Why we do not consider the divergence due to mass-shell in QED?

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Discussion Overview

The discussion revolves around the treatment of divergences in Quantum Electrodynamics (QED), specifically addressing why divergences due to mass-shell conditions are not considered in the same way as infrared (IR) and ultraviolet (UV) divergences. Participants explore the implications of mass terms and their regulation in loop calculations.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant asks why divergences in loops when \( p^{2} \to m^{2} \) are not considered, focusing instead on soft photon divergences as \( k^{2} \to 0 \) (IR divergence) and UV divergences.
  • Another participant suggests that \( p^{2} - m^{2} \) is regulated using the "iε-Term" and evaluated through Cauchy contour integrals.
  • A question is raised regarding the treatment of divergences when the mass \( m = 0 \), implying a different consideration in that case.
  • A later reply indicates that applying Cauchy's theorem when \( m = 0 \) leads to IR divergence, suggesting a connection between mass and divergence treatment.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of divergences related to mass-shell conditions versus IR divergences, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants reference specific mathematical techniques such as the "iε-Term" and Cauchy's theorem, but the discussion does not resolve the implications of these methods on the treatment of divergences.

ndung200790
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Please teach me this:
Why we do not consider the divergences in loops in QED when p^{2}→m^{2} but only consider the soft photon when k^{2}→0(IR divergence) and UV divergence?
Thank you very much in advance.
 
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p²-m² is always regulated via the "iε-Term" and is evaluated via the Cauchy contour integral
 
Why we have not the same consideration when m=0?
 
Sorry,Now I have understood.Applying Cauchy's theorem when m=0(so k^{2}→0)
we have divergence(IR divergence)
 

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