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Breit Wigner Curve

  • Thread starter Robsta
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  • #1
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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?
 
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  • #2
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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.
 

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  • #3
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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?
 
  • #4
nrqed
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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).
 

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