# Homework Help: Quark Flow diagrams - when is it Strong/EM/Weak?

1. Dec 4, 2012

### Silversonic

1. The problem statement, all variables and given/known data

(1) Indicate whether the following process (if possible) occurs due to the strong, weak or electromagnetic interation.

$\pi^{-} p → \pi^{-} \pi^{+} n$

(3) Sketch Quark flow diagrams for the following, stating whether it's due to EM, weak or strong interaction;

$K^{*+} → K^{+} \pi^{0}$

The K*+ is an excited state of the K+, it has the same quark composition.

3. The attempt at a solution

I'm having a hard time understanding whether you can know the first question is a strong or weak interaction. My answer tells me that because it's dealing only with hadrons, it is therefore the strong interaction. However, the weak interaction (W+-) is able to couple to quark pairs (e.g. $u$ and $\bar{d}$). The emission of a W boson also changes the quark flavour.

So in the first interaction, wouldn't it be possible for the $\pi^{-}$ to just be an "observer" particle, while an up quark in the proton decays via the weak interaction in to a down quark (changing the proton in to a neutron), and thus this W+ boson subsequently couple to a $u$ and $\bar{d}$ pair creating the $\pi^{+}$ meson?

The second question also has me stumped. It shows the one of the quarks of the $K^{*+}$ meson decaying via the strong interaction, emitting a gluon which subsequently produces the $\pi^{0}$. However, why can't this process happen exactly the same way with electromagnetic interaction? One of the quarks of the $K^{*+}$ decays via electromagnetic interaction, emitting a virtual photon and this virtual photon subsequently couples to a quark, antiquark pair producing the $\pi^{0}$. I don't see any violation at all? So why must it be the strong force? Hell, the same could happen with the weak interaction using the Z boson.

Last edited: Dec 4, 2012
2. Dec 5, 2012

### Simon Bridge

For the second one - you want to know why the K* cannot just decay electromagnetically to produce a photon, which then somehow becomes a quark-antiquark pair?
How would a photon do that?

3. Dec 5, 2012

### Silversonic

Can pair production not do this? I'm told that the electromagnetic propagator is able to couple to a quark and anti quark pair (just like with electrons and positron pairs).

So far we've been twisting Feynmann diagrams and replacing particles with antiparticles moving in the opposite direction in time. So maybe it's not pair-production of quarks but a manipulation of diagrams describing the photoelectric effect or annihilation?

4. Dec 5, 2012

### Simon Bridge

Have you included the different symmetries?

5. Dec 17, 2012

### Silversonic

Apologies for the necrobump. I saw what you said but I decided I'd wait a few weeks until I read over my material again to see if I'd understand it. Unfortunately I do not. When you mention symmetries you might be mentioning something outside the scope of what I've learnt. We've just learnt a very basic way of twisting Feynman diagrams to represent some interactions as others.

For instance, annihilation involves two differently charged leptons coming towards each other and annihilating each other to produce a photon. We have altered this by changing the positron into an electron moving in the opposite direction, then twisted the diagram Feynman diagram 90 degrees so it looks like an electron emitting a photon (Bremsstrahlung).

As far as I can gather from my notes, it is possible for a photon to couple in to a quark-antiquark pair. Even an official textbook tells me "In strong and electromagnetic interactions quarks and anti quarks are only created or destroyed in particle-antiparticle pairs".

So what is wrong with the possibility I have mentioned?

The only thing I could think of is that actually it is possible through Strong and Electromagnetic interactions, but as Strong has a shorter lifetime we just go with that.

A side note: My lecturer seems to conclude that if Isospin (I_3) and strangeness are conserved in an interaction, it must be Strong. However, both of these are conserved in EM too. Also, Weak CAN conserve these two (but not always). So I don't see that implication "Conservation of I_3 and S" --> "Strong interaction process". It must be down to whichever has the shortest lifetime?

Last edited: Dec 17, 2012
6. Dec 21, 2012

### vela

Staff Emeritus
In general, if a process can occur through different channels, you look at the relative strengths of the interactions.