How Does a New Lagrangian Term Affect the Fine Structure Constant?

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The discussion revolves around the implications of adding a new term to the Standard Model Lagrangian, specifically $$\frac{{\phi}F_{\mu v}F^{\mu v}}{F_{\phi}}$$, and its effect on the fine structure constant, ##\alpha##. Participants express confusion about the definitions and roles of variables like $$\phi$$, $$F_{\phi}$$, and $$A_{\phi}$$, with consensus that $$\phi$$ is indeed a field. The challenge lies in understanding how this new term influences calculations related to QED interactions and the fine structure constant. Additionally, the term is identified as non-renormalizable, complicating the analysis further. Overall, the conversation highlights the complexities of modifying the Lagrangian and its impact on fundamental constants.
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Homework Statement
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If we add to the Lagrangian of SM a term $$\frac{{\phi}F_{\mu v}F^{\mu v}}{F_{\phi}}$$

How does the fine structure constant ##\alpha## changes as $$\phi = A_\phi cos(m_{\phi} t)$$?

I am having some hard time to finding out where i should start. I remember see the strucutre constant arrising when we evaluate vertices on feynman diagrams for QED interactions... But i am too ignorant to know how to proceed..
 
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What is Fphi and Aphi?

Is phi a field?
 
malawi_glenn said:
What is Fphi and Aphi?

Is phi a field?
"What is Fphi and Aphi?" I assume they are just constants... But the question says nothing about it.
Yes, phi is a field.
 
LCSphysicist said:
"What is Fphi and Aphi?" I assume they are just constants... But the question says nothing about it.
Yes, phi is a field.
Where did you find the problem? This is a non renormalizeable term
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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