Title: EM vs QM
Date: 2019-09-23 23:00
Author: Stefan Israelsson Tampe
Sorry for the schematic math, I will post it in a nicer form later,
To want to understand QM you are not suppose to think what's the reson behind it is, but simply accept it and continue with the calculations. As I'm philosophically minded I don't accept this. Now there was some ideas that come to my attention. First an intresting approach to model maxwells equations for source terms that moves at the speed of light. The paper can be located at [Restricted Maxwell](https://www.researchgate.net/publication/320274378_Maxwell's_Equations_and_Occam's_Razor). Accompanioning this paper is a recent published book (discussed here) seam to claim that it can connect these EM theory with QED. I havent read this book yet but I still want to speculate how to model QED from this theory.
Let's begin. What's the simplest model of source terms that move with the speed of light? well that must be based on a plane wave e.g.
$$
S = C \exp(i k \cdot x), |k| = 0.
$$
If we assume the sources based on $S$ we would expect that using $A$ the potential, fields expressed as
$$\Phi = \gamma \cdot A = \Phi_0 \exp(i k \cdot x)$$, With $\Phi_0$ much slowly varying than the wave part. Therefore the value over one cycle is
$$
\int \Phi_0(x_0 + dx)\exp(ik \cdot (x-x_0)\,dx =
\int dx (x-x_0)\cdot \nabla \Phi_0(x+0)exp(ik \cdot (x+x_0)) =
\sum \frac{2\pi i}{k_i}\partial_i \Phi_0 = \sum i \hbar \partial_i \Phi_0
$$, assuming all $k_i$ is equal in magnitude.
E.g. multiplying a slowly varying field will essentially produce the derivative of it if viewed from some distance.
Now we will play a little with correctness but note that $\gamma_i = \gamma_i^-1$ and that we assume we could replace $k_i$ with $\gamma_i k_i$, then again the source terms satisfy the wave equation and also
$$
\Phi(x) ~ \sum i \hbar \gamma_i \partial_i \Phi_0
$$
But not only this, if we add an external field $\gamma \cdot B$ then we have exactly the same equation as the dirac equation with an external field but without mass.
$$
\Phi(x) ~ \sum \gamma_i (i \hbar \partial_i \Phi_0 + B_i)
$$
Note we are allowed to add a constant to the exponential function in the original source term as it will be removed taking the derivative. Now if we assume that the energy of the exponential physics is $m c$ in the small testvolume around a local and scale with the size $\Phi_0$ we get
$$
0 = \Phi_0(exp(...) - mc)) = \sum \gamma_i ((i \hbar \partial_i \Phi_0) + B_i) - m c \Phi_0
$$
Now this is a calculation on a napkin, take it for what it is worth.
