Although most particles disintegrate [the technical term is “decay”] into other particles, a few types of particles do not. Why not?
The world exhibits many types of particles — you can read about a lot of the (apparently-)elementary particles here, and there are lots of other particles which you can build out of more elementary ones, like protons and neutrons and atomic nuclei — but most of them decay in a tiny fraction of a second. I’ve explained in a previous article why most particles decay; it’s actually a form of dissipation, something we have some intuition for, from our experience of waves and vibrations. But why is it that a few types of particles do not decay at all, or at least live much longer than the 13.7 billion-year age of the (current) universe?
The only known stable particles in nature are the electron (and anti-electron), the lightest of the three types of neutrinos (and its anti-particle), and the photon and (presumed) graviton (which are their own anti-particles). The presumed graviton, too, is stable. The other neutrinos, the proton, and many atomic nuclei (and their anti-particles … I’m going to stop mentioning the anti-stuff, it goes without saying) are probably not stable but are very, very, very long-lived. Protons, for instance, are so long-lived that at most a minuscule fraction of them have decayed since the Big Bang, so for all practical purposes they are probably stable. The other rather long-lived particle is the neutron, which when on its own, outside an atomic nucleus, lives just 15 minutes or so. But neutrons inside many atomic nuclei can live far longer than the age of the universe; such nuclei provide them with a stable home. Finally, I should add that if dark matter is made from particles, then those particles, too, must be stable or very, very long-lived.
Why are these particles stable? It turns out that our world imposes some rules on particle behavior, ones not visible to us in the physics of waves and vibrations that we encounter in daily life, that prevent some particles from decaying, either rapidly or at all. The fundamental rules are “conservation laws”, statements that certain quantities in the universe never change in any physical process. (These quantities include energy, momentum, electric charge, and a few others.) There are also some approximate conservation laws, stating that certain quantities only change very rarely. Conservation laws do not appear from nowhere, imposed out of thin air by theorists; they are related to other properties of the world. For example, if the laws of nature do not change over time, then it follows (thanks to a theorem of the mathematician Emmy Noether) that energy is conserved. Meanwhile, the stability of the matter out of which we are made provides strong tests of these conservation laws, as we’ll see.
Combining these laws with the properties of particles leads to a set of simple rules that determine when particles simply cannot decay, or when they can at most decay very rarely. And these rules are (almost) entirely sufficient to explain the stability of the particles out of which we are made, and those that we interact with most often.
Here are the main ones. Their most important consequences for our universe and our lives are written in bold-face.
Rules of Nature Believed (for deep reasons) To Be Exact
1) A PARTICLE MUST DECAY TO TWO OR MORE PARTICLES.
This is why every decay that we see in nature involves two or more particles emerging from a single one. It follows simply from the laws of nature that the total energy and total momentum must stay constant in any physical process (or as physicists say, “energy and momentum are conserved.”) And Rule #1 follows directly from these conservation laws. Here’s the argument, if you are interested:
Suppose a particle of type 1 could decay to a particle of type 2 and nothing else. Let’s see there’s a contradiction: take a particle 1 and put it in front of you, sitting at rest. All of its energy is in mass energy. Now imagine it decays to particle 2. Conservation of energy says
Mass Energy of Particle 1 = Mass Energy of Particle 2 + Motion Energy of Particle 2
Since motion energy is positive, particle 2 must have mass energy less than or equal to the mass energy of particle 1. But the motion energy of particle 2 is positive, so if the mass energy of particle 2 is less than that of particle 1 that means particle 2 must be moving. But particle 1 started out at rest, so that means it had NO momentum. Particle 2 is moving, so it has SOME momentum. That’s impossible; momentum is conserved. Therefore the decay is impossible unless the two particles have equal mass. But in this case, if particle 1 could decay to particle 2, the reverse would also be true: particle 2 could decay to particle 1. Well, that’s not a decay at all; it is a mixing between the two types of particles, which is a qualitatively different phenomenon.
2) THE MASS OF A DECAYING PARTICLE MUST EXCEED THE SUM OF THE MASSES OF THE PARTICLES PRODUCED IN ITS DECAY
Total energy and total momentum are conserved in a decay, but total mass always decreases. A particle (“parent”) with a mass m1 may only decay to particles 2 and 3 (“children”) if the sum of masses of the children is less than the mass of the parent: m2 plus m3 must be less than m1. This is a simple consequence of the law of nature that the total energy must stay constant in any physical process. Here’s the argument, if you want to see it:
Imagine you are watching particle 1 at rest — just sitting there in front of you. Its energy is all mass energy m1c2. Then it decays to particles 2 and 3. Each one has mass energy AND motion energy. Since energy is conserved
Mass Energy of Particle 1 = Mass Energy of Particle 2 + Mass Energy of Particle 3 + Motion Energy of Particle 2 + Motion Energy of Particle 3
But since motion energy is always positive, we learn that the initial mass energy is more than the final mass energy, and thus m1c2 is bigger than m2c2 + m3c2, implying m1 is bigger than m2 + m3.
Since the photon is (as far as any experiment can tell) massless, there is nothing to which it can decay. That is why light waves can travel across a room, across space from the sun, and across the universe without disintegrating in flight. The same is presumably true for the graviton.
3) THE TOTAL ELECTRIC CHARGE BEFORE AND AFTER A DECAY MUST MATCH
Another conserved quantity is electric charge. A W– particle, which is very heavy and has negative electric charge -e, can decay to an electron of negative charge -e and an anti-neutrino of zero charge. But a W– cannot decay to a positron of positive charge +e and a neutrino of zero charge, because the total charge would then have changed from -e to +e. Not can a W– decay to an electron of negative charge and a positron (or “anti-electron”) of positive charge, because that combination would have total charge zero.
And since the electron is the lightest particle that has electric charge, there is nothing that it can decay to; only neutrinos, photons, gluons and gravitons are lighter, but they are all electrically neutral, so any combination of them would have zero electric charge. Any unknown particles that are lighter than the electron must also be electrically neutral, or we would easily have produced them in experiments. So the electron is stable.
4) THE TOTAL NUMBER OF “FERMIONS” BEFORE AND AFTER THE DECAY CAN CHANGE ONLY BY AN EVEN NUMBER
All particles are either fermions or bosons, as described here. The rule stated above follows from the fact that angular momentum, like energy and momentum, is conserved (which explains the tendency of spinning things like the earth to keep spinning.) The rule prevents a neutron from decaying to a proton and an electron; laws 1, 2 and 3 would be obeyed, but not 4, because all of these particles are fermions. Instead, a neutron decays to a proton, an electron and an anti-neutrino; then we have one fermion initially and three at the end, for a change of two.
There are three types of neutrinos, and it is now believed all three have mass (at least two of them do and probably all three — that’s a long story.) The lightest neutrino is the lightest known fermion, but the only known particles that are lighter, to which it could potentially decay, are bosons (the photon and the graviton.) Therefore it cannot decay at all: one cannot start with one fermion and end up with only bosons. [This neutrino could turn out to be unstable if there are even lighter fermion particles that we’ve so far missed because they interact even more weakly with ordinary matter than neutrinos do.] Incidentally, we do know neutrinos are moderately long-lived because we have observed them traveling great distances from distant star explosions, called “supernovas”.
Rules of Nature Believed (for less deep reasons) To Be Very Nearly Exact
5) THE TOTAL NUMBER OF QUARKS MINUS THE TOTAL NUMBER OF ANTIQUARKS MUST NOT CHANGE IN A DECAY
A proton contains three quarks plus many gluons and many pairs of quarks and antiquarks, so in a proton the number of quarks minus the number of antiquarks is three. A neutron also has an excess of three quarks. So a neutron, which is heavier, can potentially decay to a proton without violating rule 5 — and it does (along with an electron and an anti-neutrino).
But the proton is the lightest particle that has a more quarks than antiquarks, so it follows from this rule, along with rule 2, that it is stable. Clearly the proton cannot decay to any combination of electrons, photons, neutrinos etc. because these contain no quarks. There are some hadrons (particles made from quarks and antiquarks and gluons,) in particular pions and a few others, but these all differs from protons and neutrons in that they have equal numbers of quarks and antiquarks. Therefore a proton (which is heavier) cannot decay to any combination of pions plus non-hadrons (such as photons, electrons, neutrinos, etc.) because again the children will have the same number of quarks and anti-quarks, while the parent does not. Pions, by contrast, can decay without violating any of the rules; for example, an electrically neutral pion (which is a boson) can decay to two photons, while a positively-charged pion can decay to a neutrino and an anti-muon, a fact that is very useful in making neutrino beams.
Actually it is believed by many theorists (though not yet demonstrated experimentally) that this rule is violated by a tiny amount, and that the proton is very very very slightly unstable, with an extraordinarily long lifetime. By looking for a decade or so at huge numbers of protons (in a giant vat of water [here’s a link to the Super-Kamiokande detector]) and seeing no sign of even a single one decaying, we know the proton lifetime is at least 10,000,000,000,000,000,000,000,000,000,000,000 years. I hope I didn’t miss a zero. The age of the current phase of the universe is about 13,700,000,000 years, so there will be plenty of protons around for a long time to come.
There are other rules too, but the majority of effects we see around us follow from these few alone.
Summary, and a Couple of Questions I Haven’t Yet Answered For You
In particular, we now have all the rules needed to explain:
- why photons are stable
- why electrons are stable
- why protons are stable or very long-lived
- why at least one type of neutrino is stable or very long-lived
which is basically all you need for ordinary matter, chemistry, sunlight, and lots of other processes of daily life — except for one thing. What about the unstable neutron?!
The neutron is a very curious case. There is no rule preventing neutron decay, and indeed it does decay, after on average about 15 minutes, to a proton, an electron, and an anti-neutrino. Why is it so long-lived? This is partly because the proton and the neutron have masses that are so nearly equal. Although the neutron has mass-energy of almost a GeV, the mass-energy of the neutron is only about 0.0007 GeV larger than the sum of the mass-energies of a proton, an electron and an anti-neutrino. Decay rates become very slow when the children from a decay have masses that add up very close to that of the parent; that’s not surprising, since by rule 2 the decay rate has to decrease to zero once the children have more mass than the parent.
But the really odd thing is that if you put a neutron in certain atomic nuclei, it becomes stable! Helium, for instance, has two protons and two neutrons. Even though a neutron by itself lives a quarter of an hour, a helium nucleus will live for the age of the universe and longer. In fact this is true for the neutrons in the nuclei of all of the stable elements in the periodic table. This fact is a hugely important consequence of Einstein’s theory of relativity (and some details of the strong nuclear force), for without it all the complexity of our chemically rich world would be absent. And this remarkable story deserves an article all its own.
Oh, and by the way, if dark matter is in fact made from unknown particles, why are they stable? Nobody knows for sure, but probably the list of rules I gave you above is not enough. Most likely there is a new conservation law, exact or approximate, yet to be discovered.
88 Responses
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An international research team working on the Borexino experiment in Italy found the decaying of electrons. And now?
https://gizmodo.com/electron-lifespan-is-at-least-5-quintillion-times-the-1747606990
I am a new reader of your website, and congratulations because it’s “vicious”, but I didn’t find nothing about Borexino experiment and I could appreciate your point of view. TIA
You’d better reread the article. They didn’t conclude that electrons decayed. They put a lower bound on the half-life. Their results were consistent with no decay. (It’s impossible to measure exactly zero decay; one can only put an upper bound on the decay rate, or lower bound on the decay half-life.)
Thank you very much for your website. I greatly appreciate your making these concepts a little bit more accessible. For your rule 1, what prevents a situation where a particle at rest decays to a particle of less mass but that is still at rest? Perhaps the rest of the energy just dissipates into a non-particle form or else just completely vanishes?
There’s no such thing as energy divorced from particles. Energy conservation means that the decay particle’s energy must go into the product particles. Yes, some energy does go off in light that’s not explicitly included in the reaction as written out. (Radiative corrections.) The amount is almost always negligible.
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In a neutron star, the gravity is so strong the escape velocity is 10% or more of the speed of light. Should any neutrons break down into a proton, electron, and anti-neutrino, gravity forces them back together.
Thanks for good explanations. So in a “neutron star” neutrons are stable or not and how?
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What do you think? I wonder if ‘fundamental’ photons could decay into something more fundamental – a shower of tachyons?
In the tachyon world – tachyon particles would simply be another form of the substance that’s neither matter nor energy yet has variations in ‘tachyon energy’. Those with maximum tachyon energy would be considered ‘energyless’ (with none of our kind of energy – instead all its real ‘energy’ is in the form of tachyon energy totally travelling through tachyon space) just like how a photon is considered ‘massless’ in the energy-matter/anti-matter world. On the other hand a heavy ‘frozen’ tachyon (a tachyon having no tachyon energy) travelling in vacuum resembles a photon travelling at the speed of light – its slowest speed travelling in only ‘tachyon time’.
The maximum tachyon velocity in absolute vacuum would represent the new modern ‘universal speed limit’ with the tachyon particle as the ultimate elementary particle. Because all matter and energy is contained in the tachyon world from the pure tachyon’s frame of reference everything in our known universe currently travels at this maximum tachyon speed in absolute vacuum (yes void of tachyons) and not at the speed of light.
Would a tachyon particle also have a wave nature like a photon? I don’t know but I say it does because I think particles cause waves so I’ll stick to treating it as a particle for now.
If we change the Lorentz factor denominator term (1- (v/c)2) to (1- (v/T)2) where v is always > c but less than or equal to the max tachyon speed (T) (I wanted Tau subscript c but for this crude edit we’ll show T here). Now since the denominator is never negative and we rely on tachyons instead of photons to observe like tachyons then there’s no causality issue and no absorption/emission effect of photons going on with tachyons (we can never directly observe a tachyon using photons thus never see them travel back in time.)
Perhaps the medium of space itself within our known universe is made of mainly non-interacting tachyons propagating almost instantaneously through the known universe allowing the possibility for the perception of expanding space to travel faster than light! Space can be thought of as a sea of virtual photons (and all kinds of strange tachyons) that briefly pop into existence only to quickly decay back into specific tachyons – even creating entanglement effects.
Could certain tachyons form from near the singularity of large blackholes where super condensed photons totally lose their wave nature and fuse together to form ‘unfrozen’ tachyons that easily escape the event horizon? If so such emissions could easily stabilize a blackhole’s core for billions of years – making them invisible tachyon stars!
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I was researching the matter anti-matter balance and came across http://en.wikipedia.org/wiki/CP_violation . Some mesons decay into electrons slightly more often than positrons. Now to incorporate this imbalance into the big bang theory…
1. The big bang starts. Everything is energy, creating particle – anti particle pair, which then meet and convert back to energy. Some mesons produce extra electrons.
2. As the energy levels decrease, creation of proton / anti-proton pairs ceases, later creation of electron / positron pairs cease.
3. When electrons orbit protons to form hydrogen, the particles in these atoms are then electrically neutral and don’t attract their anti-particle, greatly increasing their lifetime. Neutral atoms composed of protons and electrons are attracted together by gravity and form the first suns.
4. Charged particles are repulsed by the same charge keeping them evenly distributed throughout the universe, and could push out the boundary of the universe. Excess electrons and anti-protons could be the dark matter and their negative electric field could be the dark energy.
5. Another decay process that is unbalanced could reduce (or increase) the number of anti-protons in the universe.
Step 4 has a serious problem. The antiprotons would occasionally collide with the protons, and annihilate them, producing products we would detect. Also the remaining electrons in particular distributed throughout the universe would cause a fog obscuring everything else.
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During any process of decay, there is a release of energy. Wouldn’t that actually be “Mass Energy of Particle 1 = Mass Energy of Particle 2 + Motion Energy of Particle 2 + Decay Energy Released” ? That would actually account for the heat released during the process of decay. This heat would likely not have measurable mass, but would be absorbed by surrounding particles. Transfer of energy.
The “release of energy” is the motion energy (kinetic energy) of the particles. The energy only becomes heat later on, when the daughter particles interact with the rest of the system.
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Thank you for this brilliant series!
I think that if the decay products are exactly stationary (in the frame of the initial particle), you get the same potential problem as one particle decaying to one particle — the process can reverse and one has instead a mixing.
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If conservation #5 can be violated, is it possible for an electron, an electron antineutrino, and a proton to be created simultaneously? (X –> e^- + v + p^+) It seems to me (a mere layperson) that this kind of production would satisfy the first four conservation laws: 1.) 1 particle into 3; 2.) assuming hypothetical particle X is heavy; 3.) neutral X into positive proton, negative electron, and neutral neutrino; 4.) 0 leptons turns into 2 leptons.
This scenario comes to mind because I know of the unanswered problem of why there is so much matter, but no antimatter in the universe. Imagine that the above transformation were slightly more common than the antimatter version (X –> e^+ + v + p^-). If this were the case, even if it were extremely rare, it would act as a ratchet in the early universe, creating more and more matter. Most generation of mass particles would be in the form of pair production, which would be unproductive due to rapid annihilation (e.g. X –> e^+ + e^- –> X). However, on occasion, you would have a triplet (X –> e^- + v + p^+) or “anti”triplet (X –> e^+ + v + p^-). If both happened at the same rate, then their products would annihilate, and, overall, the process would be unproductive. However, if one process were even slightly favored over the other, you would end up with a tiny excess of protons and electrons over the soup of matter-antimatter pairs. Since these excess protons and electrons would have no antimatter counterpart with which to annihilate, they would stick around, i.e. the process could go only in one direction, X –> e^- + v + p^+ .
Am I missing some conservation law? I understand that B – L is violated, but so what?
X -> e- + v + p+ — that’s exactly what happens when a free neutron decays, with v as the electron’s antineutrino.
I think that when looking for the asymmetry between particles and antiparticles, we should consider quarks rather than protons. The complications of combining quarks to form protons would confound the analysis.
Dear Professor,
I have a question about proton decay:
We know that the proton, with certainty, doesn’t decay very easily.
Quite unlike the free neutron, which will tend to decay when left on to it self for a bit.
And also unlike every other hadron which will decay much faster still.
We ostensibly know this because bound protons, like those in iron, or lead, or the waters of Kamiokande don’t seem to decay often enough for our experiments to pick up on it. But then, bound neutrons, like those we find in water, or lead, or iron won’t tend to decay too easily either.
How do we infer, from Super-Kamiokande or other experiments, that free protons won’t decay quickly compared to bound protons, though not fast enough to be immediately obvious over the duration of the experiment?
How do we know that free protons, in intergalactic space, or the relatively cool plasma of stars, don’t go pop a lot more often than those that have some of their mass-energy tied up in the nucleus of an atom? Can you look at a suitably big bottle of hydrogen and infer good lower bounds on (free) proton decay from there?
Perhaps it would be better to say I have a confusion about proton decay. Why does one suspect it to happen at all?The process, if it’s there, seems so dilute that it would, perhaps, better to state that proton decay has been refuted by experiment, than to imply, as you seem to do above, that the effect just hasn’t been observed… yet. But then, this would be equivalent to saying that all swans are white, because, well, that’s what we’ve seen, this far. Not good. Then, what data would you need to say, simply, nature does not allow proton decay. It seems to me that at worst, quantum fluctuations would come and bite you after the universe is cold and void from metric expansion.
In the meantime, I shall have to try to refrain from uninformed speculation while I look forward to your reply.
PS: Your blog is _thoroughly_ awesome.
Excellent question; it all has to do with looking at the energies involved very carefully, and applying energy conservation. I’m going to write an article about these things in the next week or two, but in the meantime, as a warmup, look at today’s article, http://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/the-energy-that-holds-things-together/ .
The issue is: how large is the binding energy of the system in which the proton and neutron are sitting relative to the amount of energy that will be released in the decay.
When a neutron decays to a proton, and electron and a neutrino, most of its mass-energy goes into the proton’s mass energy. The amount of energy left over is a few million electron volts. But the binding energy of a nucleus is often larger than this. And therefore the energy released in neutron decay is (in stable nuclei) insufficient to overcome the binding energy.
When a proton decays, say to a pion and a neutrino, or to any of the other possibilities that don’t violate any exact conservation laws (such as electric charge conservation), the amount of energy released is in the hundreds of millions of electron volts. But the binding energy of a proton inside an atom or molecule is a few electron volts (not a few million, just a few, period.) So that binding energy is to tiny that it is easily overcome by the energy released in the decay.
It’s the same reason why the explosion of a stick of dynamite is pretty much the same in outer space as on earth; the earth’s gravity is far too tiny to affect the large amount of chemical energy released as the dynamite explodes.
Iam very grateful for the explanations however if charge must be conserved why is their more matter than antimatter in the universe plus what makes up a photino
Charge conservation has nothing to do with the matter/antimatter asymmetry. Matter is (on average) uncharged, and so is antimatter; both atoms and anti-atoms are electrically neutral. There are just as many positively charged particles as negatively charged particles. It’s just that there are more electrons and protons than positrons and anti-protons; that puzzle isn’t about electric charge, but about the number of leptons and quarks versus anti-leptons and anti-quarks.
Photinos, if they even exist, are fundamental particles just as photons are. They aren’t made from anything more fundamental, unless photons are too — and we have no evidence for that as yet.
My last comment has also led me to reconsider law 1)…
1) A PARTICLE MUST DECAY TO TWO OR MORE PARTICLES.
In particular, what is to stop…
a) a massless particle from decaying into a single (but different) massless particle of the same energy and momentum as the original?
b) a massive particle from decaying into a single (but different) massive particle which is at rest relative to the original particle and has the same mass as that particle.
In case b) I’m picturing the classical analog of a ball that simply changes its color without any motion ensuing.
Maybe there is some uniqueness constraint at play that says there is only one type of particle with a given mass, energy, and momentum, or something…?
Also good questions. I made some sort of cryptic comment about this.
Carrying forward your analogy: if the ball can change its color from red to green, it can change it back again too. This isn’t a decay at all; it is “mixing” of states, a flipping back and forth, and it is a very important phenomenon. For example, if you try to make two particles with equal masses mix, you will discover that instead you will end up with two particles with different masses, each of which is a mixture of the particles you started with. This is an extremely famous property of quantum mechanics (although it must have a nice classical analogue that I ought to be able to explain to you, but the right example isn’t coming to mind just now.)
And when you mix two massless states, you may end up with various things. For example, if you mix two massless spin-1/2 fermions, you may in some circumstances end up with a massive spin-1/2 fermion. In fact, that’s how the Higgs field induces a mass for electrons, top quarks, and the like — once the Higgs field is non-zero, it forces this mixing to occur. See http://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-particles-if-the-higgs-field-were-zero/ .
Regarding the following implication…
2) THE MASS OF A DECAYING PARTICLE MUST EXCEED THE SUM OF THE MASSES OF THE PARTICLES PRODUCED IN ITS DECAY
=> Since the photon is (as far as any experiment can tell) massless, there is nothing to which it can decay.
The argument doesn’t quite gel with me. Your justification for 2) requires you to consider an inertial frame in which the initial particle is at rest. But this is of course impossible in the case of the photon (or any massless particle for that matter).
In any inertial frame, the initial photon has positive energy and therefore, at least within the terms of this constraint, it is conceivable that the photon could decay into two lower energy particles. For example, what is to stop the photon from decaying into two lower energy photons? In fact, what is to stop the photon from decaying into particles having positive mass, as long as the energy and momentum are conserved?
Basically, I think 2) needs a more compelling justification when massless particles are involved.
Check the algebra — you can’t do it. Just try it: start with a photon that has energy E and momentum p = E/c, and now try to make it decay to two particles of any mass, conserving energy and momentum. The only solution is if the two particles are massless and both traveling in the same direction as the photon — which is yet another mixing phenomenon (see the answer to the next question.)
Hi Matt. Thanks for these (incredibly rapid!) responses. I’ll have to ponder them some more, but in the meantime I think it would be great if you could say something explicitly about the massless case in your laws since they are technically silent about what happens if you can’t put yourself in the rest frame of the initial particle.
Thanks again. Your blog is wonderful!
Derek
P.S. Not sure where the best place is to request topics, but I would love to see a clear article about the Pauli Exclusion Principle somewhere – presumably this is the whole reason we care about the distinction between fermions and bosons (right?), yet it is really poorly explained on the web.
Hmmm… Will have to think about that (Pauli exclusion principle explanation.) I am always best at the things for which there is something intuitive we can draw upon… this isn’t one of them.
Very interesting article. There is one thing I still do not understand. I did check the algebra and it seems that a photon can “decay” into two photons flying in the same direction (but not into massive particles or into photons moving in different directions). Even so, this seems to result in light with twice the wavelength and double intensity. But clearly that is nor what is going on or the world would be a very dark place. Does this decay not happen after all? Are the two photons somehow identical with the original one? In any case, it seems quite different from any mixing phenomenon you mentioned.
Thanks for your time!
The stability of neutron is just a sign of approximate isospin symmetry, we could say that it is a naturalness principle: restore isospin, and the neutron gets the same mass that the proton, and then it is stable. A lot more puzzling is the stability of pions, via two disparate mechanisms: the anomaly for the neutral pion to gammas and, more amazing, the near equality of muon and charged pion mass. You can not invoke isospin when the mechanisms are so different, and of course you can not invoke supersymmetry when one of the particles is a composite (nor to say that you should put a kaon and some neutrinos in the game).
Um, what about the decay of a black hole? It would seem you could violate rule 5 because you can’t know how many quarks and antiquarks went into forming the black hole. Also rule 4 since You don’t know how many fermions went into the black hole.
With rule three you can know what charge a black hole has. But what if you had a black hole that was the same mass as the lightest charged particle. Would it be stable? Would it be a superposition of the black hole and the charged particle? Is that charged particle just a tiny black hole anyway?
Black holes and their decays are a whole subject, deserving of an entirely separate set of web-pages. I’m not going to get into it here… this section of the website is about standard particles, and its purpose is to explain to the non-expert why the observed properties of the known particles are what they are. Let’s put off these questions until I have time to build some black-hole pages.
Black holes conserve angular momentum, so the fermion rule is still obeyed, since the only way to get an integer amount of angular momentum is with an even number of fermions.
Thanks for this article. A couple of points:
1. The last step in the argument for Rule 1 wasn’t clear to me: “Therefore the decay is impossible unless the two particles have equal mass. But in this case, if particle 1 could decay to particle 2, the reverse would also be true: particle 2 could decay to particle 1.”
I don’t see why the second sentence follows from the first. Are we starting from the hypothesis that *any* particle can decay to *any* other(s), and then imposing rules to see which decays are left?
2. Under Rule 5, you’ve explained why protons can’t decay to other particles made of quarks and anti-quarks, but you haven’t explicitly ruled out the possibility that a proton might just disassociate into a bunch of free quarks and anti-quarks. So is there another rule needed here to say that quarks and anti-quarks aren’t allowed to run around freely on their own?
1. This is a subtle point, and I have somewhat finessed it, so you are right to complain. First, it is a fact of quantum field theory that if A -> B is allowed then B –> A is also allowed. (For instance, if particle 1 can decay to particles 2 and 3, then particles 2 and 3, if slammed together in the right way, can turn into a particle 1). So that part’s true. What is more complicated is what follows when 1 –> 2 –> 1 –> 2 –> 1 –> … is possible. This has effects that are too complicated for me to explain here, but what happens, essentially, is that instead of particles 1 and 2 of equal mass, you actually end up with combinations of 1 and 2, let’s call them 1′ and 2′, that have different masses from each other (and from 1 and 2.) So in the end, the situation you are asking about, where two particles have the same mass and could in principle decay to one another, never actually arises.
2. Good question. The reason quarks and anti-quarks can’t run around on their own is not due to a rule — a conservation law — but to the grip of the strong nuclear force, which binds them permanently inside of hadrons. The hadrons can break apart into other hadrons, but the quarks never get free on their own. Obviously the strong nuclear force, and its effect on quarks, deserves a detailed article … one which I expect to write sometime this fall.
“you actually end up with combinations of 1 and 2, let’s call them 1′ and 2′, that have different masses from each other”
Hmm, that’s surprising! At first I wondered whether this might be related at all to the phenomenon in your “if the Higgs field were zero” article, whereby the left- and right- versions of particles are “reunified” by flipping back and forth from one to the other through their repeated interactions with the Higgs field. But in that case we just end up with a single particle, so I guess the two phenomena aren’t related. Moreover, in the present case we’re just thinking about decays, which (as I understand it) are spontaneous, not mediated or triggered by any external field. (Is that right, or is there something more subtle to be said about why a particle decays at one moment rather than another?)
Nice article!
You briefly mentioned Noether, so according to Noether’s theorem a conservation law implies a symmetry.
What are the symmetries associated with rules 4 and 5?
Other way round: according to Noether, a symmetry implies a conservation law. The reverse does not follow. Rules 4 and 5 are often stated as following from Noether’s theorem, but there’s a subtlety. I want to get it precisely right before answering your question.
Look forward to reading your article explaining why neutrons are stable when they are sitting inside atomic nuclei!
Nice article. I think it might be worth giving some justification for rule 5 though as all the other rules had a justification.
Small mistake in your article: “So a neutron, which is heavier, can potentially decay to a proton without violating rule 5 — and it does (along with a positron and a neutrino).” (since that would mean no charge -> double positive charge)
Thank you! Fixed.
how a neutron decays to a proton, positron and a neutrino? Charge obviously not conserved here and the decay is not found in nature…..
It **doesn’t** decay this way; that’s why Thilo said it is a “mistake” and why I “fixed” it in the article.