Breit Wigner Curve: Finding Reaction Rate from FWHM

  • Thread starter Thread starter Robsta
  • Start date Start date
  • Tags Tags
    Curve Wigner
Click For Summary

Homework Help Overview

The discussion revolves around the interpretation of the Breit Wigner curve in relation to reaction rates and the implications of its width (FWHM) on the understanding of decay processes. Participants are exploring the connection between energy measurements and reaction rates, particularly in the context of particle physics.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants are questioning how the width of the Breit Wigner curve relates to the reaction rate and what the implications of the uncertainty principle are for determining actual lifetimes versus uncertainties. There is also an exploration of symmetry arguments related to particle decay and conservation laws.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the interpretation of the FWHM in relation to decay rates. Some guidance has been offered regarding the relationship between energy uncertainty and lifetime, but there is no consensus on the interpretation of these concepts.

Contextual Notes

Participants are navigating the complexities of decay processes and the implications of conservation laws in particle physics, with specific reference to the neutral rho meson and its decay characteristics. There are indications of missing information regarding the numerical factors that may influence the relationship between energy width and lifetime.

Robsta
Messages
88
Reaction score
0

Homework Statement


I've got a given Breit Wigner curve of the number of decays at given energies.

I've been told by several sources that the width (FWHM) of the curve gives the rate of the reaction.

I can't see however how an energy here actually translates into a rate.

Homework Equations



I understand that from the uncertainty principle $$ \Delta E \Delta t = \hbar $$ and therefore $$ \frac{\Delta E}{\hbar} = \Delta t $$ but to what does delta t refer? It seems to me like an uncertainty in the lifetime which is of course not what I'm after.

How would I find the actual lifetime rather than the error in the lifetime?
 
Last edited:
Physics news on Phys.org
Attached is a picture of the relevant curves in case it adds anything. Although I've labelled the width Gamma I am aware that it's a graph of energy of decay against frequency of decay.
 

Attachments

  • IMG_2302.JPG
    IMG_2302.JPG
    68.4 KB · Views: 909
I also have a question about the second part of the question visible in the image above. The question is "Use symmetry to explain why the neutral rho meson does not decay to two neutral pion".

My argument goes like this:

The parity of neutral rho is (-1) (since L=0). The parity of a neutral pi is also (-1). Therefore the parity of two neutral pions is therefore 1. So parity would not be conserved.
Also, the angular momentum of the neutral rho meson is 1 (L=0 S=1). And the angular momentum of the two neutral pions would be 0. So angular momentum would not be conserved.

Is the parity argument valid since parity is not necessarily conserved by the weak interaction? Is it for some reason safe to assume that it is conserved here?
 
Robsta said:

Homework Statement


I've got a given Breit Wigner curve of the number of decays at given energies.

I've been told by several sources that the width (FWHM) of the curve gives the rate of the reaction.

I can't see however how an energy here actually translates into a rate.

Homework Equations



I understand that from the uncertainty principle $$ \Delta E \Delta t = \hbar $$ and therefore $$ \frac{\Delta E}{\hbar} = \Delta t $$ but to what does delta t refer? It seems to me like an uncertainty in the lifetime which is of course not what I'm after.

How would I find the actual lifetime rather than the error in the lifetime?
The ##\Delta t## is basically the lifetime itself! (I say "basically" because there is some numerical factor).
 

Similar threads

Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K