Is Proton Decay to Positron and Photon Forbidden by Baryon Number Conservation?

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Denver Dang
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Homework Statement


Hi...

I got a problem where I'm asked to figure out if several decays is forbidden or not.

One says:

[itex]p \rightarrow e^{+} + \gamma[/itex]

I'm then thinking baryon number conservation, but I'm not quite sure if it is right.
The proton, p, is a baryon made up of 3 quarks, and the positron and photon is elemental particles, which isn't made up of any quarks. So my question, even though it sounds quite simple, just wanted to be sure, is, do they count as zero, and therefore means that I get 1 = 0, which violates the baryon number conservation, or am I mistaken ?


Regards


Homework Equations





The Attempt at a Solution

 
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Yes, that is correct, only baryons have a baryon number, for everything else such as mesons, leptons and gauge bosons it's zero. (You can also say quarks have a baryon number of 1/3, while antiquarks have one of -1/3, this way mesons also come out with 0).

Another way of looking at this specific reaction is that if the proton was unstable, the universe would look much different than it does :)
 
niklaus said:
Yes, that is correct, only baryons have a baryon number, for everything else such as mesons, leptons and gauge bosons it's zero. (You can also say quarks have a baryon number of 1/3, while antiquarks have one of -1/3, this way mesons also come out with 0).

Another way of looking at this specific reaction is that if the proton was unstable, the universe would look much different than it does :)
Ok, thank you very much :)

And I actually have another question, if you have the time.

By emitting a W--boson, you can change quarks in specific bosons, mesons.
But is it valid for every quark, or is it just some quarks that will alter by emitting a W--boson ? I mean, there has to be some rule to how the quarks change, or... ?

I have been looking in my textbook over and over, but I can't seem to find anything that answers my question.


Regards.
 
Quarks with charge -1/3 (down, strange, bottom) can become quarks with charge +2/3 (up, charm, top) by emitting a W-, preferably to the matching quark in the same family, but other transitions (for example d -> W-, c) are possible too, but with a smaller probability (for more details read up on the Cabibbo–Kobayashi–Maskawa matrix).

For up-type quarks to become down-type quarks, they need to emit a W+, everything else is the same as above.

You can remember this through conservation of charge, as W- has charge -1 and W+ has charge +1. Of course energy must be conserved too, so for example the decay of a free neutron is possible, while the decay of a free proton is not (neutron mass > proton mass).
 
Ofc... That makes sense :)

Thank you very much.