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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 $E=\frac ... 2 In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ... 0 The way you distinguish between different interactions in the universe is by looking at the mediator bosons thereof, which appear when scattering two particles under this or that interaction. Take two electrons as example and scatter them up: if you identify that photons take part to the process then it is an electromagnetic interaction; if you find out ... 1 The quantity minimized during the geometry optimization is the total energy$U$of the molecular system with nuclei being fixed. It consists of the kinetic energy of electrons, the potential energy of the Coulomb repulsions between the electrons, the potential energy of the Coulomb attractions between electrons and nuclei, and the potential energy of the ... -3 If we calculate the energy required to assemble an electron from a thinly spread out cloud of charge as e, then e is almost equal to the rest mass of the electron. (This is taking into account the experimental size of the electron from collision experiments and uniform charge density in the electron). Though this information does not explain the stability ... 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 ... 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 ... 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 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 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 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 ... 7 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 ... 1 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. ... 1 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 ... 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 $$\frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma + E_e - E_\gamma') m_e - m_e^2c^2 = E_\gamma - E_\gamma'$$ 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 ... 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 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 ... 8 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 ... 0 What's the significance of neutrino oscillations? They say something very important about the nature of neutrinos. As per this related answer, the Standard Model doesn't really cover it. Then when you ask about it, people tend to talk about neutrino mass. What you don't much hear is why do neutrinos oscillate? IMHO the answer has to be that the neutrino is ... 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 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 ... -4 I wrote a paper in which I show that the electric field of a charged particle can be described as composed of two quanta. Two types of quanta are enough to describe the electric field, the magnetic field and photons. It was shown, why electron does not fall into nucleus and why annihilation happens between particle and anti-particle. A more deep explanation ... 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 ... 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. 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 ... 0 I don't think this assumption is legit: Assume the you initially have two quarks (up and anti-up for instance) that are placed far outside of causal contact with each other for a real experimental setup. Indeed, what would have been the previous history of those two scorrelated (outside of each other's light cone) quarks, to be produced isolated? ... 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 ... 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 ... 0 A photon should be discussed as a wave, instead of a particle (as it ceases to exist in rest state). As such two waves of the same wavelength and frequency may or may not be "negative" to each other, depending on their phase difference. When their phase difference is an odd multiple of pi, they will invariably cancel out each other upon interference. 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 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 ... -4 It is important to remember that Gravity is not a force. The search for gravitons or gravitinos is misguided. As explained by Einstein, Gravity is an affect caused by space being curved, where the curvature is a linear combination of the curvature caused by the sum of the energy density provided by each object in the Universe. Einstein's GR equations lead ... 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 ... 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 ... 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 ... 0 You can make an atom with heavier generations of quark if you want to count that, for example, a top quark has the same spin and charge as an up quark and you can "construct" a proton from a charm or top quark - not sure how stable it would be. http://www.particleadventure.org/three_gen.html But as Gabriel said, Up quark, Down quark and the Electron are ... 0 The number depends on the type of atom, but all of them are mostly made of up and down quarks (which make up the proton and neutron) and electrons. 0 These are some nice answers! Wanted to add that, as you know, how strongly the quarks couple to each other (or interact with each other) is momentum-dependent. So within the nucleons (protons and neutrons) quark coupling is very strong (which is why the quarks are confined in the nucleons). Because the interquark interaction is so strong at these ... 0 These are two really great answers, so I don't feel the need to add much; only to supply that the rho can be thought of as an "excited" pion, like the Delta can be thought of as an "excited" nucleon. As the asker and @gcsantucci point out, the rho has a unit more spin than the pion. This means you could think of it as a light quark-antiquark system which has ... 0 These sorts of decays can proceed via two routes to go from the B to the final-state hadrons (pions & kaons). For example (ADS method),$B^-\to D^0K^-\to K^+\pi^- K^-B^-\to \bar{D}^0K^-\to K^+\pi^- K^-\$ The amplitudes of both of these routes will contribute to the final state with different phases, hence 'interference'. This is summarised nicely ...

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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 ...

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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 ...

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