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Are not the IR and UV divergences the same (mathematically)

  1. Aug 13, 2009 #1
    question is why speak about IR (short momentum) and UV (short distances) divergences ?

    in fact if we define [tex] \epsilon = 1/\Lambda [/tex]

    then both integrals

    [tex] \int_{\epsilon}^{\infty}x^{-k}dx [/tex] and the [tex] \int_{0}^{\Lambda}x^{k-2}dx [/tex]

    have the same rate of divergence [tex] \Lambda ^{k-1} [/tex] as the regulator 'Lambda' goes to infinity. (simply make a change of variable x=1/t )

    then if mathematically is the same to get rid off an UV or an IR divergence , and with a simple change of variable you can turn an IR divergence into an UV one then why make distinction (the logarithmic case is just another question)
  2. jcsd
  3. Aug 13, 2009 #2


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    they have VERY different physical meanings, and tell you much about whether or not your theory makes any sense. You should not mix them up!

    Many errors in the literature have come up by people blindly canceling IR divergences with counterterms, and thus getting absolute nonsense!
  4. Aug 13, 2009 #3
    But when using Dimensional regularization or analytic regularization, you do not make distinction between IR or UV don't you ??

    By the way i have some problems understanding what ' counterterms' are and how they work

    are you andrew Blechmann the one who wrote 'renormalization our great misunderstood friend' ?? .. i liked this paper much.
  5. Aug 13, 2009 #4
    This is also a thing I didn't understand well.
    You use dim reg both for UV and IR divergences, but to regulate the first you need to continuate analytically in d < 4 dimensions, while to regulate the others you need d > 4, in the same time! Or not? how do you deal with this?
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