Gradient of Time Delay
📂Electrodynamics Gradient of Time Delay Overview The gradient of time delay is as follows.
∇ t r = − 1 c
\nabla t_{r}=-\frac{1}{c} \crH
∇ t r = − c 1
Proof
Given = c ( t − t r ) \acR = c(t -t_{r}) = c ( t − t r ) t t t
∇ = ∇ ( − c t r ) = − c ∇ t r
\nabla \cR =\nabla(-c t_{r})=-c \nabla t_{r}
∇ = ∇ ( − c t r ) = − c ∇ t r
Therefore, the gradient of time delay can be computed by calculating ∇ \nabla \cR ∇
∇ = ∇ ⋅ = 1 2 ⋅ ∇ ( ⋅ ) = 1 2 ∇ ( ⋅ ) = 1 2 [ × ( ∇ × ) + × ( ∇ × ) + ( ⋅ ∇ ) + ( ⋅ ∇ ) ] = 1 [ × ( ∇ × ) + ( ⋅ ∇ ) ]
\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}
∇ = ∇ ⋅ = 2 ⋅ 1 ∇ ( ⋅ ) = 2 1 ∇ ( ⋅ ) = 2 1 [ × ( ∇ × ) + × ( ∇ × ) + ( ⋅ ∇ ) + ( ⋅ ∇ ) ] = 1 [ × ( ∇ × ) + ( ⋅ ∇ ) ]
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 ( ⋅ ∇ ) ( ⋅ ∇ ) = ( ⋅ ∇ ) r − ( ⋅ ∇ ) w = − v ( ⋅ ∇ t r )
\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*}
( ⋅ ∇ ) = ( ⋅ ∇ ) r − ( ⋅ ∇ ) w = − v ( ⋅ ∇ t r )
The second equality holds because for any vector A \mathbf{A} A ( A ⋅ ∇ ) r = A (\mathbf{A} \cdot \nabla )\mathbf{r}=\mathbf{A} ( A ⋅ ∇ ) r = A ( ⋅ ∇ ) A = ∂ A ∂ t r ( ⋅ ∇ t r ) (\abcR \cdot \nabla)\mathbf{A}=\dfrac{\partial \mathbf{A} }{\partial t_{r}}(\abcR \cdot \nabla t_{r}) ( ⋅ ∇ ) A = ∂ t r ∂ A ( ⋅ ∇ t r ) ( ⋅ ∇ ) r = (\bcR \cdot \nabla )\mathbf{r}=\bcR ( ⋅ ∇ ) r = ( ⋅ ∇ ) w = v ( ⋅ ∇ t r ) ( \bcR \cdot \nabla ) \mathbf{w} = \mathbf{v}(\bcR \cdot \nabla t_{r}) ( ⋅ ∇ ) w = v ( ⋅ ∇ t r )
Part 2. ∇ × \nabla \times \bcR ∇ × ∇ × = ∇ × r − ∇ × w = − ∇ × w = − ( − v × ∇ t r ) = v × ∇ t r
\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*}
∇ × = ∇ × r − ∇ × w = − ∇ × w = − ( − v × ∇ t r ) = v × ∇ t r
The second equality is valid because of ∇ × r = 0 \nabla \times \mathbf{r}=0 ∇ × r = 0 r \mathbf{r} r r = x x ^ + y y ^ + z z ^ \mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}} r = x x ^ + y y ^ + z z ^ A \mathbf{A} A ∇ × A = − ∂ A ∂ t × ∇ t r \nabla \times \mathbf{A} = -\frac{\partial \mathbf{A}}{\partial t} \times \nabla t_{r} ∇ × A = − ∂ t ∂ A × ∇ t r ∇ × w = − v × ∇ t r \nabla \times \mathbf{w}=-\mathbf{v}\times\nabla t_{r} ∇ × w = − v × ∇ t r
Therefore, substituting these two results into ( 1 ) (1) ( 1 )
∇ = 1 [ − v ( ⋅ ∇ t r ) + × ( v × ∇ t r ) ] = 1 [ − v ( ⋅ ∇ t r ) + v ( ⋅ ∇ t r ) − ∇ t r ( ⋅ v ) ] = 1 [ − ∇ t r ( ⋅ v ) ]
\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*}
∇ = 1 [ − v ( ⋅ ∇ t r ) + × ( v × ∇ t r ) ] = 1 [ − v ( ⋅ ∇ t r ) + v ( ⋅ ∇ t r ) − ∇ t r ( ⋅ v ) ] = 1 [ − ∇ t r ( ⋅ v ) ]
The second equality is justified by the BAC-CAB formula . To summarize,
− c ∇ t r = ∇ = 1 [ − ∇ t r ( ⋅ v ) ] ⟹ − ∇ t r ( ⋅ v ) = − ∇ t r ⟹ ∇ t r = − c − ( ⋅ v )
\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*}
⟹ ⟹ − c ∇ t r = ∇ − ∇ t r ( ⋅ v ) ∇ t r = 1 [ − ∇ t r ( ⋅ v ) ] = − ∇ t r = c − ( ⋅ v ) −
This is a general result that also applies when the charges are stationary . Since t r = t − c t_{r}=t-\frac{\cR}{c} t r = t − c
∇ t r = − ∇ c = − 1 c ∇
\nabla t_{r} = -\nabla \frac{\cR}{c}=-\frac{1}{c} \nabla \cR
∇ t r = − ∇ c = − c 1 ∇
And since ∇ = \nabla \acR = \acrH ∇ =
∇ t r = − 1 c
\nabla t_{r}=-\frac{1}{c} \crH
∇ t r = − c 1
By substituting v = 0 \mathbf{v}=\mathbf{0} v = 0
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