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    What Makes Particles Unstable ?
    By Tommaso Dorigo | November 9th 2012 07:39 AM | 18 comments | Print | E-mail | Track Comments
    About Tommaso

    I am an experimental particle physicist working with the CMS experiment at CERN. In my spare time I play chess, abuse the piano, and aim my dobson...

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    A journalist I am following on twitter just posted the question in the title of this post. I felt bound to try and give an answer with as simple concepts as I found meaningful. So, what makes a unstable particle unstable ?

    One answer is this: a particle is unstable if there is a way, not forbidden by any physical law, to convert its rest-mass into other forms of energy. One may understand this by thinking of entropy: any system left free to evolve will do so in the direction of maximum entropy. So since a single particle state is a very low-entropy system, while the decay products of its disintegration have a multitude of possible configurations and a higher entropy, the system will naturally evolve in that direction.

    Of course, the reaction must have a physically allowed way to occur: we cannot, for instance, make electric charge disappear, so an electron cannot turn into light quanta (it would also violate angular momentum conservation and a few other conservation laws, but that's beside the point).

    Another possible answer, loosely connected to the above one, is that in physics anything that is not forbidden is compulsory. If the probability of occurrence of a reaction in a finite amount of time is not zero, that reaction will occur, if the system is given enough time. Similarly, if you play the lottery enough times, you are bound to win one day. But be prepared to live for a long time before that happens...

    Maybe the most serious answer though is the following. Any quantum system is constantly subjected to quantum fluctuations, whereby virtual particles are emitted and absorbed. Take a muon: the muon is the heavy brother of the electron and, unlike the latter, it has a short lifetime (the electron is perfectly stable as far as we know). The muon constantly emits and reabsorbs photons, weak bosons, Higgs bosons: these are all the particles to which the muon "couples": the carrier of all the interactions to which the muon is subjected, by virtue of possessing some charges of the relative fields: electric charge and weak hypercharge.

    Now these virtual emissions occur for very brief instants of time, such that the materialized virtual boson can exist without borrowing too much energy from the system: the rule is that the time of the virtual particle existence, multiplied by the energy needed to materialize it, needs to be of the order of the reduced Planck constant. This is Heisenberg's Uncertainty Principle

    Note that if a muon emits a photon, even a virtual one, it remains a muon: only its energy may change in the process. However, if the muon emits a W boson, it turns into a muon neutrino (but we cannot yet speak of a real decay, until we clarify if the reaction undoes itself). Now take the virtual W boson emitted for a brief instant by the muon: its fate determines the fate of the emitting muon. If the W boson remains what it is before it gets reabsorbed by the muon neutrino, the muon will only feel a slight itch on the back and will not realize it has been turned into a muon neutrino for a brief instant of time.

    But if the W boson also in turn "quantum-fluctuates" by emitting an electron -a virtual one also-, the W "becomes" an electron antineutrino (it is incorrect to say that a boson turns into a fermion: but here I am simplyfing things. You however well understand that what happens is that the W turns into the electron-electron antineutrino pair).

    Now we're in trouble: the electron antineutrino cannot be reabsorbed by the muon neutrino, unless other quantum fluctuation reverse the whole chain in the exact inverse order (electron plus electron antineutrino merge into a W, and the W gets then reabsorbed by the muon neutrino, turning it into the original muon). So now the muon has become a muon neutrino, the W is no more, and the electron and electron neutrino are bound to live their different lives.

    In a way, it is as if you used to lend your car to your friend Joe on Thursday evenings. Every Friday morning you get back your car intact. However, one night Joe crashes on a pole, destroying the car. The irreversible accident prevents you from getting back to your original state of every Friday morning: you are no longer a car owner.

    It only remains to explain why the electron does not share the same fate of the muon in the reaction we have discussed above. Just as the muon did, the electron can also emit a virtual W boson, becoming an electron neutrino for a brief instant of time. And the W boson can also fluctuate into, say, a muon-muon antineutrino pair. However, no real accident can occur here, because at the end of these fluctuations the energy balance must return intact: and since the electron has a rest mass corresponding to a mere 0.5 MeV of energy, this is not sufficient to produce a real muon (which is 200 times heavier). The muon-muon antineutrino pair cannot, therefore, become real, and the reaction cannot develop: the virtual particle loop must close back into itself, and the electron returns to be itself.

    Note also that the virtual W can indeed turn into an electron- electron antineutrino pair and make them real, without upsetting the energy balance. In that case, though, the antineutrino must annihilate (closing the loop) with the original electron neutrino (otherwise the two extra neutrinos would be carrying away energy, and energy balance would be violated). The result is just a quantum fluctuation which is absolutely indistinguishable from the one we discussed before.

    Ian: I hope this helps!

    Comments

    Very well explained! All other physics majors should be reading this as well.

    Thanks Tomasso, great explanation for both experts and non-experts.

    Amir D. Aczel
    Very cool! Could you add some Feynman diagrams to illustrate this (I think it would be helpful here) and maybe say something about calculations using them. Thanks again!
    Amir D. Aczel
    Should the idea of a spontaneous decomposition of a particle be avoided where we can think of an alternative?

    What if a muon interacts with a higgs boson to produce a W and a neutrino:

    Higgs ~ 2Z
    Higgs ~ 4γ

    W ~ ν μ
    also W ~ 3γ
    ~ γ μ μ

    therefore ν = γ μ

    So Higgs + μ ~ 4γ μ
    ~ 3γ (γ μ)
    ~ W ν
    ie Higgs + muon ~ W + neutrino

    dorigo
    Dear Anon,

    unfortunately it is not algebra. One needs to care about all conservation laws when putting together possible and impossible reactions. The correct way to do this is through Feynman diagrams.

    Specifically though, in what you write: the muon does not need a Higgs boson intervention to turn into a W-neutrino virtual pair. It is a lowest-order weak interaction as is. If you put in a H, you are discussing a much higher-order reaction, which is therefore highly suppressed with respect to the former. Of course, suppression takes a different flavor when we discuss virtual reactions, but still the physics is overwhelmingly the one I described.

    In any case a muon plus gamma cannot be likened to a neutrino. There is no way you can turn one into the other - there is weak hypercharge to conserve.

    Cheers,
    T.
    I need to read more on weak hypercharge. But it is related to the conservation of the B-L. ν and μ- both have L=1. (Unless the wiki page is incorrect.) And L=0 for γ. So is there a problem with ν = μ- γ for conservation of B-L: 1=1?

    Yes, I know you cannot do particle physics like algebra. But I only used equivalences which I thought I was most sure of, and I couldn't write it down in my own notation without going into too much explanation. The problem for me is that the vacuum contains unseen matter/fields which are used in particle decays without attribution, and also there are antiparticles which can arrive at the interaction from the future. So you can have more matter after the interaction than before. But, also, things can annihilate meaning there can be less matter afterwards than before. It is not easy to write an algebraic equation in those circumstances. But it ought to be done where possible and all participants to the decay named.

    From your reply to another postee, you can't add steroids to any person and expect to get Sylvester Stallone. However, it seems to me that QM adds steroids to nothing and expects to get a Sylvester Stallone outputted. You shouldn't be able to add pure energy to a decay and get new matter outputted. Energy can add mass but not matter. Energy is the added steroids, it has to be added either to particles or to unseen matter lurking in the vacuum.
    If something in the vacuum nudges the decay into being, the contribution of that nudge ought to be fully attributed. What was it in the vacuum that interacted with the muon to cause it to decay? It can't be attributed merely to steroids?

    dorigo
    Well, I tried to explain what causes the decay... I don't know what to add!
    Cheers,
    T.
    That's Ok. Thanks for your replies.
    Excellent blogs!

    dorigo
    Dear Theo, Zach, and Amir,

    thanks for the encouragement!
    Amir, I unfortunately have a lot in my hands at the moment. I prefer to write
    another post than to improve this one... I am sure you understand.
    Cheers,
    T.
    rholley
    Concerning different energy levels, this has just brought to mind something I read about neutrons decaying into protons and electrons.

    At the centre of a white dwarf star, when electrons are quantum-packed much more densely, this in energetically less favourable.  At about 1.2 or 1.3 solar masses, the energy balance at the centre is tipped the other way, and neutronization sets in.

    So it may well be this, rather than the relativistically determined Chandrasekhar limit of just over 1.4 solar masses (as is generally taught), that marks the point where a white dwarf
    star collapses.

    Typically, this does not result in a neutron star.  In the commonest type, where material from a companion has been accreting on to a carbon-oxygen white dwarf, a thermonuclear reaction sets it, and the star blows apart as a supernova (Type Ia, so I am informed).

    Neutron stars are understood to result from larger mass original stars, where thermonuclear reactions have produced an iron core.  No further nucleosynthesis is possible, and the core then collapses under graviational pressure from the layers above.  If the star is not too massive, a massive emission of neutrinos occurs, causing the upper layers to undergo massive nuclear reactions leading to a type II supernova. A neutron star is formed from the core.

    If the star is even more massive, then the core collapses to a black hole, leading to a hypernova and a gamma-ray burst.

    (source: Thermal Physics by Ralph Baierlein, if I remember correctly).
    Robert H. Olley / Quondam Physics Department / University of Reading / England
    White dwarfs typically don't collapse to neutron stars at the Chandrasekhar limit. They undergo a runaway fusion reaction between oxygen atoms and carbon atoms that produces a type 1a supernova.

    rholley
    Thanks for pointing this out.  I have amended my text above accordingly.

    However, I still think that the onset of neutronization, rather than the Chandrasekhar limit, might be the point where collapse sets in.
    Robert H. Olley / Quondam Physics Department / University of Reading / England
    Thank you for that explanation Tommaso.

    That explains why particles decay but I wonder if you might shed some light on why exactly different particles have different rates of decay and also as to why some decay pathways have a higher probability than others

    dorigo
    Ah, that's another quite nice question, and very complex to answer - there's a ton of physics in it, given its generality. Phase space, coupling constants, conservation laws.

    I will think about giving some more examples in a follow-up, but not today...

    Cheers,
    T.
    Probably an obvious question, but since a muon decays to a electron pretty quick, what stops an electron from absorbing a pretty high powered photon (or something else) and turning into a muon?

    After a re-read and some thought, I guess "electron plus electron antineutrino merge into a W, and the W gets then reabsorbed by a muon neutrino, turning it into a muon" is just too unprobable to ever happen? Does the first part of this ever happen "electron plus electron antineutrino merge into a W" ?

    dorigo
    And about this other question: this is a virtual process, so it happens -with some probability dictated by coupling constants, of course, but without the problem of energy conservation: the W is of course way off-shell (mass near zero).

    As asymptotic states (real particles) electron and neutrinos could, in principle, turn into a W, for instance if you built a electron-neutrino collider you could do that. The cross section is however quite small.

    Cheers,
    T.
    dorigo
    Hi Ed,

    you should be careful. The muon decays to a muon neutrino, not to an electron. It does so by emitting a virtual W, which then materializes into an electron-electron antineutrino pair. In this reaction of course the leptonic number is conserved: lepton flavor violation has never been observed so far, it would be in fact new physics beyond the Standard Model.

    So, for your question: an electron absorbing a photon cannot turn into a muon any more than you can turn into Sylvester Stallone by taking a lot of steroids.

    Cheers,
    T.