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New answers tagged particle-physics

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It depends. In the two Higgs doublets models, there are two cases: The two doublets are independent There is a cross potential term between these doublets a la: $\phi_1^2\phi_2^2$ For case 1, the two VEVs are independent. For case 2, the two VEVs are are coupled to each other. But it is hard to tell which is inducing which.

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Fluence is the time integrated flux . Here , replace " a " for annual , by " y " for year . You obtain a more known unity : 0.3 MW-y / ...

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As about first parameter it seems that specification is talking about irradiance flux density. Neutron flux can be extracted from it like that : $$\Phi _ n = \frac {J} {E_n}$$ Where $J$ is irradiance flux density $[\text{W} /\text{m} ^2]$ and $E_n$ is neutron average kinetic energy. As about second parameter I'm not sure. In one page above in other formula ...

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The 1956 discovery of electron antineutrino from nuclear reactions is credited to Clyde Cowan and Frederick Reines in their famous Savannah River experiment. They based the expected number of neutrino events to a parity-conserving weak interaction theory. Cowan and Reines reported the observed neutrino events consistent with parity-conserving theory. However,...

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Written as $ΔE = (Δm)c^2$ the $ΔE$ is the increase in the kinetic energy of a mass being accelerated, and the $Δm$ is the corresponding increase in the inertial mass (as predicted by special relativity).

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You are fundamentally correct, but are mixing up what's necessary for each. Special relativity is both necessary and sufficient to construct the idea of rest energy, but it is merely necessary for constructing a model of conversion of rest energy to other forms of energy. In order to do that, you need particle physics, and the only way we know how to ...

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My QCD knowledge is very limited - I haven't learnt it anywhere else other than in some qualitative discussion in Thomson's particle physics textbook. So I was only seeking a qualitative answer to aid my understanding in PDFs. Thanks to @anna_v 's answer I realised I hadn't gone through Thomson's own slides of the chapter and the answer lies there all along. ...

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I copy these definitions of the scattering variables" I am not able to see what this has to do with the value of x. Does this mean when x decreases, the four momentum q of the gluon also decreases? If so, why is it the case? The definition of x is dependent on the four momentum carried by the propagator, in the diagram small $q$ vector , which is ...

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An electron and a positron can annihilate regardless of their spin states. However, their likelihood of annihilation depends on those spin states.

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So the Parton distribution functions are "essentially" the probability distributions (very suspicious how both are called pdf) of a probe particle to interact with a component inside your hadron. They depend both on the momentum at which we probe $Q^2$ and the fraction of momentum carried by said parton, the Bjorken x. They are mostly known through ...

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If you had a particle with charge but without mass, it would be subject to infinite acceleration if attracted by an opposite charge, since $\rm F=ma$. That should rule out the existence of such particles.

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Since a particle only can be said to exist if it can express its existence, its properties in interactions if it has energy and localized energy is a source of gravity and we define mass as something which exerts and feels gravity, then there cannot exist massless particles. (That is, if we define a particle as an entity which at all times has a well-defined ...

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First lets clarify one thing: Light mesons, wich is the entities envolved in this decay, have all one thing in common: their orbital angular momentum is l=0. They are, however, grouped in the pseudo-scalar mesons (with s=0 for the pions for exemple) and the vector-mesons (with s=1 for the rho and omega for example). The parity of a meson state is the product ...

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A small explanatory footnote to @anna's fine answer, which emphasizes the primacy of the LSS discovery, a major game-changer: it upended our understanding of particle interactions forever. The professionals at HSM would have more generic insights. The 1995 NP awarded to Perl & Reines was a bit special. (Cowan had died in 1974.) The background is that ...

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This belongs to the history of science, but Wikipedia is good for these questions: In 1962 Leon M. Lederman, Melvin Schwartz and Jack Steinberger established by performing an experiment at the Brookhaven National Laboratory1 that more than one type of neutrino exists by first detecting interactions of the muon neutrino (already hypothesised with the name ...

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Good answer from Penguino. Here is an example. Several items of the same type are for sale in a store with many kinds of merchandise. Customers come by at some average rate. On the average, they have a constant probability of wanting one of the items. The lifetime of an item has an exponential distribution. Examples more along the lines of a light bulb are ...

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It might be hard to find a realistic example of exponentially distributed lifetimes (EDL) in macroscopic systems. An EDL requires each member of the population to have a constant probability of failure for each time interval - and if failure doesn't occur during that interval the member's state is 'reset' to its state at the beginning of the time interval. ...

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The ground state (or vacuum state) minimizes the energy, not just the potential. A solution of the field equations with non-constant field $\phi$ (satisfying $|\phi(x)|=\phi_{0}$, so that the potential is minimized) will still have kinetic energy associated with it—making it automatically not the vacuum. That’s all there is to say about finding the vacuum ...

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These[1][2] review papers contain good introductions to fractons. Generally, there is no uncertainty principle for such systems. This is because these fractons are usually emergent particles. For example, we can think of a domain wall excitation in a 1D Ising model as a particle, but this does not have any uncertainty principle associated to it. The ...

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your question could be generalized: pion decays to odd number of photon is forbidden. That's a conclusion of Furry's theorem: even number of photons is forbidden to odd numbers of photon Here the only assumption is that strong interaction and EM process charge conjugation symmetry. As for the second process, because charge conjugation is no longer a symmetry ...

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I think it is easy to say that the charge is conserved, but might be hard to measure. The cart is before the horse in this sentence. It is because the measurements completely agree with charge conservation that charge conservation is imposed in the mathematical models describing particle data. Or lepton and baryon conservation. At elementary particle ...

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It is not clear what do you mean in statement "QED is very predictive...": QED does not contain any particles except electrons & photons. QED describes only electron-photon proccesses and when you include additional particles (for instance, muon) you obtain a different theory (strictly speaking). In Standard Model (SM) there is the following ...

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Good question! Their is lot to unpack to so I will do my best here. I think I can give you a global answer (or try to anyway), but a true answer to your question would involve knowning what process you are looking at. I am going to assume that you are thinking about $W_\pm/Z$-boson mediated interaction and not a more complicated process with $W$ boson loops ...

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since this thread got some recent recognition, I'll post the answer that I found out some time ago through some additional literature research and the one I personally was most satisfied with. As mze's answer proposed, the PREM model is used as the density profile for the matter potential and an analysis in Joachim Kopp's diploma thesis in chapter 4.2.4 ...

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The $\rho$ has the quantum numbers of $\pi\pi$ coupled to $I=1$. If they only list particles stable under the strong interaction then $\tau^-\to\rho^-\nu$ is the same as $\tau^-\to\pi^-\pi^0\nu$, which indeed has a branching ratio of 25%. The table is more helpful than that: It lists the non-resonant (non $\rho$) part of $\tau\to \pi^-\pi^0\nu$, which is ...

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Probably the easiest way to see this is to look at $W^+$ decay into a positron and a neutrino in the SM with only one generation, so CP is conserved, and C and P are violated maximally. The relevant terms in the lagrangian are proportional to $$W^+_\mu \bar \nu _R \gamma^\mu e_L + W^-_\mu \bar e _R \gamma^\mu \nu_L ~,$$ which transform under C to $$W^-_\... 3 Gauge bosons are part of the quantum mechanical models built in order to fit an enormous amount of data gathered with particle physics experiments. In that sense they have been measured in experiments by that data since the establishment as mainstream of the standard model of physics. All the particles in the table are considered measured, because with out ... 1 Yes, the reason is as simple as "there is no production channel for antineutrinos". The primary stellar fusion reaction families, the proton-proton chain and the CNO cycle (significant in stars with mass > 1.3 M_\odot) both convert some protons to neutrons, and that conversion releases a positron & an electron neutrino. There are no other ... 3 The sun starts with electrons and protons and fuses the protons into heavier nuclei. Almost all stable nuclei contain neutrons, which means that in this process, protons have to be converted to neutrons. This requires the destruction of electrons by charge conservation, which in turn requires the creation of electron neutrinos by lepton number conservation. ... 2 You must convert proton to neutron somehow (pp chain or CNO chain) and this means by charge conservation that a positron must be emitted, and by lepton number conservation to balance lepton number that anti-electron is accompanied by neutrino and not anti-nu. 1 No. The Goldhaber experiment's purpose was to test the masslessness hypothesis of neutrinos, which was proposed in 1957 by Lev Landau and Abdus Salam. The motivation for this hypothesis was that this was the simplest possible parity-violating theory for weak interactions. In it, neutrino is strictly a left-handed particle. In a parity-conserving theory one ... 1 Quantum breaking is usually used for localized condensed matter systems containing atoms in BEC state. https://www.news.ucsb.edu/2020/019817/hunt-gravitons When we usually describe this state, we describe a homogeneous field having constant charge and energy. It is very important to understand that classically this configuration is stable. At the quantum ... 1 Asking in google "quantum break" arxiv gives a number of physics preprints, This seems relevant to gravity: Quantum break in high intensity gravitational wave interactions The lowest order amplitudes for [graviton+graviton→photon+photon] lead to cross-sections of order G2, where G is the gravitational constant. These are too small to be of any ... 0 The amplitude for the decay is:$$M = \cfrac{g^{αβ}g^2 v}{2}ε^{*}_{α}(p)ε^{*}_{β}(q) $$You can obtain the vertex factor from the assosiated Lagrangian. ε here are the polarisations of the outgoing spin 1 bosons and p and q denote the momenta. You want to calculate:$$Γ = \frac{p_{f}}{32π^2 m_{H}} \int |M^2|dΩ$$For the M^2 you use the fact that W... 1 You get a pure imaginary number because the decay from rest of a Higgs into two W bosons is not allowed kinematically, for the masses you used. 1 I found this schematic in a power point presentation that may help: Particles with non-zero magnetic moment are deflected, due to the magnetic field gradient, from a straight path So it is not the north or south pole that makes the difference , but the orientations of the incoming magnetic moments interacting with the gradient of the magnetic field. In ... 2 Perusing the PDG, the fountainhead of all particle knowledge, or a half-decent HEP text, you can find out how mesons decay to answer your question. Essentially they are \bar q q quark composites, so their constituents can: Combine with more \bar q q pairs emergent out of the glue soup of QCD, and strongly decay to more mesons, like \rho \to \pi \pi, ... 2 A good example of both mass and lifetime measurements is provided by the delta baryon, which is produced in:$$ e(p, e')\Delta  which means an electron beam (energy $E \gg m_e$) impinges on a proton target (mass $M$). The scattered electron (energy $E'\gg m_e$) is detected. The invariant mass (squared) of the undetected final hadronic state is constructed ...

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A decay doesn't have a cross section associated with it, only a decay width. Most likely they are referring to the production cross section for the process $\rm X+X\to H\to W^+W^-$ at some specific collider. Note that this depends on the identity of the initial particles and their momenta. It can also depend on other properties such as the polarization of ...

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Two atoms are definitely not sufficient to tell the difference between a liquid and a solid. The difference between those states is that the structure of a solid remains effectively fixed while a liquid's changes. With a single bond, there's no reasonable way to separate those two cases. As mmesser314 pointed out in concepts, gas liquid and solid are terms ...

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It's fairly hard to answer your question due to how general it is. There's a lot of ways these measurements can be done, and which ones are used in practice highly depends which ones are the most practical for the situation. For a relatively simple to perform/understand method as an introduction, I recommend you look into bubble chamber analysis, which only ...

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Generally, "decay" refers to reactions where the initial state consists of only one object, so matter-antimatter annihilation is not decay. As an aside, whether or not a reaction "involves" all three types of "forces" (strong, weak, EM) depends on how small of a correction you're willing to look at. Generally, every interaction ...

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Well, you should clarify, what do you mean by predicting facts in QCD by means of CFT. CFT studies the correlation functions, relationships between them by means of conformal bootstrap. QCD is not a conformal theory even if we are dealing with negligible masses of fermions, due to the existence of scale - $\Lambda_{QCD}$. The theories experiencing continious ...

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It can! The question isn't very specific, so I'll only answer broadly. Of course it all depends on what you mean by high and low energies, but many naive supersymmetric models you might write down will affect low-energy physics. If your model predicts that the proton will decay quickly, or that a new particle will be created if you collide two electrons ...

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An early use of the term "high energy physics" (1950) is in the title of the Rochester Conference series. The corresponding proceedings can be consulted in the CERN Library

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To probe the world of the very very small, physicists need a probe method that has a very very short wavelength- otherwise it will not be able to resolve detail on very very tiny length scales. Since shorter wavelengths carry higher energy, the science of probing very very small distances using very very short wavelengths is known as high energy physics. ...

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In general, particle physics is not NECESSARILY high energy physics. For example quantum electrodynamics describes all interactions of light and matter, not just high energy ones. However, many of the interesting states of matter that we consider in particle physics only exist at high energies. For example, types of quarks besides up and down only exist at ...

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but either just disappeared (violating conservation of energy) or resulted in an extra-strong recoil to the nucleus. This would violate momentum conservation at the center of mass system. One has to work with four vectors, whose algebra obeys the the conservation laws. Nuclear interactions were of the first evidence for a need of special relativity in ...

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You may well be misunderstanding several things. The $N_f^2-1$ flavor charges do no count flavors. It is easiest for you to see that by imagining a world of just two flavors, u and d, namely isospin, for $N_f=2$. Then the three generators/charges effect infinitesimal flavor rotations and are just the three Pauli matrices. So you see that $\sigma_1$ just ...

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Each column of Φ is an $SU_L(2)$ doublet, since $U_L$ only scrambles rows. Each row of Φ is an $SU_R(2)$ doublet, since $U_R$ only scrambles columns. N.B. In response to your comment, $\Phi \sigma_x =\begin{bmatrix} c & a \\ d & b \end{bmatrix}$. So the first column has been interchanged with the second one, rigidly; in that sense it has been ...

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