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:


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.

  • 2
    $\begingroup$ "Jiggling" is an unhelpful analogy. Have you read en.wikipedia.org/wiki/Radioactive_decay $\endgroup$ Commented Apr 20, 2020 at 21:30
  • $\begingroup$ I agree that "jiggling" is unhelpful...at least, it didn't help me any. I have read the Wikipedia entry; I found the snow avalanche analogy equally unhelpful. I edited the question to clarify a little bit. $\endgroup$ Commented Apr 20, 2020 at 22:22
  • $\begingroup$ The reason why no classical analogy will ever help is because this is not a classical system. The fields that these nuclear states are made of extend all the way to infinity, so while it looks like that a nucleus is some well defined classical entity, in reality it's not. It is "open" to infinity. Stable atomic and nuclear states have lower energy than all other possible states and are preferred. Unstable ones are not the lowest energetic state and are therefor likely to (eventually) decay. Of course, if we accept proton decay, then no nuclear state is stable by that definition. $\endgroup$ Commented Jun 19, 2023 at 21:09

2 Answers 2


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.

  • $\begingroup$ I would have thought it is the other way around, that QM is prone to cause confusion whereas a classical analogy resolves it!! The so-called hydrodynamic quantum analogs (i.e. classical analogies) reproduce the tunneling effect, so anyone confused about QM should probably start with those. $\endgroup$
    – Steve
    Commented Apr 21, 2020 at 9:57

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.