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Dimensional regularization

  1. Jul 21, 2010 #1
    I was looking at a paper that used dimensional regularization and the following expression was derived:

    [tex]\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon} [/tex]

    Factoring out [tex]p^2(1-x)^2 [/tex]:

    [tex]\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon} [/tex]

    The part that I don't understand is that they expanded the rightmost factor in binomial expansion. [tex]\lambda^2[/tex] is smaller than [tex]p^2 [/tex] (in fact [tex]\lambda^2=p^2-m^2 [/tex]), but the 1/(1-x) changes all that when x approaches 1, making [tex]\frac{\lambda^2}{p^2(1-x)} [/tex] much greater than 1.

    Is it okay to expand the rightmost factor in binomial expansion because when x goes to 1, the factor on the left [tex][p^2(1-x)^2]^{\epsilon} [/tex] goes to zero, so it doesn't matter what's on the rightmost side at that point? If so, isn't there an error term that needs to be calculated?
  2. jcsd
  3. Jul 21, 2010 #2


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    Could be. Could you give us some more context? Do you have the citation for the paper? Cheers,

  4. Jul 21, 2010 #3
    There's a lot of pages, so I'll just link to a link of it:

    http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+DONOGHUE+AND+HOLSTEIN+and+robinett&FORMAT=www&SEQUENCE= [Broken]

    It's the second paper in the list, and you can download a copy by clicking on Scanned Version (KEK).

    The part I'm referring to is on page 49, equation (A1).

    It's an older paper, so they probably do things a little differently, but still, the result should be the same as with today's techniques, so I'm quite surprised: I haven't seen any of these techniques tried before in QFT books on QED.

    *actually, it's page 50 if you click on the scanned images (Tiff and Gif). On the actual paper, it is page 49.
    Last edited by a moderator: May 4, 2017
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