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.
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 $$ \vec{B} = \nabla_T \times \vec{A}. \label{eq:e010} $$
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:
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.
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