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The Zhemengko Equation 📂Electrodynamics

The Zhemengko Equation

Overview1

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

E(r,t)=14πϵ0∫[ρ(r′,tr)2+ρ˙(r′,tr)c−J˙(r′,tr)c2]dτ′ \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.

B(r,t)=μ04π∫[J(r′,tr)2+J˙(r′,tr)c]×dτ′ \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, trt_{r} is retarded time, and \bcR is separation vector.

Derivation

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

E=−∇V−∂A∂tB=∇×A \begin{align} \mathbf{E} &= -\nabla V-\frac{\partial \mathbf{A}}{\partial t} \\ \mathbf{B} &= \nabla \times \mathbf{A} \end{align}

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

V(r, t)=14πϵ0∫ρ(r′, tr)dτ′,A(r, t)=μ04π∫J(r′, tr)dτ′ 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 trt_{r} include r′\mathbf{r}^{\prime}, the calculation is not outright simple. The gradient of VV is, due to the chain rule, the gradient of retarded time is ∇tr=−1c\nabla t_{r}=-\dfrac{1}{c} \crH and the gradient of magnitude of separation vector is 1=−12\dfrac{1}{\cR} = -\dfrac{1}{\cR^{2}}\crH,

∇V=14πϵ0∫∇(ρ(r′, tr))dτ′=14πϵ0∫[1∇ρ(r′, tr)+ρ(r′, tr)∇(1)]dτ′=14πϵ0∫[−1∂ρ(r′,tr)∂tr∇tr−ρ(r′,tr)2]dτ′=14πϵ0∫[−ρ˙(r′,tr)c−ρ(r′,tr)2]dτ′ \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,

∂A∂t=∂∂tμ04π∫J(r′, tr)dτ′=μ04π∫1∂J(r′, tr)∂tdτ′=μ04π∫J˙(r′, tr)dτ′ \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)(3), (4)(4) into (1)(1),

E(r,t)=14πϵ0∫[ρ(r′,tr)2+ρ˙(r′,tr)c]dτ′−μ04π∫J˙(r′, tr)dτ′=14πϵ0∫[ρ(r′,tr)2+ρ˙(r′,tr)c−μ0ϵ0J˙(r′, tr)]dτ′=14πϵ0∫[ρ(r′,tr)2+ρ˙(r′,tr)c−J˙(r′, tr)c2]dτ′ \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 1c2=μ0ϵ0\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 B\mathbf{B} is more complicated to calculate because it includes the curl operator.

∇×A=μ04π[∇×(∫J(r′, tr)dτ′)] \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 (x,y,z)(\mathbf{x},\mathbf{y},\mathbf{z}), and ∫dτ′\int d\tau^{\prime} is the integration regarding (x′,y′,z′)(\mathbf{x}^{\prime},\mathbf{y}^{\prime},\mathbf{z}^{\prime}), thus they are independent.

∇×A=μ04π∫∇×J(r′, tr)dτ′=μ04π∫[1(∇×J(r′, tr))−J(r′, tr)×∇(1)]dτ′ \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) ∇×(fA)=f(∇×A)−A×(∇f)\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) - \mathbf{A} \times (\nabla f). First, let’s determine each component of ∇×J\nabla \times \mathbf{J}.

(∇×J)x=∂Jz∂y−∂Jy∂z (\nabla \times \mathbf{J})_{x}=\dfrac{\partial J_{z}}{\partial y}-\dfrac{\partial J_{y}}{\partial z}

Due to the chain rule of differentiation,

∂Jz∂y=∂Jz∂t∂t∂tr∂tr∂y \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 tr=t−ct_{r}=t-\dfrac{\cR}{c}, therefore ∂t∂tr=1\dfrac{\partial t}{\partial t_{r}}=1, ∂tr∂y=−1c∂∂y\dfrac{\partial t_{r}}{\partial y}=-\dfrac{1}{c}\dfrac{\partial \cR}{\partial y}, therefore

∂Jz∂y=−1cJz˙∂∂y \dfrac{\partial J_{z}}{\partial y}=-\dfrac{1}{c} \dot{J_{z}}\dfrac{\partial \cR}{\partial y}

Consequently,

(∇×J)x=−1cJz˙∂∂y−1cJy˙∂∂z=1c(Jy˙∂∂z−Jz˙∂∂y)=1c[J˙×(∇)]x \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,

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

And since ∇(1)=−12\nabla \left( \frac{1}{\acR} \right)=-\frac{1}{\acR^2}\acrH, with (6)(6) together when substituting into (5)(5),

B(r,t)=∇×A=μ04π∫[11c(J˙×)+J×12]dτ′=μ04π∫[J(r′, tr)2+J˙(r′, tr)c]×dτ′ \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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  1. David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p486-487 ↩︎