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2

Quark oscillations don't exist by the definition of a quark. A quark is defined as a mass eigenstate and thus it is not going to oscillate with time (energy eigenstates don't charge with time of course!). To see how this works consider the relevant quark interaction terms without any choice of basis, \begin{equation} - m _d \bar{d} d - m _u \bar{u} u - i ...


2

So over the last 48 hours I've been doing a small amount of research on my own question of why being able to detect or even image relic neutrinos would be important. Perhaps the most succinct summary can be found in these slides http://cosmo2014.uchicago.edu/depot/talk-long-andrew.pdf These suggest that there are 3 types of answer: The cosmologist's - ...


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The detector that took that image--Super Kamiokande (super-K for short)--is a water Cerenkov device. It detects neutrinos by imaging the Cerenkov cone produced by the reaction products of the neutrinos. Mostly elastic scattering off of electrons: $$ \nu + e \to \nu + e \,,$$ but also quasi-elastic reactions like $$ \nu + n \to l + p \,,$$ where the neutron ...


18

The neutrinos are coming straight at us. Indeed, their interactions with anything along the way are minimal at best. The reason the image is so big is that the angular resolution of the detector is rather poor (compared to, say, an optical telescope). This is not unexpected when it comes to neutrino telescopes. The details of how the detector work are ...


0

Martin wrote: ", hence they can only interact via the weak force, which is, as the name says, weak." Makes me wonder, is "weak" a good name? Or is this interaction better described as short, and still full of surprises? I'm thinking of {Dirac, Weyl, Majorana} spinors, rates of chiral oscillation, inertial-mass... ... which makes me wonder about the ...


3

Issue 1: The Expansion Misconception Forget everything you thought you visualized about the Big Bang. Let's start from scratch. First, picture a sheet of rubber with a grid marked on it. This sheet represents space. It is not curved, nor will it ever curve in this example. It might stretch, but that's not particularly important here either. Attach a light ...


0

Addressing part of your question: Neutrinos were not emitted prior to (or at the start of) the big bang. "Before the big bang" is a phrase that drives the philosophers nuts, but because the big bang was the beginning of space and time, it is thought that there was no "before". Also, neutrinos of any sort could not have formed until a few fractions of a ...


0

Background neutrinos can't be detected, but neutrino observatories have detected neutrinos into the high TeV range. See e.g. Icecube http://icecube.wisc.edu/ for a high energy neutrino observatory.


3

I will answer this since @rob, who provided the link that gives a summary of the proposed methods and technical difficulties, is not doing it (comments are not guaranteed to be invariant to time on this site). It is true that measuring the Cosmic Microwave Background radiation has been extremely important in the development of the model of the beginning of ...


1

In many standard theories, Neutrinos are assumed as the be most common particle in the Universe (known as cosmic neutrino background (CNB) a relic of the Big Bang). Many experiments are being carried out to detect them from different sources. Those from the CNB have been only indirectly detected, but many, originated in violent event across the universe ...


1

It is not possible, even in principle, to unambiguously detect a change in the speed of light over time, and this was true even before the SI units were redefined so as to give $c$ a defined value. The trouble here is that $c$ has units, and if a fundamental unitful number seems to change over time, there is always an ambiguity in whether the number is ...


2

The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only $\nu_L$). Namely, the desired term is (one generation suffices): $$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, ...


3

Lets analyse the Majorana condition and the Majorana mass term. A massive Majorana neutrino $\chi_j$ (a Majorana spin $1/2$ fermion) having mass $m_j>0$ can be described in a local quantum field theory (eg. the standard model) by a four component spin $1/2$ field $\chi_j(x)$ which satisfies the Dirac equation and the Majorana condition which reads: $$ ...


0

We know four fundamental forces, three of them being included in the Standard Model. At low energies (compared to the mass of $W^\pm$) the forces have strengths as follows: strong force > EM force > weak force > gravity Also, photon and neutrino are really quite different. Photon is a force carrier, while neutrino is a matter particle. Now, matter ...


10

The photon does couple directly to charged stuff, e.g. via Compton scattering. This is indirectly related to the spin, as direct interactions between fermions are hard to construct. The neutrino on the other hand does not couple immediately to any other matter particle. It requires a force-carrier. Now as it turns out the only force carriers that care about ...


7

Neutrinos having no charge means they don't participate in electromagnetic interactions, which are the strongest (at least long range). Them being leptons means they don't interact with the strong force (which is, as the name says, strong), hence they can only interact via the weak force, which is, as the name says, weak. Photons on the other hand do not ...


2

The interaction term disappears because when you integrate out the right handed neutrinos you insert in a linear combination of the other fields in their place. For example I expect that we would have something of the form, \begin{equation} G _{ ij} \sigma \bar{\nu} _L ^i \nu _R ^j \rightarrow G _{ij} G _{jk}\sigma ^2 \bar{\nu} _L ^i \nu _L ^k ...


0

The real reason is in following. Let's assume Majorana field: $$ \Psi_{M} = \Psi_{L} + \hat{C}\bar{\Psi}^{T}_{L}, \quad \hat{C} = i\gamma_{2}\gamma_{0}, \quad \Psi_{L} = \begin{pmatrix} \psi_{L} \\ 0 \end{pmatrix}. $$ By using this notation it's not hard to see that kinetic term is equal to $$ \bar{\Psi}_{M}\gamma^{\mu}\partial_{\mu}\Psi_{M} = ...


0

The short asnwer to your question is that the overall factor $\frac{1}{2}$ from the Lagrangian of a Majorana field (in the 4-component notation) $$\mathcal{L}=\frac{1}{2}(\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi -m\bar{\psi}\psi)$$ compared to the general Dirac Lagrangian is usual for self-conjugate fields and it is introduced to ensure a consistent ...



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