# [nuclear physics]Understanding what's type of interaction is

1. Jun 6, 2010

### valleyman

1. The problem statement, all variables and given/known data
My problem is on questions like "say if the following interactions are allowed or prohibited, and explain why." where the interactions are something like

$$p + p -> K^+ + K^-$$

or

$$e^- + p -> n + \nu_e$$

3. The attempt at a solution
Now the first one is not allowed because charge and barionic number are not conserved, ok, but the second is allowed and the solution says it is weak interaction. How do I recognize what type of interaction takes place? Are there any rules like those one of the conservation?
Note: my problem isn't specifically on those particular interactions I just need to understand how to recognize, in a general way, what type of interaction is
Thanks for the help,
valleyman

2. Jun 6, 2010

### vela

Staff Emeritus
Tip: Use \rightarrow to get an arrow in LaTeX.

Here's a few rules of thumb to help identify which interaction is responsible: If there's a photon, then you know it has to be an electromagnetic interaction. Similarly, if there's a neutrino, it has to be a weak interaction. If quark flavor isn't conserved, like in kaon decay, it's a weak interaction.

3. Jun 15, 2010

### valleyman

Thanks for the help those rules are very useful but there are still some things I don't understand! what if my reaction doesn't show either photons or neutrinos? And how do I check if quark flavour is conserved?
For example
$$e+e^-\rightarrow p+\overline{p}$$
this is e.m., right? But what if it is
$$e+e^-\rightarrow k^++k^-$$
Shouldn't it be e.m. anyways? How do I decide??
Thanks
valleyman

4. Jun 15, 2010

### vela

Staff Emeritus
You have to know what vertices are allowed for the various leptons and quarks, and generally you have to know the quark content of the various mesons and baryons. Electrons do not have color charge, so they will not interact through the strong force. So both reactions have to be either electromagnetic, where the electron and positron annihilate and create a virtual photon, or weak, where they annihilate and emit a virtual Z. At low energies, electromagnetism is going to dominate because the Z is so massive.

With only leptons in the initial state, the quark numbers will all be 0. Since the final state consists of a particle and its antiparticle, the quark numbers for each flavor will also be 0, so there's no problem with quark flavors.