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

[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)

Needless to say, it would be unpleasant for us if this reaction were common (all atoms would disintegrate) [...]

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 the proton, as the lightest baryon, has nowhere to go; conservation of baryon number guarantees its absolute stability.

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

 

[AdrianTheRock]

Think about conservation of momentum...

 

[nonequilibrium]

I always do <3

 

[AdrianTheRock]

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 traveling 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 + ₁H¹ → ₁H² + 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.

 

[nonequilibrium]

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.

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

 

[AdrianTheRock]

On the contrary, having by-products makes it all the more impossible. Again, if we boost into the proton's rest frame, it certainy doesn't have enough energy to transmute into a heavier particle and emit some others as well.

Interestingly, though, suppose we put the lightest particle we can think of - for convenience, we will make this an electron antineutrino, at rest - in the path of our speeding proton. This changes the situation enormously, because (provided it's going fast enough) the proton can now fire off a W⁺ boson at the antineutrino, turning itself into a neutron and the antineutrino into a positron. (More likely in practice, at very high energies, that the proton would split into a shower of hadrons - we are doing a bizarre version of deep inelastic scattering here - but we'll ignore that for the moment.)

Yes, one featherweight particle at rest makes all the difference. In a pure vacuum, I can always boost into the proton's rest frame and be left with no more than a paltry 938MeV on the energy scale. But if I do that when there's an antineutrino here, the antineutrino then ends up with loads of kinetic energy in the proton's rest frame, and I still have enough to make that neutron and positron.