Banner
    Understanding Muon Decay
    By Tommaso Dorigo | March 4th 2010 03:41 PM | 15 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...

    View Tommaso's Profile
    Yesterday somebody asked me here if I could explain how does a muon really decide when and how to decay. I tried to answer this question succintly in the thread, and later realized that my answer, although not perfectly correct in the physics, was actually not devoid of some didactic power. So I decided to recycle it and make it the subject of an independent post.

    Before I come to the discussion of how, exactly, does a muon choose when and how to decay, however, let me make a few points about this fascinating particle, by comparing its phenomenology to that of the electron.

    Muons versus Electrons

    The muon is the electron's heavier brother. It has the same charge and the same status of elementary lepton, but the one thing that makes it different from the electron (well, the other is "muon-ness", which does not count for the sake of this argument) changes most of its phenomenology. Such is the importance of mass!

    The electron mass is 0.511 MeV, the muon mass is 105 MeV: a factor of 200 in favor of the latter. Let me make a list below of the difference this makes.

    1 - The electron does not have anything lighter to decay into: being the lightest electrically charged fermion, it is condemned to live forever. The muon, instead, can turn into an electron (indeed I discuss the process in more detail below). The available energy of the reaction is the factor which dictates its lifetime: 2.2 microseconds. So the electron is eternal, while the muon is unstable. The universe would be an entirely different place if muons were as stable as electrons!

    2 - When in possess of energy in the range from a MeV to a few GeV, the electron is already quite relativistic (it travels at the speed of light: 297,000 kilometers, or only inches less than that, per second), while at the same energy the muon is much less relativistic. This fact has several important phenomenological consequences, but discussing them would bring me too far today.

    3 - The electron radiates bremsstrahlung photons much more readily than muons at the same energy. Take a W boson produced in a LHC collision: if the W decays to an electron-neutrino pair, the energetic electron will have a penchant to radiate out photons whenever it crosses the electromagnetic field of the atoms it traverses.

    The muon, with its 200 times higher mass, is subjected to a 1.6 billion times smaller energy loss by radiation at the same energy. This has important consequences for the detection of electrons and muons: the most visible one is that electrons interact in dense matter by producing a shower of secondaries, as I discussed just a few days ago here; muons, on the contrary, pass almost unhindered through large amounts of material. All particle detectors in collider physics experiments are built the way they are because of this simple difference: dense layers of material are used to detect electrons, while muons may be picked up downstream, where no other particles make it.


    So how does a muon decay ?

    We usually picture an elementary particle as a line in a Feynman graph, and we think of it as a point-like object propagating in space along a straight line. But quantum mechanics tells us otherwise: the particle should rather be pictured as a cloud with undefined boundaries. A lot is going on in its neighborhood: not only does the particle interact with the medium, constantly "talking" with the environment by emitting and absorbing bosons, the quanta of the interacting fields: the particle also emits and reabsorbs "its own stuff", virtual particles it generates by itself. The picture on the right should explain what I mean: the upper straight line is what we may think of the muon propagation; the lower graph shows a more realistic scheme, with the muon continually emitting virtual bosons (dashed lines) which occasionally turn into pairs of fermions, etcetera, ad infinitum.

    In particular, the muon emits and reabsorbs virtual W bosons of electric charge equal to its own. By doing that, it momentarily turns into a muon neutrino: electric charge and weak hypercharge get subtracted from it, to be given back as soon as the virtual W comes back from the free trip.

    Maybe I should explain what I mean by "virtual" at this junction. A virtual particle is not a ghost, or a mathematical trick! A virtual particle, in truth, is as real as a real one; the only difference is that its presence may be ignored when one describes a particle reaction in the simplest approximation.

    Now consider that W boson: however virtual, it is also authorized to temporarily split into fermion-antifermion pairs before it is claimed back by the emitting muonic line. It may thus turn into an electron- electron neutrino pair, for instance; but also other fermion pairs are possible. Then, before you know it, the pair fuses back into the W, and the W fuses back with the muon neutrino to yield the original muon. This is shown graphically on the right.

    But with electron-neutrino pairs one different thing may happen: the duo may decide to leave the scene of the virtual fluctuation for good, never fusing back into the W. They do so without violating indefinitely the energy-momentum conservation law, because the total mass of the muon neutrino, the electron, and the electron neutrino is smaller than the original muon mass. So the muon gets fooled: the quantum numbers it lent to the W boson are lost forever. It is effectively turned into a muon neutrino for good.

    Muons, as any other subatomic body, constantly emit and reabsorb virtual particles of all the kinds they couple to. The probability that a virtual W plays the disappearance trick to the muon which popped it out of the vacuum is constant every time it gets created: this creates an exponential decay law, no less than the one which applies to the stack of tokens that a unwitty player continues to bet on "red" at the roulette as he wins.

    What governs the speed of the disappearance, i.e. the muon lifetime, is a direct consequence of the likelihood of the transition discussed above, which in turn depends on the available energy of the reaction (muon mass minus mass of the final state bodies), as well as the strength of the coupling: the latter is a direct measure of the likelihood of virtual W emission.

    Taking Stock

    All in all, there is nothing particularly surprising in the description I have provided above: particle reactions readily occur if they do not violate any quantum-mechanical rule. Muons may only decay to electrons -the only particles lighter than muons which carry the same electric charge- and they do so by means of the exchange of a W boson, the carrier of the electroweak charged-current interaction.

    Once you understand how muons decay, it becomes absolutely trivial for you to figure out that a b-quark may decay to a charm quark, an electron, and a neutrino: the diagram and the qualitative explanation are the same. But the large mass of the b-quark (4 GeV, 1.2 of which get converted into the mass of the charm quark) also allows different leptons to be produced: a muon-neutrino pair, for instance. A tau-neutrino pair is also possible, but its rate is sizably smaller, since the tau lepton weighs 1.77 GeV, which means it takes away a rather large share of the available energy.

    In conclusion, the decay of elementary particles may be understood without the complicated calculations of quantum field theory. A qualitative picture may explain most of the observable features of the process. However, it is quantitative calculations what actually tell us, once compared with experimental measurements, that our qualitative picture is correct!

    Comments

    Hi Tommaso,

    This was a very nice post (coming from a particle physics student). Thanks!

    --
    Doug

    On the subject of muon decay, would anyone know how we know the mean life span of a muon "at rest" is 2.2 microseconds? At almost the velocity of light such a muon would travel about 650 meters during this time, Yet one particle accelerator where they research muons, Brookhaven Nat Lab on Long Island NY, has a diameter of only 14 meters. Hence the muon would have to revolve more than ten times. Yet muons "propagate in space along a straight line." All of which leaves me confused. If a knowlwdgeable person can help, please do.

    It depends on the energy. The muons from cosmic ray showers that are produced in the upper atmosphere are very high energy. The fact that some of them live long enough (much more than 2 microseconds!) to make it all the way to the Earth's surface (much more than 650 meters!) served as an early verification of time dilation. High energy muons are very penetrating. Cosmic ray muons have been used to explore the innards of the Great Pyramid. Also they are being discussed as an alternative to X-raying for inspection of cargo.
    Low energy muons as produced in the laboratory are quite a different matter. They can easily be slowed down and stopped in a suitable target such as a plastic scintillator. The 2 microsecond measurement of their lifetime 'at rest' results from this.
    In answer to your other question, the magnetic field in particle accelerators cause all charged particles including muons to travel in a circular path, the radius of the circle being inversely proportional to the strength of the field.

    Thank you for your informative reply, Bill. Then could it be that cosmic ray muons live longer because they have a higher energy level to begin with than low energy laboratory-produced muons, rather than because of theoretical Larmor time dilation? Like a cell phone battery that has a higher charge will last or live longer than a battery of lower charge.

    Very interesting topics, do you think you'll talk also about the adrons decay in this fashion in the future? Just to explain better the process.
    Talking about the pictures, shouldn't be an anti-electron-neutrino the particle that comes out from the W boson? (or conversely a positron and an electron-neutrino , but in this case is C-symmetrical the process?)

    dorigo
    Icy, I sometimes discuss these topics in the middle of longer posts on physics results, so yes, I think I will.
    About the process, yes, anti-neutrinos with electrons. I decided to simplify things this time.

    Cheers,
    T.
    "In conclusion, the decay of elementary particles may be understood without the complicated calculations of quantum field theory."

    Yet, Feynman diagrams (which are about all we know about quantum field theory) are still the central tool used in your post. Which suggests that, in fact, one cannot understand the decay of elementary particles without quantum field theory.

    dorigo
    I sort of agree, but I said "the complicated calculations" of QFT.
    Cheers,
    T.
    It's a nice post but it doesn't answer the most important question - why do some virtual fluctuation products end up "leaving the scene for good."

    dorigo
    That, dear P, is not different from asking why the winning lottery number was AK2340029413.
    Quantum mechanics is the random machine of the universe.

    Cheers,
    T.
    Hi T.,
    awesome article, as usual.
    You really should write a book on particle physics.. I will keep on telling you until you do it!

    Cheers

    dorigo
    Σ΄ευχαριστώ  πολύ Αλεχ! Αλλα το βιβλιο θα πρεπει να περιμενεi... 
    Hi Tommaso,
    I do not understand why you're saying: "The muon, with its 200 times higher mass, is subjected to a 40,000 times smaller energy loss by radiation at the same energy." The rate of energy loss via bremstrahlung is proportional to (E/m) at the 4th power (if the field causing radiation is orthogonal to particle speed): so, the rate at which a muon losses energy should be (200)^4=1.6 billion time less than an electron with the same energy...

    dorigo
    That is correct Leonardo, I was confusing myself with the Bethe Bloch (where you have a beta squared factor at the denominator).
    Cheers,
    T.
    Hi Tommaso, Firstable thank you for your article! Yes the electron deposits more energy in matter by a factor (E_muon/E_electron)^2 due to the electromagnetic radiation inverse to E^2. So near the rest energies (m_muon/m_electron)^2=(200)^2 = 40,000 times but at high energies like in LHC(CERN) experiments the ratio decrease and the muon may become like electrons making showers in calorimeter before the muons chambers.