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Electron positron pairs can appear whenever there exists enough energy and electromagnetic interactions. Photon photon to produce an e+e- is disfavored as far as crossections go because of the great number of vertices involved in the Feynman diagram One of the outgoing photons, if it has enough energy can be a virtual photon in a diagram creating a pair ...

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It is "easy" to show that the cross-section of the 2 reactions: $$(1)~~~\nu_{\mu} + d \to \mu^- + u~~~~~~(2)~~~\bar{\nu}_{\mu} + u \to \mu^+ + d$$ are different. The naive calculation gives a factor $\sigma_1/\sigma_2 = 3$ as shown below. The fact that the figure in the particle data group gives actually almost a factor 2 needs to take into account the ...

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Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ... 4 Nuclei have a series of discrete energy levels (somewhat analogous to electronic energy levels, but the details are, not surprisingly, different). Examples of these so-called Energy Level Diagrams can be found at, e.g., Triangle Universities Nuclear Laboratory. So, a simple alpha decay will go from one level in the parent nucleus to one level in the daughter ... 3 It means that when the neutrinos hit electrons, the electrons are moving preferentially in the same directions that the neutrinos were moving. So when we are building a water Cherenkov detector for solar neutrinos, the Cherenkov signal will be coming from the direction of the sun. This is very advantageous to suppress background and because of the daily and ... 3 Here is the elementary particle table from which all others are built up , the standard model of particle physics. Which shows the conserved quantum numbers that characterize the particles (columns and rows have quantum numbers assigned too) plus the measured masses. The quantum numbers have to "annihilate" to have an annihilation event, i.e. they should ... 3 The answer is that in general cross section of particles and antiparticles are different. For this case in specific there is one point that Dirac particle have four possible states: due two possible ranges for energy E<0, E>0, and two possible range for helicities h>0 and h<0. In the standard model of particles only the neutrino with h<0 interact ... 3 Higgs couplings of the Standard Model are flavor conserving because the separate "flavor numbers" such as$L_\mu$– the number of muons minus the number of antimuons plus the number of muon neutrinos minus the number of muon antineutrinos – are conserved in each interaction. It means that the corresponding interaction (one that is able to produce the Higgs ... 3 The UV cutoff procedure that is used to regulate Feynman diagrams must be chosen so that the region of integration involving large values of$p^\mu$(all components individually) is cut-off. A Lorentz invariant cut-off such as$p^2 < \Lambda^2$does not achieve this. However, what we can do is to Wick rotate$p^0 \to - i p^0$and then impose a cut-off ... 3 It is the classic two body versus three body decay. For a two body decay the energy in the center of mass of the two bodies is fixed for the individual decay products, a delta function for the muon, in the above example, within uncertainties. For a three body decay there is a variable spectrum for the energy of the electron. (Analogous in kinematics to ... 3 Does the invariant helicity property contribute to the the concept of a photon and an "anti-photon" being the same entity? Not really, and I think you're getting mixed up between helicity and chirality here. Take a look at this deep-water wave image by Kraaieniest. See how the red-dot test particles move in a helical-like fashion? They can't move "the other ... 3 The thing is that any reaction which produces a charged particle traveling faster than local$c$will produce Cherenkov radiation. In a neutrino experiment the two you list are the signal, and any others are the background. Someone else analyzing the same data from the same detector might not be interested in the neutrino signal and consider that ... 3 Mass is not conserved in that decay, but then there is no expectation that it would be. The "law" of conservation of mass is only an approximate rule that applies to low energy events and interactions. Chemists (well, the non-nuclear ones, anyway) talk about it, but physicists do not. The rule that does apply is the conservation of energy (mass being one ... 3 I'm going to expand on dmckee's answer because this used to puzzle me in my younger days, and I think it's a fascinating part of modern physics. It's tempting to think of a particle like a muon as a chunk of matter whizzing around - something like a tiny billiard ball. However the physicist's description of a particle is much, much stranger. Quantum field ... 3 According to the Review of Particle Physics (Section 33.7.4 of the 2014 edition) there are two main causes of radiation damage for electronic devices: Bulk damage due to displacement of atoms from their lattice sites. This leads to increased leakage current, carrier trapping, and build-up of space charge that changes the required operating voltage. ... 3 Ionizing radiation loses energy in matter by creating electron-ion pairs. Suppose you have an 1 MeV charged particle stopping in a silicon crystal. The first ionization energy for free silicon atoms is about 8 eV. The ionization energy will be a little different for silicon atoms on the lattice, but not grossly so: your 1 MeV charged particle is going to ... 2 There is no reason you can't have nuclear or nucleon excitation; or meson production reactions in the mix. Reaction like $$\nu + ^{16}\!\mathrm{O} \to \nu + ^{16}\!\mathrm{O}^* \,$$ with a subsequent decay of the excited nucleus or $$\nu + n \to e^- + \Delta^+ \,,$$ or $$\nu + p \to e^- + p + \pi^+ \,,$$ are accessible. The three I have exhibited will ... 2 In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ... 2 You say energy conservation gives $$E_{\gamma}' + E_e' = E_{\gamma}$$ I think it should be $$E_{\gamma}' + E_e' = E_{\gamma} + E_e$$ where$E_e = m_e c^2$. This would make your second to last equation $$\frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma + E_e - E_\gamma') m_e - m_e^2c^2 = E_\gamma - E_\gamma'$$ 2 How many real elementary particles (not hypothetical) make up an atom or can be in an atom? This is tricky, because of the inclusion of the word "real". Let's say we're talking about a helium atom, and we're talking about how many different types of elementary particles there are. The helium atom is comprised of protons, neutrons, and electrons. OK, now ... 2 I don't know nearly enough QFT to address the background or implications of your question. However, I'd basically answer yes to your first two questions, but it depends a little on your definition. A single phonon mode is not localized in space. However a wave packet can in principle be built up of a small range of frequencies, giving a fairly well defined ... 1 In superfluid helium-4, the phonon excitation spectrum includes a mode which has the same energy and momentum as a neutron with a speed of about 440 m/s (wavelength$\lambda \approx 9\,Å$). You can create a neutron beam which contains only 9 Å neutrons by starting with cold neutrons and being clever with diffraction from crystals. If you send these ... 1 are the decay products produced totally probabilistic or can we "tune" an accelerator to increase the probability of a particular decay channel occuring? Once a particle is produced, it decays completely independently of its production. The only dependence on production occurs when, for instance, a particle is produced in an entangled state and ... 1 Here is the phi mass 1019.445±0.020 MeV/c2. Here is the K mass 493.667±0.013 MeV/c2 , times two 987 1019-987~32 MeV/c2 are left over to be shared as kinetic energy of the two kaons. It is a matter of phase space. The two pions of the rho are ~280 MeV/c, leaving a lot of phase space to facilitate the decay, i.e. give larger probability because of larger ... 1 This view would not be accepted by physicists today. Charged particles have mechanical mass, momentum, and energy (rest and kinetic) and the fields have energy and momentum. Total energy is conserved. Total momentum is conserved. Are there cases where it can be sensible to imagine field momentum as an additional mechanical momentum? Sure, consider the ... 1 The tau decay hadronically thorugh neutrino + Wboson (greater Branching ration). And then the W boson decay in one charged hadron (narrow jets). In QCD things are a little different, gluons/quarks for example doesn't have neutrino on his decays and then multiplicity on final state is larger, more hadrons produced (wider jets). And gluons can produce more ... 1 Usually the first step in deriving the reps of Poincaire is to go to the rest frame of the particle. This amounts to choosing a basis where$P^0$acting on the state is nonzero, and where the eigenvalue a of$P^i$are zero. We can do this if the momentum is timelike, that is if the eigenvalue of$P_\mu P^\mu$is negative (in -+++ signature). Furthermore the ... 1 The$kT$comes from the Bose-Einstein statistics. The photons are governed by this statistic, nothing suspicious is here. The power three appears when we go form the variable$E$to the variable$E/kT\$ in the integral.

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The only neutrino interactions known to be detected are those you provided: The charged current interaction and the neutral current interaction, respectively. These interactions can also be differentiated with respect to flavor. In IceCube, for example, the charged-current interaction from a muon neutrino produces a "track" in the detector along the path of ...

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Percentage Depth Dose Curves (or PDDs) are used to determine how many Monitor Units (MU) a treatment machine needs to give (or how long the machine needs to be on) to deliver a particular dose to a particular depth. The depth at which the PDD curve peaks is referred to as dmax. Treatment machines are often calibrated so that 1MU = 1 cGy at dmax in water. ...

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