The proton has never been observed to decay into other particles. So it is possible that it is an absolutely stable particle and will never decay. However, there are some grand unified theories that argue that the proton does decay and physicists have been carrying out experiments to detect them decaying. The catch is that these theories predict an extremely long lifetime for the proton, greater that 1031 years! How does one do such an experiment?
The way to do it is to take advantage of the probabilistic nature of quantum mechanics. When we say that the average lifetime of a particle is some value, what that means is that there is a range of times in which the particle can decay, starting with zero (i.e., immediately) as the lower bound and going up all the way to infinite times, and one can assign probabilities to any given time interval. Short-lived particles have probability distributions that peak at the low end, while those of long lived particles peak at higher values.
Of course the chance that any given proton will decay in the lifetime of an experimenter is extremely small. So what one does is observe a vast number of protons and thus increase the odds of seeing one decay. So if the proton lifetime is 1031 years, by observing a collection of 1031 protons, one could expect to see one decay per year. Of course, the technological challenges are immense.
A new paper reports on results of an experiment using a 50,000 ton tank of water surrounded by detectors that looked for proton decay predicted by one particular theoretical model. They looked for decays for 17 years but saw none. This null result means that the minimum lifetime of the proton has been nudged even more upwards and now stands at 5.9×1033 years for this particular decay pathway.
One aspect of proton decay that fascinates me is the role for it or some process like it in explaining the imbalance of matter and anti-matter (the baryogenesis problem). As I understand it, even a very slow proton decay process could have a corresponding anti-proton decay that proceeds much faster, explaining how all the initial anti-protons decayed to electrons early in the history of the universe. But if there is NO proton decay, then there should be NO anti-proton decay either.
It will happen, but it would be unseemly if proton decay were to be observed so soon after the discovery of the Higgs. I give it another 15 years.
Maybe they’re just shy.
@1:
Do you have a reference? My understanding is that, unless CPT symmetry is broken, the lifetimes of antiprotons, etc would be exactly the same as the corresponding matter particles. And if CPT is broken, so is Lorentz invariance.
Damn those baby-boomer protons! When will they step aside and let the younger particles have a chance?
Rob,
To get an asymmetry in the proton-antiproton decay lifetimes, you need only CP-violation.
Mano, to get matter-antimatter asymmetry, you need all of the Sakharov conditions
1) B (baryon number) violation
2) C and CP violation
3) departure from thermal equilibrium
And these are necessary, not sufficient, conditions.
I think the lifetime of the antiproton is a separate issue, and its equality to the proton lifetime is a consequence of CPT invariance. That’s not to say (small) violations should be discounted.
“So if the proton lifetime is 10^31 years, by observing a collection of 10^31 protons, one could expect to see one decay per year.”
I’m not a physicist, but I have to say I don’t understand this. It seems to me that what we need to know here isn’t the lifetime, it’s the expected variation around that lifetime, as well as the average birthdate (so to speak) of the protons in question. Now, it’s quite possible that the numbers do work out that nicely, but I don’t see why they should. (Or is that lifespan the expected value of a uniform distribution?)
psweet @8: The lifetime is defined according to the variation, in terms of the decay width Γ, where ℏ is the reduced Planck constant
τ = ℏ/Γ
The ‘birthdate’ of the proton is irrelevant. If you have N protons at time t=0 (i.e. whenever you start your stopwatch), the number remaining after time t is
Nexp(-t/τ)
Ah, okay, I guess I’m too used to thinking of organisms, where lifespan does work that way.
Thanks.
Shouldn’t we call an “anti-proton” a conton?
Observer effect: a watched proton never decays.