📂Electrodynamics

Moving Point Charges and the Electric Fields They Create

Overview ¹

The electromagnetic field created by a moving point charge is as follows.

$$\begin{align*} \mathbf{E}(\mathbf{r}, t) &= \frac{q}{4\pi\epsilon_{0}} \frac{\cR} {( \bcR\cdot \mathbf{u} )^3 } \left[(c^2-v^2)\mathbf{u} +\bcR\times (\mathbf{u} \times \mathbf{a} ) \right] \\ \mathbf{B} (\mathbf{ r}, t) &=\frac{1}{c} \crH\times \mathbf{ E } (\mathbf{ r}, t) \end{align*}$$

Description

An introduction to the induction process for electric fields.

Induction

The electric and magnetic fields created by a moving point charge can be obtained using the Liénard-Wiechert potentials.

$$V(\mathbf{r}, t)= \frac{1}{4\pi \epsilon_{0}} \frac{qc}{ (\cR c -\bcR\cdot \mathbf{v})} ,\quad \mathbf{A}(\mathbf{r}, t) = \frac{ \mathbf{v} } {c^2} V(\mathbf{r}, t)$$

Additionally, the electromagnetic field can be obtained by the following equation.

$$\mathbf{ E} = -\nabla V -\frac{\partial \mathbf{ A} }{\partial t},\quad \mathbf{B}=\nabla \times \mathbf{A}$$

Let’s derive $\nabla V$ and $\dfrac{\partial \mathbf{A}}{\partial t}$ in turn. First, if we calculate $\nabla V$,

$$\nabla V=\frac{qc}{4 \pi \epsilon_{0}} \nabla \dfrac{1}{(\cR c - \bcR \cdot \mathbf{v}) }$$

Since $\dfrac{ d}{dx} \left( \dfrac{1}{f(x)} \right)=\dfrac{-1}{ [f(x)]^2} f^{\prime}(x)$,

$$\begin{equation} \nabla V = \frac{qc}{4\pi\epsilon_{0}} \frac{-1}{ (\cR c -\bcR \cdot \mathbf{v} )^2}\nabla (\cR c -\bcR \cdot \mathbf{v} ) \end{equation}$$

Therefore, we shall calculate $\nabla \cR$ and $\nabla (\bcR \cdot \mathbf{v})$. First, if $\nabla ( \bcR \cdot \mathbf{v} )$ is calculated according to Multiplication Rule 2,

$$\begin{equation} \nabla (\bcR \cdot \mathbf{v}) = (\bcR \cdot \nabla ) \mathbf{v}+(\mathbf{v} \cdot \nabla ) \bcR + \bcR\times (\nabla \times \mathbf{v}) +\mathbf{v}\times (\nabla \times \bcR) \end{equation}$$

First Term

$$\begin{align*} (\bcR \cdot \nabla) \mathbf{v} &= \left( \cR_{x} \frac{\partial}{\partial x} + \cR _{y} \frac{\partial}{\partial y}+\cR_{z}\frac{\partial }{\partial z} \right)\mathbf{v} \\[1em] &= \cR_{x}\frac{d \mathbf{v} }{dt_{r}}\frac{\partial t_{r}}{\partial x} +\cR_{y}\frac{d \mathbf{v} }{dt_{r}}\frac{\partial t_{r}}{\partial y} + \cR_{z}\frac{d \mathbf{v} }{dt_{r}}\frac{\partial t_{r}}{\partial z} \\[1em] &= \frac{d \mathbf{v} }{ dt_{r}} \left( \cR_{x}\frac{\partial t_{r}}{\partial x} + \cR_{y}\frac{\partial t_{r}}{\partial y}+\cR_{z}\frac{\partial t_{r}}{\partial z}\right) \\[1em] &= \mathbf{a} ( \bcR \cdot \nabla t_{r}) \end{align*}$$

Here, $\mathbf{a}$ is the acceleration of the particle (point charge) at the retarded time.

Second Term

Since $\bcR=\mathbf{r}-\mathbf{w}$,

$$\begin{equation} (\mathbf{v} \cdot \nabla ) \bcR=(\mathbf{v}\cdot \nabla)\mathbf{r} -(\mathbf{v}\cdot \nabla ) \mathbf{w} \end{equation}$$

For an arbitrary vector $\mathbf{A}=(A_{x}, A_{y}, A_{z})$,

$$\begin{align*} (\mathbf{A} \cdot \nabla ) \mathbf{r} &= \left( A_{x}\frac{\partial}{\partial x}+A_{y}\frac{\partial}{\partial y}+A_{z}\frac{\partial}{\partial z} \right) ( x \hat{\mathbf{x}} + y \hat{\mathbf{y}} +z\hat{ \mathbf{z} }) \\ &= A_{x}\hat{\mathbf{x}}+A_{y} \hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}} \\ &= \mathbf{A} \end{align*}$$

And

$$\begin{align*} (\mathbf{A} \cdot \nabla ) \mathbf{w}(t_{r}) &= \left( A_{x}\frac{\partial}{\partial x}+A_{y}\frac{\partial}{\partial y}+A_{z}\frac{\partial}{\partial z} \right) \mathbf{w} \\[1em] &= A_{x}\frac{\partial \mathbf{w} } {\partial x}+A_{y}\frac{\partial \mathbf{w} }{\partial y}+A_{z}\frac{\partial \mathbf{w} }{\partial z} \\[1em] &= A_{x}\frac{\partial \mathbf{w} }{\partial t_{r}}\frac{\partial t_{r}} {\partial x}+A_{y}\frac{\partial \mathbf{w} }{\partial t_{r}}\frac{\partial t_{r}}{\partial y}+A_{z}\frac{\partial \mathbf{w} }{\partial t_{r}}\frac{\partial t_{r}}{\partial z} \\[1em] &= A_{x} \mathbf{v}\frac{\partial t_{r}} {\partial x}+A_{y}\mathbf{v}\frac{\partial t_{r}}{\partial y}+A_{z}\mathbf{v}\frac{\partial t_{r}}{\partial z} \\[1em] &= \mathbf{v} \left( A_{x} \frac{\partial t_{r}} {\partial x}+A_{y}\frac{\partial t_{r}}{\partial y}+A_{z}\frac{\partial t_{r}}{\partial z}\right) \\[1em] &= \mathbf{v}(\mathbf{A} \cdot \nabla t_{r}) \end{align*}$$ Thus, $(3)$ is

$$(\mathbf{v} \cdot \nabla ) \bcR=\mathbf{v} -\mathbf{v}(\mathbf{v}\cdot \nabla t_{r} )$$

Third Term

First, the following holds.

$$\begin{align*} \nabla \times \mathbf{v} &= \left( \frac{\partial v_{z}}{\partial y} - \frac{\partial v_{y}}{\partial z} \right)\hat{\mathbf{x}} + \left( \frac{\partial v_{x}}{\partial z} - \frac{\partial v_{z}}{\partial x} \right)\hat{\mathbf{y}} + \left( \frac{\partial v_{y}}{\partial x} - \frac{\partial v_{x}}{\partial y} \right)\hat{\mathbf{z}} \\[1em] &= \left( \frac{\partial v_{z}}{\partial t_{r}} \frac{t_{r}}{\partial y} - \frac{\partial v_{y}}{\partial t_{r}} \frac{t_{r}}{\partial z} \right)\hat{\mathbf{x}} + \left( \frac{\partial v_{x}}{\partial t_{r}} \frac{t_{r}}{\partial z} - \frac{\partial v_{z}}{\partial t_{r}} \frac{t_{r}}{\partial x} \right)\hat{\mathbf{y}} + \left( \frac{\partial v_{y}}{\partial t_{r}} \frac{t_{r}}{\partial x} - \frac{\partial v_{x}}{\partial t_{r}} \frac{t_{r}}{\partial y} \right)\hat{\mathbf{z}} \\[1em] &= \Big( a_{z} (\nabla t_{r})_{y} - a_{y} (\nabla t_{r})_{z} \Big)\hat{\mathbf{x}} + \Big( a_{x} (\nabla t_{r})_{z} - a_{z} (\nabla t_{r})_{x} \Big)\hat{\mathbf{y}} + \Big( a_{y} (\nabla t_{r})_{x} - a_{x} (\nabla t_{r})_{y} \Big)\hat{\mathbf{z}} \\[1em] &= -\mathbf{a} \times \nabla t_{r} \end{align*}$$

Therefore

$$\bcR \times ( \nabla \times \mathbf{v} )= -\bcR\times (\mathbf{a} \times \nabla t_{r})$$

Fourth Term

Since $\bcR=\mathbf{r}-\mathbf{w}$ and $\nabla \times \mathbf{r}=0$,

$$\nabla \times \bcR = \nabla \times \mathbf{r} -\nabla \times \mathbf{w}=-\nabla \times \mathbf{w}$$

Using the result of $\nabla \times \mathbf{v} = -\mathbf{a} \times \nabla t_{r}$ calculated above, we can know $\nabla \times \mathbf{w} = -\mathbf{v} \times \nabla t_{r}$. Therefore,

$$\mathbf{v}\times (\nabla \times \bcR)=\mathbf{v} \times (- \nabla \times \mathbf{w})=\mathbf{v} \times (\mathbf{v} \times \nabla t_{r})$$

Now, by substituting the above calculation results into $(2)$,

$$\begin{align} \nabla (\bcR \cdot \mathbf{v} ) &= \mathbf{a} ( \bcR \cdot \nabla t_{r}) + \mathbf{v} -\mathbf{v}(\mathbf{v} \cdot \nabla t_{r}) -\color{blue}{\abcR\times(\mathbf{a} \times \nabla t_{r} )} +\color{green}{\mathbf{v} \times (\mathbf{v} \times \nabla t_{r})} \nonumber \\[1em] &= \mathbf{a} ( \bcR \cdot \nabla t_{r}) + \mathbf{v} -\mathbf{v}(\mathbf{v} \cdot \nabla t_{r}) -\color{blue}{\left[\mathbf{a} ( \abcR \cdot \nabla t_{r}) - \nabla t_{r}(\abcR \cdot \mathbf{a} ) \right]}+ \color{green}{\left[ \mathbf{v}(\mathbf{v} \cdot \nabla t_{r} ) -\nabla t_{r} (\mathbf{v} \cdot \mathbf{v}) \right] } \nonumber \\[1em] &=\mathbf{v} + \nabla t_{r}(\bcR \cdot \mathbf{a}) -\nabla t_{r}( \mathbf{v} \cdot \mathbf{v}) \nonumber \\[1em] &= \mathbf{v}+( \bcR \cdot \mathbf{a} -v^2) \nabla t_{r} \end{align}$$

The second equality holds due to the BAC-CAB rule.

Conclusion

Since $\cR=c(t-t_{r})$, $\nabla \cR = -c\nabla t_{r}$ applies. By substituting this and $(4)$ into $(1)$,

$$\begin{align} \nabla V &= \frac{qc}{4\pi\epsilon_{0}} \frac{-1}{ (\cR c -\bcR \cdot \mathbf{v} )^2}\nabla (\cR c -\bcR \cdot \mathbf{v} ) \nonumber \\[1em] &= \frac{qc}{4\pi\epsilon_{0}} \frac{-1}{ (\cR c -\bcR \cdot \mathbf{v} )^2} \Big[ c\nabla \cR -\nabla (\bcR \cdot \mathbf{v} ) \Big] \nonumber \\[1em] &=\frac{qc}{4\pi\epsilon_{0}} \frac{-1}{ (\cR c -\bcR \cdot \mathbf{v} )^2} \Big[ -c^2\nabla t_{r} -\mathbf{v}-(\bcR \cdot \mathbf{a} -v^2)\nabla t_{r} \Big] \nonumber \\[1em] &=\frac{qc}{4\pi\epsilon_{0}} \frac{1}{ (\cR c -\bcR \cdot \mathbf{v} )^2} \Big[ \mathbf{v} + (c^2 -v^2+\bcR \cdot \mathbf{a} )\nabla t_{r}\Big] \end{align}$$

And the gradient of the retarded time is as follows.

$$\nabla t_{r} = \frac {-\bcR }{\cR c -\bcR \cdot \mathbf{v} }$$

Substituting this into $(5)$,

$$\begin{align*} \nabla V &= \frac{qc}{4\pi\epsilon_{0}} \frac{1}{ (\cR c -\bcR \cdot \mathbf{v} )^2} \left[ \mathbf{v} + (c^2 -v^2+\bcR \cdot \mathbf{a} )\nabla t_{r}\right] \\ &= \frac{qc}{4\pi\epsilon_{0}} \frac{1}{ (\cR c -\bcR \cdot \mathbf{v} )^2} \left[ \mathbf{v} + (c^2 -v^2+\bcR \cdot \mathbf{a} ) \frac {-\bcR }{\cR c -\bcR \cdot \mathbf{v} } \right] \\ &= \frac{qc}{4\pi\epsilon_{0}} \frac{1}{ (\cR c -\bcR \cdot \mathbf{v} )^3} \Big[ (\cR c -\bcR \cdot \mathbf{v})\mathbf{v} - (c^2 -v^2+\bcR \cdot \mathbf{a} ) \bcR \Big] \end{align*}$$

Finally, if we derive $\dfrac{\partial \mathbf{A}}{\partial t}$, it is done. Upon calculation, the result is as follows.

$$\frac{ \partial \mathbf{A}}{ \partial t }=\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\cR\mathbf{a}/c-\mathbf{v})+ \frac{\cR}{c}\mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right]$$

By defining as $\mathbf{u}\equiv c \crH-\mathbf{v}$ and organizing,

$$\begin{align*} \mathbf{E}(\mathbf{r},t) &= -\nabla V - \frac{ \partial \mathbf{A}}{ \partial t} \\ &= \frac{q}{4\pi\epsilon_{0}}\frac{\cR}{(\bcR\cdot \mathbf{u})^{3}}\left[ (c^{2}-v^{2})\mathbf{u} + \mathbf{u}(\bcR\cdot\mathbf{a})-\mathbf{a}(\bcR\cdot\mathbf{u}) \right] \\ &= \frac{q}{4\pi\epsilon_{0}}\frac{\cR}{(\bcR\cdot \mathbf{u})^{3}}\left[ (c^{2}-v^{2})\mathbf{u} + \bcR \times (\mathbf{u}\times \mathbf{a}) \right] \end{align*}$$

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