You've jumped to the wrong conclusion.That interpretation is never going to work. This is obvious when you compare the equations of motion. The equation of motion of a charged particle in a given electromagnetic field is given by the Lorentz force:
$$On the other hand the trajectory of a particle in a curved space is given by the geodesic equation I told you about in the other thread:
m \frac{\mathrm{d}^{2}\bar{x}}{\mathrm{d}t^{2}} \,=\, q \bigl[ \bar{E} \,+\, \bar{v} \times \bar{B} \bigr] \,. \qquad (1)
$$
$$
\ddot{x}^{i} \,+\, \Gamma^{i}_{jk} \dot{x}^{j} \dot{x}^{k} \,=\, 0 \,. \qquad (2)
$$
Start from scratch and think about the electromagnetic field is curved space. The electron has an electromagnetic field, which is isotropic. It also has magnetic dipole moment, which should signal rotation to you. So think in terms of frame-dragging, and picture the particle as an extended entity like this:In order to interpret the electromagnetic field as curved space you'd have to show it predicts the right trajectories for charged particles, which means showing that you could interpret equation (1) as a special case of equation (2). But it's clear you can't: these equations describe qualitatively different behaviour that can't be reconciled.
This is a flat two-dimensional depiction, but it ought to suffice. Now imagine another particle just like it, and think of them in terms of dynamical vortices of stress-energy. If they have no initial relative motion, two similar vortices will move linearly apart. We draw radial "electric field lines" to depict this linear motion. However if they have some initial lateral motion, the vortices will rotate around one another. We draw concentric "magnetic field lines" to depict this rotational motion.
Equation 1 is inappropriate because there are no "point charges". A charge is a region of space where the curvature is total. Equation 2 isn't appropriate because it isn't some big bland large-scale curvature.For example, in equation (1) the acceleration is an affine (constant + linear) function of the velocity, while in equation (2) the acceleration is a quadratic function of velocity. In equation (1) the acceleration is also different for different particles (it explicitly depends on the charge/mass ratio) which also contradicts equation (2).
No. Nobody has told me that.Also, I'm sure someone must have told you at some point that the idea of curvature only in space immediately breaks Lorentz invariance, which is a known symmetry of electrodynamics.
It's true because this is what Maxwell was saying, only he didn't get it quite right. He talked about molecular vortices, he didn't know about electrons. He had the vortices down as being in the space the particles moved through rather than the particles themselves. Yes, it isn't an accurate description of quantum electrodynamics, but I think it's a fairly good description of the underlying reality. A start at least.And all of this is true because... you say so? This isn't an accurate description of quantum electrodynamics, if that's what you were aiming for.
It's classical.What does that even mean? In quantum physics the wavefunction is not a wave in a classical field. Part of the point of my brief explanation was to make this clear: in quantum electrodynamics, the whole electromagnetic field gets its own wavefunctional. (Or at least, you can attribute it with a wavefunctional. As I said QFT textbooks don't describe it this way for practical reasons. They use the state/operator formalism instead.)
The photon and neutrino aren't static, set them aside. Most of the particle zoo are ephemera. Look at the lifetimes.+1 for the photon (free photons don't decay into anything). +6 for the neutrinos and their antiparticles, which as far as I know are too light to decay into anything else. + possibly a few more I'd have to look up in my old particle physics notes, due to chirality. What's your point?
OK.Er, yes it is, at least in the sense I was using the term "field configuration". The only reason I included that example was to clarify the sense in which I was using the term "field configuration".
Yes, I've read The Refractive Index in Electron Optics and the Principles of Dynamics by Ehrenberg and Siday.I already know about the Aharonov-Bohm effect. You know the effect appears even if you treat the electromagnetic field as classical, right?