# Weber Electrodynamics

## Weber's Force Equation

On this webpage, we present several forms that the Weber force equation can take. We show that Ampere's force equation is obtainned by imposing the defining properties of an Amperian current element, and that the Lorentz force equation is a special case.

### General Definition

The force on the test charged body, $$q_{Ti}$$, due to the source charged body, $$q_{Sj}$$, is given by $$\vec{F}_{TiSj} = - \hat{r}_{TiSj} \dfrac{d}{d r_{TiSj}}U_{TiSj}.$$ where $$U_{TiSj}$$ is the Weber potential energy function defined as $$U_{TiSj} \equiv \dfrac{q_{Ti} q_{Sj} }{4 \pi \epsilon_0} \dfrac{1}{r_{TiSj}} \left( 1 - \dfrac{\dot{r}_{TiSj}^2}{2 c^2} \right).$$

Explicitly, the force equation can be written as $$\vec{F}_{TiSj} = q_{Ti} \left( 1 - \dfrac{1}{2 c^2} \dot{r}_{TiSj}^2 + \dfrac{1}{c^2} r_{TiSj} \ddot{r}_{TiSj} \right) \dfrac{q_{Sj}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2}.$$

The way we have writting Weber's force equation suggests the following form $$\vec{F}_{TiSj} = q_{Ti} \gamma_W \vec{E}_{TiSj}^{(C)},$$

where we define $$\gamma_W \equiv \left( 1 - \dfrac{1}{2 c^2} \dot{r}_{TiSj}^2 + \dfrac{1}{c^2} r_{TiSj} \ddot{r}_{TiSj} \right)$$ to be the "Weber coupling coefficient", and $$\vec{E}_{TiSj}^{(C)} \equiv \dfrac{q_{Sj}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2}$$ is the Coulomb field.

Another way that we can write Weber's force equation is $$\vec{F}_{TiSj} = q_{Ti} \vec{E}_{TiSj}$$ where we now define the "Weber electrodynamics function" to be $$\vec{E}_{TiSj} \equiv \gamma_W \vec{E}_{TiSj}^{(C)}.$$

### Including Mass Energy

An interesting observation is that we can use the equation $$E = m c^2$$ to write the force equation as $$\vec{F}_{TiSj} = \dfrac{q_{Ti} q_{Sj}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \dfrac{1}{E} \left( E - \dfrac{1}{2}m \dot{r}_{TiSj}^2 + m r_{TiSj} \ddot{r}_{TiSj} \right) .$$

In this form, the $$\dfrac{1}{2}m \dot{r}_{TiSj}^2$$ term is a kinetic energy term, and the $$m r_{TiSj} \ddot{r}_{TiSj}$$ term is a potential energy term. These energies are subtracted and added, respectively, to the mass energy $$E$$ and the whole thing is normalized by the mass energy $$E$$. It is not clear what $$m$$ refers to here.

### Ampere's Force Equation

Another form that the Weber force equation can be put into is \begin{align} \begin{split} \vec{F}_{TiSj} = &\dfrac{q_{Ti} q_{Sj}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \\ & - \dfrac{q_{Ti} q_{Sj}}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 v_{Ti} v_{Sj} \left( 2 \hat{v}_{Ti} \bullet \hat{v}_{Sj} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right) \right) \\ &+ \dfrac{q_{Ti} q_{Sj}}{8 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 v_{Ti} v_{Ti} \left( 2 \hat{v}_{Ti} \bullet \hat{v}_{Ti} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right) \right) \\ &+ \dfrac{q_{Ti} q_{Sj}}{8 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 v_{Sj} v_{Sj} \left( 2 \hat{v}_{Sj} \bullet \hat{v}_{Sj} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right) \right) \\ &+ \dfrac{q_{Ti} q_{Sj} \mu_0}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \vec{r}_{TiSj} \bullet \vec{a}_{Ti} - \dfrac{q_{Ti} q_{Sj} \mu_0}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \vec{r}_{TiSj} \bullet \vec{a}_{Sj} . \end{split} \end{align}

In this form, we see the angular relationships through the dot products involving unit vectors. (Recall that $$\hat{a} \bullet \hat{b} = \cos{\theta}$$ where $$\theta$$ is the angle between the unit vectors $$\hat{a}$$ and $$\hat{b}$$.) These angle relationships, in the 2nd term, are the same as the angle relations in Ampere's force equation.

Ampere's force equation, for the force between two Amperian current elements, not between two discrete charged bodies, can be written as $$d\vec{F}_{1,2} = -\dfrac{I_1 dl_1 I_2 dl_2}{4 \pi} \mu_0 \dfrac{\hat{r}_{1,2}}{r_{1,2}^2}\left(2 \cos\left(\varepsilon\right) - 3 \cos\left( \alpha \right) \cos\left( \beta \right) \right) ,$$ where the angle $$\varepsilon$$ is the angle between the direction of Amperian current element $$I_1 dl_1$$ and the direction of Amperian current element $$I_2 dl_2$$, $$\alpha$$ is the angle between the direction of Amperian current element $$I_1 dl_1$$ and the line connecting $$I_1 dl_1$$ to $$I_2 dl_2$$, and $$\beta$$ is the angle between the direction of Amperian current element $$I_2 dl_2$$ and the line connecting $$I_1 dl_1$$ to $$I_2 dl_2$$. (See the books Ampere's Electrodynamics, by Assis and Chaib, Apeiron, 2015, Newtonian Electrodynamics, by Graneau and Graneau, World Scientific Pub. Co., 1996, and Ampere-Neumann Electrodynamics of Metals, by Graneau, Hadronic Press, 1994.)

In particular, this Ampere force equation is similar to the term $$- \dfrac{q_{Ti} q_{Sj}}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 v_{Ti} v_{Sj} \left( 2 \hat{v}_{Ti} \bullet \hat{v}_{Sj} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right) \right)$$ of Weber's force equation in form and angle dependency.

We can obtain Ampere's force equation from the Weber force equation by imposing the definition of an Amperian current element. An Amperian current element (A.C.E.) is

1. a charge neutral current element, having two equal but oppositely charged bodies,
2. such that we can approximate the location of each of the two charged bodies as being at the same location.

We will divide the sums up into 5 different sets, writing the force equation as $$\vec{F}_{TS} = \vec{F}^{(1)} + \vec{F}^{(2)} + \vec{F}^{(3)} + \vec{F}^{(4)} + \vec{F}^{(5)},$$ where \begin{align} &\vec{F}^{(1)} \equiv \sum\limits_{i\in \{+,-\}} \sum\limits_{j\in \{+,-\}} \dfrac{q_{Ti} q_{Sj}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2}, \\ &\vec{F}^{(2)} \equiv \sum\limits_{i\in \{+,-\}} \sum\limits_{j\in \{+,-\}} - \dfrac{q_{Ti} q_{Sj}}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 \vec{v}_{Ti} \vec{v}_{Sj} \left( 2 \hat{v}_{Ti} \bullet \hat{v}_{Sj} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right) \right), \\ &\vec{F}^{(3)} \equiv \sum\limits_{i\in \{+,-\}} \sum\limits_{j\in \{+,-\}} \dfrac{q_{Ti} q_{Sj}}{8 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 \vec{v}_{Ti} \vec{v}_{Ti} \left( 2 \hat{v}_{Ti} \bullet \hat{v}_{Ti} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Ti} \right) \right), \\ &\vec{F}^{(4)} \equiv \sum\limits_{i\in \{+,-\}} \sum\limits_{j\in \{+,-\}} \dfrac{q_{Ti} q_{Sj}}{8 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \mu_0 \vec{v}_{Sj} \vec{v}_{Sj} \left( 2 \hat{v}_{Sj} \bullet \hat{v}_{Sj} - 3 \left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right)\left( \hat{r}_{TiSj} \bullet \hat{v}_{Sj} \right) \right), \\ &\vec{F}^{(5)} \equiv \sum\limits_{i\in \{+,-\}} \sum\limits_{j\in \{+,-\}} \bigg( \dfrac{q_{Ti} q_{Sj} \mu_0}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \vec{r}_{TiSj} \bullet \vec{a}_{Ti} - \dfrac{q_{Ti} q_{Sj} \mu_0}{4 \pi} \dfrac{\hat{r}_{TiSj}}{r_{TiSj}^2} \vec{r}_{TiSj} \bullet \vec{a}_{Sj} \bigg) . \end{align}

For $$\vec{F}^{(1)}$$, we write out the sums, and use the approximation that the two charged bodies of an Amperian current element are at the same location, to get $$$$\vec{F}^{(1)} = \dfrac{q_{T+} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} + \dfrac{q_{T+} q_{S-}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} + \dfrac{q_{T-} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} + \dfrac{q_{T-} q_{S-}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2}.$$$$ Since the Amperian current elements are charge neutral, we have \begin{align} &q_{T-} = - q_{T+}, \\ &q_{S-} = - q_{S+}. \end{align} We now have $$\vec{F}^{(1)} = \dfrac{q_{T+} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} - \dfrac{q_{T+} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} - \dfrac{q_{T+} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2} + \dfrac{q_{T+} q_{S+}}{4 \pi \epsilon_0} \dfrac{\hat{r}_{TS}}{r_{TS}^2}.$$ The sum is therefore zero $$\vec{F}^{(1)} = \vec{0}.$$

We will skip $$\vec{F}^{(2)}$$ for now and come back to it at the end of this derivation.

For $$\vec{F}^{(3)}$$ we have $$$$\begin{split} \vec{F}^{(3)}= &\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T+} v_{T+} \left( 2 \hat{v}_{T+} \bullet \hat{v}_{T+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right) \right) \\ &+\dfrac{q_{T+} q_{S-}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T+} v_{T+} \left( 2 \hat{v}_{T+} \bullet \hat{v}_{T+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right) \right) \\ &+\dfrac{q_{T-} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T-} v_{T-} \left( 2 \hat{v}_{T-} \bullet \hat{v}_{T-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right) \right) \\ &+\dfrac{q_{T-} q_{S-}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T-} v_{T-} \left( 2 \hat{v}_{T-} \bullet \hat{v}_{T-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right) \right). \end{split}$$$$ Writing this in terms of all positive charges gives $$$$\begin{split} \vec{F}^{(3)}= &\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T+} v_{T+} \left( 2 \hat{v}_{T+} \bullet \hat{v}_{T+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right) \right) \\ &-\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T+} v_{T+} \left( 2 \hat{v}_{T+} \bullet \hat{v}_{T+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T+} \right) \right) \\ &-\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T-} v_{T-} \left( 2 \hat{v}_{T-} \bullet \hat{v}_{T-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right) \right) \\ &+\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{T-} v_{T-} \left( 2 \hat{v}_{T-} \bullet \hat{v}_{T-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{T-} \right) \right), \end{split}$$$$ which is seen to be zero $$$$\vec{F}^{(3)}= \vec{0}.$$$$

For $$\vec{F}^{(4)}$$ we write $$$$\begin{split} \vec{F}^{(4)}= &\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S+} v_{S+} \left( 2 \hat{v}_{S+} \bullet \hat{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right) \right) \\ &+\dfrac{q_{T+} q_{S-}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S-} v_{S-} \left( 2 \hat{v}_{S-} \bullet \hat{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right) \right) \\ &+\dfrac{q_{T-} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S+} v_{S+} \left( 2 \hat{v}_{S+} \bullet \hat{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right) \right) \\ &+\dfrac{q_{T-} q_{S-}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S-} v_{S-} \left( 2 \hat{v}_{S-} \bullet \hat{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right) \right) . \end{split}$$$$ Writing this equation in terms of positive charges gives $$$$\begin{split} \vec{F}^{(4)}= &\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S+} v_{S+} \left( 2 \hat{v}_{S+} \bullet \hat{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right) \right) \\ &-\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S-} v_{S-} \left( 2 \hat{v}_{S-} \bullet \hat{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right) \right) \\ &-\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S+} v_{S+} \left( 2 \hat{v}_{S+} \bullet \hat{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S+} \right) \right) \\ &+\dfrac{q_{T+} q_{S+}}{8 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 v_{S-} v_{S-} \left( 2 \hat{v}_{S-} \bullet \hat{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right)\left( \hat{r}_{TS} \bullet \hat{v}_{S-} \right) \right) . \end{split}$$$$ This is easily seen to be zero $$$$\vec{F}^{(4)}= \vec{0}.$$$$

For $$\vec{F}^{(5)}$$ we write $$$$\begin{split} \vec{F}^{(5)}= & \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T+} - \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S+} \\ &+ \dfrac{q_{T+} q_{S-} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T+} - \dfrac{q_{T+} q_{S-} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S-} \\ &+ \dfrac{q_{T-} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T-} - \dfrac{q_{T-} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S+} \\ &+ \dfrac{q_{T-} q_{S-} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T-} - \dfrac{q_{T-} q_{S-} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S-} . \end{split}$$$$ Writing this equation in terms of positive charges gives $$$$\begin{split} \vec{F}^{(5)}= & \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T+} - \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S+} \\ &- \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T+} + \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S-} \\ &- \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T-} + \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S+} \\ &+ \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{T-} - \dfrac{q_{T+} q_{S+} \mu_0}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \vec{r}_{TS} \bullet \vec{a}_{S-} . \end{split}$$$$ And this is seen to be zero $$$$\vec{F}^{(5)}= \vec{0}.$$$$ That is, regardless of the accelerations of the individual charged bodies in either of the current elements, the accelerations contribute nothing to the overall force on the test current element due to the source current element.

We are now left with only the $$\vec{F}^{(2)}$$ sums, $$$$\begin{split} \vec{F}_{TS} = \vec{F}^{(2)}= & - \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \left( 2 \vec{v}_{T+} \bullet \vec{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) \right) \\ & - \dfrac{q_{T+} q_{S-}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \left( 2 \vec{v}_{T+} \bullet \vec{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \right) \\ & - \dfrac{q_{T-} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \left( 2 \vec{v}_{T-} \bullet \vec{v}_{S+} - 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) \right) \\ & - \dfrac{q_{T-} q_{S-}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \left( 2 \vec{v}_{T-} \bullet \vec{v}_{S-} - 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \right) . \end{split}$$$$ Writing this in terms of the positive charges, and collecting similar terms together, we have $$$$\begin{split} \vec{F}_{TS} = & - \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T+} \bullet \vec{v}_{S+} + \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T+} \bullet \vec{v}_{S-} \\ & + \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T-} \bullet \vec{v}_{S+} - \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T-} \bullet \vec{v}_{S-} \\ &+ \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) - \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \\ &- \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) + \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right)\left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) . \end{split}$$$$ This can be written as $$$$\begin{split} \vec{F}_{TS} = & \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T+} \bullet \left(- \vec{v}_{S+} + \vec{v}_{S-} \right) \\ & + \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \vec{v}_{T-} \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \\ &+ \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right) \left( \left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) - \left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \right) \\ &+ \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right) \left( - \left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) + \left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \right) . \end{split}$$$$ Simplifying gives $$$$\begin{split} \vec{F}_{TS} = & -\dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 2 \left(\vec{v}_{T+} - \vec{v}_{T-} \right) \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \\ &+ \dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 3 \left( \left( \hat{r}_{TS} \bullet \vec{v}_{T+} \right) - \left( \hat{r}_{TS} \bullet \vec{v}_{T-} \right) \right) \left( \left( \hat{r}_{TS} \bullet \vec{v}_{S+} \right) - \left( \hat{r}_{TS} \bullet \vec{v}_{S-} \right) \right), \end{split}$$$$ which can be further simplified to $$$$\begin{split} \vec{F}_{TS} = -\dfrac{q_{T+} q_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \bigg(& 2 \left(\vec{v}_{T+} - \vec{v}_{T-} \right) \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \\ &- 3 \hat{r}_{TS} \bullet \left( \vec{v}_{T+} - \vec{v}_{T-} \right) \hat{r}_{TS} \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \bigg). \end{split}$$$$ We now define Amperian differential current elements to be \begin{align} &I^{(A)}_{T}d\vec{l}_{T} \equiv dq_{T+} \left(\vec{v}_{T+} - \vec{v}_{T-} \right) = \lambda_{T+} \left(\vec{v}_{T+} - \vec{v}_{T-} \right) dl_T, \\ &I^{(A)}_{S}d\vec{l}_{S} \equiv dq_{S+} \left(\vec{v}_{S+} - \vec{v}_{S-} \right) = \lambda_{S+} \left(\vec{v}_{S+} - \vec{v}_{S-} \right) dl_S, \end{align} where $$\lambda_{T+}$$ and $$\lambda_{S+}$$ are line charge density functions for the Apmerian current elements. Using these definitions, and the differential form of the force equation, given by $$$$\begin{split} d^2\vec{F}_{TS} = -\dfrac{dq_{T+} dq_{S+}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \bigg(& 2 \left(\vec{v}_{T+} - \vec{v}_{T-} \right) \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \\ &- 3 \hat{r}_{TS} \bullet \left( \vec{v}_{T+} - \vec{v}_{T-} \right) \hat{r}_{TS} \bullet \left( \vec{v}_{S+} - \vec{v}_{S-} \right) \bigg), \end{split}$$$$ we have $$$$d^2\vec{F}_{TS} = -\dfrac{I^{(A)}_{T} I^{(A)}_{S}}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \bigg( 2 d\vec{l}_{T} \bullet d\vec{l}_{S} - 3 \left( \hat{r}_{TS} \bullet d\vec{l}_{T} \right) \left( \hat{r}_{TS} \bullet d\vec{l}_{S} \right) \bigg),$$$$ and $$$$d^2\vec{F}_{TS} = -\dfrac{I^{(A)}_{T}dl_T I^{(A)}_{S} dl_S}{4 \pi} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \mu_0 \bigg( 2 d\hat{l}_{T} \bullet d\hat{l}_{S} - 3 \left(\hat{r}_{TS} \bullet d\hat{l}_{T} \right) \left( \hat{r}_{TS} \bullet d\hat{l}_{S} \right) \bigg).$$$$ With \begin{align} &\cos{(\epsilon)} \equiv d\hat{l}_{T} \bullet d\hat{l}_{S} \\ &\cos{(\alpha)} \equiv \hat{r}_{TS} \bullet d\hat{l}_{T} \\ &\cos{(\beta)} \equiv \hat{r}_{TS} \bullet d\hat{l}_{S} \end{align} we can write $$$$\boxed{d^2\vec{F}_{TS} = -\dfrac{I^{(A)}_{T}dl_T I^{(A)}_{S} dl_S}{4 \pi} \mu_0 \dfrac{\hat{r}_{TS}}{r_{TS}^2} \bigg( 2 \cos{(\epsilon)} - 3 \cos{(\alpha)} \cos{(\beta)} \bigg), restrictions: A.C.E. \label{eq:e6.77BB} }$$$$ which is Ampere's force equation for the force on an Amperian current element $$I^{(A)}_{T}d\vec{l}_T$$ due to another Amperian current element $$I^{(A)}_{S}d\vec{l}_S$$.

The restrictions for the applicaility of this equation to Amperian current elements only is indicated by the "restrictions: A.C.E." writen with the equation. This must always be kept in mind. Ampere's force equation is not applicable for situations involving individual charged bodies that do not form Amperian current elements.

### Lorentz Force Equation

Next, we find that the Weber force equation can be written into the form \begin{align} \begin{split} \vec{F}_{TiSj} =& q_{Ti} \vec{E}_{TiSj}^{(C)} + q_{Ti} \vec{v}_{TiSj} \times \vec{B}_{TiSj} + q_{Ti} \dfrac{1}{c^2} \vec{v}_{TiSj} \left( \vec{v}_{TiSj} \bullet \vec{E}_{TiSj}^{(C)} \right) \\ &- q_{Ti} \dfrac{3}{2} \dfrac{1}{c^2} \left( \hat{r}_{TiSj} \bullet \vec{v}_{TiSj} \right)^2 \vec{E}_{TiSj}^{(C)} + q_{Ti} \dfrac{1}{c^2} \left(\vec{r}_{TiSj} \bullet \vec{a}_{TiSj} \right) \vec{E}_{TiSj}^{(C)} \end{split} \end{align}

Note that the first two terms $$q_{Ti} \vec{E}_{TiSj}^{(C)} + q_{Ti} \vec{v}_{TiSj} \times \vec{B}_{TiSj}$$ look like the Lorentz force equation. But we have to remember that we are dealing with Weber quantities and not Maxwell quantities. So, these terms are not exactly the Lorentz force equation. Still the Lorentz force equation is buried within these terms.

Recall the definition of the relative velocity $$\vec{v}_{TiSj} \equiv \vec{v}_{Ti} - \vec{v}_{Sj} .$$

And recall that the magnetic induction function can be written as $$\vec{B}_{TiSj} = \dfrac{-q_{Sj}}{4 \pi \epsilon_0 c^2 } \dfrac{\vec{v}_{TiSj} \times \hat{r}_{TiSj}}{r_{TiSj}^2}.$$

We then see that Maxwell's magnnetic induction function $$\vec{B}_{TiSj}^{(M)} = \dfrac{q_{Sj}}{4 \pi \epsilon_0 c^2 } \dfrac{\vec{v}_{Sj} \times \hat{r}_{TiSj}}{r_{TiSj}^2}.$$ is contained within.

Picking out the appropiate terms, we have $$\vec{F}_{TiSj} = q_{Ti} \vec{E}_{TiSj}^{(C)} + q_{Ti} \vec{v}_{Ti} \times \vec{B}_{TiSj}^{(M)} + (Other \; Terms)$$

Therefore, the Lorentz force equation is compatible with Weber electrodynamics, and is not compatible with Maxwell electrodynamics. Experiments confirming the Lorentz force equation is a confirmation of Weber electrodynamics, not a confirmation of Maxwell electrodynamics, provided a reason for dropping the "Other Terms" can be justified.