Gradient of Time Delay
Overview
The gradient of time delay is as follows.
$$ \nabla t_{r}=-\frac{1}{c} \crH $$
Proof
Given $\acR = c(t -t_{r})$ and since $t$ is independent of spatial variables,
$$ \nabla \cR =\nabla(-c t_{r})=-c \nabla t_{r} $$
Therefore, the gradient of time delay can be computed by calculating $\nabla \cR$.
$$ \begin{align} \nabla \cR &= \nabla \sqrt{\bcR \cdot \bcR} \nonumber \\ &= \frac{1}{2\sqrt{\bcR\cdot \bcR}} \nabla (\bcR \cdot \bcR ) \nonumber \\ &= \frac{1}{2\cR} \nabla (\bcR \cdot \bcR ) \nonumber \\ &= \frac{1}{2\cR} \Big[ \bcR \times (\nabla \times \bcR ) + \bcR \times (\nabla \times \bcR) + (\bcR \cdot \nabla)\bcR +(\bcR \cdot \nabla)\bcR\Big] \nonumber \\ &= \frac{1}{\cR} \Big[\bcR\times (\nabla \times \bcR) + (\bcR \cdot \nabla) \bcR \Big] \end{align} $$
The fourth equality holds due to the Product Rule (b). Now proceeding with the remaining calculations one by one,
- Part 1. $(\bcR \cdot \nabla)\bcR$
$$ \begin{align*} (\bcR \cdot \nabla ) \bcR &= (\bcR \cdot \nabla ) \mathbf{r} - ( \bcR \cdot \nabla) \mathbf{w} \\ &= \bcR - \mathbf{v} ( \bcR \cdot \nabla t_{r}) \end{align*} $$
The second equality holds because for any vector $\mathbf{A}$, $(\mathbf{A} \cdot \nabla )\mathbf{r}=\mathbf{A}$1 and $(\abcR \cdot \nabla)\mathbf{A}=\dfrac{\partial \mathbf{A} }{\partial t_{r}}(\abcR \cdot \nabla t_{r})$2 makes $(\bcR \cdot \nabla )\mathbf{r}=\bcR$ and $( \bcR \cdot \nabla ) \mathbf{w} = \mathbf{v}(\bcR \cdot \nabla t_{r})$ true.
- Part 2. $\nabla \times \bcR$
$$ \begin{align*} \nabla \times \bcR &= \nabla \times \mathbf{r} -\nabla \times \mathbf{w} \\ &= -\nabla \times \mathbf{w} \\ &= -(- \mathbf{v} \times \nabla t_{r} ) \\ &= \mathbf{v} \times \nabla t_{r} \end{align*} $$
The second equality is valid because of $\nabla \times \mathbf{r}=0$. Each component of $\mathbf{r}$ is independent of the other components, which is a matter of course. If you are not sure what this means, set it to $\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}$ and calculate it directly. The third equality holds because for any vector $\mathbf{A}$, $\nabla \times \mathbf{A} = -\frac{\partial \mathbf{A}}{\partial t} \times \nabla t_{r}$3 makes $\nabla \times \mathbf{w}=-\mathbf{v}\times\nabla t_{r}$ true.
- Part 3. Conclusion
Therefore, substituting these two results into $(1)$ gives
$$ \begin{align*} \nabla \cR &= \frac{1}{\cR} \Big[ \bcR -\mathbf{v}(\bcR \cdot \nabla t_{r}) + \bcR \times (\mathbf{v} \times \nabla t_{r}) \Big] \\ &= \frac{1}{\cR} \Big[ \bcR -\mathbf{v}(\bcR \cdot \nabla t_{r}) + \mathbf{v}(\bcR \cdot \nabla t_{r})-\nabla t_{r}(\bcR \cdot \mathbf{v}) \Big] \\ &= \frac{1}{\cR} \Big[ \bcR -\nabla t_{r} (\bcR \cdot \mathbf{v} )\Big] \end{align*} $$
The second equality is justified by the BAC-CAB formula. To summarize,
$$ \begin{align*} && -c \nabla t_{r} = \nabla \cR &= \frac{1}{\cR} \Big[ \bcR -\nabla t_{r} (\bcR \cdot \mathbf{v} )\big] \\ \implies && \bcR - \nabla t_{r}(\bcR \cdot \mathbf{v}) &= -\cR \nabla t_{r} \\ \implies && \nabla t_{r} &= \frac{-\bcR}{\cR c - (\bcR \cdot \mathbf{v})} \end{align*} $$
This is a general result that also applies when the charges are stationary. Since $t_{r}=t-\frac{\cR}{c}$,
$$ \nabla t_{r} = -\nabla \frac{\cR}{c}=-\frac{1}{c} \nabla \cR $$
And since $\nabla \acR = \acrH$,
$$ \nabla t_{r}=-\frac{1}{c} \crH $$
By substituting $\mathbf{v}=\mathbf{0}$ into the above result, it is equivalent.
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