Bob_for_short

You are right, of course. Let us return to the spherical magnet configuration. The radial magnetic field looks weird to me. Maybe one cannot get a complete cancellation of azimuthal fields inside and it will be something complex, with closed lines.

Being a student, I heard of Maxwell multi-poles. For example, you take two opposite-sign charges Q at some distance R. Their dipole moment is Q*R. Then you decrease the distance R and increase the charge values so that the dipole moment does not change. In the limit R = 0 you have an electric dipole with practically closed lines.

A dielectric bar, like a magnet bar but charged with rho(x) = k*(|x|-L)*sgn(x), -L < x < L.

So a "fundamental" dipole is different from an electric dipole? An electric dipole can also be made with continuous charge distribution, without a gap. OK, Is it possible to construct a ball like the OP asked? I think we can take a ball, cut it in 8 equal parts with three mutually perpendicular planes, magnetize each piece along its axis of symmetry (from the center to the surface) and make them stick together with a glue.

Magnetic field lines are the same as electric dipole lines: they start at one pole and finish at another one.

No, the magnetic field in such a configuration will be confined within the body. It is just like two equal electrical charges, one at the center, the other one is on the sphere. No field outside the body. However the magnetic field inside will be radial. I think it is well possible to make.

Projection [math]< \Phi | \Psi >[/math] or the amplitude measurement is not an evolution or a collapse.

Yes, that is right. A(x,t) is not a constant vector in space and time so both E and B may be different from zero. In particular, if A(x,t) = a⋅exp(ikx-iωt), you have a plane EMW.

A little wind without closet will do.

Steven Hawking has Science at his hand to explain everything, so we don't need other scientists.

It is not the SWE that should predict the projections but the angular momentum operator. SWE still holds if the interaction does not depend on spin variables. The Pauli equation is reduced to SWE if there is no magnetic field, for example. Each spinor component obeys SWE then. The same statement is valid for KGWE.

No! A charge does not emit any potential V( r )! V( r ) is a potential energy of interaction of two charges and r is their relative distance. There is no other meaning of the potential energy.

The interference picture is not destructive in all the space. In some regions the fields cancels each other, in other they add. The energy of EMF is an integral over the whole space from the energy density (E2 + B2). It remains constant despite the bright and dark region may travel in space.

In my pet theory this bla-bla about virtual photons surrounding an electron is represented in a concrete formula (solution): the electron is a part of quantum oscillators. The whole system is called an "electronium". When you push the electron, the state of oscillators change - they get excited that corresponds to appearing real photons. On the other hand I need a Coulomb interaction term too. It is another, independent property of charges (apart from emitting photons) and is not reduced to exchange of "virtual" photons. I my model each electronium has its own oscillators, like each atom has its own electrons and energetic levels. So the charge attraction/repulsion is another feature of charges. In the standard QED this question is quite vague due to initially decoupling the "mechanical" and electromagnetic properties of a change and considering the coupling by the perturbation theory. Factually nobody has written the solution for a real electron surrounded by the virtual photons. Such a solution has to include the "mechanical" and the "wave" variables. I advances such a solution from physical and mathematical reasoning as an ansatz in http://arxiv.org/abs/0811.4416 and http://arxiv.org/abs/0806.2635

Photons from A do not disappear in an accelerated reference frame. The radiated EMF energy density is positive and cannot be canceled with a variable change. The emitted photons can be "registered" by a photomultiplier or reflected by a mirror, so their presence is easily visible in an accelerated RF.

Yes, we can and we make. By increasing the projectile energy in the Bremsstrahlung, for example. A regular, visible light becomes gamma-ray flux in a fast-moving reference frame due to Doppler shift (increase) of frequency. But normally nuclear gamma-rays sources in laboratories emit much less photons than there is in a visible light. So gamma-rays consist of rare "particles" (hard photons).

This interpretation is much better because it treats the electron classically and photon quantum-mechanically. Yes, [math]\Lambda_C[/math] determines the energy loss of the incident photon (which leads to increasing its wavelength). No, only problems with such an interpretations (runaway exact solutions). No, because it does not predict the discrete spin projections. If you take into account the magnetic field energy (together with electrostatic), you may arrive at larger "classical" electron radius proportional to the Compton length.

Of course, it would. Accelerators are designed in a way to get the desired beam path. It is exactly A who is "manipulated".

The paths in the integral do not depend on A. The phase factors do. You cannot manipulate with paths because they are given. You cannot change the integral value because it is unique, like 1 + 2 = 1 +1 + 1. You always hear the sum = 3.

Yes, CED describes lasers in the sense of emitting and propagating the EMF. However it does not deal with photons. The number of photons in a coherent beam is uncertain anyway. What is certain is the wave phase. If you read QED, you may find a chapter "Radiation of a classical current" or so (see QED by Akhiezer, Berestetski or other sources with coherent light description). No. The classical electron radius is a dimensional parameter appearing, for example, in the Tompson formula. The other its meanings are misleading. The electron Compton wave-length, appeared first in the Compton formula, has nothing to do with the electron size. Remember, in the Compton formula derivation the electron is considered as a classical relativistic particle with mass m. The h-bar comes from the photon property E = h-bar*omega (different from the energy of a classical EMW). After recombining the dimensional parameters to get the Compton length, you characterize the photon as a classical wave with a classical wavelength (no h-bar involved) which was not the initial assumption. Photon is not characterized with a wavelength (but energy) and the electron is not quantum in this problem. Bohr radius is meaningful and includes the electron charge. This is a real size amongst three of them, if you like.

The value of alpha has nothing to do with your question because in radiative processes the "small" parameter is alpha*ln(omega/m) or so. For soft radiation it may be large. Another important thing is the number of photons. Classical Electrodynamics describes well such cases. There exists a CED in media, with phenomenological epsilon and mu. Generally, CED describes average values, and as long as the number of photons is large (the fluctuations are relatively small), one can use CED.

Keeping in mind quantum electromagnetic, with a stable medium.

For a health reason, I cannot follow your reasoning about protons. Atoms are also compound systems. Atom-atomic collisions may be approximately described with a potential interaction. The cross section is given in http://arxiv.org/abs/0806.2635, formula (11). I did not get the point but the interference pattern will remain the same in my opinion. If the screens, slits, and detectors are some boundary conditions, then the material properties do not matter.

Quantum mechanics answers these questions. If I knew what you want to know...

The energy emitted by my emitter is entirely absorbed by my absorber so no energy is left to create other universes.