Yukawa Interaction: What Is It & Why Not Part of Standard Model?

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Discussion Overview

The discussion centers around the Yukawa interaction, its definition, and its absence from the Standard Model of particle physics. Participants explore the theoretical implications and historical context of the Yukawa potential, particularly in relation to the strong force.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant inquires about the Yukawa interaction and its exclusion from the Standard Model, suggesting a lack of clarity on the topic.
  • Another participant references the Fourier transform of the Coulomb potential, proposing a mathematical relationship to the Yukawa potential.
  • There is a question about the meaning of the parameter α and confirmation that r represents radius.
  • A participant notes that the Yukawa potential was historically used to model the strong force, highlighting its exponentially decaying term and its implications for force behavior at short distances.
  • It is mentioned that the current understanding of the strong force involves gluons, suggesting that the Yukawa potential may not fully capture the nature of the strong force.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement on the Yukawa interaction and its relevance to modern physics. Some acknowledge its historical significance while others suggest that the current framework provides a more accurate description of the strong force. The discussion remains unresolved regarding the full implications of the Yukawa interaction.

Contextual Notes

Participants express uncertainty about specific parameters and their roles in the Yukawa potential, indicating a need for further clarification on definitions and mathematical relationships.

AleksanderPhy
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Hello There I may have some mistakes so correct me when I wrote something wrong.My question is:what's the yukawa's interaction and why it is not part of standard model?
 
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I remembered having seen that once it was to compute the Fourier transform of the coulomb potential it is like $$V(r)=\frac{e^{-ar}}{r}$$ then you put a to zero after the transform.
 
Thank you very much(;But what is α
PS! Am I correct that r is radius?
 
AleksanderPhy said:
Thank you very much(;But what is α
PS! Am I correct that r is radius?
Right.
a is just some adjustable parameter. There are other threads on the topic too maybe worth to look at them.
 
Thank you so much it helped me lot.
 
The Yukawa potential was used early on to model the strong force I believe. The exponentially decaying term gave rise to an exponential weakening of the force (strong force is short distance). But we have a better picture now of the strong force (what with the gluons and stuff). The Yukawa potential itself is just a potential with some functional force, whether nature conforms to this picture is a separate issue.
 
Thx same words:biggrin:
 

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