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Strong and weak interactions in nuclear physics

by rayman123
Tags: interactions, nuclear, physics, strong, weak
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rayman123
#1
Jan4-11, 03:31 AM
P: 152
1. The problem statement, all variables and given/known data

How do we determine whether we have a strong or a weak interaction? (for the following processes)

For example we have a reaction
[tex] K^{-}+p\Rightarrow \Xi^{-}+K^{+}[/tex]

or another example
[tex]K^{+}\Rightarrow \pi^{+}+\pi^{0}[/tex]

thanks!


2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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vela
#2
Jan4-11, 04:53 AM
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PF Gold
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Start by looking up the quark content of the individual particles.
rayman123
#3
Jan4-11, 07:38 AM
P: 152
for the first one is
[tex] u\overline{s}+uud\Rightarrow dss+u\overline{s}[/tex]
but I still do not know how from this I can determine the type of interaction

vela
#4
Jan4-11, 04:21 PM
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PF Gold
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Strong and weak interactions in nuclear physics

That can't be right. You have the same quarks for K+ and K-.
rayman123
#5
Jan5-11, 02:59 AM
P: 152
according to my physics handbook Carl Nordling both these kaons have the same content (the same quarks).....maybe it is an error??? Or maybe it follows some kind of a notation system?
From particle data group I found for [tex] K^{+} (u \overline{s})[/tex] and for [tex] K^{-} (\overline{u}s)[/tex]
vela
#6
Jan5-11, 03:36 AM
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PF Gold
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They're a particle-antiparticle pair. Your handbook probably just lists what the quarks are for the particle. Just swap quark for antiquark and vice versa to figure out what the antiparticle is. So now you have

[tex]\overline{u}s+uud \Rightarrow dss+u\overline{s}[/tex]

You just have to check if the process is consistent with the various interactions. For example, strong interactions will not change quark numbers. If quark numbers aren't conserved in a process, you can rule out the strong interaction. Another approach is to draw the Feynman diagram for the process. You just need to know what vertices are allowed.

A few shortcuts you can take:

1. A photon only interacts electromagnetically, so if a photon is present, the process is electromagnetic.
2. Similarly, the neutrino only interacts via the weak force, so if a neutrino is present, it's a weak interaction.
3. Leptons don't carry color charge, so if a lepton is involved, it can't be a strong interaction.
rayman123
#7
Jan11-11, 02:02 PM
P: 152
I have checked and strangeness is conserved, baryon nr is conserved, charge nr is conserved, isospin is conserved the only thing which is not conserved is the projection of the isospin. Does this rule the strong interaction out and we are left with the possible weak interaction?
vela
#8
Jan11-11, 02:30 PM
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PF Gold
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It looks like I3 is conserved to me. Why did you decide it wasn't?
rayman123
#9
Jan11-11, 02:38 PM
P: 152
hm maybe I made mistake...Yes you are right, I got on the left hand side of the reaction: -1/2,+1/2
on the left hand side: -1/2, +1/2
It is just my physics handbook which is a bit confusing....but I think I figured out how the system works now
so I guess we can write that because there is not violation of any of above numbers and the interaction is strong
rayman123
#10
Jan11-11, 02:39 PM
P: 152
I have another question, can [tex] I_{3}[/tex] be violated in case of the weak interaction?
vela
#11
Jan11-11, 02:56 PM
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Yeah, it looks like a strong interaction to me.
vela
#12
Jan11-11, 02:57 PM
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PF Gold
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Quote Quote by rayman123 View Post
I have another question, can [tex] I_{3}[/tex] be violated in case of the weak interaction?
Yes, the weak interaction can change quark flavors, so isospin isn't always conserved.


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