Tag Info

New answers tagged

3

Neutrinos are produced in the atmosphere all the time as a consequence of cosmic ray interactions, and they mostly then fly right through the planet. So there is no problem with a source. In fact the "flying right through" bit is one of the problems: even the full diameter of a planet simply doesn't intercept a large enough fraction of the beam to make ...


4

Neutrino flavor is defined as agreeing with the flavor of the charged lepton participating in the interaction, so that the neutrino in the reaction $$ \nu + A \to \mu + X \,, $$ is defined to be a muon neutrino and the one in $$ \nu + n \to e + p $$ is a electron neutrino by definition. We have no way of knowing the alleged flavor of a neutrino ...


2

The de Broglie wavelengths of freely propagating particles (i.e. forget interactions) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse of the scale factor. As neutrinos have a non-zero mass (perhaps 0.1 eV - see Battye & Moss 2014, ...


3

It's possible to detect neutrinos in whichever flavor they are oscillating through, so that won't necessarily cause a "dropped packet" problem. The answer is, technically, yes, there is no physical law preventing the use of neutrinos as a communication medium. It has been demonstrated that we can cause the emission and detection of neutrinos. For example, ...


1

In the quantum mechanical regime of elementary particle interactions there is a number of conserved quantum numbers and quantities: energy, momentum,angular momentum, and all the relevant quantum numbers that define the particles of the standard model. Experiments are scattering experiments , of two particle scattering and the results are compared with ...


3

The existance of the neutrino was deduced by Pauli from studying beta decay. Beta decay apparently produces a proton and an electron from the decay of a neutron. However this process violates several conservation laws, including conservation of angular momentum, and the way Pauli resolved this was to suggest the decay also produces a neutrino that wasn't ...


1

So when talking about "blowing up the nucleus" with a neutrino you are referring to a process we call Deep Inelastic Scattering. Deep refers to penetrating individual nucleons and interacting on the quark level, where as inelastic refers to the production of new particles (mostly hadrons). A more useful quantity to discuss in this interaction, other than ...


2

Nuclear binding energies are of order a few to 9 MeV per nucleon. So to "smash" apart the nucleus would take energies of this order. The mass of the particle involved is not directly relevant to how much energy they they may have. Photons could have GeV energies but they are massless. It just means your neutrinos are travelling at even closer to the speed ...



Top 50 recent answers are included