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I Shift of momenta cures IR divergence?

  1. Sep 26, 2017 #1

    CAF123

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    Consider the following integral $$\int \frac{d^4k}{k^2}$$ It is UV divergent but is it IR finite or IR divergent? The integrand is singular as ##k \rightarrow 0## so this suggest an IR divergence but this is no longer the case if I make a shift of the loop momenta by say ##p_1## and write the same integral as $$\int \frac{d^4k}{(k+p_1)^2}$$

    Usually we say an IR divergence can be cured by addition of a mass in the denominator (as then the integrand won't be singular as k goes to 0) but isn't the IR divergence also cured by simply making a lorentz transformation on the momenta (assuming ##p_1^2 \neq 0##) ? I don't understand this result so where is the failure in the reasoning?
     
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  3. Sep 26, 2017 #2

    mfb

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    Don't you get a divergence at k=-p1 now?
     
  4. Sep 26, 2017 #3

    Orodruin

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    Yes.
     
  5. Sep 26, 2017 #4

    CAF123

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    What if ##p_1## is large numerically? Then the divergence at ##k=-p_1## is not in the IR?

    With a mass M, e.g in the integral $$\int \frac{d^4k}{k^2-M^2},$$ we have a divergence at ##k=\pm M##. The mass is said to cure the IR divergence (because integrand no longer singular as ##k\rightarrow 0##) but there is still this ##k=\pm M## divergence in place that is cured by the Feynman prescription.

    So if ##k=\pm M## is not an IR divergence then why is ##k=-p_1## one?
     
  6. Sep 26, 2017 #5

    mfb

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    It is still there, and renaming it won't help.
    Substituting k -> k-p1 doesn't change the integral at all.
    That is a different type of divergence (note the difference of squares instead of a squared difference), and there are ways to deal with it.
     
  7. Sep 26, 2017 #6

    CAF123

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    Ok,so if I took the IR limit (k->0) in the integral ##\int d^4k/(k+p_1)^2## then I lose all k dependence from the integrand and so the result is like ##\int d^4k## which is infinite. Is it in this sense that the IR divergence is still clear?

    Are these 'ways' you mention the wick rotation to Euclidean 4 momentum etc?
     
  8. Sep 29, 2017 #7

    vanhees71

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    This integral is IR save since ##\mathrm{d}^4 k \propto |k|^3## (taking it as a Euclidean/Wick rotated QFT integral).
     
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