What's Up with Weak Decays? (Standard Model Part 6)
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- čas přidán 20. 02. 2022
- In this chapter, we discuss particle decays in order to introduce a brand new interaction into the standard model: the weak interaction.
Full Standard Model Playlist: • Standard Model
I love how you explain these terms without losing all of the tecnichal information.
The W+ boson is dual to the W- boson.
Leptons and quarks come in pairs -- duality!
"Always two there are" -- Yoda.
Awesome video!! 😊
We missed your videos. :)
Love 😘 your video 👍
Great Vid
Welcome back! I hope you're doing well.
Around 9:30 what does "directly observed" really mean? Don't we always just observe decay products? How is the high energy interaction different than observing the decay with "borrowed" energy? Is it more decay products associated with a single event? (and then doing "statistics" to it)
Good to be back! Great questions as always. I definitely get what you are saying, and you are right, we don't directly "see" a W boson in a detector. The answer has to do with the structure of the amplitudes for a process to occur. Basically, what happens in this case is that the amplitude changes behavior when the initial state energies are high enough to produce the heavy intermediate states. Essentially, it becomes far more likely for certain events to occur when the system no longer has to "borrow" energy to produce the intermediate states, so when we scan, e.g., an electron-positron collision over energy levels, you can typically see a dramatic increase in the number of decay products corresponding to, in this case, weak interactions. Now, what this behavior looks like depends on the intermediate state that is produced. I believe that for the original discovery of the W boson, they actually looked for W pair production (so, really, they needed even *higher* energies, >2*MW). In the signal (basically, the number of decays you expect from this process against the energy), you expect to see a "hill" which starts at twice the mass of the W boson.
Now, of course take the whole hand-waving argument of the system "borrowing energy" to produce an intermediate state with a grain of salt. This argument relies heavily on a somewhat literal interpretation of what we see in the Feynman diagrams, which we know isn't really a correct thing to do. Virtual states don't really physically exist: in fact, for energies below MW, no matter how closely you looked, you will never see an intermediate W boson (though, in principle, the closer you look, the higher energies you need). However, above MW, you could see that a W boson is physically produced if you had a good enough detector.
Perhaps the better way of saying what is going on is that the leading-order amplitudes for these processes feature heavy propagators which give a structure like E^2/(E^2 - MW^2) where E is related to the energy of the external state. So you see that, at low external momenta, you essentially get a E^2/MW^2 suppression of the amplitude, whereas when E ~ MW, the amplitude becomes very large.
@@zapphysics Hmmm let me try to understand this: what it means to "borrow" energy to make a heavy particle briefly is that the energy is unaccounted for by the total energy of the incoming particles beams? while on the direct detection of short-lived particle, the energies of incoming particles are large enough to make the intermediate state without having to refer to quantum uncertainty? When the intermediate particle *does* exist, the pattern of particle path in detectors will be different than if it doesn't, every experimental parameter being the same?
@@zapphysics This video also reminded me of the thoughts i have on these topics:
1. A weird thing to notice: even though W+- particles act as force-carrying bosons, they also act as charges to another force-carrying boson (photon)!
2. Kind of weird to me that everytime weak interaction is mentioned, it never seems to have anything to do with motion, and rather only is involved in transformation of particles to other particles (photons can make charges move in different ways in many length scales as we know, while gluons "glue" (attract) quarks together and spawning the nuclear force, which attracts nucleons together). Hmmm do the weak bosons have any repulsion/attraction property whatsoever?
3. On the point of virtual particles being a literal interpretation of feynman diagrams, i also gravitate more to that idea. If they're actually "particles", are they localized in space? can we pick a single virtual particle out like we can resolve an atom in a crystal lattice with X-ray diffraction? And the explanation on how particle exchange seems to be able to attract two charges (while having all the momentum just right) doesn't seem to be that satisfying to me. I'm beginning to think that our theories don't really need "particles" for them to work. Maybe QFT can result in something... else
@@geoffrygifari3377 Hopefully I can answer all of your questions! For your initial question, I think your general understanding is correct. The only thing I will say is that it isn't necessarily the behavior of the particles that we see in the detector that changes when we reach initial state energies sufficient to produce real intermediate states, but it usually has to do with the rate that a reaction happens. When a state no longer has to "borrow" energy, usually these rates increase dramatically.
As for your other questions/comments, I'll answer them one-by-one:
1.) Excellent observation of the somewhat strange fact that the W+/- is charged under electromagnetism. This is a pretty tricky topic, but it has to do with the fact that the weak interaction is actually deeply intertwined with electromagnetism. In fact, these two interactions can be unified into a single, electroweak interaction. At the risk of spoiling future videos, this electroweak interaction actually behaves remarkably similarly to QCD, most notably the gauge bosons associated with the interaction can interact with each other. Now, the symmetry associated with the electroweak interaction is spontaneously broken by the Higgs mechanism, resulting in some of the gauge bosons picking up a mass. The resulting gauge bosons which remain massless combine together to form the photon. However, the ability for the electroweak gauge bosons to interact does not go away and hence, we are left with weak bosons which interact with photons.
2.) Yes, absolutely there is a force associated with the weak interaction (in fact, there are a couple)! One can take a classical limit of these interactions and one finds a force very similar to the electric interaction between two charges. The reason it isn't as noticeable or as talked about is because it gains an additional exponential suppression like exp(-M*r) where M is the mass of the gauge boson and r is the distance between the particles. Since the weak gauge bosons are *so* massive, this force is hugely suppressed, especially when it may compete with electromagnetism or the strong interaction, except at extremely small separations. However, this is exactly how neutrinos (which don't interact with either the strong or electromagnetic interactions) are detected in experiments like Super Kamiokande: a neutrino comes into the detector and interacts with, say, an electron. This both causes a recoil (due to the weak force) and also changes the neutrino into a charged particle which can then be detected. As a note, there is another part of the weak interaction, known as the neutral-current interaction, which is basically identical to the electromagnetic interaction, but with a very heavy boson in place of the photon.
3.) Here, I will warn you to not take the idea of virtual particles too literally. In fact, this is truly a so-called "semi-classical" interpretation. This means that we are extrapolating the rules of classical mechanics to get an approximation of quantum mechanics. In other words, when we do calculations with Feynman diagrams we are not doing the full quantum-mechanical calculation. In quantum mechanics, we can really only ever know what we put into an experiment and what we get out of the experiment. Trying to talk about what happens in between in *true* quantum mechanics is pretty much non-sense. However, we can use virtual particles and Feynman diagrams as a mathematical tool to get a good approximation to the true quantum result in certain circumstances.
You are exactly correct that our theories do not need particles in order to work: we can, of course, just talk about quantum fields. It just so happens that, in non-interacting QFTs, one can truly define particles as localized excitations of these quantum fields. In our usual approximation of perturbation theory, we essentially treat the fields as free fields with small corrections due to the interactions. Therefore, we can still typically talk about things in terms of particles to a very good approximation. However, this assumes two things: there are no so-called global effects which wouldn't be seen by this particle picture and that the interactions are weak. If the interactions become too strong (as is the case for low-energy QCD), then this whole perturbative picture completely breaks down and it no longer makes sense to talk about individual particles. This is why it makes no sense to talk about a free quark in the same way as a free electron. For such strongly-coupled theories, one must either truly use the full field theory if possible (which generally is not the case outside of some instances with extreme amounts of symmetry) or a different type of approximation such as an effective field theory, e.g. chiral perturbation theory for low-energy QCD or lattice QFT.
I hope this mostly answered your questions!
@@zapphysics Very cool. Does this mean that electrons draw neutrinos to them kinda like the electron is drawn to a proton? Or is it the W boson that attracts them? Also, how do Z bosons fit into the picture? Thanks for making the complicated simple!
I examined the diagram. So do down quarks lose their negative electric charge by emitting w bosons? How do they gain a positive electrical charge after turning into up quarks? Where does this positive electric charge come from?
Great video! One question though. At 8:18, shouldn't the particle in the top right be a positron instead of an electron? Shouldn't the anti-muon decay into a positron, a neutrino, and an antineutrino?
Oh! That “a 4 particle interaction replaced with two three particle interactions, introducing a new heavier intermediate particle” surprised me and seems quite pleasant.
When one says that when quantizing gravity one gets a non-renormalizable theory, would one suspect something similar to happen beyond the energy levels where the effective theory works? Or, would it not apply in that case due to the reasons for it being non-renormalizable being different in some way?
(Hoping my question isn’t assuming something confused)
@drdca it is a great question and very insightful! This is really one of the main reasons why people sort of assume that there is a quantum theory of gravity and not that quantum mechanics needs to be "general relativized" (for a lack of a better term). One can absolutely try to include gravitational interactions with the standard model particles, and what one finds is exactly something with the structure of an effective theory with a "cutoff scale" of the Planck mass (~10^19 GeV). One can even do effective quantum gravity calculations using this effective theory (and people do!). But since it so strikingly has the structure of an EFT, the expectation is that there should be some sort of new physics (maybe not a new heavy particle, it could be something else) at energies above this scale.
Could the symmetry be in a n gatively curved space?
Griping, but did not explain why a neutron turns into a proton (or downess goes upness) or why this process occurs in two otherwise completely separate families of particles i.e Leptons and Baryons?
Alt yazi çeviri ayarlayamadim
1:41 what does it mean to "not participate in strong interactions"?
Good question! The strong interaction is basically just a force that only certain particles feel. For example, this is what keeps the nucleus of an atom together despite the fact that there are lots of positively-charged protons living very close together feeling a huge electric repulsion. For the muon to not participate in strong interactions just means that it doesn't feel this force. This is pretty much in exactly the same way that a neutron doesn't feel the electromagnetic force since it is electrically neutral. When a particle doesn't participate in strong interactions, that is a good indicator that it is not made up of quarks (since the strong interaction and QCD are very deeply related) and can't be classified as a hadron!
When first introducing beta positive decay to students (p+ --> n0 e+ v) with little to no particle physics background, I struggle to find an explanation that makes sense. Do you have an intuitive way of describing this without deeply involving the weak force?
@Spencer Wenzel I feel like the inclusion of "intuitive" makes this a bit of a loaded question lol, but I can try to help out a bit. I would say that the most "correct" way to go about it is to recognize that particles can typically be grouped together into families. For example, the proton and neutron form a family (really, it is the up and down quark here), the electron and (electron) neutrino form a family, the muon and (muon) neutrino form a family, etc. Weak interactions (specifically, the charged-current interactions) tie together particles within these families, so whenever you have a proton in a weak interaction, it has to come paired with a neutron. Whenever you have an electron, you have to have a neutrino. So on. Figuring out whether it is an electron or positron in the final state just comes down to charge conservation. As a note: if you want to be more specific, these "families" are determined by a quantity known as weak isospin.
So, if we know that we start with a neutron in the initial state, we know that we have to have a proton due to the connection through weak decays. But, this means that the initial state has zero charge and the final state has a charge of +e, so we have to balance that charge with an electron to have a net-zero charge in the final state. Again, since it is a weak interaction, this electron has to come paired with an anti-neutrino (seeing why it is an anti-neutrino is a bit tricky, though. Basically, in an interaction, you should always have the same number of baryons and leptons in the initial and final states. Anti-baryons/leptons count as "negative" baryons/leptons, so since we have no leptons in the initial state, we need to have one lepton and one anti-lepton. I will say this is a little misleading since baryon and lepton numbers are not truly conserved, but for all intents and purposes for this discussion, they are).
Now, we can do the same exercise for a proton within a nucleus decaying. We know from the connection through weak interactions that the final state better have a neutron, but this leaves us with an initial state with +e and an electrically neutral final state. To get this to work out, we need a positron in the final state, which is always going to come paired with a neutrino in a weak interaction.
If you don't really like this explanation, and if they tend to grasp beta-minus decay alright, you could go the route that beta-plus decay is essentially just beta-minus decay in reverse. The only caveat here is that you need some "particle-physics math" in order to re-organize your decay. The sort of quick and dirty way to justify how this all works is to remember that mathematically speaking, a particle travelling forward in time is the same object as an anti-particle travelling backwards in time (note, I don't _love_ this explanation, but it should get the job done). This essentially means that we can take an incoming particle and replace it with an out-going anti-particle and vice versa. In other words, in a particle physics process like A B -> C D, we are allowed to shuffle things around, e.g., B->anti-A C D, as long as everything is allowed by other conservation laws.
So, we can start with beta-minus decay n -> p+ e- anti-v and reverse everything in time to get anti-n -> p- e+ v, and now just swap the anti-neutron into the final state and the anti-proton into the initial state to get p -> n e+ v. Like I said, this is a bit quick and dirty, but it should work.
Let me know if you have any other questions or if you want clarifications, etc.!
@@zapphysics @ZAP Physics Thankyou for the response! I have never thought of explaining beta positive decay as beta negative decay in reverse. I also will find the ideas of families useful in my explanation. I appreciate it. If I could ask a few clarifying question, that would be great.
I find my students can grasp beta negative decay easily because the rest mass of the particles before the decay are greater than that of the particles after the decay. However, for beta positive decay, the rest mass of the particles before the decay is less than that of the particles after. You made mention to this in your video how a free proton decaying to a neutron would not conserve energy. If you could clarify a few points, it would be greatly appreciated
1) Is the mass before beta positive decay in the binding energy of the nucleus? So it would be:
proton + binding energy --> neutron + posititron + electron neutrino
in this formulation, is mass-energy conserved?
2) What specifically happens in beta positive decay? I am going to take a wild guess but am probably very wrong. Does a proton emmit a W+ boson, using the binding energy of the nucleus, and then then leaves the neutron while the W+ boson decays into a positron and neutrino?
I took mostly math courses in university and am trying to understand particle and quantum physics through youtube videos mostly. Any help would be great.
@Spencer Wenzel I would be more than happy to help out and clarify these points! I think that the key thing to reinforce is that, in nuclear decay, one can't really treat the protons and neutrons in the nucleus the same way as one would for free protons and neutrons because they are constantly interacting and are tied up in a bound state. The idea that interacting particles in bound states behave very differently from their free counterparts should be very familiar: just think of the electrons in an atom. The electrons in an atom are confined to a discrete set of states where they have this "cloud-like" behavior whereas free electrons can have any energy they want and often behave like localized point particles.
To directly address your questions:
1.) You are pretty much correct here, though I would say that instead of just having the binding energy on the left-hand side, I would say "change in binding energy." Again, these nucleons are bound up into nuclei (aside from the case where you are just looking at free neutron decay), both in the initial and final state, and neither of these nuclei will have the same mass as just the sum of all of the free proton and neutron masses individually. If you look at the masses of unstable isotopes which beta decay, you'll find that they will always be heavier than their daughter nucleus plus the electron mass (the neutrino mass is basically zero for all intents and purposes here), so there will actually be some _extra_ energy leftover in the final state. This extra energy just goes into the kinetic energy of the decay products. Actually, a very interesting bit of history is that this book-keeping of energies is exactly how the neutrino was first predicted!
Now, actually calculating the binding energy is very tricky since nuclear physics is incredibly messy. There are some approximations which give a pretty good result, and I think that the easiest to understand is the liquid drop model, which you can read about here: en.wikipedia.org/wiki/Semi-empirical_mass_formula
For beta decays, the volume and surface areas of the nuclei don't really change between the initial and final states, so this won't really account for any of the mass difference in this case (alpha decay is a different story). The main effects essentially come from two sources: the first is pretty easy to understand in that the electrical repulsion between protons in the nucleus wants to blow the nucleus apart, so the strong nuclear force which binds the nucleons together has to put in more energy to keep the nucleus bound, and this extra energy translates to more mass for the nucleus just by E=mc^2. The other major effect comes from the fact that the nucleons are fermions, just like the electron, so no two similar nucleons are allowed to be in the same quantum state by the Pauli exclusion principle. The effect is the same as for electrons in an atom: the nucleons are confined to "shells" of discretized energies. So, say we have a nucleus with 5 neutrons and 3 protons. By the same rules as for atoms, you see that two neutron shells are filled with one neutron forced to be in a higher-energy third shell, while only one proton shell is filled with an extra proton in the second shell. Here, you see that if we converted that extra neutron into a proton, it could live in the second proton shell and nothing would live in the higher-energy third shell at all, so the total energy of the nucleus would be reduced. This would give a preference for this nucleus to beta-minus decay (and in fact, if you look it up, you see that Lithium-8 does indeed beta-decay into Beryllium-8). There is also a third effect that essentially prefers that the shells are filled, exactly the same way that atoms prefer to have filled electron shells.
How these effects actually balance is pretty complicated, but if we account for the fact that the nuclear force is significantly stronger than the electromagnetic force, we could guess that, as long as we don't have many protons and the Coulomb energy doesn't get too high, the main effect at play is the extra energy coming from mis-matched numbers of protons and neutrons, leading to nucleons living in higher-than-necessary shells. However, when we start adding more and more protons, this Coulomb energy starts to really play a role, so we will need to try to space them out using neutrons so that the repulsion gets a bit smaller. This, combined with the fact that the energy difference between shells gets smaller as we add more and more nucleons means that heavier nuclei should have more protons than neutrons. And this is exactly what we see (en.wikipedia.org/wiki/Beta_decay#/media/File:Table_isotopes_en.svg): for light nuclei, the preference is for the same number of protons and neutrons, while for heavier nuclei, the preference is to be neutron-dominated.
All of this is to say that there are definitely ways to predict whether a nucleus will beta-plus or beta-minus decay and that the effect has much more to do with the interactions between nucleons in the nucleus than the actual mass of the individual nucleons.
2.) I think your understanding here is much better than you give yourself credit for! Your picture is basically correct up to some small nit-picky details. One thing that I will say is that a _real_ intermediate W boson in never produced in beta decay. In fact, if you look at the mass of the W boson, you'll see that it by itself is heavier than the entire nucleus for a good chunk of the atoms on the periodic table. So just emitting a W boson in this process is extremely forbidden by conservation of energy. This is where the idea of "borrowing" energy using the uncertainty principle comes into play: the system can borrow enough energy to produce an intermediate W boson, as long as it gives it back extremely quickly. The initial and final states conserve energy, but the intermediate state doesn't. This is one of the main reasons that the weak interaction is an incredibly short-range force.
Now, I will say that this picture is what we would call a "semi-classical" one: it really is an approximation using classical calculations of interactions modified by quantum mechanics in that classically forbidden intermediate states are allowed to exist. In true quantum mechanics, we can't know anything about the intermediate states. All we know is what we send into an experiment and what we measure coming out. So perhaps the "most correct" thing to say is just that the reaction occurs due to the fact that it is energetically favorable (the final state is lighter than the initial state) and it can be facilitated through interactions with a quantum "weak field" in the same way that 2->2 electron scattering is facilitated through interactions with a quantum electromagnetic field. However, most people just use the semi-classical description because it gives a pretty clear picture of the interaction in our head (and it turns out to be a _really_ good approximation in this case).
Sorry if that was a bit long-winded, but I hope it was helpful! Let me know if any other questions come up!
@@zapphysics thank-you for the response. It answered a question I had for months and answered questions I didn't know I had. Thank-you for making these videos, they are right on the edge of my zone of proximal development and they have been invaluable in helping me understand physics 👍
Is it true that in order to see a lone quark, one would have to put so much energy into the collision that a new quark pair would be created and so one can't theoretically observe a lone quark? If that is true, then doesn't that mean we can't scientifically confirm the existence of quarks? Isn't a consequence of this that if one had an alternate theory that didn't include quarks, was equally obscure to direct confirmation, and made the same mathematical predictions that neither of the two theories would any longer be science because in order to have science one must have a unique prediction and the ability to observe directly the theorized phenomenon? I mean to ask, are quarks really hard science? I don't see it yet, and people go on about strings and other fantasies... It makes me sad for the state and future of science.
You know, the ancient Egyptians started developing medicine, but then it seems their method of discovery became contaminated with religious beliefs and their progress stopped entirely. I see a similar thing happening to physics, where it mostly is just fantasy that Michio Kaku can sell popular books about now. People are talking about the 'science of consciousness' and so on.. it is all quite sad and embarrassing.
In the same way you can confirm the existence of protons and neutrons by creating a nuclear collision at such a high energy that the nuclei don't interact as a whole but the nucleons do, you can create a higher energy collision in which the quarks interact with each other and not a nucleon as a whole. This is, if I am not mistaken, what basically quark-gluon plasma is. We have observed it in accelerators and what we've seen agrees with the existence of smaller particles in nucleons. Does that put your mind to rest?