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
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  $\mathrm{d} \Omega_{CM}$ for a $1\rightarrow 2$ particle decay?

+ 1 like - 0 dislike
1629 views

The differential solid angle is described in e.g. Srednicki's QFT text but only for the case of scattering. Because in the case of scattering it's defined with respect to the incoming three-momentum ${\bf k_1}$ and it's scattered three-momentum ${\bf k_1'}$. This definition would not make much sense for a particle decay (I can't at least see how it would easily carry over).

What about the case of particle decay? What is $$\mathrm{d} \Omega_{CM}$$ in the case of particle decay? Any good references where it is properly defined?

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
asked Apr 13, 2014 in Theoretical Physics by Love Learning (165 points) [ no revision ]
It is a solid angle, but its definition depends on the particular case. How many particles you have in the final state? Check "Elementary Particles and Their Interactions", Quang Ho-Kim.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger
The title mentions 1->2 particle decay. Do you have any idea about the answer? Thanks for replying.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
Ok, sorry. See the answer and comment below. There is nothing special about this angle.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger

1 Answer

+ 0 like - 0 dislike

Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments:

$$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega |\mathcal{M}|^2=\dfrac{|\vec{p_1}|}{8 \pi M^2}|\mathcal{M}|^2,$$

where $\mathcal{M}$ is the amplitude of the process, $M$ is the mass of the initial particle and $|\vec{p_1}|=|\vec{p_2}|$ is the final 3-momentum of one of the particles.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger
answered Apr 13, 2014 by JSchwinger (0 points) [ no revision ]
Yes I know it's $4\pi$ in this case, but I mean, how is it defined. Because in the case of scattering it's defined with respect to the incoming three-momentum ${\bf k_1}$ and it's scattered three-momentum ${\bf k_1'}$. This definition would not make much sense for a particle decay (I can't at least see how it would carry over in any sense).

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
It is the angle of one of the particles with respect to a given axis. Since the amplitude does not depend on this angle, you have only $4\pi$. Think in terms of the general formula for the phase-space: $$\mathrm{d}\phi_f \propto \frac{\mathrm{d}^3 p_1}{2 E_1}\frac{\mathrm{d}^3 p_2}{2 E_2} \delta^4(p_1+p_2-p).$$ We integrate the momentum of one of the particles to get rid of the distribution $\delta^3$ and at the end we'll have only a 3D integral. To do this integral, you have to define an angle.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger

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$\hbar$ysics$\varnothing$verflow
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
...