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Expansions of Bose Functions

  1. Mar 4, 2016 #1
    1. The problem statement, all variables and given/known data
    To study the thermodynamic behavior of the limit $$z\rightarrow1$$ it is useful to get the expansions of $$g_{0}\left( z\right),g_{1}\left( z\right),g_{2}\left( z\right)$$

    $$\alpha =-\ln z$$ which is small positive number. From, BE integral,
    $$g_{1}\left( \alpha \right) =-ln\left( 1-z\right) =-ln\alpha+\dfrac {\alpha } {2}-\dfrac {\alpha ^{2}} {24}+O({\alpha ^{4}})$$
    and hence

    $$g_{0}\left( \alpha \right) =-\dfrac {\partial } {\partial \alpha }g_{1}\left( \alpha \right)=\dfrac {1} {\alpha }-\dfrac {1} {2}+\dfrac {\alpha } {12}O({\alpha ^{3}})$$

    [Source: A.Khare, Fractional Statistics and Quantum Theory, Two Dimensional Bose Gas, p.113]

    Could anyone help me to derive this expressions? I can't figure out what does it mean writing this functions in the powers of α.

    2. Relevant equations
    https://en.wikipedia.org/wiki/Polylogarithm (BE integral)

    3. The attempt at a solution

     
  2. jcsd
  3. Mar 9, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 9, 2016 #3

    Charles Link

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    Homework Helper

    Suggestion: Write ## z=exp(-\alpha) ## and you get a term ## ln(exp(\alpha)-1) ## plus another term ## ln(exp(\alpha))=\alpha ## In the first term (inside the ## ln ## ) expand ## exp(\alpha) =1+\alpha+(\alpha)^2/2 +... ## and subtract the 1. Then factor out ## \alpha ## and you get ## ln(\alpha)+ln(1+\alpha/2+..) ## The expansion of ## ln(1+u)=u ## for small ## u ##.
     
  5. Mar 14, 2016 #4
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