Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why does local gauge invariance suggest renormalizability?

+ 6 like - 0 dislike
1586 views

I'm reading Gauge Field Theories: An Introduction with Applications by Mike Guidry and this particular remark is not obvious to me:

A tempting avenue is suggested by the QED paradigm, for if a local gauge invariance could be imposed on the weak interaction phenomenology we might expect the resulting theory to be renormalizable. [Guidry, section §6.5, p. 232]

Is there an obvious argument for this "local gauge invariance suggests renormalizability" remark? I should add that I still tend to get lost in the streets of renormalization when unsupervised, i.e. I'm not familiar enough with the entire concept to have any real intuition about it. (references on renormalizability that might help are of course also welcome)

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user Wouter
asked Apr 28, 2014 in Theoretical Physics by Wouter (30 points) [ no revision ]
retagged May 4, 2014
I suspect that even though its badly phrased. They just mean that they are working with a large cutoff, as in the SM, and so the nonrenormalizable terms are small

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user JeffDror
I'd agree with Jeff's assessment, it seems badly phrased. A priori, a gauge symmetry by no means indicates renormalizability.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user JamalS

1 Answer

+ 0 like - 0 dislike

This statement is related to the fact that renormalizability of a theory depends of the mass dimension of the coupling constants in the Lagrangian. Couplings with zero or positive mass dimensions lead to renormalizable theories. As a consequence, writing down only terms with appropriate mass dimensions is required in order to construct a theory that is renormalizable.

In quantum electrodynamics, all operators consistent with both (local) gauge and Poincaré symmetry that are at most of mass dimension 4 automatically satisfy the above criterion. One could understand the statement in the reference in this way. Of course, this does not hold for term of higher dimension.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user Frederic Brünner
answered Apr 28, 2014 by Frederic Brünner (1,130 points) [ no revision ]
Imposing only gauge invariance and Poincare symmetry we can still have other nonrenormalizable terms such as $\bar{\psi}\psi \bar{\psi}\psi$,$\bar{\psi}\psi F_{\mu\nu} F^{\mu\nu}$, etc.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user JeffDror
@JeffDror: I have edited my answer.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user Frederic Brünner

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...