# 2 Q's about statements in Griffiths' book (regarding proton decay)

1. Mar 8, 2012

### nonequilibrium

Hello,

I'm taking a particle physics class and we're using Griffiths' book "Introduction to Elementary Particles". I was reading in it but two statements in it (on the same page, for reference p33 in the second edition) struck me as weird, and as I would greatly appreciate if anyone could clarify them for me:

Statement A
(regarding the non-decay of protons)
We know free neutrons have a finite lifetime, but they are stable in nuclei (due to the Pauli principle, if I understand correctly). Why wouldn't the same logic apply to protons (well I suppose I get why it doesn't work for hydrogen)?

Statement B
(after introducing the law of conservation of baryon number)
But if that is the logic, isn't it possible to just send a very high energy photon onto a proton to get a neutron (to make up for the relatively low rest mass of the proton)? Or for example can't a really fast moving proton (= high kinetic energy) decay into another (necessarily more massive) baryon (with by-products, to make the mechanical conservation laws work out)? Then again that wouldn't make sense for any particle, from a relativistic stand point (i.e. the decay can't happen from the proton's reference frame, not having the kinetic energy), however it seems there is nothing that forbids the decay from happening in the certain reference frame in which the proton is going near the speed of light(?)*

* this last remark is applicable to any particle and isn't proton-specific

2. Mar 8, 2012

3. Mar 8, 2012

### nonequilibrium

I always do <3

4. Mar 9, 2012

My point was that if a fast-moving proton were to transmute somehow into another particle, that particle would have to have the same momentum, and therefore to conserve the total energy the resulting particle cannot be more massive than the original proton. It is not possible for particle interactions to occur in some (inertial) frames of reference but not others. An interaction either can happen in any frame, or none. The choice of frame simply reflects the velocity at which the observer is travelling relative to the "action".

It's easier to think about interaction possibilities in the centre-of-mass frame of reference where, in this case, the proton would have zero momentum, so only its rest mass would be available for creating daughter particles.

It is indeed possible to send a fast proton into another one and make a more massive hadron, the hadron in question being a deuterium nucleus. This is exactly what happens in the sun:

1H1 + 1H11H2 + e+ + $\nu$e

There is still some hope for proton decay - various grand unification theories predict it - but if it does happen it must do so very slowly as experimental results have now put lower limits on the proton lifetime that are of the order of 1034 years - see http://en.wikipedia.org/wiki/Proton_decay.

5. Mar 9, 2012

### nonequilibrium

Yes, that's why I included "with by-products". Then the reasoning doesn't work.

6. Mar 9, 2012