# Weber Electrodynamics

## Coverting Maxwell Into Weber

On this webpage, we demonstrate one way to convert Maxwell's electrodynamics into Weber's electrodynamics. Recall that Maxwell's electrodynamics has two implicit restrictions: $$\partial \rho_S / \partial t = 0$$ and $$\vec{v}_{Ti} = \vec{0}$$. By removing these two restrictions, and adding two terms to the scalar potential function, we arrive at Weber's electrodynamics.

Again we state that we are working under the restrictions:

1. All relational speeds are significantly less than the speed of light.
2. Time retardation (time delay) can be ignored.

### The Conversion Steps

First, we need to change the definition of the current density function. In Maxwell's electrodynamics, this is (we use a superscript $$(M)$$ to indicate Maxwell quantities) $$\vec{J}^{(M)} \equiv \rho_S \; \vec{v}_S .$$

Note that this definition for current density ignores the current due to the velocity of the test charged body (the detector). That is, consider a source charged body and a test charged body, both initially at rest in the arbitrarily chosen coordinate system, and so at rest with respect to one another. Now, move the source charge away from the test charge with a constant velocity. This is the usual concept for the current. But now, reset so that both charges are at rest with respect to one another. This time, we move the test charged body with constant velocity but in the opposite direction to the direction that we moved the source charged body. Again, it appears, to the test chrged body (to the detector) that there is a current. And the detector can not tell if the source charged body is "moving" or if the detector is "moving" (constant velocity). All that matters is that there is a relative motion between the source charged body and the test charged body.

In Weber's electrodynamics we therefore define the current density function to be $$\vec{J} \equiv \rho_S \; \vec{v}_{TiS} = \rho_S ( \vec{v}_{Ti} - \vec{v}_{S} )$$

Note that when the Maxwell restricition $$\vec{v}_{Ti} = \vec{0}$$ is applied we recover the Maxwell definition by $$- \vec{J} \rightarrow \vec{J}^{(M)} \;\; restriction: \vec{v}_{Ti} = \vec{0}.$$ The "extra" minus sign is due to the way we have chosen to define $$\vec{v}_{TiS}$$.

We now use this definition of current density function in the definition of the vector potential function to transition from the Maxwell definition to the Weber definition, $$\vec{A}^{(M)} \equiv \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \vec{J}^{(M)} \dfrac{1}{r} d\tau \rightarrow \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \dfrac{-\vec{J}_{TiS}}{r_{TiS}} dVol_S \equiv \vec{A}_{Ti}$$ which then propagates to the definition of the magnetic induction function $$\vec{B}^{(M)} \equiv \nabla \times \vec{A}^{(M)} \rightarrow \nabla_T \times \vec{A}_{Ti} \equiv \vec{B}_{Ti}$$

The above transitions removes the $$\vec{v}_{Ti} = \vec{0}$$ restriction of Maxwell's electrodynamics.

The second step in transitioning from Maxwell into Weber electrodynamics is to add two terms to the scalar potential function definition $$\Phi^{(M)} \equiv \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \dfrac{1}{r} d\tau \rightarrow \dfrac{1}{4 \pi \epsilon_0} \iiint \rho_S \dfrac{1}{r}_{TiS} \left( 1 + \dfrac{\dot{r}_{TiS}^2}{2 c^2} + \dfrac{1}{c^2} r_{TiS} \ddot{r}_{TiS} \right) \; dVol_S \equiv \Phi$$

The third step is to remove the $$\partial \rho_S / \partial t = 0$$ restriction of Maxwell's electrodynamics. Recall that in the continuous sources case, Weber's electrodynamics has an additional term in the decomposition of the electrodynamics function, $$\vec{E}_{Ti}$$, that depends on the time derivative of the charge density function. This doesn't appear in Maxwell's electrodynamics because of this restriction on the charge density function. The transition is $$\vec{E}^{(M)} \equiv - \nabla \Phi^{(M)} - \dfrac{\partial}{\partial t} \vec{A}^{(M)} \rightarrow - \nabla_T \Phi_{Ti} - \dfrac{\partial}{\partial t} \vec{A}_{Ti} - \vec{R}_{Ti}^{(E)} \equiv \vec{E}_{Ti}$$ where $$\vec{R}_{Ti}^{(E)} \equiv \dfrac{1}{4 \pi \epsilon_0 c^2 } \iiint \left( \dfrac{\partial \rho_S}{\partial t} \right) \dfrac{\vec{v}_{TiS}}{r_{TiS}} \; dVol_S .$$

The final step is to use the Weber force equation $$\vec{F}_{Ti} = q_{Ti} \vec{E}_{Ti}$$ and not to use the Lorentz force equation. The Lorentz force equation is just two terms of the full Weber force equation. When using the Lorentz force equation some justification for dropping the remaining terms from the complete Weber force equation needs to be given.

This completes the conversion of Maxwell's electrodynamics into Weber's electrodynamics for the case of continuous sources. However, as noted on the webpage describing problems with Maxwell's electrodynamics, the discrete sources case is the more fundamental case.