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MSbar Renormalisation + Gamma Functions

  1. Oct 28, 2011 #1
    Hi all,

    I am currently reading and calculating stuff on Next-To-Leading-Order calculations and I came across a paper that I want to calculate by myself and try to get the same results.

    There is one part, though, where I have absolutely no idea how this works. It comes in the context of renormalisation which is a topic I know something about but in which there remain still a lot of mysteries for me:

    In this paper, the author calculated the NLO Vertex function, which I evaluated also:

    [itex]\tilde{\Gamma} = \frac{\alpha}{4\pi} C_F \left(\frac{-q^2}{4\pi \mu^2}\right)^{\frac{\epsilon}{2}} \frac{\Gamma^2(1+\frac{\epsilon}{2})\Gamma(1-\frac{\epsilon}{2})}{\Gamma(2+\epsilon)} \left[-\frac{2}{\epsilon^2}(2+\epsilon)^2\right][/itex]

    Then comes the text
    "The renormalisation constant Zlambda is defined through the relation
    [itex]Z_\lambda^{-1} =1 + \tilde{\Gamma_{UV}}[/itex]
    and of course depends on the way the ultraviolet divergent part GammaUV is isolated. Within the MSbar scheme, it can easily be ascertained to be

    [itex]Z_\lambda = 1+\frac{\alpha}{4 \pi \Gamma(1+\frac{\epsilon}{2})}C_F \left(\frac{1}{4\pi}\right)^{\frac{\epsilon}{2}} \frac{8}{\epsilon}[/itex]


    I have no clue how this solution can be derived. I know that you can do Zlambda = 1/(1+Gamma) = 1 - Gamma to order alpha but thats as far as I get. I know how to expand the Gamma function in terms of epsilon and gamma,but the easy-peasy way I learnt it in Peskin SChroeder only has one Gamma function with one epsilon and one gamma and one 4pi part, where the MSbar scheme explicitely says how to absorb this. But now I have four Gamma functions, which are all basically convergent, and I end up with only one Gamma function left (I would have assumed that I expand all of them) and this denominator Gamma function somehow is different from the denominator Gamma function you start with.

    Can anyone give me a hint how you would actually start rewriting the gamma functions and how you in fact would then apply the msbar scheme formally to get the UV part?
     
  2. jcsd
  3. Oct 31, 2011 #2

    Avodyne

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    Science Advisor

    The denominator gamma function (and the power of 4pi) are only there to convert from MS to MS-bar. To leading order in e=epsilon, Gamma(1+e/2)=exp(-g e/2), where g is Euler's constant.
     
  4. Nov 1, 2011 #3
    Ah I see. I never heard about this method of shifting between these two schemes.

    I think then I understand how the rest works- Thank you!
     
    Last edited: Nov 1, 2011
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