New answers tagged neutrinos
I've found an historical detailed description in this book: Controversy and Consensus: Nuclear Beta Decay 1911-1934 by Carsten Jensen , From 1911 to 1934, 23 years, a lot of ping-pong with the experiments and theories went forth and back. I will not try to resume the history and the book deserves a reading. My textbooks, aged, only mention the winning ...
The lower mass limits of neutrinos is not 0eV. They have to have a mass, since we can observe neutrino oscillations. This is something the standard model did not get right. Looking at PDG (2014) the boundaries seem to be 0 < m < 2ev.
The swarzchild radius is the distance a spherical mass has to be shrunk before it becomes a black hole. This is r = (2G/c²) m. 2G/c² is approximately equal to 1.48×10⁻²⁷ m/kg Estimates for the mass of the neutrino vary somewhat. One estimate is from around 0.2 eV to 2 eV. Picking 1eV for simplicity, the swarzchild radius is approximately 1.48×10⁻²⁷ m/kg ...
It's a whole lot of questions wrapped up in two sentences. I suggest you look at the standard model. (see below) - and this one doesn't have the higgs, but it was hard to find one the right size to post. All mass is a result of an interaction between particles with rest mass and the Higgs field or Higgs Boson if you like, but I think Higgs field is ...
They have weak hyperchage, flavor, mass and spin (though the mass states are not flavor states with various consequences). How many more "innate properties" do you want?
To begin, lets go over the basics again. Any ensemble of two body decays in which the parents and children have the same masses in each event has a delta-function energy spectrum, or violates at least one of energy- or momentum-conservation. The fact that the beta decay spectrum is broad and continuous implies that at least one of the pre-conditions is ...
It depends on the probability of interaction. This probability is computed using Fermi's golden rule, and it involves the strength of the interaction and the number of allowed final states. Weaker interactions means higher probabilities of going through the atom. Some examples: Neutrinos only feel the weak interaction, so their probability of going ...
Any sufficiently fast particle can go through the atom since the repulsing force is finite and you can prepare a projectile with a high enough energy.
Most sub-atomic particles can. In the Rutherford gold foil experiment, alpha particles (helium nuclei) often went through atoms. Beta particles (high speed electrons) can go through paper. There are more than a billion neutrinos going through you every single day.
HyperLuminal asked: "Does that mean that electrons are infinitely stable?" Think about Dirac's model of an electron, which includes left and right handed contributions. Now add the (Nobel-worthy) Brout-Englert-Higgs idea, that the left-handed bit interacts with a condensate of weak hypercharge, while the right-handed bit does not. This suggests a ...
The statement is true for decays, where lifetimes can be measured. It is not true for interactions though. A suicidal electron meeting a positron has a good probability to disappear, together with the positron, into two gamma rays, at low energies. Electron-positron annihilation It is intriguing that this is not true for neutrinos. If an electron ...
This is not exactly true. It is believed that net charge is conserved, but there is a weak process called electron capture, where an electron is captured by a nucleus, (usually from an inner "orbital" so there is a spectroscopic signature), a neutrino is emitted and a proton changes to a neutron. So therefore your textbook is wrong!
Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference ...
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