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Cuts of a Feynman diagram and the massless limit

  1. Sep 6, 2015 #1


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    Consider a ##j## point all massive leg one loop polygonal Feynman diagram ##P## representing some scattering process cut on a particular mass channel ##s_i##. Invoking the relevant Feynman rules and proceeding with the integration via dimensional regularisation for example gives me an expression for ##\text{Cut}_{s_i}P\left(p_1^2, \dots, p_j^2, \epsilon\right)##.

    I am wondering if someone could explain why it is or is not possible to send the mass corresponding to the channel that was cut to zero if I wanted to consider a massless limit?

    Many thanks!
  2. jcsd
  3. Sep 11, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Sep 16, 2015 #3
    Looks like you are talking infrared divergence. If you sent all the masses to zero, the amplitude is divergent as energy goes to zero, thus you have to apply a cutoff. The argument for QED is, by considering there is measuring energy threshold for any apparatus, the final result would not depend on any cutoff but the apparatus threshold.
  5. Oct 30, 2015 #4


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    Sorry for the late reply and for just coming back to this thread now. I have yet to take a proper course in the path integral formulism in QFT, so what I say may be a bit incorrect. As far as I understand it, dimensional regularisation is a regularisation procedure in which there is no need to introduce a cut off scale?

    If we consider a 4 propagator cut of the massive box, then taking four prop cuts reduces its weight to zero and thus its expansion in the regulator starts off as const + A##\epsilon##. I am not sure if this is related to the fact that the 4 prop cut of the all massless box can be obtained by simply sending all masses to zero in the 4 cut of the massive box? At least my prof gave me the impression you could, but I am still understanding why.

    This thread is a month old and Rainbowend only has two posts so might be away. If anyone else has anything to say, please do, thanks :)
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