34

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms.

Qmechanic
  • 201,751
Justin
  • 743

7 Answers7

25

Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2$\mu$s. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter particles that it could decay into while conserving the total charge of $-e$ and total spin of $\tfrac{1}{2}$.

Whether a particle decays or not depends on whether there are any lighter particles for it to decay to. A muon weighs about 105.7 MeV while an electron weighs about 0.511 Mev. So a muon can transform into an electron and have 105.2 Mev left over to go into the two neutrinos and the kinetic energies of all the particles. An electron can't transform in to a muon unless it can find the extra 105.2 MeV from somewhere. If we supply the extra energy, for example in the LEP Collider, then electrons can and indeed do "decay" into muons.

John Rennie
  • 355,118
  • 16
    Also, depending on your application, 2μs can be a very long time. :) – vsz Sep 22 '15 at 18:56
  • 1
    Thank you for your comprehensive answer. To be honest I was not expecting anything like the response i got for this little question but is still leaves me confused. I have not been convinced as to why these heavier are not just complex quantities of destruction. A fleeting mirage of momentum and interaction that appears as a flash in existence. – Justin Sep 23 '15 at 04:56
  • If you can consider as an analogy, smashing a balloon full of water against a wall, and for the purposes of this analogy the fundamental particles of the balloon are bits if rubber and water molecules. We would not propose of a new object being a split balloon with a body of liquid waters protruding from it would we? But because of the realm of quantum scale everything seems more granular and staccato so those intermediate fleeting states appear more important. But maybe that's philosophically misleading – Justin Sep 23 '15 at 05:06
  • 1
    The fundamental particles of the Standard Model, e.g. electrons, muons, quarks, neutrinos etc are not composites. They are not made up from smaller components. You cannot smash a muon - you can only destroy it by removing energy from the muon field and transferring that energy somewhere else. On the other hand, if you're talking about protons, neutrons etc then they are composites in a way analogous to your balloon. Perhaps you could clarify what you are asking about. – John Rennie Sep 23 '15 at 05:30
  • Doh! I realise how bad that analogy was. I was talking about fundamentals i guess its hard to phrase what i mean but you are hitting on it when you talk of energy fields. So new analogy. How about "burning a balloon full of petrol." do we count the partially exploded balloon in our standard model? To me, anergy fields are at play here. Is this what string theory is? – Justin Sep 23 '15 at 05:47
  • 3
    All fundamental particles are excitations in a quantum field, and quantum fields can exchange energy with each other. When a muon decays the muon field transfers energy to the electron and neutrino fields. NB quantum fields are rather abstract things called operator fields - they are not an energy field. – John Rennie Sep 23 '15 at 05:53
  • Hey John. Which fundamental particles are stable? i was looking for a list but I could not find one. Mostly about masses and spins – Justin Sep 23 '15 at 05:59
  • 2
    The only particles that don't spontaneously transform into other particles are the electron and the up and down quarks (and their antiparticles). But be careful about calling these stable - add a bit of energy and they will also transform into other SM particles. Possibly also the electron neutrino, though the neutrinos are a bit of a special case as all three transform into each other. – John Rennie Sep 23 '15 at 06:07
  • Clearly over short enough time scales, unstable particles are stable (muons over yoctosecond time scales are pretty stable: E18. On a second time scale, U-238 has an E18 half life). How about pressure: If you had a sufficiently high pressure environment, could unstable particles become stable? (ie, not enough "room" for the decay products somehow) Or are the decay products in this case the wrong kind for pressure-based stability? – Yakk Sep 23 '15 at 14:39
  • @JohnRennie Is there really such a thing as a muon field, or could it be something like a combination of the electron field and the Higgs field or something? – Michael Sep 23 '15 at 20:28
  • @Michael: yes there is a muon field. There are fields for all the fundamental particles. No the muon field isn't a combination of the electron field and the Higgs field, or indeed any other combination. – John Rennie Sep 24 '15 at 04:57
  • @Yakk actually, yes, in some sense. I don' feel that realistic environment can make a muon (or other unstable elementary particle) stable, but for unstable composite particles example is easy to find. Neutron is unstable, while iron nucleus is stable (and so are neutrons inside it). Or a more spectacular example would be a neutron star. – Fedxa Sep 24 '15 at 18:52
  • @Fedxa ah! So if we have an area every 'nearby' state occupied by electrons, it might take 'too much' energy for the muon to decay. But, given how heavy the muon is, that is a lot of electron states that need to be filled. A white dwarf is basically a huge atom (as far as the electrons are concerned -- all the "orbitals" are full, and the pressure is keeping it inflated): a muon in one might be "far enough" down the white dwarf's potential well (muons not being excluded by said electrons) that it would have difficulty decaying? – Yakk Sep 24 '15 at 19:33
  • Naturally, someone has already worked this out: http://physics.stackexchange.com/questions/135478/stable-muon-density-inside-a-white-dwarf-star It even has a cool aplet. Punch line seems to be that white dwarfs aren't "deep" enough, and muons are too heavy to be trapped. – Yakk Sep 24 '15 at 19:35
  • @Yakk Yes, the idea is more or less like this. But White dwarfs do not provide dense enough envirtonment, see https://en.m.wikipedia.org/wiki/Fermi_energy#White_dwarfs . The states with energies of the order of muon mass are not occupied. – Fedxa Sep 24 '15 at 20:18
12

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles?

Here I will only consider elementary, non composite particles. All the hadronic resonances are composite particles of quark antiquark combinations as well as the neutron .

The standard model of particle physics has a relatively small number of elementary point (0 dimensional) particles and is a model of a large number of observations encapsulated in their simplest form.

enter image description here

The Standard Model of elementary particles (more schematic depiction), with the three generations of matter, gauge bosons in the fourth column, and the Higgs boson in the fifth.

This model incorporates a group symmetry ( SU(3)xSU(2)xU(1) ) which identifies it. The basic hypothesis is that at large energies the symmetry is unbroken, i.e. that there exists no Higgs field to give masses to the particles entering in the table, and thus at those energies all these particles are stable, since there cannot be a lower energy state to which they can decay. They have a separate existence in the group symmetries due to the quantum numbers which give a unique niche for each particle.

After the electroweak symmetry is broken at about 246 GeV and below , the Higgs vacuum expectation value, the particles acquire the masses we have measured in the laboratory and the composites display the beautiful group representations that gave rise to the standard model to begin with.

It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms.

The standard model says that the unstable particles are an artifact of the low energy in which one observes them, at high enough energies they are no longer unstable because they have zero mass.

The standard model took years to develop. When I started in 1963, different models,( four fermi interactions, vector meson dominance, Regge poles,eightfold way) were in fashion, which due to the accumulation of data slowly morphed into the standard model. It is the different quantum numbers, conserved by different interactions, that give the particular group structures observed and finally the particle table.

It is interesting to note though, that string theories, which can embed the standard model of particle physics, posit one kind of "particle", instead of a point , a one dimensional string where the particles are vibrational levels of the string which display the standard model group representations.

anna v
  • 233,453
  • Hi Anna, please consider reopening this meta thread - it is an excellent example of how these things are meant to work and should be something we can point to. Maybe flag it for a moderator to tag as status-completed? (sorry to bother you here, by the way.) – Emilio Pisanty Sep 22 '15 at 17:41
  • @EmilioPisanty OK, will do. I thought it might not be useful since the post was openedby votes. – anna v Sep 22 '15 at 18:32
  • I think it is useful as a record of how things can and do happen: posting on meta does get questions reopened when it is appropriate, and rather quickly too. – Emilio Pisanty Sep 22 '15 at 18:45
  • What kind of experimental conditions would allow one to observe stable "high energy" muons? I guess that would be best as a distinct question, but that elaboration might make this answer more concrete in its application to the OP's question. – Yakk Sep 23 '15 at 14:41
  • @Yakk Cosmological at about 10^-10 seconds http://cms.web.cern.ch/content/story-universe when the electroweak symmetry breaks.also see http://hyperphysics.phy-astr.gsu.edu/hbase/forces/unify.html – anna v Sep 23 '15 at 15:16
  • That's a really good answer I can understand. But does this mean that at colossal energies we will see a colossal amount of new particles in the standard model? – Justin Sep 23 '15 at 22:14
  • Not for the standard model, it has a definite structure measured at our labs. Projected at very high energies it still retains it except for the symmetry breaking. There exist extensions of the SM still, they would start with a definite structure at low energy densities and retain it except for the mass at high. Other models, such as ones using string theory, do predict more and more string excitations with energy, but then the only elementary part is the string, the particles are just excitations that happen to fit the standard model at low energies. – anna v Sep 24 '15 at 02:52
  • Thanks Anna, that starting to make sense. So Anna how is string theory coming along? – Justin Sep 24 '15 at 06:16
  • IMO it is doing OK. We have to wait for the new data of the LHC to see whether there is supersymmetry, or resonances due to strings, or effects of large extra dimensions ( which last is not very probable).http://theory.tifr.res.in/conferences/kanha/Kanha09_Antoniadis.pdf – anna v Sep 24 '15 at 07:13
5

I think the most direct answer to this would be the fact that a heavier particle can decay into many different lighter particles for different reactions. The abundance of occurence of these relations are const. Again the same heavy particle can be created in multiple types of collisions of various different lighter particles. Thus we cannot say that the heavier particle is just "composite particle".

Although there are some particles of this category you mentioned. For instance "composite fermions" which are bound state of a electron and flux quanta. But not all.

mpv
  • 5,220
Ari
  • 2,841
4

Inmediately is not really true, there is some proportionality.

Sorry I am not answering directly about "the standard model", this is quarks and leptons. But they will fit the general pattern, you will see. Let me first consider all the "particles" listed in the particle data group file.

Most of the particles decaying via photons have a half-live about inverse the cube of its mass, times a constant.

Most of the particles decaying via beta have a half-live about inverse the fifth power of its mass, time a constant.

See here the picture. Horizontal is mass, vertical is decay width, which is about the inverse of halflife.

Mass vs Decay Width

The neutron is a lot more stable, so it is not pictured here, it is down in the plot then.

The "resonances" and "excited states" are in the huge clud with very small lifetimes, or wide decay widths. So even if W, Z or the top decay faster than any resonance, they decay slow enough compared with its mass. Still, even a resonance is more stable that the inverse of its mass, $ t_{1/2} > \hbar / mc^2$, the grey line above in the drawing

Even for the Higgs (the decay width has not been really measured directly yet, same that for Upsilon 1,2 and 3, but anyway we can guess or calculate them) you can see that it is stable enough compared to its mass. Also, there is still some unmeasured excited states of Bs which should be between the two lines, and it seems it is not going to get a name, just Bs* will be. So it is something of a thumb rule, but it looks acceptable when you plot everything. Amusingly most textbooks seem not to have space for a such nice log plot.

So, at the end, what about the standard model fermions? Well, as charge is preserved, and accidentally also baryon and lepton number, they only decay via the weak force. You can see in fact the "mu" exactly in the low corner of the plot, and a imaginary $m^5$ line across it will cross the tau lepton too. Thus they are all more stable that the yellow line in the plot, except the top, which is middle between both, due to the restauration of electroweak symmetry. So for the fermions, As for W and Z, and Higgs, they also conform with the intuitive scale. Say that, they are all more "fundamental" and we would surely consider them particles in its own right even if they were not as stable as they are. But they are very stable.

arivero
  • 1,943
2

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles?

Nobody has an issue calling the electron a particle. Ditto for a neutron. It's stable in a nucleus, and the fact that a free neutron decays in circa 15 minutes doesn't much matter. It's similar for a muon, which lasts for a couple of microseconds, because that's quite a long time in particle physics. Muons leave tracks, as do other non-stable particles. And because we can see those tracks, I don't think many people have an issue calling them particles.

Things start to get interesting when we move on to shortlived baryons where you hear the word "resonance". I'm afraid to say I always think of a twanged ruler when I hear that. Something that doesn't last, that's more of an action or an event than a thing. Particularly when it comes to the Z-boson which lasts for circa 3 x 10$^{-25}$ seconds. Nobody has actually seen a Z-boson, or a track. Its existence is inferred. It's similar for some other particles: "Experiments don’t detect the Higgs boson directly – instead, its existence is inferred by looking at the particles left behind when it decays".

It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms

I guess particle physicists call them particles, and everybody else goes along with that. But maybe somebody like anna v can give a better answer.

John Duffield
  • 11,097
2

You're question is interesting because it is connected to the notion of elementary particle. As mentioned by anna v, the elementary particles (fermions) of the standard model have very specific properties under the symmetry of the standard model ($SU(2)_L\times U(1)_Y \times SU(3)_c$): they lie in the fundamental representation of the group, which in familiar language means that they are the building block of all other particles (baryons, mesons etc). This definition does not imply that elementary particles are necessarily stable. An electron, a muon and a tau (3 elementary particles, leptons) have exactly the same property under a transformation of the symmetry group. Nothing distinguishes one from the others except their mass. Heavier particles can decay into the lighter ones.

Paganini
  • 1,823
  • This really is the nub of the matter, i.e. the irreducibility and the fundamentality of the representations as I (a nonspecialist) understand it. – Selene Routley Sep 23 '15 at 13:27
1

Your question seems quite general, but perhaps you're confused about what "decay" is. When we say something "decays" we don't always mean that it's somehow "breaking up" into it's constituent parts. In fact, we hardly ever do. The heavier particles aren't really "transient interplay of the stable forms", unless I misunderstand, and that isn't something that it makes sense to say.

Matt
  • 1,758
  • 1
    I have commented on Johns answer that may clarify how i see it. And i understand that the laws of conservation cannot be broken. How can fundamental particles have constituent parts? So i'm not including neutrons. By interplay i mean of energy more I guess – Justin Sep 23 '15 at 05:19