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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

What is the g-factor of the W boson?

+ 3 like - 0 dislike
153 views

Is there any experimental data available about the g-factor of the W boson?

Is there any theoretical prediction about its g-factor?

asked Mar 25 in Experimental Physics by Fred [ no revision ]

2 Answers

+ 4 like - 0 dislike

According to the Lagrangian of the Standard Model, the g-factor of the W-boson is given at tree level by $g=2$.

Experimental results are compatible with this result, see eg p4 of https://arxiv.org/abs/hep-ex/0407042 for results from LEP.

There are radiative corrections to the tree level value which can be computed in the Standard Model: $g=2+\dots$, but, as far as I understand, the precision of the experiments is still insufficient to be sensible to these corrections.

answered Mar 25 by 40227 (4,660 points) [ revision history ]
+ 0 like - 0 dislike

The g-factor is not a proper particle property, but depends on interactions involved and the "state of particle", in which it is calculated. In other words, the radiative corrections are state- (or process-) dependent.

answered Mar 26 by Vladimir Kalitvianski (22 points) [ no revision ]

The reference interactions are of course those of the electroweak QFT.

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$\hbar$ys$\varnothing$csOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...