# Weber Electrodynamics

## Correction To Ampere's Law

We wish to verify Ampere's Law (not Ampere's force equation) that is one of Maxwell's four differential equations. But we find that a correction is required. We choose the simplest, least restrictive, approach to correcting Ampere's Law.

We will work in the SI system of units.

### Setting Up The Mathematics

First, we will distinguish between the coordinates of the sources by using a subscript "S", and the coordinates of the detector (observation, or test) position by using a subscript "T". Then, for example, the $$\nabla$$ vector operator, when referring to the detector position, will be written as, in a Cartesian coordinate system $$\nabla_T \equiv \dfrac{\partial}{\partial x_T}\hat{x} + \dfrac{\partial}{\partial y_T}\hat{y} + \dfrac{\partial}{\partial z_T}\hat{z}.$$

In Maxwell electrodynamics Ampere's Law is written as $$\nabla_T \times \vec{B} = \mu_0 \vec{J}_f + \mu_0 \epsilon_0 \dfrac{\partial}{\partial t}\vec{E}, \label{eq:e001}$$ where $$\vec{J}_f$$ is the free current density function.

We use the scalar $$\Phi \equiv \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \dfrac{1}{r_{TS}} d\tau_S$$ and the vector $$\vec{A} \equiv \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \vec{J} \dfrac{1}{r_{TS}} d\tau_S \label{eq:e002}$$ potential function definitions, where $$r_{TS}$$ is the distance from the source differential volume element $$d\tau_S$$ to the detector (observation, or test) position, $$\rho$$ is the charge density function, and $$\vec{J}$$ is the current density function defined by $$\vec{J} \equiv \rho \vec{v}_S$$ and $$\vec{v}_S$$ is the "average drift velocity" of the source charges passing through the differential volume element $$d\tau_S .$$

The electric field can be decomposed into the gradient of the scalar potential function and the time derivative of the vector potential function as $$\vec{E} = -\nabla_T \Phi - \dfrac{\partial}{\partial t} \vec{A}.$$ This can be written as $$\vec{E} = \vec{E}^{(C)} - \dfrac{\partial}{\partial t} \vec{A}.$$ where $$\vec{E}^{(C)} \equiv -\nabla_T \Phi = \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \dfrac{\hat{r}_{TS}}{r_{TS}^2} \; d\tau_S \label{eq:e006}$$ is the Coulomb field.

The magnetic induction field can be written as $$\vec{B} = \nabla_T \times \vec{A}. \label{eq:e010}$$

### Attempting Verification Of Ampere's Law

Taking the curl of both sides of Eq. (\ref{eq:e010}), $$\nabla_T \times \vec{B} = \nabla_T \times \left( \nabla_T \times \vec{A} \right) ,$$ and using the relation $$\nabla \times \left( \nabla \times \vec{V} \right) = \nabla \left( \nabla \bullet \vec{V} \right) - \nabla^2 \vec{V} ,$$ we have $$\nabla_T \times \vec{B} = \nabla_T \left( \nabla_T \bullet \vec{A} \right) - \nabla_T^2 \vec{A} .$$

Using Eq. (\ref{eq:e002}) for the vector potential function, and interchanging the order of the derivative and integration operations, we have $$\nabla_T \times \vec{B} = \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \nabla_T \left( \nabla_T \bullet \dfrac{\vec{J}}{r_{TS}} \right) \; d\tau_S - \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \nabla_T^2 \dfrac{\vec{J}}{r_{TS}} \; d\tau_S , \label{eq:e020}$$ where the current density function is defined as $$\vec{J} \equiv \rho \vec{v}_S .$$

We will make use of the following realtions where $$f$$ is a scalar function and $$\vec{V}$$ is a vector function: \begin{align} & \nabla \bullet \left( f \vec{V} \right) = f \left( \nabla \bullet \vec{V} \right) + \left(\nabla f \right) \bullet \vec{V} , \\ & \nabla^2\left( f \vec{V} \right) = \vec{V} \nabla^2 f + 2 \left(\nabla f \bullet \nabla \right) \vec{V} + f \nabla^2 \vec{V} . \end{align}

Noting that spatial derivatives of the source velocity, $$\vec{v}_S$$, with respect to the test body coordinates, $$\vec{r}_T$$, are zero, and that the charge density function $$\rho$$ does not depend on the test body (detector) coordinates, so that $$\nabla_T \bullet \vec{J} = 0$$ and $$\nabla_T^2 \vec{J} = \vec{0}$$ we can write Eq. (\ref{eq:e020}) as $$\nabla_T \times \vec{B} = \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \nabla_T \left( \vec{J} \bullet \nabla_T \dfrac{1}{r_{TS}} \right) \; d\tau_S - \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \vec{J} \nabla_T^2 \dfrac{1}{r_{TS}} \; d\tau_S .$$

We now use the relations \begin{align} & \nabla_T \dfrac{1}{r_{TS}} = \dfrac{-\hat{r}_{TS}}{r_{TS}^2} , \\ & \nabla_T^2 \dfrac{1}{r_{TS}} = - 4 \pi \delta \left( \vec{r}_T - \vec{r}_S \right) , \end{align} where $$\hat{r}_{TS}$$ is the unit vector from the differential source element to the test body, to write \begin{equation} \begin{split} \nabla_T \times \vec{B} = &\dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \nabla_T \left( \rho \vec{v}_S \bullet \left( \dfrac{-\hat{r}_{TS}}{r_{TS}^2} \right) \right) \; d\tau_S \\ &- \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \vec{J} \left( - 4 \pi \delta \left( \vec{r}_T - \vec{r}_S \right) \right) \; d\tau_S . \end{split} \end{equation}

Performing the second term's integration and then reordering the terms gives $$\nabla_T \times \vec{B} = \dfrac{1}{ \epsilon_0 c^2} \vec{J}_f - \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \rho \nabla_T \left( \dfrac{\vec{v}_S \bullet \hat{r}_{TS}}{r_{TS}^2} \right) \; d\tau_S ,$$ where $$\vec{J}_f \equiv \vec{J} \left(\vec{r}_T, t \right) ,$$ is the free current density function.

Using the relation, for vector functions $$\vec{a}$$ and $$\vec{b}$$ $$\nabla \left( \vec{a} \bullet \vec{b}\right) =\vec{a}\times\left(\nabla\times\vec{b}\right)+\vec{b}\times\left(\nabla\times\vec{a}\right) + \left(\vec{a}\bullet\nabla\right)\vec{b} + \left(\vec{b}\bullet\nabla\right)\vec{a} ,$$ and remembering that spatial derivatives of the source velocity, $$\vec{v}_S$$ , with respect to the test body coordinates are zero, we have \begin{equation} \begin{split} \nabla_T \times \vec{B} = &\dfrac{1}{ \epsilon_0 c^2} \vec{J}_f \\ &- \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \rho \left( \vec{v}_S \times \left( \nabla_T \times \dfrac{\vec{r}_{TS}}{r_{TS}^3}\right) + \left( \vec{v}_S \bullet \nabla_T \right) \dfrac{\vec{r}_{TS}}{r_{TS}^3} \right) \; d\tau_S . \end{split} \end{equation}

Next, we use the relations $$\nabla_T \times \dfrac{\vec{r}_{TS}}{r_{TS}^3} = \vec{0} ,$$ and $$\left( \vec{v}_S \bullet \nabla_T \right) \left( \dfrac{\vec{r}_{TS}}{r_{TS}^3} \right) = \dfrac{1}{r_{TS}^3}\vec{v}_S - 3 \dfrac{\hat{r}_{TS}}{r_{TS}^3}\left( \hat{r}_{TS} \bullet \vec{v}_S \right) ,$$ to obtain $$\nabla_T \times \vec{B} = \dfrac{1}{ \epsilon_0 c^2} \vec{J}_f - \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \rho \left( \dfrac{1}{r_{TS}^3}\vec{v}_S - 3 \dfrac{\hat{r}_{TS}}{r_{TS}^3} \left(\hat{r}_{TS} \bullet \vec{v}_S \right) \right) \; d\tau_S . \label{eq:e029}$$

Let us now consider the time derivative of the Coulomb field $$\vec{E}^{(C)}$$ defined in eq. \ref{eq:e006}. We obtain \begin{equation} \begin{split} \dfrac{\partial}{\partial t}\vec{E}^{(C)} = & \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \left( \dfrac{\partial}{\partial t} \rho \right) \dfrac{\hat{r}_{TS}}{r_{TS}^2} \; d\tau_S \\ &+ \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \rho \left( \dfrac{-3}{r_{TS}^3} \left( \vec{v}_T \bullet \hat{r}_{TS} \right) \hat{r}_{TS} + \dfrac{1}{r_{TS}^3} \vec{v}_T \right) \; d\tau_S \\ &- \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \rho \left( \dfrac{1}{r_{TS}^3} \vec{v}_S - 3 \dfrac{\hat{r}_{TS}}{r_{TS}^3} \left( \hat{r}_{TS} \bullet \vec{v}_S \right) \right) \; d\tau_S . \end{split} \label{eq:e030} \end{equation}

Comparing Eq. (\ref{eq:e030}) with Eq. (\ref{eq:e029}), we see that in order to obtain an equation close to Maxwell's equation Eq. (\ref{eq:e001}), Ampere's Law, we are now required to impose two restrictions, not obvious for Maxwell's electrodynamics in general.

First, the charge density function, $$\rho\left(\vec{r}_S,t\right)$$, must be constant in time. This removes the first term of Eq. (\ref{eq:e030}).

Second, the test body position $$\vec{r}_T$$ (detector or observation position) must be stationary in the arbitrary coordinate system chosen $$\left( \vec{v}_T = \vec{0} \right)$$. This removes the second term of Eq. (\ref{eq:e030}).

With these two restrictions $$\left( \partial\rho / \partial t = 0, \vec{v}_T = \vec{0} \right)$$, Eq. (\ref{eq:e030}) becomes $$\dfrac{\partial}{\partial t}\vec{E}^{(C)} |_{restrictions} = - \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \rho \left( \dfrac{1}{r_{TS}^3} \vec{v}_S - 3 \dfrac{\hat{r}_{TS}}{r_{TS}^3} \left( \hat{r}_{TS} \bullet \vec{v}_S \right) \right) \; d\tau_S . \label{eq:e031}$$

Using Eq. (\ref{eq:e031}) in Eq. (\ref{eq:e029}), and the relation $$1/c^2 = \mu_0 \epsilon_0$$, results in $$\nabla_T \times \vec{B} = \mu_0 \vec{J}_f + \mu_0 \epsilon_0 \dfrac{\partial}{\partial t} \vec{E}^{(C)}, restrictions: \partial \rho / \partial t = 0, \vec{v}_T = \vec{0} . \label{eq:e032}$$

This is the mathematically correct form for Ampere's Law.

The alternative, to use $$\vec{E}$$ instead of $$\vec{E}^{(C)}$$, would require additional restrictions to be imposed. To see this, recall that $$\vec{E}$$ can be written in terms of $$\vec{E}^{(C)}$$ $$\vec{E} = \vec{E}^{(C)} - \dfrac{\partial}{\partial t} \vec{A}.$$ Then the time derivative of $$\vec{E}$$ is written as $$\dfrac{\partial}{\partial t}\vec{E} = \dfrac{\partial}{\partial t}\vec{E}^{(C)} - \dfrac{\partial^2}{\partial t^2} \vec{A} . \label{eq:e0034}$$

Performing the time derivatives indicated in Eq. (\ref{eq:e0034}) gives (dots over quantities indicate derivatives with respect to time) \begin{equation} \begin{split} \dfrac{\partial}{\partial t}\vec{E} = & \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \dot{\rho} \dfrac{\hat{r}_{TS}}{r_{TS}^2} \; d\tau_S \\ &+ \dfrac{1}{ 4 \pi \epsilon_0 } \iiint \rho \left(\dfrac{-3}{r_{TS}^3}\left(\vec{v}_T \bullet \hat{r}_{TS} \right)\hat{r}_{TS} + \dfrac{1}{r_{TS}^3}\vec{v}_T \right) \; d\tau_S \\ & -\dfrac{1}{ 4 \pi \epsilon_0 } \iiint \rho \left(\dfrac{-3}{r_{TS}^3}\left(\vec{v}_S \bullet \hat{r}_{TS} \right)\hat{r}_{TS} + \dfrac{1}{r_{TS}^3}\vec{v}_S \right) \; d\tau_S \\ & -\dfrac{1}{ 4 \pi \epsilon_0 c^2} \iiint \left( \ddot{\rho} \vec{v}_S \dfrac{1}{ r_{TS} } + 2 \dot{\rho} \vec{a}_S \dfrac{1}{r_{TS}} - 2 \dot{\rho} \vec{v}_S \dfrac{1}{r_{TS}^2} \dot{r}_{TS} \right) \; d\tau_S \\ &- \dfrac{1}{ 4 \pi \epsilon_0 c^2 } \iiint \left( \rho \dot{\vec{a}}_S \dfrac{1}{r_{TS}} -2\rho \vec{a}_S \dfrac{1}{r_{TS}^2} \dot{r}_{TS} \right) \; d\tau_S \\ &- \dfrac{1}{ 4 \pi \epsilon_0 c^2 } \iiint \left( 2\rho \vec{v}_S \dfrac{1}{r_{TS}^3} \dot{r}_{TS}^2 -\rho \vec{v}_S \dfrac{1}{r_{TS}^2} \ddot{r}_{TS} \right) \; d\tau_S . \end{split} \label{eq:e035} \end{equation}

Instead of simplifying the situation, Eq. (\ref{eq:e035}) contributes additional terms that do not cancel and would require imposing still further restrictions to eliminate these undesired terms.

Therefore, Eq. (\ref{eq:e032}), repeated here $$\nabla_T \times \vec{B} = \mu_0 \vec{J}_f + \mu_0 \epsilon_0 \dfrac{\partial}{\partial t} \vec{E}^{(C)}, restrictions: \partial \rho / \partial t = 0, \vec{v}_T = \vec{0} ,$$ is the closest result to Maxwell's equation Eq. (\ref{eq:e001}), Ampere's Law, obtainable from directly performing the calculations indicated with the least number of imposed restrictions.

This has demonstrated that Maxwell's electrodynamics has two implicit restricitons:

1. $$\partial \rho / \partial t = 0$$,
2. $$\vec{v}_T = \vec{0}$$.

As a consequence of the the second restriction, the Lorentz force equation $$\vec{F} = q_T \vec{E} + q_T \vec{v}_T \times \vec{B}$$ must be written as $$\vec{F} = q_T \vec{E}$$ because $$\vec{v}_T = \vec{0}$$.

That is, the Lorentz force equation is not compatible with Maxwell's ellectrodynamics. But don't worry, the Lorentz force equation is compatible with Weber's force equation.