# Tag Info

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Beam instabilities driven by collective effects The from the beam dynamics point of view, we pretty soon encounter instabilities given by collective effects. The most relevant ones, in case of a single bunch, are the beam-beam interaction, which takes place when two bunches cross each others probing their self-fields, and the wakefields/impedance/image ...

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how do these cathode rays cross the anode? Either through the perforations or past the edges. Some won't make it through, but the ones that do will be observed. The electric field is strongest in the region between the anode and cathode. While some of the field does leak out to the sides and behind them, it is significantly smaller. The electrons are ...

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You could rephrase the question as: what problems will I encounter if I gradually increase the pressure? A colleague of mine who used to design cathode ray tubes explained that the main reason for pursuing a high vacuum is lifetime. With some residual pressure, the tube would still work, but any (positive) ions that are formed due to gas-electron collisions ...

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Well, "low pressure" could be considered almost any CRT built by man, since it is very, very hard to achieve a true, absolute vacuum. Cathode rays were systematically observed in Crookes tubes at 10^-6 atm, which is definitely not a vacuum, but can also be considered "low pressure".

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Elementary particles are classified into two groups: Bosons & Fermions. Fermions comes with two families: quarks and leptons. Leptons come with three generations (till date no fourth generation leptons observed). Same is true for quarks as well.The first generation consists of electron $e^{-}$ and electron-neutrino $\tau_{e}$. Standard way of ...

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Fermion number is conserved in the standard model, Fermion propagation paths never end in a Feynman diagram. The examples given are beyond the standard model. However, in various Grand Unification models Fermion number violation is predicted. Neutrinoless Double beta decay in based on SO(10) which is a GUT model

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Another attractive feature of this conjecture is that it is similar to another conjecture related to hadrons that is known to be true: that the spin of a hadron is equal to the sum of the spins of the quarks in the hadron (which come in discrete half integer increments), even though non-quark partons in the hadron have non-zero spins that "magically" cancel ...

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Let´s go: [1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition. We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations: From a cross-section of longitudinal (scalar) ...

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I would answer this question differently. From other perspective, a electron gets its charge by the only generator that is not broken after the S.S.B of the SU(2)xU(1) gauge group. In this case $$Q = \frac{1}{2} Y + T_{3}$$ Where $Y$ is the hypercharge eingevalue and $T_{3}$ is the eigenvalue related to the SU(2) diagonal generator. So, as every ...

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The solution you quote actually doesn't conserve momentum. You can use that $p^2$ is a Lorentz invariant and solve $(p_\nu+p_p)^2=(p_e+p_n)^2$, considering the left hand side in the lab frame and the right hand side in the CM frame. You can find $E_\nu$ and check that now momentum is conserved. Anyway, as mentioned in the previous comment, ...

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As you say, loop corrections change the value of the quartic coupling $\lambda$ is modified by loop corrections. If the renormalization group changed its value so $\lambda \gg 1$, the perturbative interpretation of the theory would break apart. Note that the condition $\lambda < 1$ (or $\lambda/(4\pi) < 1$) is only a hint of the perturbativity, the ...

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The parton model assumes that nucleons are composed of three valence quarks that share the momentum of the nucleon in an approximately equitative way. That means that the valence quark pdf's have a peak around $x = 1/3$, and have much lower values at $x \to 0$ and $x \to 1$. In addition to that, you have interactions between quarks (nowadays we know that ...

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I am very bad at drawingin "paint", but the process can go as antineutrino +neutrino to Z0 , Z0 to s antis (or up antiup), a gluon vertex from the quark to an up antiup (or strange antistrange quark) the parenthesis are the alternate diagram. , So it is not forbidden, it has two weak vertices and so very small cross section.

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Your first diagram is wrong, since there is no vertex in a Lorentz invariant theory where three fermions and a vector meet. However, I don't see why you say the interaction is forbidden. It would surely be insanely suppressed since amplitudes are extremely weak, but I don't see a problem with the diagram (for instance): Notice that quarks mix, so the ...

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The problem with this process is, that there is no term in the standard model which allows a coupling of a fermion a W-boson and a quark-antiquark pair. Even with the neutrino having no charge, you have to take this arrow you draw on the fermion line seriously, so the fermion flow is violated at this interaction points. Furthermore you can only couple ...

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An electron is a fundamental particle(Lepton) and is different from both protons and neutrons, which are not fundamental(i.e they are made from even more smaller particles called quarks). For an electron, the charge it has is an intrinsic property, that is it is a part of its description along with mass and spin. Coming to neutrons and protons, even though ...

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1 How does an electron get its charge? This is the elementary particle table . The electron is an elementary particle and its charge is an observable attribute that , together with its other quantum numbers and mass, classify it as an electron. And how can it maintain that charge for very long (infinite) periods of time? Observations ...

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Scattering experiments can be used to determine the size of a particle. The results for an extended object are different than that of a point particle. But all of these scattering experiments depend on getting the probe particle "close" to the scattering object. In the case of electrons, that means launching the probe with enough energy to overcome the ...

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The Spin of a particle is better understood from a group-theoretic point of view. It is just telling you how a particle, i.e. an asymptotically free state of your theory, transforms under the symmetry of your theory, Lorentz symmetry. Well, actually under its double cover as Weinberg explains in his first book, that is why we are allowed to have spinors. ...

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I'd say that this claim is specific for the current experiments at the LHC. We collide protons there, and the protons are made of quarks and gluons -- strongly interacting stuff. You can even say that there are already $b$-quarks in the proton. So, when the protons collide, this strongly interacting stuff produce events that are similar to the genuine $h\to ... 3 In order to get a good mass accuracy using the gamma gamma channel one needs to measure well the gamma energy in the electromagnetic calorimeter which can easily contain all the energy. The four vectors have measurement errors but not missing energy. b and b_bar decay weakly to a number of particles including neutrinos and the subsequent decays end on ... 4 The notation$pp \rightarrow t\bar{t}$doesn't literally mean the two protons disappear leaving just a top and antitop. It means a top and antitop are created as well as a shower of other debris. Because protons are composite objects a 13TeV proton-proton collision is an exceedingly messy business. At that energy the quarks are resolved so it's really a ... 2 Apologies for evincing magisterial cluelessness about what your diagrams represent and what you'd want to achieve, but I'd array the standard facts on tetraquarks avoiding Young diagrams, although they are self evident in the Dynkin labelling, which I also give, next to the tensor labelling. They may be useful to what you appear to be after--but I can't ... 3 The process by which particles are created after inflation is known as "reheating". One needs a coupling between the inflaton and another particle's field for this to happen. Generally it occurs at the end of inflation when the inflaton is oscillating in the well of its potential and the expansion rate falls below the interaction rate between the inflaton ... 2 My 2 cents on it is that in QM (be it "standard" QM or QFT) one describes only the state of a particle. Having said that, the most general state for a single particle is indeed a wave packet. Now, if you localise certainly a particle at some point in time, then later on it will be associated with a spreading wave packet because of Heisenberg indeterminacy ... 6 A particle is not a wavepacket. And there are no particle states for interacting theories. We define particle states in QFT by expanding the free field into its Fourier modes and using these modes as creation/annihilation operators for particle states - the mode of momentum$p$creates the particle state$\lvert p\rangle$with momentum$p$. The Hilbert ... 2 Spin is connected to the intrinsic magnetic dipole moment of the particle, this is what makes the particle capable of interacting with an external magnetic field. Namely, the intrinsic dipole magnetic moment$\vec\mu$of a particle with spin$\vec{S}$can be found through: $$\vec\mu = g\left(\frac{q}{2m}\right)\vec{S}$$ where$g$is the g-factor, and$q$... 0 A can be photon, Z or gluon. B can be u or d quark. 0 As AccidentalFourierTransform has pointed out, there is no change of flavour in A, because the$u$and$\bar{u}$are annihilating, so it is a photon. For the$\pi^0$decay, B is a virtual$u$or$d$. 0 The idea behind this is not only drawing pretty interesecting lines; but to have a larger symmetry which includes all of the standard model (SM). To each simple group is attached a coupling 'constant': in the SM we have three of these because$G_{SM} = SU(3)_C \times SU(2)_L \times U(1)_Y$. The values of these coupling 'constants' vary with energy and this ... 1 Neutral charged objects are made of charges. Although net charge is zero, imperfections in spatial arrangement of charges, causes them to interact with electromagnetic fields. Edit: 1) Neutrino's are formed during nuclear reactions. Once created, they sustain their direction of spin which does not change during course of time. 2) Photons don't change their ... 3 The FRG can be thought of as a modern version of Wilson RG, although the technical details are of course very different. But all in all, if one could do all calculations exactly, these different versions would all be the same. Now, about these technical differences. In Wilson RG (and in Polchinski's functional version) one work with a low energy action for ... 1 In one sense, this should not be a coincidence, because the mass and the charge radius are actually determined by more fundamental quantities (quark masses, strong force coupling constant). So given that those quantities have their particular values, in theory it inexorably determines what "mass times charge radius" is going to be. What should be a ... 1 "Perhaps a sagging potential is one of ratios rather than sizes." Your conclusion is correct, but the idea is not restricted to the sagging well. This is a neat exercise in scaling, and it all comes from the choice of parametrization of the potential for given length$L$. To see this better in the present case, just rewrite the hamiltonian in terms of ... -1 If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of$h\nu$into two photons of energy$h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe. 1 No. Feynman diagram calculations are much more complicated; the$\alpha^n$scaling is just one piece. For example, consider the amplitude for$N \gamma \to (N+1) \gamma$, where$N$photons interact to turn into$N+1$photons. The lowest-order Feynman diagrams contain one electron loop connecting all the photons together, so the cross section, according to ... 2 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. Let me review its antecedents as you seem to be aware of the phenomenon, when you have all quarks, ups and downs, mix, and not just the downs, ... 0 An electron being a ball of uniform mass and charge is not consistent with its observed gyromagnetic ratio. The charge must be pushed out and the mass must be pushed comparatively inwards to satisfy the existing ratio of about 2. See Classical proof of the gyromagnetic ratio$g=2$1 At lowest order, color screening comes from virtual quark/antiquark pairs, just like charge screening comes from electron/positron pairs in QCD. The diagram/effect is also referred to as 'vacuum polarization' and is shown below. Charge antiscreening comes from virtual gluon pairs; its diagram is below. 2 In an imaginary world where the leptons and the neutrinos are all massless, they would exist only in the left-handed helicity version. In this imaginary world the decay of the pion in lepton plus neutrino would be prohibited: you could not conserve both the total spin (zero) and the total momentum, in that this would imply emitting two particles both with ... 2 First, let's clear up some terminology: the usual statement "Majorana fermions are their own antiparticles" is correct, but confusing because the words we usually use to describe neutrinos are made for Dirac fermions. If neutrinos had no mass at all, there would be two independent types of neutrino: a left-handed and a right-handed neutrino. These particles ... 0 The PDGlive webpage lists their fit as $$\text{BR}(D^0 \to \pi^- e^+ \nu_e) = (28.9\pm0.8)\times10^{−2}$$ Looking at the experimental analyses from which this combination was obtained, e.g. this CLEO analysis, we find that inclusion of charge-conjugate states is implied throughout this report In other words, I believe that PDG lists the branching ... 2 The color of the photon is related to its frequency$f$, which can be related to the energy of the photon by the expression$E = hf$, where$h$is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ... 0 One way to think about it is similar to a chemical equilibrium. Electron-hole pairs are spontaneously generated every now and then from random thermal fluctuations, and when an electron collides with a hole they annihilate with each other (some fraction of the time). The frequency with which electrons and holes collide is$np$. In steady-state, this ... 0 "Why is$np$always equal to$n_i^2$?" Well, first of all, the easy way to answer your question "Why is$np$always equal to$n_i^2$?" would be simply to notice that$np$is independent of the Fermi level$E_F$, and thus independent on the fact that the semiconductor is doped or not. In the case of a non-degenerate semiconductor (i.e. when$E_F$is ... 0 The process is discussed at the parton level – both in the initial form and the desired form – so the conversion cannot depend on PDFs. Now, the Mandelstam variable$t$is equal to $$t = (p^\mu_1 - p^\mu_3)^2 = m_1^2+m_3^2 -2 E_1 E_3 +2 \vec p_1\cdot \vec p_3$$ in the "mostly minus" metric convention. The masses of particles are fixed and the total energy ... 2 Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter$E_f$, the Fermi level (assume a fixed temperature). Provided$E_f$is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form$\propto e^{E_f}$for electrons and$\propto e^{-E_f}$for ... 1 There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient$7.181\times 10^{-16}$, when converted from MeV to eV, should give$7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives$10^{30}$, not just$10^{15}$. In this way, the OP has to add a$10^{-15}$factor to his result. That makes his result$10^{-3}\$ ...

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I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...

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