I believe this example shows how microscopic em theory is just an extension of macroscopic em theory. The cancellation of the tangential electric field is standard practice in macroscopic conductors. Does this mean that a charge is just some sort of conductor? No, that’s a circular argument and gets us no closer to the truth (conductor = point charges in a material, point charge = conductor).

It might be possible to eliminate mass as a concept using the above example. Note how the classical radius depends on the mass of the point charge. Conventionally, we would say that the mass regulates the acceleration due to the incident force, a = F/m. Let’s look at this example from a purely em viewpoint and start with a spherical charge of a given radius but no mass. The only physical law in this new em universe is that there can be no tangential electric field on a charge’s surface and that the charge will accelerate as needed to maintain that law. How it accelerates is unknown, just like it is in standard physics. How far can we get in constructing this em universe? Where does the theory break down?

]]>Now, we know from observation that an electron can orbit a proton at a given frequency without net radiation, which seems to be impossible in electromagnetic theory when you view the electron as a collection of charge. However, what happens when the electron is modeled as a charged sphere with an extremely high permittivity/permeability that can rotate? Let’s look at the case where the increase in permittivity and permeability are proportional, so that there is no reflection at the sphere’s surface, just a decrease in the speed of light in the sphere. When this charged surface electron travels around a proton, its acceleration will produce two spherical electromagnetic waves: one moving outwards and one moving inwards. Since we’re doing a lot of handwaving already, let’s assume the following: the sphere isn’t rotating, the body carries the inward em wave with it as it moves in a circle, and the time for the inward em wave to traverse the body equals the time taken for charge to travel halfway around the proton. Well, that now-outward traveling wave will have an opposite sign compared to the new em wave generated by the electron’s circular path. Basically, it will reduce the subsequent outward wave due to the orbit. Is this how an electron is able to maintain a stable orbit around a proton?

]]>When the Big Bang happened billions of years ago, everything was determined by the initial configuration. What we see, what we hear, how the particles in our neurons are going to behave due to those stimuli, how our brains are going to decide what to do based on those neurons, etc., all that was determined billions of years ago. The fact that you stumbled upon this blog and are reading this sentence right now was, you guessed it, determined billions of years ago. This is the dreaded “free will” debate.

]]>

What effect will this emitted em wave have on a free electron that’s initially at rest? As earlier, let’s place the free electron along the axis of the hydrogen atom **and assume axis of rotation remains constant** so that the free electron sees an amplitude-modulated, chirped, circularly polarized em wave. The electron is accelerated in a circular path by the rotating electric field, with an orbital radius proportional to the magnitude of the electric field. Simultaneously, the electron is forced downstream by v x B. This longitudinal acceleration is proportional to the rate of change of the amplitude and frequency of the em wave. Note that if both the amplitude and frequency are constant, the longitudinal acceleration will vanish.

What is the overall path of the electron during the interaction? It’s a helix whose radius varies with the amplitude and frequency of the em wave, i.e. from zero to some maximum then back to zero again. The length of the helix depends on the amplitude and frequency of the em wave. It’s important to note that an initially-at-rest electron will return to rest following the passage of the em wave if it doesn’t interact with a third party while under the influence of the em wave.

Some important notes for this simplified example:

- The emitted em wave always has the same amplitude and frequency envelope for a given set of hydrogen orbits.
- The emitted em wave’s amplitude decreases as 1/r with distance from the hydrogen atom
- The free electron’s path is a limited helix
- The free electron’s position is changed by a fixed amount that depends on the distance from the source
- The maximum amount of energy available to the free electron decreases with distance
- Only a tiny amount of the emitted energy will be intercepted by the free electron.
- The free electron will only retain energy from the em wave if it interacts with a third party

Due to #2 and #6, at any reasonable distance from a single hydrogen atom, the classical em wave is billions of times too weak to transfer a photon’s worth of energy to the free electron. Clearly, a classical explanation of the photon effect must depend on energy already contained in the target electron’s environment.

What possible local mechanism will mimic receipt of a photon of energy by the free electron? There is no known classical mechanism … but that doesn’t mean that one doesn’t exist. In addition, we never experiment with single, isolated electrons but always with collections of them. That enables third party interactions. If the initial conditions are right, individual electrons could be ejected from those collections while under the influence of em waves. That would be where probability and statistics come into the model, not at the low-level individual interaction level.

With so many unknowns in the classical world, it seems premature to discard determinism and switch to the quantum world with all its “weirdness”. From a software perspective, I consider such “weirdness” a bug, not a feature.

]]>1. A photon is a “particle” of em energy that travels through space. There are several problems with this approach. How can a particle have a frequency? That requires a spatial extent. In addition, how would a particle interfere with itself in a two-slit experiment?

2. A photon is some sort of em wave “packet” that’s limited in space. Constructing a reasonable packet will always fail because em waves obey superposition, so there’s no way to keep it together over long distances. You would have to go nonlinear with the em wave equations, which implies some sort of medium interaction. And just like #1, how do you explain the two-slit experiment? Finally, you still need to have the unrealistically high electric field in the em wave.

3. A photon isn’t a “thing” but rather a classical em wave. Space is filled with very strong classical em waves, presumably produced by all the other particles in the universe, that mimic a vacuum background energy spectrum. The photon’s classical em wave, with its 1/r dependency, forces the electron into a motion that interacts with the “seething vacuum”, resulting in a net transfer of a photon’s worth of energy. This is the Stochastic model. While this approach has some merit and is the least fanciful, it still relies on very strong em waves (the unrealistic “seething vacuum”).

Model #1 seems to be the favored approach right now. Unfortunately, it’s the least physical and drags in statistics, with all its voodoo, as a fundamental part of physics. A recipe for disaster. Fundamental physics should always be determinate. The statistical part comes in at a higher level; when dealing with large numbers of imprecisely defined determinate systems.

]]>The S2 orbit is at 2.119 x 10^-10 m, where the proton’s electric field is 3.207 x 10^10 V/m and the electron’s orbital frequency is 8.216 x 10^14 Hz. The electron falls to the S1 orbit at 5.297 x 10^-11 m, where the electric field is 5.131 x 10^11 V/m and the orbital frequency is 6.573 x 10^15 Hz. In the process of falling from S2 to S1, the electron emits a photon with an energy of 10.204 eV, corresponding to a frequency of 2.467 x 10^15 Hz.

So far, so good. The emitted photon’s frequency is neatly tucked in between the two orbital frequencies (about 29% of the way from S2 to S1). We get a nice physical picture of an electron falling from one non-radiating orbit to a second non-radiating one, emitting an em wave with an intermediate frequency. Classically, we would say that this intermediate frequency is some sort of mean or average value since the frequency would increase from S2 to S1, i.e. the photon would be a “chirped” em wave with an amplitude that goes from zero to some maximum then back to zero again.

Now for the fundamental problem. The emitted photon flies off into space and interacts with an isolated, stationary electron that happens to be sitting along the axis of the emitting hydrogen atom, so that the incident em wave is circularly polarized (CP). What happens? Well, QM says that **all** the photon energy is absorbed by the electron and then re-emitted. Let’s calculate how strong the electric field needs to be for an isolated electron to have absorbed all that energy. The CP wave forces the electron into a circular orbit at the frequency of the photon. If we calculate the electron’s orbital kinetic energy and equate that to the photon’s energy, we get an orbital radius of 1.222 x 10^-10 m and an electric field of 1.669 x 10^11 V/m. This is an impossibly high value for the electric field of an em wave and, worse, according to QM it is independent of the distance between the hydrogen atom and electron.

That is the fundamental problem with quantum mechanics and the reason why QM will never have an adequate physical model.

]]>