Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
The meaning of the "physical" electron charge in Peskin (Chap 7)
Reply to thread
Message
[QUOTE="niss, post: 6331100, member: 674727"] [B]Homework Statement:[/B] Why is the "physical" electron charge e_0 / (1-Pi(0)) and not e_0 / (1-Pi(q^2)) [B]Relevant Equations:[/B] (physical charge) = e = sqrt(Z_3) e_0 = sqrt(Z_3) (bare charge) On p. 246 in the Peskin QFT textbook, below is stated where Z[SUB]3[/SUB] is defined as the residue of the q[SUP]2[/SUP] = 0 pole, explicitly as $$Z_3=\frac{1}{1-\Pi(0)}$$ and e is the bare charge. In advance, the exact photon two point function is calculated as $$\frac{-ig_{\mu\nu}}{q^2(1-\Pi(q^2))}$$ Though the photon propagator should be divided by $$1-\Pi(q^2)$$ , why the defined "physical" charge is divided by $$1-\Pi(0)$$ ? Why only the low-q[SUP]2[/SUP] scattering is being considered here? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
The meaning of the "physical" electron charge in Peskin (Chap 7)
Back
Top