New answers tagged

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Because we have observed processes where the photon number is not conserved. For example positronium can decay into 2 or 3 (or more) photons. This means that it is not possible to assign a global conserved charge to photons. For neutrinos we can assign lepton number and so far we have not observed a process that would violate total lepton number (lepton ...


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Short answer: to accurately model reality. Long answer: The weak interaction has several peculiar properties: The $W$ bosons are vector bosons (so the weak theory is likely a gauge theory) The $W$ bosons have electric charge The $W$ bosons have mass. (The $Z$ boson hadn't been observed experimentally; it was a prediction of the SM) The $W$ bosons couple ...


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Detailed answer The thing is that you cannot really force in an a priori interpretation of the gauge group with which you extend your existing theory. You can only decide on the symmetries of your theory. So keeping things very general you start by trying to gauge the symmetry group $SU(2)$. This has three generators and so gives rise to three independent ...


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It is one of the experimental observations that led to the standard model of particle physics. The model has symmetries ( SU(3)xSU(2)xU(1) ) that build up the representations and these allow only for integer multiples charges, i.e. are consistent with observations. For example, why is there no meson existing of two up quarks, giving a charge of 4/3? ...


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This is just very sloppy language on the paper's part. As you say, gauge bosons are very real and their existence has physically measurable consequences (otherwise, why would we ever waste time talking about them?). (By the way "photons and electrons" are not good examples of non-gauge particles, because photons are also gauge bosons :) .) The paper just ...


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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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The question that you have asked have some vague arguments as well as some partially true facts regarding Standard Model (SM). First, Yes SM describes physics up to some energy scale which is 14 TeV. On the other hand, if we accept Plank energy ($~10^{18}$GeV) as a fundamental energy scale, then we can possibly expect new beyond the SM energy scale. A ...


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You have drawn a Feynman diagram. Feynman diagrams are iconic shorthand for integrals over the variables of the problem. The calculation gives the probability for the reaction to happen, in this case the decay of a neutron . The observables are the four vectors of the initial (neutron) and final particles. The integral is over the variables . Here is a ...


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It can be stated this way: In this particular diagram, the W boson is in a state named off-shell i.e. we say that this boson is virtual. Virtual particles are allowed to have any mass value. They can't although violate charge conservation at the vertex. This 80 GeV mass of the W boson, is for a real W boson, which is on the on-shell state. The real ...


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Newton's third law is a statement that momentum is conserved, so it is equivalent to the law of conservation of momentum. Conservation of momentum follows from a fundamental symmetry (of the action) called space shift symmetry and as far as we know this applies to all our physical theories. So Newton's law is still valid but has to be treated with some care ...


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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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There exist no free gluons, it is called confinement: Since gluons themselves carry color charge, they participate in strong interactions. These gluon-gluon interactions constrain color fields to string-like objects called "flux tubes", which exert constant force when stretched. Due to this force, quarks are confined within composite particles called ...


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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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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 ...


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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 ...


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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 ...


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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 $d$.


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It's a good question, and the answer is surprisingly simple and physical. There is indeed no fundamental objection to having a superposition of particles of different charge. But it turns out this is not stable. This is basically due to wavefunction collapse, or more sophisticatedly, due to decoherence. Imagine having a single particle in a superposition ...


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Experimentally the charge distribution of protons and neutrons has been measured as a function of the radius. So the different charge content of the two nucleons does affect the distributions. As the other answer states this is the regime where only quantum chromodynamics models can attempt to describe the wavefunctions of the quarks within the ...


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This is not a meaningful question. No, really, it isn't. Quarks don't exist as free charged objects on which we could take the classical limit and consider "forces" on them. They are confined, and occur only as constituents of bound states. In quantum mechanics, it doesn't make sense to ask whether the constituents of a bound state "repel" or "attract" each ...


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The electric charges of the states $\:|uuu⟩,|ddd⟩,|sss⟩\:$ are $\:+2,−1,−1\:$ respectively. More exactly these baryon states are the baryons $\:Δ++,Δ−,Ω−\:$. If $\:|X⟩\:$ or $\:|Y⟩\:$ would represent a baryon what would be the electric charge of this particle ? And electric charge is one of many quantum numbers. This problem is pointed out by @Cosmas Zachos ...


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A basic postulate in elementary particle theories is CPT invariance. Also the weak interaction is the only fundamental interaction that breaks parity-symmetry, and similarly, the only one to break CP-symmetry. ...... The laws of nature were long thought to remain the same under mirror reflection, the reversal of one spatial axis. The results of an ...


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In the absence of Yukawa couplings (only kinetic terms), the SM has the global flavor symmetry: $$G_{y=0} = U(N_f)^5=U(3)^5$$ Because there are 5 distinct representations in the SM (3 for quarks: $u_R$, $d_R$, $Q_L$; and 2 for leptons: $e_R$, $L_L$). However, $U(N) \sim SU(N)\times U(1)$, so the group can also be written as: $$G_{y=0} = SU(3)^5 \times ...


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I often see $\mathrm{SU}{(3)}_\text{flavor}$. However, I have seen $$\mathrm{U}(3)_\mathrm L \times \mathrm{U}(3)_\mathrm R = \mathrm{SU}(3)_\mathrm L \times \mathrm{SU}(3)_\mathrm R \times \mathrm{U}(1)_\text{vector} \times \mathrm{U}(1)_\text{axial}$$ where the last one is broken by the quantum anomaly. See slide 14 in this lecture summary of Theoretical ...


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If we identify a force as a scattering process, i.e. with a mediator of some interactions, then this need not be a vector boson of course. One can speak of "Higgs" force for instance if the process under consideration is mediated by the the Higgs (which is a scalar). There are also numerous cases where the interaction is mediated by a fermion. Therefore, ...


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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, ...


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Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...



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