How Can We Visualize Proton Decay and Its Implications in Particle Physics?

[VreemdeGozer]

Hello there,

I was just reading some articles on Wikipedia.org and found an article on proton decay. I will quote a part of it:

...allowing protons to decay via the Higgs particle, magnetic monopoles or new X bosons.

How would that be visualized? I've got a basic understanding of particle physics, but these concepts seem rather advanced, yet, I insist on wanting to understand them.. Can anybody help me out?

Thanks in advance.

 

[mathman]

I can't answer your question directly. However you should be aware of the fact that as of now there is no evidence of proton decay. Experiments have shown that protons (if decayed) would have a half-life > 10³³ years.

The concept is part of some theories which try to complete the standard model of particle physics.

 

[AdrianTheRock]

In some Grand Unified Theories there are particles called X bosons which can change the identities of particles in much the same way as the familiar W bosons change (eg) electrons into neutrinos. X bosons, however,

• can change quarks into leptons (or vice versa), but
• are very massive, say ≈ 10¹⁶ GeV.

At a quark level, such a reaction might be something like

u → X + e⁺
X + u → anti-d​

(I have left off the quarks' charges in the above for brevity.) To account for conservation of the charges, the X boson in this example would have to have an electric charge of -1/3 and a single colour charge the same as that of the initial u quark, say red. If we say the second up quark, that absorbs the X, is blue, then the anti-d would be anti-green. (Alternatively, the X could be absorbed by a d quark to produce an anti-u.)

Within the proton as a whole, we have now converted one u to a positron and the other to an anti-d. The overall effect of this interaction is therefore

p⁺ → e⁺ + π⁰​

As mathman says, the fact we have not yet observed a single proton decay despite having established some large experiements to look for precisely this means that the mass of the X bosons would have to be at least of the order stated above.

 

[VreemdeGozer]

I'm aware of it being a hypothetical process, I was just curious how exactly it would have happened (because there's no conservation of lepton and quarknumber). Can an X-boson break this law in some way under some condition?

EDIT:
The way it looks to me is that the only real laws of conservation are that of:
- Charge
- Color charge
- Impulse
- Energy

In a universe where proton decay is not possible, doesn't the conservation of color charge imply the conservation of quarknumber (and possibly, with some expansion, leptonnumber)?

 

[Bill_K]

In most GUT theories, B and L are not conserved separately but the combination B - L is conserved.

Color is conserved because the hypothetical X boson itself carries color. For example take the proton decay p -> e⁺ + π⁰. In terms of quarks suppose the proton was p = (up-red, up-blue, down-green). Then in terms of quarks what happened was up-red + up-blue -> X -> e⁺ + antid-antigreen, and meanwhile the down-green goes along unchanged. The resulting pion is formed from (d-green, antid-antigreen).

 

[VreemdeGozer]

Hm, that's not quite what I meant, but that's interesting. I've never heard of B-L conservation. In the given process, how would B-L be conserved?

I'm aware of the color conservation, but I meant that as an alternative to quark and leptonnumber conservation, there could be some kind of expanded way of defining the color charge to also include leptons, which might get around the problem too.. But B-L is probably a lot more elegant.

 

[ophase]

For example, according to RPV SUSY model proton should decay via superpartner of s-quark or b-quark which corresponds to bosonic states.
https://upload.wikimedia.org/wikipedia/commons/b/b9/Proton_decay_GUT_simple.svg

So X maybe these superpartners. But their vertex factors which are proportional to lambdas are quite small and depend on these boson masses. Super Kamiokande set some limits on lambdas as below:

λprime_(11k)λdouble_prime_(11k) < 8x10^27x (MassOfBosonX)^2 / 100 GeV

If the energy scale is high enough, proton life time should be shorter. So it's relevant for early universe or Planck scale.

 

[Bill_K]

That's exactly the same process I described above in words. Thanks for the picture.

 

[VreemdeGozer]

That picture explained it pretty well, thanks for that! But I don't see how B-L is conserved in the process:

u+u+d: B=1; L=0; B-L=1
X+d: B=1/3 + ?; L=?; B-L=1
e⁺+anti-d+d: B=0; L=-1; B-L=1

What are the values of B and L in the middle step?

 

[ophase]

That picture explained it pretty well, thanks for that! But I don't see how B-L is conserved in the process:

u+u+d: B=1; L=0; B-L=1
X+d: B=1/3 + ?; L=?; B-L=1
e⁺+anti-d+d: B=0; L=-1; B-L=1

What are the values of B and L in the middle step?

Superparticles have also the same baryon and lepton numbers of their superpartners. In the middle step we have B=-1/3 and L=0. (minus bcs it's an anti-particle)
Yes! Proton decay violates B-L conversation as well as R-parity conversation.

B-L conversation is good for low energies. But for high energies at about Planck scale, one has to take into account unperturbative radiative processes too. Because they are also Lorentz invariant. Here is the original potential term for such processes

http://binary-services.sciencedirect.com/content/image/1-s2.0-S037026931101416X-si37.gif

As you can see all terms violate B or L conversation with 1 quantum number. The second and the third terms above are responsible for the proton decay. It's not a solution to forbid these kind of processes. Have you ever thought who made the law of B-L conversation? B-L conversation is just a phenomena comes directly after Standard Model. The best thing to do is to investigate their limits at high scales. Because Baryogenesis or Leptogenesis mostly depend on these kind of processes.

 

[AdrianTheRock]

That picture explained it pretty well, thanks for that! But I don't see how B-L is conserved in the process:

u+u+d: B=1; L=0; B-L=1
X+d: B=1/3 + ?; L=?; B-L=1
e⁺+anti-d+d: B=0; L=-1; B-L=1

What are the values of B and L in the middle step?

Actually, looking at the above again, the middle step above doesn't have all the particles. Based on the version of the process I'd used in my example, ie

u → X + e⁺
X + u → anti-d​

the particles after the first "stage" of the virtual process are u+X+d+e⁺, so we have B=2/3, L=-1, so B-L = 5/3 ... unless we assign a B value of -2/3 to the X, noting that in this example B-L does revert to 1 after the totality of the interaction.

There may still in general be processes at this level which don't conserve B-L - personally, I have nowhere near the level of understanding of these theories needed to comment specifically. But it does strike me that B-L conservation is really just saying that the number of excess leptons over antileptons in the universe - in everyday reality, the number of electrons that don't have positron partners anywhere - must stay the same as the number of excess protons that don't have matching antiprotons. Violation of B-L conservation is no barrier to baryogenesis or leptogenesis - all it actually says is that when these processes occur they must create equal numbers of excess baryons and leptons, eg an electron for every proton.

 

[VreemdeGozer]

I found that just as striking as you did, then I learned of the Rishon model (and the expended Harari-Steinberg model). There's quite a few neat papers on it on the net.
http://www.weizmann.ac.il/home/harari/files/Nuclear_PhysicsB_Vol204.pdf (1982 version)
http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-2310.pdf (1979 version)

Of course, due to my insufficient understanding of all the mathematical mumbojumbo, I could only pick up a few things. It does seem to explain the matter-antimatter problem neatly, though.

Ophase, could you elaborate on that formula please? I'm not really familiar with those formulae.

 

[ophase]

Sorry, i didn't have time to elaborate formulas at all. But i'll do it in the future, nice sharing.
I am not sure of that if R letter of "R-parity" expression comes from the "Rishon value". Is that true?
If not what's the origin of the letter R in R-parity?

 

[VreemdeGozer]

As far as I know, no, R does not come from Rishon or Rishon value.

I found this redirect, but it's not really helping, I guess. ( Unless the R comes from matteR. :P )
http://en.wikipedia.org/w/index.php?title=Matter_parity&redirect=no

EDIT:
I spotted this:

Different branching ratios between the X boson and its antiparticle (as is the case with the K-meson) would explain baryogenesis.

On the WIKIPEDIA page on X and Y bosons.

How would it explain baryogenesis?