📂Electrodynamics

The Zhemengko Equation

Overview¹

When the distribution of continuous charge changes over time, the electric field is as follows.

$$ \mathbf{E} (\mathbf{r},t)=\frac{1}{4\pi \epsilon_{0}} \int \left[ \frac{ \rho (\mathbf{r}^{\prime}, t_{r}) }{\cR ^2} \crH + \frac{ \dot{\rho}(\mathbf{r}^{\prime}, t_{r})}{c\cR}\crH-\frac{\dot{\mathbf{J}}(\mathbf{r}^{\prime},t_{r}) }{c^2 \cR} \right]d\tau^{\prime} $$

When the distribution of continuous current changes over time, the magnetic field is as follows.

$$ \mathbf{B}( \mathbf{r}, t) = \dfrac{\mu_{0}}{4\pi} \int \left[ \frac{\mathbf{J}(\mathbf{r}^{\prime},t_{r})}{\cR^2} + \dfrac{ \dot{\mathbf{J}}(\mathbf{r}^{\prime}, t_{r}) } {c\cR} \right]\times \crH d\tau^{\prime} $$

These two equations are collectively known as the Jefimenko’s equations. Here, $t_{r}$ is retarded time, and $\bcR$ is separation vector.

Derivation

The electric and magnetic fields can be determined by the following equation.

$$ \begin{align} \mathbf{E} &= -\nabla V-\frac{\partial \mathbf{A}}{\partial t} \\ \mathbf{B} &= \nabla \times \mathbf{A} \end{align} $$

Here, $V$ and $\mathbf{A}$ are the retarded potentials changing over time, as shown below.

$$ V(\mathbf{r},\ t)=\dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{ \rho (\mathbf{r}^{\prime},\ t_{r}) }{ \cR } d\tau^{\prime},\quad \mathbf{A}( \mathbf{r},\ t) = \dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR}d\tau^{\prime} $$

Since $\cR$ and $t_{r}$ include $\mathbf{r}^{\prime}$, the calculation is not outright simple. The gradient of $V$ is, due to the chain rule, the gradient of retarded time is $\nabla t_{r}=-\dfrac{1}{c} \crH$ and the gradient of magnitude of separation vector is $\dfrac{1}{\cR} = -\dfrac{1}{\cR^{2}}\crH$,

$$ \begin{align} \nabla V &= \dfrac{1}{4\pi\epsilon_{0}} \int \nabla \left( \dfrac{ \rho (\mathbf{r}^{\prime},\ t_{r}) }{ \cR } \right) d\tau^{\prime} \nonumber \\ &= \dfrac{1}{4\pi\epsilon_{0}} \int \left[ \dfrac{1}{\cR}\nabla \rho (\mathbf{r}^{\prime},\ t_{r}) + \rho (\mathbf{r}^{\prime},\ t_{r}) \nabla \left( \dfrac{1}{\cR} \right) \right] d\tau^{\prime} \nonumber \\ &= \frac{1}{4\pi\epsilon_{0}} \int \left[ -\dfrac{1}{\cR}\dfrac{\partial \rho (\mathbf{r}^{\prime}, t_{r})}{\partial t_{r}} \nabla t_{r} -\rho (\mathbf{r}^{\prime},t_{r}) \dfrac { \crH} {\cR ^2} \right] d\tau^{\prime} \nonumber \\ &= \frac{1}{4\pi\epsilon_{0}} \int \left[ -\dfrac{\dot{\rho}(\mathbf{r}^{\prime}, t_{r})}{c \cR} {\crH} - \dfrac{\rho (\mathbf{r}^{\prime},t_{r})}{\cR ^2} \crH \right] d\tau^{\prime} \\ \end{align} $$

And by calculating the time derivative of the vector retarded potential,

$$ \begin{align} \dfrac{\partial \mathbf{A}}{ \partial t} &= \dfrac{\partial}{\partial t} \dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR}d\tau^{\prime} \nonumber \\ &= \dfrac{\mu_{0}}{4\pi} \int \dfrac{1}{\cR} \dfrac{\partial \mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\partial t} d\tau^{\prime} \nonumber \\ &= \dfrac{\mu_{0}}{4\pi} \int \dfrac{\dot{ \mathbf{J}}(\mathbf{r}^{\prime},\ t_{r})}{\cR} d\tau^{\prime} \end{align} $$

Since variables of time and space are independent, we can change the order of integration. By substituting $(3)$, $(4)$ into $(1)$,

$$ \begin{align*} \mathbf{E} (\mathbf{r},t) &= \dfrac{1}{4\pi\epsilon_{0}} \int \left[ \dfrac { \rho (\mathbf{r}^{\prime},t_{r})} {\cR ^2}\crH +\dfrac{\dot{\rho}(\mathbf{r}^{\prime}, t_{r})}{c\cR} \crH \right]d\tau^{\prime} -\dfrac{\mu_{0}}{4\pi} \int \dfrac{\dot{ \mathbf{J}}(\mathbf{r}^{\prime},\ t_{r})}{\cR} d\tau^{\prime} \\ &= \dfrac{1}{4\pi\epsilon_{0}} \int \left[ \dfrac { \rho (\mathbf{r}^{\prime},t_{r})} {\cR ^2}\crH +\dfrac{\dot{\rho}(\mathbf{r}^{\prime}, t_{r})}{c\cR} \crH -\mu_{0}\epsilon_{0}\dfrac{\dot{ \mathbf{J}}(\mathbf{r}^{\prime},\ t_{r})}{\cR} \right]d\tau^{\prime} \\ &= \dfrac{1}{4\pi\epsilon_{0}} \int \left[ \dfrac { \rho (\mathbf{r}^{\prime},t_{r})} {\cR ^2}\crH +\dfrac{\dot{\rho}(\mathbf{r}^{\prime}, t_{r})}{c\cR} \crH -\dfrac{\dot{ \mathbf{J}}(\mathbf{r}^{\prime},\ t_{r})}{c^2\cR} \right]d\tau^{\prime} \end{align*} $$

The last equality holds due to $\dfrac{1}{c^2}=\mu_{0} \epsilon_{0}$. This represents Coulomb’s law when the charge density changes over time. If the charge distribution is constant over time, this is the same as Coulomb’s law studied in electrostatics.

The magnetic field $\mathbf{B}$ is more complicated to calculate because it includes the curl operator.

$$ \nabla \times \mathbf{A} = \dfrac{\mu_{0}}{4\pi} \left[ \nabla \times \left( \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR} d\tau^{\prime} \right)\right] $$

Here, $\nabla \times$ is the differentiation regarding $(\mathbf{x},\mathbf{y},\mathbf{z})$, and $\int d\tau^{\prime}$ is the integration regarding $(\mathbf{x}^{\prime},\mathbf{y}^{\prime},\mathbf{z}^{\prime})$, thus they are independent.

$$ \begin{align} \nabla \times \mathbf{A} &= \dfrac{\mu_{0}}{4\pi} \int \nabla \times \dfrac{ \mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR} d\tau^{\prime} \nonumber \\ &= \dfrac{\mu_{0}}{4\pi} \int \left[ \dfrac{1}{\cR}\big( \nabla \times \mathbf{J}(\mathbf{r}^{\prime},\ t_{r})\big) - \mathbf{J}(\mathbf{r}^{\prime},\ t_{r}) \times \nabla \left(\dfrac{1}{\cR}\right) \right]d\tau^{\prime} \end{align} $$

The last equality holds by the product rule (e) $\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) - \mathbf{A} \times (\nabla f)$. First, let’s determine each component of $\nabla \times \mathbf{J}$.

$$ (\nabla \times \mathbf{J})_{x}=\dfrac{\partial J_{z}}{\partial y}-\dfrac{\partial J_{y}}{\partial z} $$

Due to the chain rule of differentiation,

$$ \dfrac{\partial J_{z}}{\partial y}=\dfrac{\partial J_{z}}{\partial t} \dfrac{\partial t}{\partial t_{r}}\dfrac{\partial t_{r}}{\partial y} $$

Here, since $t_{r}=t-\dfrac{\cR}{c}$, therefore $\dfrac{\partial t}{\partial t_{r}}=1$, $\dfrac{\partial t_{r}}{\partial y}=-\dfrac{1}{c}\dfrac{\partial \cR}{\partial y}$, therefore

$$ \dfrac{\partial J_{z}}{\partial y}=-\dfrac{1}{c} \dot{J_{z}}\dfrac{\partial \cR}{\partial y} $$

Consequently,

$$ \begin{align*} (\nabla \times \mathbf{J})_{x} &=-\dfrac{1}{c} \dot{J_{z}}\dfrac{\partial \cR}{\partial y}-\dfrac{1}{c} \dot{J_{y}}\dfrac{\partial \cR}{\partial z} \\ &= \frac{1}{c} \left( \dot{J_{y}}\dfrac{\partial \cR}{\partial z}-\dot{J_{z}}\dfrac{\partial \cR}{\partial y}\right) \\ &= \frac{1}{c} \left[ \dot{\mathbf{J}} \times (\nabla \cR) \right]_{x} \end{align*} $$

Also, since $\nabla ( \acR)=\acrH$,

$$ \begin{equation} \nabla \times \mathbf{J} = \frac{1}{c} \left( \dot {\mathbf{J}} \times \crH \right) \end{equation} $$

And since $\nabla \left( \frac{1}{\acR} \right)=-\frac{1}{\acR^2}\acrH$, with $(6)$ together when substituting into $(5)$,

$$ \begin{align*} \mathbf{B}(\mathbf{r},t)=\nabla \times \mathbf{A} &= \dfrac{\mu_{0}}{4\pi} \int \left[ \dfrac{1}{\cR} \frac{1}{c} \left( \dot {\mathbf{J}} \times \crH \right) + \mathbf{J} \times \frac{1}{\cR^2}\crH\right]d\tau^{\prime} \\ &= \dfrac{\mu_{0}}{4\pi} \int \left[\frac{ \mathbf{J} (\mathbf{r}^{\prime},\ t_{r}) }{\cR^2} + \frac{ \dot {\mathbf{J}} (\mathbf{r}^{\prime},\ t_{r}) }{c\cR} \right]\times \crH d\tau^{\prime} \end{align*} $$

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