# Why is this decay allowed and another not?

Tags:
1. Mar 26, 2015

### unscientific

Consider $\Sigma^0 \rightarrow \Lambda + \pi^0$. (Not Allowed)

According to griffiths, this strong interaction is not allowed by 'Conservation of Energy'. I'm not sure why, as this simply shows an up and anti-up quark coming together, producing a gluon, where mass of gluon is mass of up and anti-up quark combined.

Consider $p + p \rightarrow p + p + p + \bar p$. (Allowed)

Why is this possible by the strong interaction then?

2. Mar 26, 2015

### The_Duck

Consider this process in the rest from of the initial $\Sigma^0$ particle. What is the total energy of the initial $\Sigma^0$ particle? What is the minimum possible total energy of a final state containing a $\Lambda^0$ particle and a $\pi^0$ particle?

This is not a diagram that contributes to $p + p \to p + p + p + \bar p$. In your diagram, one of the protons is unaffected and might as well not be there. So effectively the process you have drawn is $p \to p + p + \bar p$, which like your first example is not allowed because of energy conservation. (Also there is a more technical reason why this diagram is not allowed, which is that a single gluon cannot produce a proton-antiproton pair because of color conservation. You need at least two gluons).

Here is a possible diagram for $p + p \to p + p + p + \bar p$:

3. Mar 26, 2015

### Orodruin

Staff Emeritus
Also, the reason a proton-proton collision can create an additional proton-antiproton pair is that the pp system is not a bound system and the protons can have a significant kinetic energy in the CoM frame (if they do not have enough, the process is forbidden).

4. Mar 26, 2015

### dukwon

Put simply, the first one is a (forbidden) decay, whereas the second one is a collision

5. Mar 31, 2015

### unscientific

That totally clears it up. Thanks!