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Does that mean that electrons are infinitely stable? The neutrinos of the three leptons are also listed as having a mean lifespan of infinity.

Qmechanic
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HyperLuminal
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    Also, there is perhaps only one single electron that keeps travelling back (as positron) and forth trough time ;) – Hagen von Eitzen Apr 27 '15 at 21:32
  • That only works for pair production/anialation events, @HagenvonEitzen. The weak interaction in particular torpedos that, as is explained in the rest of the anecdone you are referencing (as told by Feynman about a telephone conversation. It was not his notion; such claims are misquiting.) – JDługosz Apr 28 '15 at 17:11
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    This is exactly why I hate electrons. "They keep spinning around forever". Right, sure, perpetual motion, like I believe that. Electrons are evil. I think they are just made up and don't really exist, like gremlins. Oh, no wait, they are not really spinning because they are actually a cloud, no wait they are really waves. It's just BS piled on BS. – Ambrose Swasey Apr 29 '15 at 12:02
  • @HagenvonEitzen, positrons don't travel backwards through time. Only the theoretical tachyon does that. – psusi Apr 30 '15 at 23:06
  • Technically, the neutrinos continually decay into each other: https://en.wikipedia.org/wiki/Neutrino_oscillation – Zo the Relativist Jul 27 '15 at 19:42

4 Answers4

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Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that.

We'll go to your own rest frame--any decay you can do has to occur in all reference frames, and it's easiest/most limiting to talk about the electron's rest frame. In this frame, you have no kinetic energy, only rest mass energy equal to about 511 keV. So whatever you decay to has to have less rest mass than that--you might decay to a 300 keV particle, and give it 100 keV of kinetic energy, but you can't decay to a 600 keV particle. (There's no way to offset this with kinetic energy--no negative kinetic energy.) Unfortunately, every other charged lepton and every quark is heavier than that. So what options does that leave us? Well, there are massless particles (photon, gluon, graviton). There are also the neutrinos, which are all so close to massless that it took until very recently for anyone to tell that this was not the case. So you can decay to neutrinos and force carriers, maybe. Except then you run into a problem: none of these have any electric charge, and your decay has to conserve charge. You're stuck.

tl;dr: Electrons are the lightest negatively charged particle and therefore cannot decay into lighter particles without violating charge conservation.

zeldredge
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    Bravo! Great visual layman answer. I have a question though: Electrons move at an incredible (actually, unpredictable) speed. Wouldn't that affect what it could turn into? – HyperLuminal Apr 27 '15 at 17:42
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    No, because it has to happen in all frames. So if I'm riding along at the same speed as the electron--if I'm in a reference frame where the electron isn't moving, the "rest frame"--it needs to happen just the same. I might not know what that frame is, but I know that it exists. Because of that, I know that decays that I observe have to occur in that rest frame, and can use energy conservation in that frame to constrain the decay. (This is very different in, for instance, collisions, where the "rest frame" is that of the center of mass, not of the lone particle.) – zeldredge Apr 27 '15 at 17:45
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    Of course, this assumes that the electron lives in isolation. If there are other particles in the universe, they can give the electron some energy and/or charge, enabling it to change. – Ypnypn Apr 28 '15 at 04:31
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    If we consider that the electron is not a particle but really a probability density, can we use that to overcome the idea of a reference frame? Since in that case there's no single reference frame... I imagine it's an infinite number each with some probability. – user541686 Apr 28 '15 at 06:47
  • @Mehrdad it has to work in all reference frames, so if you consider an infinite number of them, you still have to take in account the 'worst' one which has no kinetic energy. You're free to take energy from outside (i.e. being struck by another particle at high speed) but that's not decay. – Peteris Apr 28 '15 at 10:51
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    Maybe this is a stupid question, but isn't there negative gravitational energy? Could you give up -89 keV of gravitational energy and decay to a 600 keV particle? – Konrad Höffner Apr 28 '15 at 14:38
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    @KonradHöffner negative changes in gravitational PE come when you move toward a massive body. The situation you're talking about would require something like an electron at 100km altitude "decaying" into an electron plus another particle at 1km altitude. Which is a cool idea to think about, but as far as we know physics is local, and when a particle decays, its decay products emerge in the same spot, so there is never any change in GPE. – David Z Apr 28 '15 at 14:49
  • BTW -- The totalitarian principle is usually attributed to Gell-Mann. – dmckee --- ex-moderator kitten Apr 28 '15 at 15:48
  • @DavidZ What if there's an extremely powerful magnetic gradient, like near the surface of a neutron star? So, like, one nanometer is enough distance to give up 89 keV? – Rag Apr 28 '15 at 16:45
  • @BrianGordon according to our current understanding, there's no way a particle can disappear from one location and its decay products appear in another location, no matter how close those two locations are. – David Z Apr 28 '15 at 17:17
  • So electrons last forever basically? – HyperLuminal Apr 28 '15 at 22:03
  • Oh yea, what about the rest of the lepton neutrinos? Could you give another answer? – HyperLuminal Apr 28 '15 at 22:04
  • Electrons will last forever unless they encounter another particle. As for the neutrinos, I think the relevant quantity is lepton number instead of charge, but it's a stranger question due to neutrino oscillations and I don't feel comfortable giving a firm answer about it. – zeldredge Apr 28 '15 at 22:29
  • There are force carriers with an electric charge, the $W^+$ and $W^-$ weak charged bosons. http://en.wikipedia.org/wiki/W_and_Z_bosons – Scott Centoni Apr 30 '15 at 20:29
  • They're not massless. – zeldredge Apr 30 '15 at 20:33
  • @zeldredge Good, the mass was my follow up question :P – HyperLuminal May 02 '15 at 12:09
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The statement is true for decays, where lifetimes can be measured.

It is not true for interactions though. A suicidal electron meeting a positron has a good probability to disappear, together with the positron, into two gamma rays, at low energies.

e+e-

Electron-positron annihilation

It is intriguing that this is not true for neutrinos. If an electron neutrino meets an anti-electron neutrino, the corresponding Feynman diagram would have two Z0s. As the Z0 is very massive, the annihilation/disappearance of the neutrinos could not happen at low energies, in contrast to what happens to electron/positrons.

Peter Mortensen
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anna v
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    Doesn't that diagram show an electron emitting a pair of gamma rays and turning around to go back in time? – Mark Apr 28 '15 at 01:13
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    @Mark If you take the x axis as the time axis, yes. with the y axis the time axis it is a prescription for getting the cross-section of an electron and a positron annihilating into two gammas. Feynman diagrams are iconic prescriscriptions/shorthand of calculations to be done. – anna v Apr 28 '15 at 02:58
  • @Mark http://physics.stackexchange.com/q/17521/44080 – J... Apr 28 '15 at 11:15
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    The electron wouldn't be suicidal if it hadn't spent its entire life being negative. – Darth Wedgius Apr 28 '15 at 22:05
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    Can't a particle decay pass through states with arbitrarily high energy (in the form of virtual particles) as long as the final collection of particles has the same amount of energy as the initial collection? – Tanner Swett Apr 29 '15 at 15:17
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    @Tanner Yes. So neutrinos theoretically can anihilate into photons. And havier neutrinos can decay into ligter ones. They just are so light that the decay takes too much time to be observed in practice. – BartekChom Apr 29 '15 at 15:42
  • @TannerSwett In the case of low energy neutrino antineutrino the virtual Z0 have to connect to another particle, but the weak coupling constant is very small and the crossection not measurable though calculable. The zero mass of the photon is that which allows for the above diagram on shell ( the only off shell is the electron between the two vertices) – anna v Apr 29 '15 at 16:55
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This is not exactly true. It is believed that net charge is conserved, but there is a weak process called electron capture, where an electron is captured by a nucleus, (usually from an inner "orbital" so there is a spectroscopic signature), a neutrino is emitted and a proton changes to a neutron. So therefore your textbook is wrong!

Rabi
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  • Maybe I stated "lifespan" incorrectly. I mean if there is nothing disturbing it, will it have an infinite lifespan? – HyperLuminal Apr 27 '15 at 17:43
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    Well there are plenty of processes that destroy an electron. Positron-electron annihilation will do it as well. This isn't usually what we mean by a lifetime, though. – zeldredge Apr 27 '15 at 17:45
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    But in the neutron isn't the electron just in 'cold storage' so to speak? It hasn't really decayed, but rather built up the neutron by combining its mass and charge with a proton. By the weak force, eventually the neutron will decay and the electron will be on its way once more. – docscience Apr 27 '15 at 19:15
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    @docscience Neutrons in stable nuclei are stable and will not decay. Also neutron is not composed of electron and proton. The electron really ceased to exist after the capture. There is nothing like a "cold storage" of particles. When electron is caltured by a proton, there is really one electron less in the universe. If a neutron decays, a new electron is born. – mpv Apr 27 '15 at 19:25
  • @mpv I'm familiar with the fact that neutrons inside the nucleus are more stable than those outside (experiment shows this), and there are theories that propose why this is so, but I don't believe we know why with any high degree of confidence. Your statement, "When electron is ca[p]tured by a proton, there is really one electron less in the universe." is also a proposition, right? – docscience Apr 28 '15 at 00:01
  • @mpv ...Show me experimental evidence that strongly supports this proposition. And assuming you could somehow track the particle how do propose to test that it's a 'new' electron that's born and not just the same one you started with? Isn't "cold storage' just as good a proposition? – docscience Apr 28 '15 at 00:01
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    @doc The theories that have had staggering success predicting properties of existing particles and the existence of of originally unknown particles are quite clear on the matter: the lepton number associated with a captured electron goes with the neutrino leaving zero leptons behind with the neutron. And while I'm not aware of any neutrino observations associated with electron capture we have all the other neutrino--nucleon processes. By contrast, confining an electron inside a neutron would cost more energy than is available. So no, "cold storage" is not a good alternative. – dmckee --- ex-moderator kitten Apr 28 '15 at 00:40
  • @dmckee But there wouldn't be any "cost of energy" once the state of the neutron is achieved within the nucleus. Don't the nuclear forces themselves maintain the more stable state of equilibrium to prevent decay of the neutron? And isn't that why if the neutron gets knocked out of the nucleus its life expectancy is severely shortened? What experiment can refute the existence of a stable proton-electron "system" within the nucleus vs a neutron fundamental particle? How do we know we are looking at another particle vs a system of particles? – docscience Apr 28 '15 at 14:05
  • @docscience You're focusing too much on the "how to disprove it". You first have to have "why believe it happens in the first place". If electrons are just an excitation of the electron field, the question "is it the same electron" simply doesn't make sense. But yeah, I imagine you could calculate how much of a electromagnetic moment the neutron would have if that were the case, and if it's not too tiny to measure, measure it. But first you'd have to show why the quark nature of neutrons is a bad description - the burden of proof is on your side. What is it chromodynamics is wrong about? – Luaan Apr 29 '15 at 07:55
  • @Luaan Todd Platt published an excellent editorial in Science, Oct 1964: "Strong Inference". His argument basically that Science should embrace the approach of multiple hypotheses rather than any one favorite hypothesis. Furthermore that one can never 'prove' a hypothesis but rather refute or falisfy competing hypotheses. It's a dated however historical and sound advice. – docscience Apr 29 '15 at 13:28
  • @docscience I'm totally on board with that, and it's the basis of scientific thinking. But you still should overcome reasonable doubt before you add a new hypothesis as an alternative - if I say neurons are communicating not just by electric impulses, but also by quantum gravity, it's an alternate hypothesis you can refute by eventually finding out that nope, there's no effect from quantum gravity. But it is more complex than the alternatives, without giving any new predictions or explaining any known discrepancies. It's correct scientific approach, but not efficient. – Luaan Apr 29 '15 at 13:35
  • @docscience How to disprove it should always be a part of scientific inquiry, and yes, we don't have the source codes to the universe so we can't proove any scientific law, but that's not the part I'm objecting to. It's just that by default, the simplest explanation wins. If you have an explanation that better fits the facts, that's awesome - but if you only have an alternative explanation to be alternative, with no way to distinguish between the two and yours is more complicated... why do you believe it in the first place? There had to be something to make you believe it, right? – Luaan Apr 29 '15 at 13:37
  • @Luaan In my case for this particular question it was just lack of sufficient information. So I posed the question here: http://physics.stackexchange.com/questions/178865/what-experiments-have-or-can-refute-the-existence-of-an-electron-particle-sys/178990#178990 Check it out. It really raised a ruckus. But it helped reshape my world view. – docscience Apr 29 '15 at 13:50
  • @docscience It's a good question and the answers are very good as well. I was only pointing out your rationalist mistake, not attacking or supporting your hypothesis or QCD exactly (the same as you, I'm not a particle physicist). Rational misfirings like this are just something humans do by default, so it's necessary to fight it wherever we go :)) As a programmer, I've taken the life advice of "fail often, fail fast" to heart - sure, spending thirty years proving your mistake is scientific, and the results might very well be important. But what if you could fail 30 years earlier? – Luaan Apr 29 '15 at 14:02
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HyperLuminal asked:

"Does that mean that electrons are infinitely stable?"

Think about Dirac's model of an electron, which includes left and right handed contributions.

Now add the (Nobel-worthy) Brout-Englert-Higgs idea, that the left-handed bit interacts with a condensate of weak hypercharge, while the right-handed bit does not.

This suggests a simple extension of the standard model: an SU(2) confinement able to hold together a fermion's left and right hand parts. Think of quark confinement, but at a shorter range.

Regarding "are electrons infinitely stable?", if such fractions of fermions can be associated, then nature may have a process for dissociating them... gamma-ray bursts?

For those working on so-called "WIMP miracles" to explain dark matter, it's this sort of electroweak connection that looks interesting; pre-electronic, pre-photonic, massive pre-fermions evolving into things that can emit photons (and hence be detected).

nnunn
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