# Tag Info

38

Yes, neutrinos "hit" electrons all the time inside the sun, on their way to getting out, which results into the resonant conversion of their flavor, predicated on the changing effective index of refraction. They interact with electrons, and protons, and neutrons, etc... through their favorite interaction, the weak , not electromagnetic interaction. (They can ...

33

Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract". In the electroweak case, the charges in question are weak isospin and weak hypercharge. For weak isospin, there are two isospin charges (or flavors), up and down, and their associated anti-charges, anti-up and anti-down. ...

26

Conservation of energy and the electron-degenerate pressure. For the neutron to decay you must have $$n \to p + e^- + \bar{\nu}$$ or $$n + \nu \to p + e^- \quad.$$ In either case that electron is going to stay around, but in addition to the neutrons being in a degenerate gas, the few remaining electrons are also degenerate, which means that adding a ...

26

Kaons and sigmas contain strange quarks, so the "ground state" particles must decay by changing quark flavour. The strong and electromagnetic interactions cannot change flavour, but the weak interaction can, hence they can only decay weakly. The strong interaction can easily produce $s\overline{s}$ pairs, which can subsequently pair up with lighter quarks ...

26

Can neutrinos “hit” electrons? Yes, e.g., Image credit

25

I think this question contains a misconception unfortunately caused by popular science descriptions of the Standard Model. The question seems to assume there needs to be some concrete source that particles "get" mass from, as if mass is a resource like money and the Higgs field is giving it out. But that's not right. In a generic field theory there is no ...

23

The interaction mediated by a massive boson will produce interaction potential of Yukawa type: $$V(r) = - \frac{\alpha_\chi}{r} e^{-mr},$$ As is well known, a scalar exchange leads to a universally attractive potential, whereas a vector exchange will attract particles to antiparticles, but repel pairs of particles or pairs of antiparticles. For Z-boson ...

23

Since the spin of the charged $\pi$ is $0$, the spins of the daughter particles need to add up to $0$ as well, i.e., their spins need to be anti-parallel. That's nothing else than the conservation of angular momentum. Assuming the anti-neutrino to be massless, it is always right-handed. Right-handed means that the momentum vector and the spin vector are ...

21

$c$ is not first and foremost the speed of light. It is first and foremost the universal speed limit of a cause-effect relationship - if $A$ influences $B$ (in the same inertial frame) causally, and if $B$ is a distance $d$ from $A$, then the minimum time that must elapse before the influence can reach $B$ is $d/c$. Since the interactions you name are ...

18

The weak force is one of the fundamental forces, but it doesn't have a strong attractive or repulsive effect on particles, and the term "weak interaction" is often preferred. For details on the attractions and repulsions associated with the weak interaction please see this answer by Alfred Centauri. All of the fundamental fermions (i.e., quarks and ...

16

Neutron decay is not an electromagnetic phenomenon at all. It is governed by the weak nuclear force. This is well supported by fact that the lifetime of the neutron fits neatly into weak universality, by the energy distribution of neutron decay products, and by the fact that anti-neutrino initiated "inverse beta decay" can be shown to be proportional to the ...

16

The problem with "weak charges" is that electroweak symmetry is spontaneously broken. Before the symmetry breaking, electroweak symmetry is described by an $SU(2)_L \times U(1)_Y$ gauge group.This amounts to three charges: weak hypercharge $Y$ for $U(1)_Y$ and weak isospin (total isospin $T$ and third component $T_3$) for the $SU(2)_L$. Some examples of ...

16

It's isn't terribly meaningful to say that Photons made of the electromagnetic force or that the W and Z bosons made of the weak force, though it's certainly true that these particles are the gauge bosons associated with these forces. However the main reason the photon is stable is because it is massless. An isolated massless particle cannot decay into ...

14

You can estimate the neutron lifetime using dimensional analysis. Beta decay is correctly described by the well known four-fermion Fermi theory, so the amplitude must be proportional to the coupling $G_F\approx10^{-5}\text{GeV}^{-2}$ (the Fermi constant). The decay rate is proportional to the squared amplitude: $$\Gamma\propto G_F^2\thinspace.$$ $\Gamma$ ...

14

In a neutron star there are mostly "free" neutrons and the question then is why don't they all beta-decay into electrons and protons? Well, some of them do, but the point is that when the electron (or proton, there are equal numbers of each) numbers build up then they become degenerate (meaning no more than two electrons [spin-up and spin-down] can occupy ...

13

There is no analogy of "voltage", "current", "charge" or "flux" to electromagnetism for the weak force, at least none that would be helpful. The reason for this is that all of these are classical concepts, while the notion of the weak force is completely quantum. Taking the classical limit just makes it vanish because the classical force law of forces with ...

12

Some background As it seems that you aren't terribly familiar with the nitty-gritty mathematics of quantum mechanics (and believe me, I'm not either), perhaps I can bridge your curiosity to a more intuitive understanding of what the weak force is. This may not be a fully-fledged answer, however my thoughts are too long to fit within a comment. As Stéphane ...

11

Here is the particle table for exchange bosons. You will see that the massive intermediate bosons are not assigned a parity. Parity is an operator. To have a definite value the state must be an eigenvalue of this operator. In the case of the massive weak interaction mediating bosons no such eigenvalue exists because in the standard model they carry both ...

11

No, there are no weak or strong waves in the sense as there are for electromagnetic or gravitational waves. The electromagnetic and gravitational waves are classical objects, they are possible vacuum solutions to the classical equation of motion for the field strength of the respective force, and can be radiated by objects charged under the respective force....

11

Electroweak instantons violate baryon number (and lepton number) by three units (all three generations participate in the 't Hooft vertex). This is explained in 't Hooft's original paper. As a result, the proton is absolutely stable in the standard model. The lightest baryonic state that is unstable to decay into leptons is $^3$He. The deuteron is unstable ...

11

One has to realize that words in physics have a definite meaning. This meaning, for words used in physics, is dependent on the specific mathematical model that uses it. The mathematical model for particle physics is called the standard model All matter and energy emerges from interactions and composites of these particle. The model encapsulates the large ...

11

If by "force" you mean a change of energy and momentum but not the type or number of particles involved, then weak neutral currents would be the answer. These were predicted by Salam, Glashow & Weinberg, and observed by the Gargamelle experiment in 1973. There a particle was observed to start moving, after interacting with an (unobserved) neutrino. The ...

10

You're mixing a few things up here. When you say "three" for the strong force, you're counting the number of colors of quarks, but when you guess "three" for the weak force, you're counting the number of force carriers. These are two different things. For example, if you counted the number of gluons (the force carriers for the strong force), you'd get ...

10

Your question is founded on misunderstandings. The theory of what goes on inside the nucleus is neither simple nor intuitive. "Meson" is simply a name for any particle that is a bound state of a quark and an anti-quark, just like "baryon" is a name for a bound state (like a proton or a neutron) of three (anti-)quarks. It does not denote a particular ...

9

All couplings in QFT are measured in Lorentz-Heaviside rationalized natural units. That is, for instance, for the electric charge, $$\alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} \approx 1/137 .$$ In these units $\epsilon_0=1$, so the elementary electric charge is simply  e = \sqrt{4\pi\alpha} ~\sqrt{ \hbar c} \approx 0.30282212 \sqrt{ \hbar c}...

9

Half life is, by definition, the amount of time until half of an infinitely large sample would decay. That's precisely equivalent (according to the frequentist interpretation of probability, if that matters to you) to the time until an individual particle's probability of decay reaches one half. The half life is a theoretical quantity that doesn't depend on ...

9

The cosmic rays consist of all sorts of particles ranging from heavy protons to little to no mass neutrinos. There are trillions of trillions of neutrinos passing through the earth at any given time. While the proton, neutrons and other 'very social' particles get trapped by the many layers of the crust, these hermit neutrinos can stream through the matter ...

8

Let us clear up a few basic concepts. Physics is about studying the way nature is . To start with, this could be descriptive, as it was for centuries before Newton. Descriptive does not answer "why" but "how" questions, in the same way that a map does not really answer "why" the landscape is like that, but describes as ...

8

To begin, let's 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 ...

8

Short answer: mostly no. Slightly longer answer: the interaction of neutrinos with the molecules of your body take several forms, but they all come down to ionizing dose. However, there are many larger sources of ionizing dose in your like. Things like the Potasium-40 in the food you eat, radon and Carbon-14 in the air you breath, cosmic rays, and on and on....

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