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IR and UV divergences

  1. Apr 10, 2009 #1
    perhaps is a dumb quetion but,

    given a IR divergent integral (diverges whenever x tends to 0)

    [tex] \int_{0}^{\infty} \frac{dx}{x^{3}} [/tex]

    then using a simple change of variables x=1/u the IR integral becomes an UV divergent integral

    [tex] \int_{0}^{\infty} udu [/tex] which is an UV divergent integral (it diverges whenever x tends to infinity)

    then why we call IR or UV divergences if they are essentially the same thing ??
  2. jcsd
  3. Apr 11, 2009 #2


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    They are not the same thing. An IR divergence is one that arises from integrals over low momentum or energy, and a UV divergence is one that arises from integrals over high momentum or energy.
  4. Apr 11, 2009 #3
    yes of course, but from the mathematical point of view a change of variable would turn an IR divergence into a UV one, for a mathematician both functions or divergences would be the same since from the cut-off we can define a function [tex] \epsilon = 1/ \Lambda [/tex]

    with this epsilon tending to 0
  5. Apr 11, 2009 #4


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    But the physical variable is not the same, that is why we call them different
  6. Apr 11, 2009 #5
    of course is not the same taking the integral

    [tex] \int_{0}^{\infty} d\lambda f( \lambda ) [/tex]

    or taking the integral [tex] \int_{0}^{\infty} dp f(p) [/tex]

    in the first integral we integrate over the wavelength (meters) whereas in the second we integrate over moment (kg.m/second) but for a mathematician both singularities would seem the same.
  7. Apr 11, 2009 #6

    Vanadium 50

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    But it's physicists that are using these terms, and they do distinguish between divergent in E and in 1/E.
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