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
5,354 answers , 22,792 comments
1,470 users with positive rep
820 active unimported users
More ...

  Decay width average in the isospin invariant limit

+ 3 like - 0 dislike
1297 views

Suppose we have the following experimental values for $\eta' \rightarrow \eta \pi \pi$ decay width:

$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$

$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0430 \pm 0.0022$

We are investigating this decay in the isospin invariant limit and want to compare our results with experimental rates. So we should average over these 2 values to find the experimental decay width in this limit. It is written in one paper that we should average

$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$

2$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0860 \pm 0.0044$

in a specific way (which is not my question). I don't know why we should twice the second value and then average?

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user soodeh
asked Apr 29, 2014 in Theoretical Physics by soodeh (15 points) [ no revision ]
Could you please provide a reference regarding the paper you mention?

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user Frederic Brünner
Yes, of course. Phys. Rev. D60, 034002 (page 3)

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user soodeh

1 Answer

+ 2 like - 0 dislike

The average is taken this way because their theoretical prediction does not distinguish between $\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-}$ and $2\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0}.$ See for example equation 2.1. The experimental value, however, is different for both decay channels. Therefore, in order to make a meaningful comparison, they choose to compare to the average of both values. Since the theoretical equivalence include the factor of $2$, the average should also include it.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user Frederic Brünner
answered Apr 29, 2014 by Frederic Brünner (1,130 points) [ no revision ]
thank you so much.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user soodeh

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$\hbar$ysicsOv$\varnothing$rflow
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
...