Which force causes decay in resonance states?

  • Thread starter Thread starter Hypersquare
  • Start date Start date
  • Tags Tags
    Decay Force
Hypersquare
Messages
14
Reaction score
0
Hey now!

I have just calculated the lifetime of a resonance state using its total width. I want to work out which force this decay is due to. How would one go about doing this?

Thanks
 
Physics news on Phys.org
To know the interaction responsible for a particular decay you have to know what the decay is, i.e. what particle(s) the particle you're looking decays into. One particle can decay in many different ways. However, since you use the word "resonance", this might be a hint that the decay is due to the strong interaction, since this is usually part of what one means by the word "resonance".

What is the specific particle and decay you're looking at?
 
Thanks fo the response Kloptok!

No such information is given. I am just told that the total width of delta 1232 resonance i 118 Mev.
 
http://pdglive.lbl.gov/Rsummary.brl?nodein=B033&inscript=Y&sub=Yr&return=BXXX010

The dominant decay is (uud) -> (udd) + (u anti-d)
No quark flavour changes are involved, therefore it can occur via the strong interaction. The broad width of the resonance is another sign of this.

Another observed decay is delta1232 -> p gamma. As you can see from the photon, the electromagnetic interaction is involved.
 
Last edited by a moderator:
You can find ideas here.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
9
Views
193
Replies
1
Views
1K
Replies
11
Views
3K
Replies
2
Views
2K
Replies
1
Views
1K
Back
Top