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39

To maintain lepton number as a conserved quantity. Consider, in detail, what's going on in a beta decay (well, I'm going to ignore the nuclear context). The reaction is then $$n \longrightarrow p^+ + e^- + \nu \,,$$ where you should take the symbol $\nu$ to mean some neutrino (without prejudice about matter-type or anti-matter-type for the moment). There ...

20

Short answer: only the down-type quarks mix by the CKM matrix, by convention and without loss of generality. The lepton sector is completely analogous to the quark sector where the up-type quarks play the role of the neutrinos and the down-type quarks play the role of the charged leptons. Long answer: Let's recall some facts about the standard model. You ...

16

Your question touches the question of ontology in particle physics. Historically we are used to be thinking of particles as tiny independent entities that behave according to some laws of motion. This stems from the atomistic theory of matter, which was developed some two thousand years ago from the starting point of what would happen if we could split ...

14

No, muons can't decay into quarks because quarks are confined; the final product cannot be quarks, but rather composite particles made of quarks, such as mesons and baryons. The lightest mesons are the pions, which are already heavier than the muon, so any such decay is forbidden by energy conservation. On the other hand, the extremely heavy tau can and ...

12

Ill answer the second part of your question about effective mass and quasiparticles, since I see that CuriousOne has answered the rest better than I could have. In a metal or semiconductor, the electron is not in the same free state it would be in a vacuum. It is bound to (although delocalised within) a lattice of positive ions. So its dispersion ...

10

A fermion is any particle characterized by Fermi–Dirac statistics and obeying the Pauli exclusion principle. So for example quarks are fermions, as are Helium-3 atoms. A fermion does not have to be an elementary particle. I'm not even sure that it has to be spin $\tfrac{1}{2}$, though I can't think of any fermions that aren't. A lepton is a spin $\tfrac{1}{... 8 When energy in a given reaction is much larger than the masses of all leptons, these masses can be neglected. For example if you consider$W$bosons decays, then $$M_W \gg m_\tau.$$ Tau is the heaviest among the charged leptons, so if you can neglect mass of the tau, you also can neglect masses of other leptons. But now you neglected the only quantity that ... 8 Absolutely they can exist. In fact, physicists often creat muonic hydrogen to study things like the structure/size of the proton with more accuracy. The reason we don't see muonic/tauonic atoms in nature is that these particles decay very quickly, whereas the electron, being the lightest of the three generations of leptons, has an essentially infinite ... 6 This question can be split into two parts: How do three generations affect the way the universe runs within the standard model, and how do they affect beyond standard model physics? It is likely that the three generations are important in beyond standard model physics and without them particle physics would be very different. However we do not have any ... 6 Somewhat reluctantly, I decided to add to the otherwise excellent technical answers above, since none confronted the fundamental false premise of your question, "why do charged leptons not mix?". Of course they do. Charged weak currents merely link generations. Let me review its antecedents as you seem to be aware of the phenomenon, when you have all quarks,... 6 It is as @dmckee says, the weak eigenstates are the same as the mass eigenstates for the leptons, therefore the corresponding Cabibo angle is zero. Without going into the mathematics this plot says all: The Cabibbo angle represents the rotation of the mass eigenstate vector space formed by the mass eigenstates . θC = 13.04°. The CKM matrix is an ... 6 There exists a standard model of cosmology, i.e. accepted as the current status of research, called the Big Bang. This makes extensive use of the known interactions of particle physics encapsulated in the standard model., and assumes a singularity at the beginning of the universe where the energy seen in the universe now was originally generated. I like ... 5 No, leptons are not coloured. The reason the Pauli principle requires colour is that there are particles made of 3 quarks that would otherwise all be in the same state, so we have to have this other quantum number, as not to violate Pauli (poor Pauli, people always trying to violate him). Even if there was some state where some leptons were all packed ... 5 The Standard Model includes 12 elementary known as fermions that respect the Pauli exclusion principle. They include six quarks (up, down, charm, strange, top, bottom), and six leptons (electron, electron neutrino, muon, muon neutrino, tau, tau neutrino) (ref) All leptons are fermions, but not all fermions are leptons. 5 Honest answer for your question is ''No''! As in the standard model (SM), the number of fermion generations appears as an arbitrary parameter, meaning that a mathematically consistent theory can be built up using any number of fermion generations. Therefore, In order to answer the question perhaps we may need to beyond the standard model. In the quest for ... 5 Mass of neutron mass of proton The mass difference 939.6-938.3 is 1.3 MeV and baryon number is conserved anyway, so the neutron can only decay into a proton after and the excess energy is 1.3 MeV . The mass of the muon is 107 MeV and the mass of the tau 1777MeV. Energetically only the electron with its small mass of 0.5 MeV is available. 5 Yes. In fact, the first process you generally learn about in quantum field theory is the annhilation of an electron/positron pair to produce a muon/antimuon pair. 5 A left-handed up quark and a left-handed anti-down quark can annihilate into a W boson because of the nature of the weak interactions. Right-handed quarks do not interact via W boson because W boson is only left-handed, so they will not annihilate each other. On the other hand, electron and an anti-muon can not annihilate each other in tree level because it ... 5 You are contrasting the Weak interactions to the Strong interactions, where the breaking is spontaneous (and large) as contrasted to explicit (and small), respectively. ($e^{-}, \nu_{e}$) are a doublet under weak isospin. This means the the corresponding SM lagrangian is invariant under an SU(2) gauge group -- you could rotate the fields under such a ... 5 Kinetic energy does not matter. In its own frame, the neutron does not have enough mass/energy to decay in a proton and a muon. 5 The decay width$\Gamma=K m^5 $is an approximation that holds when all final state particles are much lighter than the decaying particle. For the muon, there is only one decay mode(the muon is not massive enough to create the lightest meson). So$\Gamma_\mu\approx K m_\mu^5 $. The tau has two leptonic decay modes, and one hadronic. The u and d quarks are ... 5 "Only the weak force can change quark flavours" is probably what you heard. This is not the same as "the weak force is only involved when quark flavours change". All fermions couple to the weak force. In fact, neutrinos only couple to the weak force. If an interaction involves neutrinos, then the weak force has to be involved. 4 Electric charge is a "special" kind of physical property because it corresponds to a very simple physical effect. But that's not true of most physical properties. The lepton number doesn't have any force associated with it, the way electric charge does, because it's not a coupling constant. Lepton number is just a mathematical expression of what it means to ... 4 Let me address your questions one by one. Why is it said that lepton number is conserved in Standard Model (SM)? How do I know that lepton number is an Abelian charge? The SM Lagrangian is invariant under the fermion transformations, $$\psi \to e^{iL\theta}\psi$$ where$L$is assigned such that$e^-$,$\mu^-$and$\tau^-$leptons and lepton-neutrinos ... 4 Part 1: The branch of string theory which actually tries to match experiment is called string phenomenology. The state of the art in string phenomenology is that, starting from different forms of string theory (heterotic string theory, M-theory, F-theory...), it is possible to define space-time geometries, arrangements of branes, background fluxes... such ... 4 In fact, the Yukawa Lagrangian is (more or less) only the term$\mathcal{L}_Y = -g \bar{\psi}\psi \phi$. The (massless) Dirac Lagrangian for fermions and Klein-Gordon Lagrangian (plus potential) for the Higgs are not shown in your formula. The main difference between the Yukawa Lagrangian and the simpler$-g \bar{\psi}\psi \phi\$ is that the Standard Model ...

4

This question showed up in my particle physics exam and I'm not sure of the answer. I said no, since I can't see any reason why they must be equal. There are no conserved quantities between the lepton and quark generations (for example there is no rule such as a top quark can only decay into a tau lepton or something), so there is no actual reason why ...

4

The strong force has each nucleon made up of three valence quarks and a quark antiquark gluon sea holding everything within the nucleon ( proton or neutron) , it is the Quantum Chromodynamic force in the standard model of particle physics The "nuclear strong force" is a spill over force from the strong force , in a similar way that the Van Der Waals forces ...

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