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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 ...

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Every nucleon has what are called sea quarks in it, in addition to the valence quarks that define the nucleon as a proton or neutron. Some of those sea quarks, especially the strange quarks, have some secondary relevance in practical terms regarding how the residual strong nuclear force between protons and neutrons in an atomic nucleus is calculated from ...

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Particles are called elementary if they are not made up of other particles. However, interactions can change an elementary particle into another kind of elementary particle. Quarks and leptons are currently believed to be elementary. (This could change if we could observe particles interacting at higher energies than, say, the LHC can achieve.) However the ...

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"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.

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There are all sorts of quarks around us all the time, but they only exist for short periods of time. The up and down quarks are what are known as first generation quarks. They have a relatively small rest mass of $1.7\,\mathrm{MeV}$ and $4.1\,\mathrm{MeV}$ respectively. Comparing this with the 2nd generation of quarks, the strange and charm quarks with ...

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It is a brute experimental fact, already apparent within the first year or two of the discovery of the structure of nucleons, in the late 60s, at SLAC. Deeply-inelastically scattering electrons off nucleons, one could probe the charge content of the proton and the neutron, and, through Feynman's parton model, understand its kinematics. By 1971, Feynman had ...

5

First note that it's the $\Delta^+$ particle that is $uud$. The $\Delta^0$ is $udd$ like the neutron. The $\Delta^+$ particle is an excited state of the proton, that is the proton is the ground state of two up and one down quarks and the $\Delta^+$ particle is the first excited state. Likewise the $\Delta^0$ particle is the first excited state of the ...

5

In QCD, much as in E&M, there is a chromo-magnetic dipole-dipole contact interaction $\propto S\cdot {S}'$, which will obviously differ between spin 1/2 and 3/2 baryons. Bag-modelers have used it to estimate the mass difference between nucleons and deltas with tolerable accuracy.

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We have absolutely no idea. Within the Standard Model there is no explanation, it just is. There have been speculative attempts to explain this kind of structure, but my impression is that they mostly focus on explaining why each family is much heavier than the last, not on the small differences within each family. Within the first family, you can provide ...

5

The reason for this is the so-called weak interaction (see Wikipedia - Weak interaction - Interaction types. With very low probabilities (that's why it is named "weak") it causes reactions for example like these: $$d \to u + W^- \tag{1}$$ $$u \to d + W^+ \tag{2}$$ $$d + W^+ \to u \tag{3}$$ $$u + W^- \to d \tag{4}$$ $$W^- \to e^- + \overline{\nu}_e \tag{5}$$...

5

Lets start with the basic definitions: The elementary charge, usually denoted by e or sometimes $q_e$, is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron, which has charge −1 e Charge is a quantity measured in the laboratory, and in coulombs, is $1.60217662 × 10^{-19}$ ...

4

In this entry one finds the Feynman diagrams of the triple vertex : The experimental measurements confirm the existence of these diagramsm and thus the ttH coupling . The results of statistically independent searches for Higgs bosons produced in conjunction with a top quark-antiquark pair and decaying to pairs of W bosons, Z bosons, photons, τ leptons, ...

4

Note that in the 3-dimensional complex space spanned by basis $\boldsymbol{\lbrace}\boldsymbol{u}\overline{\boldsymbol{u}},\boldsymbol{d}\overline{\boldsymbol{d}},\boldsymbol{s}\overline{\boldsymbol{s}}\boldsymbol{\rbrace}$, this basis is replaced by $\boldsymbol{\lbrace}\boldsymbol{\pi^{0},\boldsymbol{\eta},\boldsymbol{\eta}^{\prime}}\boldsymbol{\rbrace}$ ...

4

In the standard model, the quark masses are proportional to their coupling to the Higgs field; but the magnitudes of those couplings (called yukawa couplings) are unexplained. A theory which goes beyond the standard model, by explaining why the values of the yukawa couplings are what they are, has the potential to explain why the tilt of the quark masses ...

4

Momentum, not mass. The sentence you are misreading refers to highly energetic protons with kinetic energies dozens of times higher than the mass of the nucleon or antinucleon in question, the realm of perturbative QCD. The valence partons involved carry a fraction x of the momentum of the struck nucleon —see the formal definition in the PDG review — of the ...

4

Take a look at the pseudoscalar mesons, a well understood family, since they comprise the pseudogoldstone bosons of spontaneous chiral symmetry breaking in 3 light flavor QCD. Flavor SU(3) arranges the octet in this pattern, and the singlet is $\eta'$, off the picture, since it has no isospin or strangeness, so U or V spin, to have any of its properties ...

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They do form, but don’t last long. Delta baryons can have three up quarks (for the $\Delta^{++}$) or three down quarks (for the $\Delta^-$). These baryons are unstable and last only a few trillionths of a trillionth of a second.

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This is a question that entails the ability to manage QCD at low energies and this is an active field of research yet as we are not able to do it, unless for lattice computations. It can be considered as part of the more general problem of the determination of the phase diagram of QCD (see my answer here). The idea of chiral symmetry breaking in strong ...

3

The history is as you have described: “Color quantum numbers ... were discovered as a consequence of the quark model classification, when it was appreciated that the spin S = ​3⁄2 baryon, the Δ++, required three up quarks with parallel spins and vanishing orbital angular momentum. Therefore, it could not have an antisymmetric wave function, (due to the ...

3

You are asking about deep inelastic scattering. Originally, neutrons and protons were thought to be the same particle, the only difference (other then EM charge and mass) was called isospin. Protons and neutrons behave almost identically under the influence of the nuclear force within the nucleus. The concept of isospin, in which the proton and ...

3

They do! All three types of neutrinos are around us in great abundance and they are absolutely part of "ordinary matter" (as opposed for example to dark matter). The reason neutrinos do not combine with other particles to form something like atoms is because they do not have any electric charge. Yes. The standard reference website with the most up-to-date ...

3

For inelastic scattering, the cross section can be modeled like this: $$\left( \frac{\text{d}^2\sigma}{\text{d}\Omega\text{d}E} \right) = \left( \frac{\text{d}\sigma}{\text{d}\Omega} \right)_\text{Mott}\, \left[ W_2(Q^2,\nu) + 2W_1(Q^2,\nu) \tan^2 \frac{\theta}{2} \right]$$ where the first term corresponds to the electric part and the second one (angle ...

3

Let $|0\rangle$ be the vacuum state, and let $q_C(f)$ denote the operator that, when applied to any state, adds another quark with color $C$ and "spatial wavefunction" $f$. The statement that quarks are fermions means that these operators anticommute with each other. Now consider the three-quark state  |f,g,h\rangle := \sum_{\pi}(-1)^\pi q_{\pi(R)}(f)q_{...

3

Nucleons are complicated quark-gluon systems. If you smear out the energy density and model a nucleon as a sphere of mass around $10^{-27}$ kilograms and a radius of around $10^{-15}$ meters, you will find that the gravitational time dilation, which differs from 1 by something on the order of $GM/c^2R\sim 10^{-39}$, is completely negligible. If you think ...

3

As in your profile you say you are a ninth class student in India, it means that your physics background is still at understanding the classical physics level. The strong nuclear force and the strong force of particle physics belong to the quantum mechanical studies, which I doubt it is taught in your year. In classical physics force can be defined as the ...

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It depends on whether or not the system corresponds to a $SU (3)$ singlet representation (a one dimensional irreducible representation). This is the color confinement postulate: (here somewhat sloppily) particles we can observe are in a singlet of $SU (3)$. You can show by tensorial decomposition that the mesons and the baryons contain a singlet ...

2

One reason that a $\pi_0$ can’t decay into one photon is that a massive particle decaying into a single massless particle violates the conservation of momentum. This is obvious if you think about it in the rest frame of the pion. Another reason is that this would violate charge-conjugation symmetry. The C-parity of a neutral pion is $+1$, while the C-parity ...

2

Individual quarks cannot be seen in isolation in bubble chambers. By placing a bubble chamber in a powerful magnetic field, charged particles can be distinguished from their antiparticles because their trajectories will be bent in opposite directions.

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