87

It doesn't matter. Suppose two electrons approach each other, exchange a photon, and leave with different spins. Are these "the same electrons" as before? This question doesn't have a well-defined answer. You started with some state of the electron quantum field and now have a different one; whether some parts of it are the "same" as before are really up to ...


73

Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays. Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are ...


71

There is lots of experimental evidence that the electromagnetic field exchanges energy with atoms in discrete chunks, and if we call these chunks photons then photons exist. Which is all very well, but my guess is that you’re really interested to know if the photon exists as a little ball of light speeding through space at $c$, and if so then, well, that’s a ...


60

That a particle decays into other particles is completely disjoint from it having substructure/being fundamental or composite. Some examples: A highly energetic photon may "decay" into an electron and a positron in the presence of another object that takes the excess momentum. That doesn't mean a photon is a composite of electron and positron. A free ...


51

Under special relativity nothing can be incompressible: consider any object of nonzero size and finite mass in its rest frame; when you apply a force to it on one side it will start moving. If it were completely incompressible, the other end would start moving simultaneously. Since the ends are spatially separated, there is a frame in which the other end ...


48

Indeed, we can't know for sure if muons are elementary are not. In this sense, the situation is like the 1950's, where we had a zoo of mesons and hadrons, but didn't yet know they were made of quarks. We wouldn't have direct confirmation of quarks for decades to come, like how we have no indications of muon compositeness now. Despite this, the quark model ...


42

As far as we know the electron is a point particle - this is addressed in the question Qmechanic suggested: What is the mass density distribution of an electron? However an electron is surrounded by a cloud of virtual particles, and the experiments in the links you provided have been studying the distribution of those virtual particles. In particular they ...


37

Seeing = detecting photons that happen to interact with your retina. You can't see photons when they are just travelling nearby. Take lasers for example. When someone is using laser pointer, the only reason you see the beam is that photons collide with dust and air particles and therefore their direction is changed. For example into your eye. Otherwise you ...


34

In quantum field theory, an elementary particle doesn't have one precise location and size in space. The quantum of an electron field in free space has different extent compared to the electron around a hydrogen atom, for example (i.e. it's harder to bounce an electron off a free electron than off a hydrogen atom). While in one very real way, an electron's ...


33

The gauge boson associated with the magnetic field is the photon. Electric and magnetic fields are in effect different views of the same thing, i.e. the electromagnetic field, and the gauge boson for the electromagnetic field is of course the photon. Consider you are looking a static charge, which obviously has just a static electric field. But now suppose ...


28

Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed ...


25

Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2$\mu$s. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter ...


24

The best place to seek evidence that decay doesn't equal compositness is in particle creation. Because if decay meant compositeness, then creation would require you to get the constituents together. When you bash two nucleons together at high enough energy you get a lot of junk coming out. Some of that junk is lepton particle-antiparticle pairs, and many of ...


24

One who is familiar with the history of particle physics, and physics in general, knows that physics is about observations fitted with mathematical models. This review examines the limits on size we presently accept for the fundamental particles which presently are at the foundation of the present standard model of particle physics. This analysis of what "...


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 ...


22

The shape of a distribution of charges is described in terms of multipole expansion, which you can think of as similar to Fourier expansion but in two dimensions. The total charge gives you the "monopole term", whose interaction is spherically symmetric. If there's an offset between the center of the mass distribution and the center of the charge ...


22

Scattering experiments can be used to determine the size of a particle. The results for an extended object are different than that of a point particle. But all of these scattering experiments depend on getting the probe particle "close" to the scattering object. In the case of electrons, that means launching the probe with enough energy to overcome the ...


22

How elementary are you talking. All mesons are unstable, a proton is possibly the only stable baryon – its half life is at least $10^{32}$ years. If you consider quarks, it's a tricky answer. Strange and heavier quarks decay via the Weak interaction –but so do down quarks, not in a proton, but certainly in a neutron ($n\rightarrow p + e^...


21

No. The decay products of a certain particle are not equivalent to its constituents. This is evident especially in the context of fundamental particles: quarks can decay into other particles, but that does not mean that a quark is not elementary (see my answer to this question). Nuclei are made of neutrons and protons, which in turn consist of quarks and ...


19

That's an interesting question, even though it might be biased by the definition of forces, and on what particles they apply. For instance, if you want to describe the force that exists between photon (even though direct photon-photon scattering has not been observed yet), it is mainly due to electron loops, so in that case the `force' is fermionic. On a ...


19

Gravitons are hypothetical, but they're far less hypothetical than most of the other particles which theorists speculate about (such as axions, magnetic monopoles, strings, sterile neutrinos, and the like). That probably sounds a little strange. Let me explain. We don't have a complete theory of quantum gravity. But we do actually have an extremely ...


19

The neutron is in no way "composed" of a proton and an electron. It can decay to a proton, electron, and an antineutrino. But that doesn't mean that these three particles literally co-exist inside the neutron at the beginning. Instead, the decay involves some real transmutation of elementary particles. The only thing that one can say because of the decay is ...


19

The electron is a fundamental point particle. It does not have sub-particles “inside”. However, its quantum interactions with other particles should give it a small electric dipole moment, according to the Standard Model of particle physics. (It is a very difficult calculation but there are estimates for it.) Some people like to picture this in their minds ...


18

Spin is not about stuff spinning. (Confusing, I know, but physicists have never been great at naming things. Exhibit A: Quarks.) Spin is a purely quantum mechanical phenomenon, it cannot be understood with classical physics alone, and every analogy will break down. It has also, intrinsically, nothing to do with any kind of internal structure. (Non-...


18

We study it using deep inelastic scattering. It was this type of experiment that first revealed the proton had an internal structure, and if the electron has an internal structure it will be this type of experiment that reveals it. No experiment has yet discovered evidence for an internal structure in the electron, however that doesn't prove the electron ...


17

As you can guess from the answers which vary widely from "yes" to "no," your question touches on a very sensitive topic for those who follow science. The answer will quickly boil down to frustratingly phrased questions like "what is real" because what you ask is sufficiently tricky. In the world of philosophy, science is categorized as part of "empiricism."...


17

There is a particle mediating the electromagnetic interaction: the photon. In the quantum version of electromagnetism (which is a particular example of a quantum field theory), the existence of mediator boson particles for forces is implied. The following may be worth mentioning: We say "electromagnetic" (and not "electric" or "magnetic") because this is ...


16

In quantum mechanics, spin is a kind of angular momentum that is intrinsic to a particle. Like normal angular momentum, when we go to the quantum world there are two numbers that describe the spin of a body, which we label $s$ and $m_s$. $s$ can be integers or half integers, depending on what you're trying to describe, and $m_s$ is an integer or half integer ...


15

Gravitons are the particles you get from quantizing General Relativity. Since we don't know yet how to correctly quantize GR (or whether trying to quantize it is actually the right way to go forward; for all we know it might just be an effective theory where the more fundamental theory has to be quantized instead), we cannot know for sure whether the ...


15

As the other answers state, the individual nuclei have a probability of decay and this happens randomly, as they sit there. You are correct though in wondering about a trigger, because at the atomic level that is exactly what happens with lasing, induced-emission = induced-decay. Spontaneous decay is random, controlled by the quantum mechanical individual ...


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