electricity - Why do charges enter the capacitor in the first place?

I have been studying Capacitors for the past year now and the one thing I don't understand is how a charge is stored on the capacitor. Essentially, a circuit with a capacitor is an incomplete circuit right?

So why do the electrons start to gather up on one of the plates. Its like they are tricked into thinking its a complete circuit and get trapped once they realise it isn't! What is this sorcery!

What is happening to make the electrons do that and if it simply is electrons moving and then turning back or getting stuck, would a switch have the same effect? (Assuming the switch is open).

Hopefully someone can finally fill the major gap in my knowledge because everything else I read doesn't explain this.

Answer

Imagine taking a disc of some conductor and putting it in an electric field (with the plane of the disc at right angles to the field lines). The free electrons in the conductor will move in response to the field, so the face at the positive side gets a net negative charge and the other face gets a net positive charge. You get a transient current through the body of the disk when you turn on the electric field and a transient current in the other direction when you turn it off.

Now split the disk through the middle, i.e. turn it into the two plates of a capacitor, and connect a wire between the two outside faces, then switch on the external electric field again. Once more the disk gets polarised, but this time you get a transient current flowing from one side of the (now split) disk to the other through the wire. You end up with the same charge separation as before, but the electrons flowed through the piece of wire instead of through the body of the disk.

Finally, connect a battery in the middle of the piece of wire. This produces an electric field between the two sides of our split disk (i.e. capacitor), and just like the externally applied field it will polarise the disk and electrons will flow through the wire from one face of the disk to the other.

That's why when you connect a battery to a capacitor you get a transient current as the capacitor polarises.

standard model - Highest and Lowest $SU(3)_F$ states

For the finite dimensional $(p,q)$-irreducible representation of $SU(3)_F$, we can label the states as $\mid T_3,Y\rangle$. Where $T_3$ is the third component of isospin and $Y$ is the hypercharge. How can I obtain the highest and Lowest weight states? I mean what are the $T_3$ and $Y$ values?

nuclear physics - Anti-neutrons, anti-quarks, isospin: What is observed and what is derived?

I would be a little more restrained with the existence of antineutrons. First at all - if I understood right - the existence of antiquarks is hypothetical. If one not agree with this please refer to experimental data which shows their observation.

Second one has to show that the neutron is not able to decay in electron, neutron and anti-neutrino as well as in the anti-particles. If one will use the isospin he has to go back in history and to explain how and why the isospin was discovered. Or was he invented?

Third a neutron-antineutron collision - after all what we have seen in this kind of collisions with protons-antiprotons and electrons-positrons- has to lead to pure energy in the form of photons. Did we get this data in some experiment?

Answer

You didn't understand any of these questions right. Antiquarks and their bound states, including the antineutrons, are produced and observed as easily as bread and butter. Lots of details experiments with e.g. antineutrons have been performed, e.g.

Scattering of antineutrons with hydrogen

http://www.sciencedirect.com/science/article/pii/037026939290998J

(this one was done more than 20 years ago) and all these experiments agree with the theoretical predictions. Millions of antiquarks are produced at the LHC each second (when it's running), too.

There's not a tiny doubt that every particle has an antiparticle. For most particles, the antiparticle is different from the original particle. Only "truly neutral" particles such as photons, gravitons, and Higgs bosons (but not neutrons!) have antiparticles that are identical to the original particle. All the antiparticle species to the known particle species have been observed, too.

The neutron always or virtually always decays to a proton, an electron, and an electron antineutrino. There's no doubt about it – this fact can be calculated from the Standard Model and it may experimentally verified, too.

Also, the antineutron can't decay to the same products as the neutron (or vice versa). That would violate the conservation of the baryon number and the conservation of the isospin in the processes based governed by the strong interaction. These conservation laws are "laws in our theories" but we only believe all these laws because there is an overwhelming experimental support for all these things.

It's impractical to measure the neutron-antineutron annihilation because both particles are (equally) unstable. But totally analogous annihilation of antineutrons and protons (see the paper above) – with some light charged products aside from neutral pions – are almost the same thing and indeed, virtually all the rest mass gets converted to pure energy just like in the case of any annihilation.

All the things you doubt – and hundreds of much more advanced, detailed, and accurate insights of a similar kind – are completely indisputable and experimentally verifiable, often very directly.

general relativity - What is the status of existing measurements of the speed of gravity?

In replying to a recent question I stated:

Gravitational waves have not been yet experimentally observed so as to have their velocity measured.

Which after the fact prompted me to try and verify it. I found a "speed of gravity" claimed measurement, which has been in a controversial discussion. In the last paragraph of this wiki article I found:

In September 2002, Sergei Kopeikin and Edward Fomalont announced that they had made an indirect measurement of the speed of gravity, using their data from VLBI measurement of the retarded position of Jupiter on its orbit during Jupiter's transit across the line-of-sight of the bright radio source quasar QSO J0842+1835. Kopeikin and Fomalont concluded that the speed of gravity is between 0.8 and 1.2 times the speed of light, which would be fully consistent with the theoretical prediction of general relativity that the speed of gravity is exactly the same as the speed of light.[18]

There has been criticism, described in the paragraph.

Has this matter been resolved ? is the measurement valid?

The speed of gravity has also been calculated from observations of the orbital decay rate of binary pulsars and found to be consistent to the speed of light within 1% ( same link).

This is the speed that gravitational fields transfer information, and though the number itself depends on the theoretical framework solving for the system parameters it is true that just the existence of gravitational damping in the binary system would imply that the speed cannot be infinite.

Along these lines I was wondering whether studying the second star in a binary system where the first has gone nova would not give a cleaner and model independent measurement of the speed of transfer of information gravitationally. In this day and age where everything is digital one should be able to have data on this from now on.

---

LIGO has officially (Feb 11,2016) announced the observation of gravitational waves , at the same time consistent with the speed of light being c, and thus measuring it. See the webcast for a summary.

Answer

It is difficult to design empirical tests that specifically check propagation at c, independently of the other features of general relativity. The trouble is that although there are other theories of gravity (e.g., Brans-Dicke gravity) that are consistent with all the currently available experimental data, none of them predict that gravitational disturbances propagate at any other speed than c. Without a test theory that predicts a different speed, it becomes essentially impossible to interpret observations so as to extract the speed.

Has this matter been resolved ? is the measurement valid?

Kopeikin was wrong. See:

The speed of gravity has also been calculated from observations of the orbital decay rate of binary pulsars and found to be consistent to the speed of light within 1% ( same link).

This is not really correct. GR predicts that low-amplitude gravitational waves propagate at c. The binary pulsar observations are in excellent agreement with GR's predictions of energy loss to gravitational waves. That makes it unlikely that GR's description of gravitational waves is wrong. However, it's not a direct measurement of the speed of gravitational waves. There is no way to get such a measurement without a viable theory that predicts some other speed.

Along these lines I was wondering whether studying the second star in a binary system where the first has gone nova would not give a cleaner and model independent measurement of the speed of transfer of information gravitationally.

The fact that a star goes nova doesn't produce any abrupt change in its gravitational field. If mass is expelled, there will be a gradual change.

statistical mechanics - Bose-Einstein condensate for general interacting systems

There is Bose-Einstein condensate (BEC) for non-interacting boson systems. Can we prove the existence of BEC for interacting systems?

special relativity - Photon Emission/Absorbsion from the Photons Perspective

First some assumptions.

1) Photons travel at the speed of light. 2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel. 3) From the photon's reference it is emitted and absorbed at the same time and in the same place.

Now imagine a setup with 1 emitter and 2 detectors (E, D1, D2). E emits photons toward the center of D1. D2 is directly behind D1 in the direction in which E emits (D1 is in the line-of-site of E, D2 is hidden from E behind D1.

E -> D1 D2

From our frame of reference photons are emitted and always absorbed by D1.

From the photon's reference, aren't the emitter and detectors sitting in the same place in its perceived 2D space? Wouldn't it then be equally valid for the photon to be absorbed by D2, or both D1 and D2? Maybe neither? Does contracted spacetime maintain some sort of z-order?

It's been my understanding that when things happen doesn't remain constant between reference frames, but the effects, what actually happens does.

Is there a paradox here or am I looking at it wrong? Is it ever valid to ask what happens in an external frame of reference?

Answer

Let's start with your assumptions:

1) Photons travel at the speed of light.

Right. No problem with that.

2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel.

Wrong. The photon doesn't have any kind of reference frame. To appreciate why, imagine you're travelling at the speed of light. We know you can't actually do this, but just imagine it. Some would say that in your reference frame everything happens at once. But I know your course, and I can nudge an asteroid into your path: BLAM! It's game over for you. You didn't see the asteroid because you don't see anything when you're moving at the speed of light. Light would have to move faster than light for that, and it doesn't. So you don't see everything happening at once, you don't see anything. Or think anything either. You're a popsicle.

3) From the photon's reference it is emitted and absorbed at the same time and in the same place.

Wrong. Like I said, the photon doesn't have any reference frame. Its wavelength isn't length-contracted to zero either. It gets emitted, it moves through space at the speed of light with its E=hc/λ wave nature, then it gets absorbed. And these three events didn't happen at the same time. Note that I measure my time using my optical clock, which has light moving inside it. The elapsed time relates to how much the light has moved.

Now imagine a setup with 1 emitter and 2 detectors (E, D1, D2)... ...From the photon's reference, aren't the emitter and detectors sitting in the same place in its perceived 2D space?

No. It doesn't perceive anything. Just like you didn't perceive that asteroid.

Wouldn't it then be equally valid for the photon to be absorbed by D2, or both D1 and D2?

No.

Does contracted spacetime maintain some sort of z-order?

Space doesn't change one iota just because you changed your speed through space. Nor do the things in space. Instead you change, along with your measurements of space and time and things.

Is there a paradox here or am I looking at it wrong?

You're looking at it wrong I'm afraid.

Is it ever valid to ask what happens in an external frame of reference?

Yes, that's no problem. The problem comes when you think there is a frame of reference when actually there isn't. And you are not the only person who has this problem.

quantum mechanics - Momentum operator acting on a bound state doesn't return an eigenvalue although kinetic energy operator does. Why?

We know that $[\hat{H}, \hat{P}]=i\hbar\frac{\mathrm{d}V}{\mathrm{d}x}$, therefore $\hat{H}$ and $\hat{P}$ commute if $\frac{\mathrm{d}V}{\mathrm{d}x}=0$, which is true for $V=0$. In that case the operators share the same eigenstates. As a matter of fact, both of the operators should return an eigenvalue if they operate on an eigenstate.

Now if we take a plane wave state $|\psi\rangle = Ae^{ikx}$, we see the above argument holds: \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &=\hbar k|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &=\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}

But if we go to the particle in a box problem, then $|\psi\rangle =\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$, where $k=\frac{n\pi}{a}$, so \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &= -i\hbar\frac{n\pi}{a}\sqrt{\frac{2}{a}}\cos\left(\frac{n\pi}{a}x\right) \neq\hbar\frac{n\pi}{a}|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &= -\frac{\hbar^2}{2m}\left(\frac{n\pi}{a}\right)\left(-\frac{n\pi}{a}\right) \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right) =\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}

So in a bound state, $\hat{H}$ returns an eigenvalue but $\hat{P}$ doesn't, although the commutation condition is satisfied. (For the harmonic oscillator potential, I am not surprised by a similar result as the commutation criteria are not satisfied in the first place). Why is this?