just for the 0. order it is the same one)
  5. Dec 1, 2008 #4


    User Avatar
    Homework Helper

    In the first term, I multiplied the exponential through to get

    [tex]}e^{\beta k} + e^{\beta k} \beta k \dfrac{1}{2} \left(\dfrac{m_{\pi}}{k}\right)^2} \right) - 1[/tex]

    I then moved the -1 over next to the lone exponential and factored out the [itex]e^{\beta k} - 1}[/itex] term to get

    [tex]\left(e^{\beta k}-1\right) \left(1 + \frac{e^{\beta k} \beta k \dfrac{1}{2} \left(\dfrac{m_{\pi}}{k}\right)^2}{e^{\beta k} -1} \right)[/tex]

    The second factor can now be binomially expanded (since you have 1 over that). The term with the [itex](e^{\beta k} -1)^{-1}[/itex] is just the zeroth order term. The second term from that expansion is the first term in the second order integral I wrote in my last post.

    For the second term in the expression quoted at the top of this post, you could do the same expansion in the denominator as I just did, but that will generate terms of order higher than m_\pi^2, so to second order I just dropped those terms, which is effectively like setting m_\pi = 0 in the denominator of that term.

    Now, that's second order. Since I didn't try th expansion any further, third order mystifies me a little, as it's not clear to me that any third order terms will come out of this expansion (since it's in terms of m_\pi^2 s). It certainly bugs me a little. I suspect it may be a typo and meant to be 4th order, as there's no 1st order term. I may have to think about that some more.
    Last edited: Dec 1, 2008
  6. Dec 3, 2008 #5


    User Avatar
    Homework Helper

    Okay, I thought about it some more, and I've decided that I currently have very little clue how they get the terms beyond second order. I keep trying, and the integrals I have to evaluate to get the coefficients keep diverging, no matter how I try to do the expansion (and I've tried a few ways now). The 'cleanest' way I've thought of so far is to change variables to get the integral into the following form (s = \beta m_\pi):

    [tex]I(s) = \int_{s}^{\infty}du~\frac{(u^2-s^2)^{3/2}}{e^u - 1}[/tex]

    and then either: expanding the numerator using the bionomail theorem; approximating the lower limit as 0 for the zeroth and second order terms seems to recover what we have above; at fourth order I get a divergent integral doing that. The other method is to expand I(s) in a power series about s = 0; however, the expansion once again hits a brick wall at third and fourth order, where the coefficients seem to diverge.

    So, for now, I'm afraid I'm simply at a loss as to how to get the higher order terms. =/
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook