What is the physical basis behind the half life of nuclear decay?

Question

I'm trying to understand the physical basis of decay half-life (understanding there may be different bases for different types of decay...) and I came across this article:

https://www.lanl.gov/museum/news/newsletter/2017/2017-05/science-question-decay.php

Quoting from that article:

A nucleus consists of protons and neutrons held together by powerful forces. Certain combinations are more stable than others. It has to do with ratios and “gluons,” but I will try to avoid that part of the explanation. Within one element—and to be careful, let's assume we are talking about one nuclide (or one isotope of one element)—every atom has exactly the same combination of protons and neutrons. It would appear that all the nuclei are identical, but the particles have some combination of “jiggles” inherent in them, like siblings in the back seat of a car.

For a nuclide with a long half-life, most of the time the total jiggling stays inside the limit of the binding force that holds the nucleus together: the nucleus doesn’t break apart. Every once in a long while, however, the jiggles might line up or form a resonance that sends the nucleus across the limit of its cohesion, and the nucleus splits.

What are the "jiggles" that this author is referring to? I assume that it refer to oscillations of some type...what is oscillating though?. Can someone point me to references for the "ratios and gluons" that form the part of the explanation that the author avoided?

I speculate that perhaps there's something behind this explanation like the formation and annihilation of gluons (for a strong-force mediated decay) or W,Z bosons (for a weak-force mediated decay)...but I haven't been able to confirm my suspicions.

Answer 1

The "jiggling" refers to the fact that when a particle is confined in some small volume (here, the nucleus), it must gain momentum uncertainty in accordance with the Pauli exclusion principle. Nevertheless, a stable nucleus of a given isotope is a stationary state: it is unchanging, and all ground state nuclides of the same A and Z in the universe are identical at all times. Something that has momentum (aka: motion), that is in a stationary (aka: unchanging) state is purely a quantum mechanical concept.

It appears the author is trying to describe alpha decay. Classically, alpha decay makes no sense, since the nuclear binding (aka: some ratios of gluons and stuff) is too strong (and increases with range up to a cutoff) to allow alphas to get far enough away for the electric repulsion to kick them out: even if they jiggle. Quantum tunneling resolves that problem.

Radioactive decay is deeply quantum mechanical, and any classical analogy will cause confusion for someone who knows a little physics. Of course, the URL says it's from the museum, and the description is good enough for visitors.

Answer 2

Classically, a radioactive nucleus ought to be stable because there is a big energy barrier to anything in it being ejected. But one consequence of quantum uncertainty is the ability of something to tunnel through the energy barrier and appear on the other side, in an even lower-energy state than it started. The decay of one subatomic particle into two others is governed by a similar mechanism. Thus, radioactive decay is a way of getting rid of excess energies.

In a situation where such quantum tunnelling is possible, its statistically uncertain nature leads to a characteristic "half-life" time period at which point there is a fifty-fifty chance of it having occurred. Each specific situation within a radioactive nucleus has a different relation between the energy barrier and the particle's uncertainty, and hence a different half-life.

Gluons are not really involved as everything with them happens so quickly, but some radioactive mechanisms do involve a boson as intermediary. For example an excited quark may decay into a lower-energy quark of different flavor and spit out a boson, which in turn decays into an electron and a neutrino.