Homework Help: Strong, weak, electromagnetic interactions

1. Nov 17, 2007

genloz

1. The problem statement, all variables and given/known data
I had a whole list of reactions and had to show whether or not they could occur... I understood most of the reasons, bar the following, but was more confused about whether or not a reaction was strong or weak... I know that if the reaction involves a lepton its weak, and if it involves a photo, it's electromag, but baryons alone don't define whether a reaction is strong or weak:
For example:

$$\Sigma^{-} \rightarrow n + \pi^{-}$$
I understand strangeness is not conserved, but why is it a weak interaction if it only involves hadrons?

$$e^{+} + e^{-} \rightarrow \mu^{+} + \mu^{-}$$
Why is this an electromag reaction when only leptons are involved and charge is conserved on both sides and no photons are involved? Shouldn't it be a weak?

$$K^{-} \rightarrow \pi^{-} + \pi^{0}$$
This only involves hadrons and strangeness isn't conserved, so should this be an impossible strong reaction?
I thought weak decays only didn't conserve partity, but the answers say this is a possible weak decay, does that mean weak decays don't need to conserve strangeness either? Also why is it weak rather than strong?

$$\pi^{-} + p \rightarrow \Lambda^{-} + \K^{0}$$
I thought this would be prevented due to lack of strangeness conservation, but it's allowed via the strong interaction, why?

Thanks very much!!

2. Nov 17, 2007

malawi_glenn

$$\Sigma^{-} \rightarrow n + \pi^{-}$$

Have you checked isospin and 3-component of isospin? It can be a weak, if it is done by a W boson. The general rule you are referring to that if neutrinos are emitted, then it is weak, dont go the reverse way, it is not equivalence. All neutrinos comes from weak, but not all weak produces neutrinos. So you must give up this thinking, same holds for the EM interaction.

$$e^{+} + e^{-} \rightarrow \mu^{+} + \mu^{-}$$
can proceed both by virtual photon and Z-boson, i.e it can do both weak and EM, but weak is ... weak = unlikley, so the observed muons comes from EM interaction in like 99.99999999999999999999999999999999999% of the cases.

$$K^{-} \rightarrow \pi^{-} + \pi^{0}$$
Strangeness is not conserved right? So it is a weak.
Also: Check isopspin and 3-component of isospin.

$$\pi^{-} + p \rightarrow \Lambda^{-} + K^{0}$$

I have never heard of Lambda minus baryon.. you must mean lamda zero (charge is not concerved otherwise)
Lamda-0 has one strange quark, and K-0 has one anti-strange quark, so total strangeness is conserved.
Please check if you wrote this reaction correct before proceeding altough.

Last edited: Nov 17, 2007
3. Nov 17, 2007

genloz

pi- I:1 I3:1 Tz: ??
pi0 I:1 I3:0 Tz: ??
n I:1/2 I3:1/2 Tz:uud 1/2-1/2-1/2:-1/2
K- ubar s I: 1/2+0:1/2 I3:-1/2+0: -1/2 Tz: ??
Sigma- dds I:1/2+1/2+0:1 I3:-1/2+-1/2+0:-1 Tz: -1/2-1/2-1/2
Tz: u c t neutrinos 1/2
Tz: d s b charged leptons -1/2

So
[tex]\Sigma^{-} \rightarrow n + \pi^{0}{/tex]
I: 1 -> 1/2 + 1 not conserved
I3: -1 -> 1/2 + 0 not conserved
Tz: not sure

[tex]K^{-} \rightarrow \pi^{-} + \pi^{0}{/tex]
I: 1/2 -> 1 + 1 not conserved
I3: -1/2 -> 1 + 0 not conserved
Tz: not sure

Whcih still doesn't make sense...

4. Nov 17, 2007

malawi_glenn

You dont know what isospin and third component of isospin is? Look it up. i cant help you with that.

I dont understand what you have written... just a mess, neutrinos dont have isospin etc..

Aslo check the thing i wrote to you regarding last reaction with that lamda minus of yours.