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The main reason is that lightest common particle/anti-particle pair is the electron/positron. When they collide, the overwhelmingly most likely result is emitting photons. The electron and positron weigh 511 kev, so when they annihilate, there is 1.02 MeV of energy. So each photon has 511 kev. Photons with this energy are gamma rays. Other photons like ...


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It is possible. The rule is simple: the total energy (which is at least mc^2, m the sum of both masses) will be converted into other particles (photons or particle-anti-particle pair). So it is possible for a photon to pick up only a very few energy (for example by having a lot of photons, which is possible, even dough very unlikely). I think, it is ...


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The amount of energy to emit is enormous ($2mc^2$ plus kinetic energy), so the emitted energies per photon will be at least the rest mass of the electron, which is half GeV (not counting some extreme and unlikely redshifts due to the collision point moving with respect to the observer). And that's well in the gamma spectrum.


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To expand on a @ACuriousMind's comment, consider an electron-positron pair such that their center of mass is at rest. Since energy and momentum are conserved in the annihilation process, the resulting photons will be emitted in opposite directions with an energy $E = m_e = 511\ \text{keV}$ each (units such that $c = 1$). If the pair is instead in motion, one ...


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In the field of PET (positron emission tomography) people tend NOT to call the product of the annihilation "gamma rays", but rather "annihiliation photons". While that may be a subtle distinction, the view is that a "gamma ray" is emitted by a nucleus, while a "photon" is a more general term used for a quantum of electromagnetic energy. But according to ...


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At tree level, a matter-antimatter annihilation reaction doesn't just produce gamma rays, nor can you exclude neutrinos in the final state. Even the simplest such reaction can—given enough energy—produce a variety of particle-pairs. However, those pairs are subject to two subsequent processes: If the particles are not stable, they will decay towards ...


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As is discussed in this answer, the "rest frame" of the cosmic neutrino background would be very similar to that defined by the cosmic microwave background if neutrinos were very light (say $<0.1$ eV). The Sun would be moving with respect to this frame at around 370 km/s. But if neutrinos were more massive(say getting on for 1-2 eV) then they are ...


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I just want to add something to the answers above, because I did not understand this initially. As far as clustering goes, the crucial point is how slow the neutrinos are travelling with respect to the characteristic escape velocities of galaxies (600 km/s) and clusters (2000 km/s). If you assume a rest mass of 0.1 eV, use the 1.95K temperature and the ...


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As explained in this paper, the dominant effect is due to gravitational interactions, which can yield overdensities up to a factor of $10^3$ for neutrino masses of the order of 1 eV. The clustering of relic neutrinos can be modeled well using the collision free Boltzmann equation (Vlasov equation) where the densities evolve under the influence of the ...


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I thought about this a little more since I prompted your question and there are a couple of other complications worth noting. This may be more like a comment than an answer, but it's too long for a comment. First, and directly addressing your question: the cosmic neutrino background is already present before the gravitational well of the star forms. If ...


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It requires that they lose energy somehow (to drop hyperbolic orbits into periodic ones). There are two basic mechanisms available: gravitational scattering and weak scattering. In both cases we expect the interaction to be elastic, but that doesn't mean the neutrino has as much kinetic energy in the star's frame afterward the interaction as before: it ...


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Short answer is we don't know, there may be right handed neutrinos and left handed antineutrinos that either don't interact through the weak force or do so extremely less than neutrinos. Sterile neutrinos a theoretical candidate for these particles.


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Your post reminds me of this paper by Wigmans (also via arXiv) which I learned about during a colloquium in the distant past. Wigmans points out a narrow and interesting region in the parameter space for neutrinos in the mass region 0.1-1 eV. A longer paper (arXiv) by the same author; you'll enjoy mining citations. Such neutrinos, redshifted to the CνB ...


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As far as theory goes, the Cosmic Neutrino Background (CvB) was created within the first second after the Big Bang, when neutrinos decoupled from other matter. Nevertheless, while the universe was still hot neutrinos stayed in thermal equilibrium with photons. Neutrinos and photons shared a common temperature until the universe cooled down to a point where ...


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Yes, you can in a vector theory. But in the SM you cannot do that because of the chiral structure of the model: $l_L \sim \mathbf{2}$ but $e_R\sim (e^c)_L\sim \mathbf{1}$. So you need a scalar which transforms as $\mathbf{2}$.


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Consider the average beta decay, which at nucleon level looks like $$ n \longrightarrow p + e^- + \bar{\nu} \,. \tag{1}$$ The distribution of electron energies (as measured in proton's frame) is controlled by the phase space of the products. We observe an electron energy spectrum consistent with these physics. What you propose is essentially that this ...


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This argument might conceivably work for weak decays, though I believe there is evidence to the contrary. This comes to mind, but I won't 100% swear it's quite what you're after. Peterh mentions in a comment that weak decay rates (e.g. in supernova afterglows) appear to be independent of the local density of dark matter. There is no reason to believe the ...


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The experimental detection of slow neutrinos is indeed a big problem, but one that is very important. The cosmic neutrino background is at a temperature of around 2K and likely to consist of non-relativistic neutrinos for plausible neutrino rest masses - with a density of around 340 cm$^{-3}$ (all flavours). It is at this low temperature for precisely the ...


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Strictly speaking, it is indeed incorrect that neutrinos travel at "close to the speed of light". As you said, since they have mass they can be treated just like any other massive object, like billiard balls. And as such they are only traveling at nearly the speed of light relative to something. Relative to another co-moving neutrino it would be at rest. ...


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The cross-section of a nucleon to a solar (few MeV) neutrino is about $10^{-46}$ m$^2$ and higher than the cross-section for leptonic reactions (see here). The mean density of the Earth is 5500 kg/m$^3$, and this is essentially made up of protons and neutrons with a number density of $3.3\times 10^{30}$ m$^{-3}$. The mean free path is $(n\sigma)^{-1} \...


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The basic assumption in the Big Bang cosmolocigal model is that the CMB departures from a uniform black body radiation observed in the map All-sky map of the CMB, created from 9 years of WMAP data. are a relic of the density of matter at the time of the decoupling of photons, about 380000 years after the BB. In contrast, neutrino decoupling, if ...



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