The Liénard-Wiechert potentials are given as:
Φ
(
r
→
,
t
)
=
1
4
π
ϵ
0
q
c
(
R
c
−
R
→
⋅
v
→
)
A
→
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r
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t
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=
v
→
c
2
Φ
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r
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t
)
Φ
(
r
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t
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=
1
4
π
ϵ
0
q
c
(
R
c
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R
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⋅
v
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)
A
→
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r
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t
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=
v
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2
Φ
(
r
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,
t
)
{:[Phi( vec(r)”,”t)=(1)/(4piepsilon_(0))(qc)/((Rc-( vec(R))*( vec(v))))],[ vec(A)( vec(r)”,”t)=(( vec(v)))/(c^(2))Phi( vec(r)”,”t)]:} \begin{aligned}
& \Phi(\vec{r}, t)=\frac{1}{4 \pi \epsilon_0} \frac{q c}{(R c-\vec{R} \cdot \vec{v})} \\
& \vec{A}(\vec{r}, t)=\frac{\vec{v}}{c^2} \Phi(\vec{r}, t)
\end{aligned} Φ ( r → , t ) = 1 4 π ϵ 0 q c ( R c − R → ⋅ v → ) A → ( r → , t ) = v → c 2 Φ ( r → , t )
These equations can now be used to calculate the
(
E
→
,
B
→
)
(
E
→
,
B
→
)
( vec(E), vec(B)) (\mathbf{\vec{E}, \vec{B}}) ( E → , B → ) due to a point charge in accelerated motion, using the equations:
E
→
=
−
∇
→
Φ
−
∂
A
→
∂
t
B
→
=
∇
→
×
A
→
E
→
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−
∇
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−
∂
A
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t
B
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∇
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×
A
→
{:[ vec(E)=- vec(grad)Phi-(del( vec(A)))/(del t)],[ vec(B)= vec(grad)xx vec(A)]:} \begin{aligned}
& \vec{E}=-\vec{\nabla} \Phi-\frac{\partial \vec{A}}{\partial t} \\
& \vec{B}=\vec{\nabla} \times \vec{A}
\end{aligned} E → = − ∇ → Φ − ∂ A → ∂ t B → = ∇ → × A →
Calculation of
E
→
E
→
vec(E) \mathbf{\vec{E}} E → :
E
→
=
−
∇
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Φ
−
∂
A
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∂
t
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∇
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∂
A
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∂
t
{: vec(E)=- vec(grad)Phi-(del( vec(A)))/(del t):} \begin{aligned}
\vec{E}=-\vec{\nabla} \Phi-\frac{\partial \vec{A}}{\partial t} \\
\end{aligned} E → = − ∇ → Φ − ∂ A → ∂ t
Here,
∇
→
Φ
=
q
c
4
π
ϵ
0
∇
→
[
(
R
c
−
R
→
⋅
v
→
)
−
1
]
=
q
c
4
π
ϵ
0
(
−
1
)
(
R
c
−
R
→
⋅
v
→
)
2
∇
→
(
R
c
−
R
→
⋅
v
→
)
⇒
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
∇
→
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R
→
⋅
v
→
)
−
c
∇
→
R
]
∇
→
Φ
=
q
c
4
π
ϵ
0
∇
→
(
R
c
−
R
→
⋅
v
→
)
−
1
=
q
c
4
π
ϵ
0
(
−
1
)
(
R
c
−
R
→
⋅
v
→
)
2
∇
→
(
R
c
−
R
→
⋅
v
→
)
⇒
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
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2
[
∇
→
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R
→
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→
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−
c
∇
→
R
]
{:[ vec(grad)Phi=(qc)/(4piepsilon_(0)) vec(grad)[(Rc-( vec(R))*( vec(v)))^(-1)]],[=(qc)/(4piepsilon_(0))((-1))/((Rc-( vec(R))*( vec(v)))^(2)) vec(grad)(Rc- vec(R)* vec(v))],[=> vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc-( vec(R))*( vec(v)))^(2))[ vec(grad)( vec(R)* vec(v))-c vec(grad)R]]:} \begin{aligned}
\vec{\nabla} \Phi & =\frac{q c}{4 \pi \epsilon_0} \vec{\nabla}\left[(R c-\vec{R} \cdot \vec{v})^{-1}\right] \\
& =\frac{q c}{4 \pi \epsilon_0} \frac{(-1)}{(R c-\vec{R} \cdot \vec{v})^2} \vec{\nabla}(R c-\vec{R} \cdot \vec{v}) \\
\Rightarrow \vec{\nabla} \Phi & =\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\vec{R} \cdot \vec{v})^2}[\vec{\nabla}(\vec{R} \cdot \vec{v})-c \vec{\nabla} R]
\end{aligned} ∇ → Φ = q c 4 π ϵ 0 ∇ → [ ( R c − R → ⋅ v → ) − 1 ] = q c 4 π ϵ 0 ( − 1 ) ( R c − R → ⋅ v → ) 2 ∇ → ( R c − R → ⋅ v → ) ⇒ ∇ → Φ = q c 4 π ϵ 0 1 ( R c − R → ⋅ v → ) 2 [ ∇ → ( R → ⋅ v → ) − c ∇ → R ]
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
∇
→
(
R
→
⋅
v
→
)
−
c
∇
→
R
]
,
…
(
1
)
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
∇
→
(
R
→
⋅
v
→
)
−
c
∇
→
R
]
,
…
1
{: vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc-( vec(R))*( vec(v)))^(2))[ vec(grad)( vec(R)* vec(v))-c vec(grad)R]”,” dots(1):} \begin{aligned}
\vec{\nabla} \Phi=\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\vec{R} \cdot \vec{v})^2}[\vec{\nabla}(\vec{R} \cdot \vec{v})-c \vec{\nabla} R], & \ldots\left(\mathrm{1}\right)\\
\end{aligned} ∇ → Φ = q c 4 π ϵ 0 1 ( R c − R → ⋅ v → ) 2 [ ∇ → ( R → ⋅ v → ) − c ∇ → R ] , … ( 1 )
Since retarded time is given by:
t
r
=
t
−
R
c
⇒
R
=
c
(
t
−
t
r
)
t
r
=
t
−
R
c
⇒
R
=
c
t
−
t
r
{:[t_(r)=t-(R)/(c)],[=>R=c(t-t_(r))]:} \begin{aligned}
t_r & =t-\frac{R}{c} \\
\Rightarrow R & =c\left(t-t_r\right)
\end{aligned} t r = t − R c ⇒ R = c ( t − t r )
This gives
∇
→
R
=
−
c
∇
→
t
r
∇
→
R
=
−
c
∇
→
t
r
vec(grad)R=-c vec(grad)t_(r) \vec{\nabla} R=-c \vec{\nabla} t_r ∇ → R = − c ∇ → t r
Solving for 1st term in equation 1:\
Using the product rule:
∇
→
(
X
→
⋅
Y
→
)
=
(
X
→
⋅
∇
→
)
Y
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+
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Y
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⋅
∇
→
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X
→
+
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→
×
(
∇
→
×
Y
→
)
+
Y
→
×
(
∇
→
×
X
→
)
⇒
∇
(
R
→
⋅
v
→
)
=
(
R
→
⋅
∇
→
)
v
→
+
(
v
→
⋅
∇
→
)
R
→
+
R
→
×
(
∇
→
×
v
→
)
+
v
→
×
(
∇
→
×
R
→
)
)
,
…
(
2
)
∇
→
(
X
→
⋅
Y
→
)
=
(
X
→
⋅
∇
→
)
Y
→
+
(
Y
→
⋅
∇
→
)
X
→
+
X
→
×
(
∇
→
×
Y
→
)
+
Y
→
×
(
∇
→
×
X
→
)
⇒
∇
(
R
→
⋅
v
→
)
=
(
R
→
⋅
∇
→
)
v
→
+
(
v
→
⋅
∇
→
)
R
→
+
R
→
×
(
∇
→
×
v
→
)
+
v
→
×
(
∇
→
×
R
→
)
)
,
…
2
{:[ vec(grad)( vec(X)* vec(Y))=( vec(X)* vec(grad)) vec(Y)+( vec(Y)* vec(grad)) vec(X)+ vec(X)xx( vec(grad)xx vec(Y))+ vec(Y)xx( vec(grad)xx vec(X))],[=>grad( vec(R)* vec(v))=( vec(R)* vec(grad)) vec(v)+( vec(v)* vec(grad)) vec(R)+ vec(R)xx( vec(grad)xx vec(v))+ vec(v)xx( vec(grad)xx vec(R)))”,”dots(2)]:} \begin{aligned}
\vec{\nabla}(\vec{X} \cdot \vec{Y})=(\vec{X} \cdot \vec{\nabla}) \vec{Y}+(\vec{Y} \cdot \vec{\nabla}) \vec{X}+\vec{X} \times(\vec{\nabla} \times \vec{Y})+\vec{Y} \times(\vec{\nabla} \times \vec{X}) \\
\Rightarrow{\boldsymbol{\nabla}}(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})=(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{v}}+(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{R}}+\overrightarrow{\boldsymbol{R}} \times(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{v}})+\overrightarrow{\boldsymbol{v}} \times(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{R}})),\ldots\left(\mathrm{2}\right)\\
\end{aligned} ∇ → ( X → ⋅ Y → ) = ( X → ⋅ ∇ → ) Y → + ( Y → ⋅ ∇ → ) X → + X → × ( ∇ → × Y → ) + Y → × ( ∇ → × X → ) ⇒ ∇ ( R → ⋅ v → ) = ( R → ⋅ ∇ → ) v → + ( v → ⋅ ∇ → ) R → + R → × ( ∇ → × v → ) + v → × ( ∇ → × R → ) ) , … ( 2 )
1st term in eq. (2):
(
R
→
⋅
∇
→
)
v
→
=
(
R
x
∂
∂
x
+
R
y
∂
∂
y
+
R
z
∂
∂
z
)
v
→
(
t
r
)
=
R
x
d
v
→
d
t
r
∂
t
r
∂
x
+
R
y
d
v
→
d
t
r
∂
t
r
∂
y
+
R
z
d
v
→
d
t
r
∂
t
r
∂
z
=
a
→
(
R
→
⋅
∇
→
t
r
)
(
R
→
⋅
∇
→
)
v
→
=
R
x
∂
∂
x
+
R
y
∂
∂
y
+
R
z
∂
∂
z
v
→
t
r
=
R
x
d
v
→
d
t
r
∂
t
r
∂
x
+
R
y
d
v
→
d
t
r
∂
t
r
∂
y
+
R
z
d
v
→
d
t
r
∂
t
r
∂
z
=
a
→
R
→
⋅
∇
→
t
r
{:[( vec(R)* vec(grad)) vec(v)=(R_(x)(del)/(del x)+R_(y)(del)/(del y)+R_(z)(del)/(del z)) vec(v)(t_(r))],[=R_(x)(d( vec(v)))/((d)t_(r))(delt_(r))/(del x)+R_(y)(d( vec(v)))/((d)t_(r))(delt_(r))/(del y)+R_(z)(d( vec(v)))/((d)t_(r))(delt_(r))/(del z)],[= vec(a)(( vec(R))*( vec(grad))t_(r))]:} \begin{aligned}
(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla}) \vec{v} & =\left(R_x \frac{\partial}{\partial x}+R_y \frac{\partial}{\partial y}+R_z \frac{\partial}{\partial z}\right) \vec{v}\left(t_r\right) \\
& =R_x \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t_r} \frac{\partial t_r}{\partial x}+R_y \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t_r} \frac{\partial t_r}{\partial y}+R_z \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t_r} \frac{\partial t_r}{\partial z} \\
& =\vec{a}\left(\vec{R} \cdot \vec{\nabla} t_r\right)
\end{aligned} ( R → ⋅ ∇ → ) v → = ( R x ∂ ∂ x + R y ∂ ∂ y + R z ∂ ∂ z ) v → ( t r ) = R x d v → d t r ∂ t r ∂ x + R y d v → d t r ∂ t r ∂ y + R z d v → d t r ∂ t r ∂ z = a → ( R → ⋅ ∇ → t r )
2nd term in eq. (2):
(
v
→
⋅
∇
→
)
R
→
=
(
v
→
⋅
∇
→
)
r
→
−
(
v
→
⋅
∇
→
)
w
→
(
v
→
⋅
∇
→
)
R
→
=
(
v
→
⋅
∇
→
)
r
→
−
(
v
→
⋅
∇
→
)
w
→
( vec(v)* vec(grad)) vec(R)=( vec(v)* vec(grad)) vec(r)-( vec(v)* vec(grad)) vec(w) (\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{R}}=(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{r}}-(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{w}}\\ ( v → ⋅ ∇ → ) R → = ( v → ⋅ ∇ → ) r → − ( v → ⋅ ∇ → ) w →
As,
R
→
=
r
→
−
w
→
(
t
r
)
R
→
=
r
→
−
w
→
(
t
r
)
{: vec(R)= vec(r)- vec(w)(t_(r)):} \begin{aligned}
\overrightarrow{\boldsymbol{R}}= \overrightarrow{\boldsymbol{r}}- \overrightarrow{\boldsymbol{w}}\boldsymbol{(t_r)}\\
\end{aligned} R → = r → − w → ( t r )
Here,
(
v
→
⋅
∇
→
)
r
→
=
(
v
x
∂
∂
x
+
v
y
∂
∂
y
+
v
z
∂
∂
z
)
(
x
i
^
+
y
j
^
+
z
k
^
)
=
v
x
i
^
+
v
y
j
^
+
v
z
k
^
=
v
→
(
v
→
⋅
∇
→
)
r
→
=
v
x
∂
∂
x
+
v
y
∂
∂
y
+
v
z
∂
∂
z
(
x
i
^
+
y
j
^
+
z
k
^
)
=
v
x
i
^
+
v
y
j
^
+
v
z
k
^
=
v
→
{:[( vec(v)* vec(grad)) vec(r)=(v_(x)(del)/(del x)+v_(y)(del)/(del y)+v_(z)(del)/(del z))(x hat(i)+y hat(j)+z hat(k))],[=v_(x) hat(i)+v_(y) hat(j)+v_(z) hat(k)= vec(v)]:} \begin{aligned}
(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{r}} & =\left(v_x \frac{\partial}{\partial x}+v_y \frac{\partial}{\partial y}+v_z \frac{\partial}{\partial z}\right)(x \hat{\boldsymbol{i}}+y \hat{\boldsymbol{j}}+z \hat{\boldsymbol{k}}) \\
& =v_x \hat{\boldsymbol{i}}+v_y \hat{\boldsymbol{j}}+v_z \hat{\boldsymbol{k}}=\overrightarrow{\boldsymbol{v}}
\end{aligned} ( v → ⋅ ∇ → ) r → = ( v x ∂ ∂ x + v y ∂ ∂ y + v z ∂ ∂ z ) ( x i ^ + y j ^ + z k ^ ) = v x i ^ + v y j ^ + v z k ^ = v →
Proceeding in a similar manner as for the 1st term in eq. (2):
(
v
→
⋅
∇
→
)
w
→
=
v
→
(
v
→
⋅
∇
→
t
r
)
(
v
→
⋅
∇
→
)
w
→
=
v
→
v
→
⋅
∇
→
t
r
( vec(v)* vec(grad)) vec(w)= vec(v)( vec(v)* vec(grad)t_(r)) (\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{w}}=\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right) ( v → ⋅ ∇ → ) w → = v → ( v → ⋅ ∇ → t r )
Hence, we obtain
(
v
→
⋅
∇
→
)
R
→
=
v
→
−
v
→
(
v
→
⋅
∇
→
t
r
)
(
v
→
⋅
∇
→
)
R
→
=
v
→
−
v
→
v
→
⋅
∇
→
t
r
( vec(v)* vec(grad)) vec(R)= vec(v)- vec(v)( vec(v)* vec(grad)t_(r)) (\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}}) \overrightarrow{\boldsymbol{R}}=\overrightarrow{\boldsymbol{v}}-\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}}\cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right) ( v → ⋅ ∇ → ) R → = v → − v → ( v → ⋅ ∇ → t r )
3rd term in eq. (2):
∇
→
×
v
→
=
(
∂
v
z
∂
y
−
∂
v
y
∂
z
)
i
^
+
(
∂
v
x
∂
z
−
∂
v
z
∂
x
)
j
^
+
(
∂
v
y
∂
x
−
∂
v
x
∂
y
)
k
^
=
(
d
v
z
d
t
r
∂
t
r
∂
y
−
d
v
y
d
t
r
∂
t
r
∂
z
)
i
^
+
⋯
=
−
a
→
×
∇
→
t
r
∇
→
×
v
→
=
∂
v
z
∂
y
−
∂
v
y
∂
z
i
^
+
∂
v
x
∂
z
−
∂
v
z
∂
x
j
^
+
∂
v
y
∂
x
−
∂
v
x
∂
y
k
^
=
d
v
z
d
t
r
∂
t
r
∂
y
−
d
v
y
d
t
r
∂
t
r
∂
z
i
^
+
⋯
=
−
a
→
×
∇
→
t
r
{:[ vec(grad)xx vec(v)=((delv_(z))/(del y)-(delv_(y))/(del z)) hat(i)+((delv_(x))/(del z)-(delv_(z))/(del x)) hat(j)+((delv_(y))/(del x)-(delv_(x))/(del y)) hat(k)],[=((dv_(z))/((d)t_(r))(delt_(r))/(del y)-(dv_(y))/((d)t_(r))(delt_(r))/(del z)) hat(i)+cdots],[=- vec(a)xx vec(grad)t_(r)]:} \begin{aligned}
\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{v}} & =\left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right) \hat{\boldsymbol{i}}+\left(\frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x}\right) \hat{\boldsymbol{j}}+\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right) \hat{\boldsymbol{k}} \\
& =\left(\frac{\mathrm{d} v_z}{\mathrm{~d} t_r} \frac{\partial t_r}{\partial y}-\frac{\mathrm{d} v_y}{\mathrm{~d} t_r} \frac{\partial t_r}{\partial z}\right) \hat{\boldsymbol{i}}+\cdots \\
& =-\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{\nabla}} t_r
\end{aligned} ∇ → × v → = ( ∂ v z ∂ y − ∂ v y ∂ z ) i ^ + ( ∂ v x ∂ z − ∂ v z ∂ x ) j ^ + ( ∂ v y ∂ x − ∂ v x ∂ y ) k ^ = ( d v z d t r ∂ t r ∂ y − d v y d t r ∂ t r ∂ z ) i ^ + ⋯ = − a → × ∇ → t r
Therefore,
R
→
×
(
∇
→
×
v
→
)
=
−
R
→
×
(
a
→
×
∇
→
t
r
)
R
→
×
(
∇
→
×
v
→
)
=
−
R
→
×
a
→
×
∇
→
t
r
vec(R)xx( vec(grad)xx vec(v))=- vec(R)xx( vec(a)xx vec(grad)t_(r)) \overrightarrow{\boldsymbol{R}} \times(\vec{\nabla} \times \overrightarrow{\boldsymbol{v}})=-\overrightarrow{\boldsymbol{R}} \times\left(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right) R → × ( ∇ → × v → ) = − R → × ( a → × ∇ → t r )
4th term in eq. (2): (Same as 3rd term in eq. 2)
∇
→
×
R
→
=
∇
→
×
r
→
−
∇
→
×
w
→
=
0
−
[
−
v
→
×
∇
→
t
r
]
=
v
→
×
∇
→
t
r
∇
→
×
R
→
=
∇
→
×
r
→
−
∇
→
×
w
→
=
0
−
−
v
→
×
∇
→
t
r
=
v
→
×
∇
→
t
r
{:[ vec(grad)xx vec(R)= vec(grad)xx vec(r)- vec(grad)xx vec(w)],[=0-[- vec(v)xx vec(grad)t_(r)]],[= vec(v)xx vec(grad)t_(r)]:} \begin{aligned}
\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{R}} & =\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{r}}-\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{w}} \\
& =0-\left[-\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right]\\
& =\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r
\end{aligned} ∇ → × R → = ∇ → × r → − ∇ → × w → = 0 − [ − v → × ∇ → t r ] = v → × ∇ → t r
Therefore,
v
→
×
(
∇
→
×
R
→
)
=
v
→
×
(
v
→
×
∇
→
t
r
)
v
→
×
(
∇
→
×
R
→
)
=
v
→
×
v
→
×
∇
→
t
r
vec(v)xx( vec(grad)xx vec(R))= vec(v)xx( vec(v)xx vec(grad)t_(r)) \overrightarrow{\boldsymbol{v}} \times(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{R}})=\overrightarrow{\boldsymbol{v}} \times\left(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right) v → × ( ∇ → × R → ) = v → × ( v → × ∇ → t r )
Then eq. (2) becomes,
∇
→
(
R
→
⋅
v
→
)
=
a
→
(
R
→
⋅
∇
→
t
r
)
+
v
→
−
v
→
(
v
→
⋅
∇
→
t
r
)
−
R
→
×
(
a
→
×
∇
→
t
r
)
+
v
→
×
(
v
→
×
∇
→
t
r
)
∇
→
(
R
→
⋅
v
→
)
=
a
→
R
→
⋅
∇
→
t
r
+
v
→
−
v
→
v
→
⋅
∇
→
t
r
−
R
→
×
a
→
×
∇
→
t
r
+
v
→
×
v
→
×
∇
→
t
r
{:[ vec(grad)( vec(R)* vec(v))= vec(a)( vec(R)* vec(grad)t_(r))+ vec(v)- vec(v)( vec(v)* vec(grad)t_(r))],[- vec(R)xx( vec(a)xx vec(grad)t_(r))+ vec(v)xx( vec(v)xx vec(grad)t_(r))]:} \begin{aligned}
\vec{\nabla}(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})= & \overrightarrow{\boldsymbol{a}}\left(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right)+\overrightarrow{\boldsymbol{v}}-\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right) \\
& -\overrightarrow{\boldsymbol{R}} \times\left(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right)+\overrightarrow{\boldsymbol{v}} \times\left(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right)
\end{aligned} ∇ → ( R → ⋅ v → ) = a → ( R → ⋅ ∇ → t r ) + v → − v → ( v → ⋅ ∇ → t r ) − R → × ( a → × ∇ → t r ) + v → × ( v → × ∇ → t r )
Using BAC-CAB rule:
A
→
×
(
B
→
×
C
→
)
=
B
→
(
A
→
⋅
C
→
)
−
C
→
(
A
→
⋅
B
→
)
R
→
×
(
a
→
×
∇
→
t
r
)
=
a
→
(
R
→
⋅
∇
→
t
r
)
−
∇
→
t
r
(
R
→
⋅
a
→
)
v
→
×
(
v
→
×
∇
→
t
r
)
=
v
→
(
v
→
⋅
∇
→
t
r
)
−
v
2
∇
→
t
r
A
→
×
(
B
→
×
C
→
)
=
B
→
(
A
→
⋅
C
→
)
−
C
→
(
A
→
⋅
B
→
)
R
→
×
a
→
×
∇
→
t
r
=
a
→
R
→
⋅
∇
→
t
r
−
∇
→
t
r
(
R
→
⋅
a
→
)
v
→
×
v
→
×
∇
→
t
r
=
v
→
v
→
⋅
∇
→
t
r
−
v
2
∇
→
t
r
{:[ vec(A)xx( vec(B)xx vec(C))= vec(B)( vec(A)* vec(C))- vec(C)( vec(A)* vec(B))],[ vec(R)xx( vec(a)xx vec(grad)t_(r))= vec(a)( vec(R)*( vec(grad))t_(r))- vec(grad)t_(r)( vec(R)* vec(a))],[ vec(v)xx( vec(v)xx vec(grad)t_(r))= vec(v)( vec(v)* vec(grad)t_(r))-v^(2) vec(grad)t_(r)]:} \begin{aligned}
\overrightarrow{\boldsymbol{A}} \times(\overrightarrow{\boldsymbol{B}} \times \overrightarrow{\boldsymbol{C}}) & =\overrightarrow{\boldsymbol{B}}(\overrightarrow{\boldsymbol{A}} \cdot \overrightarrow{\boldsymbol{C}})-\overrightarrow{\boldsymbol{C}}(\overrightarrow{\boldsymbol{A}} \cdot \overrightarrow{\boldsymbol{B}}) \\
\overrightarrow{\boldsymbol{R}} \times\left(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right) & =\overrightarrow{\boldsymbol{a}}\left(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla} t_r\right)-\vec{\nabla} t_r(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}) \\
\overrightarrow{\boldsymbol{v}} \times\left(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right) & =\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right)-v^2 \overrightarrow{\boldsymbol{\nabla}} t_r
\end{aligned} A → × ( B → × C → ) = B → ( A → ⋅ C → ) − C → ( A → ⋅ B → ) R → × ( a → × ∇ → t r ) = a → ( R → ⋅ ∇ → t r ) − ∇ → t r ( R → ⋅ a → ) v → × ( v → × ∇ → t r ) = v → ( v → ⋅ ∇ → t r ) − v 2 ∇ → t r
Therefore,
∇
→
(
R
→
⋅
v
→
)
=
a
→
(
R
→
⋅
∇
→
t
r
)
+
v
→
−
v
→
(
v
→
⋅
∇
→
t
r
)
−
a
→
(
R
→
⋅
∇
→
t
r
)
+
∇
→
t
r
(
R
→
⋅
a
→
)
+
v
→
(
v
→
⋅
∇
→
t
r
)
−
v
2
∇
→
t
r
⇒
∇
→
(
R
→
⋅
v
→
)
=
v
→
+
(
R
→
⋅
a
→
−
v
2
)
∇
→
t
r
∇
→
(
R
→
⋅
v
→
)
=
a
→
R
→
⋅
∇
→
t
r
+
v
→
−
v
→
v
→
⋅
∇
→
t
r
−
a
→
R
→
⋅
∇
→
t
r
+
∇
→
t
r
(
R
→
⋅
a
→
)
+
v
→
v
→
⋅
∇
→
t
r
−
v
2
∇
→
t
r
⇒
∇
→
(
R
→
⋅
v
→
)
=
v
→
+
R
→
⋅
a
→
−
v
2
∇
→
t
r
{:[ vec(grad)( vec(R)* vec(v))= vec(a)( vec(R)*( vec(grad))t_(r))+ vec(v)- vec(v)( vec(v)*( vec(grad))t_(r))],[- vec(a)( vec(R)*( vec(grad))t_(r))+ vec(grad)t_(r)( vec(R)* vec(a))+ vec(v)( vec(v)* vec(grad)t_(r))-v^(2) vec(grad)t_(r)],[=> vec(grad)( vec(R)* vec(v))= vec(v)+( vec(R)* vec(a)-v^(2)) vec(grad)t_(r)]:} \begin{aligned}
\vec{\nabla}(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}) & =\overrightarrow{\boldsymbol{a}}\left(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla} t_r\right)+\overrightarrow{\boldsymbol{v}}-\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}} \cdot \vec{\nabla} t_r\right) \\
& -\overrightarrow{\boldsymbol{a}}\left(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla} t_r\right)+\vec{\nabla} t_r(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}})+\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{v}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right)-v^2 \overrightarrow{\boldsymbol{\nabla}} t_r \\
\Rightarrow \vec{\nabla}(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}) & =\overrightarrow{\boldsymbol{v}}+\left(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}-v^2\right) \vec{\nabla} t_r
\end{aligned} ∇ → ( R → ⋅ v → ) = a → ( R → ⋅ ∇ → t r ) + v → − v → ( v → ⋅ ∇ → t r ) − a → ( R → ⋅ ∇ → t r ) + ∇ → t r ( R → ⋅ a → ) + v → ( v → ⋅ ∇ → t r ) − v 2 ∇ → t r ⇒ ∇ → ( R → ⋅ v → ) = v → + ( R → ⋅ a → − v 2 ) ∇ → t r
Hence, using above results in eq. (1):
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
v
→
+
(
R
→
⋅
a
→
−
v
2
)
∇
→
t
r
+
c
2
∇
→
t
r
]
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
v
→
+
R
→
⋅
a
→
−
v
2
∇
→
t
r
+
c
2
∇
→
t
r
{: vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc- vec(R)* vec(v))^(2))[ vec(v)+( vec(R)* vec(a)-v^(2))( vec(grad))t_(r)+c^(2)( vec(grad))t_(r)]:} \begin{aligned}
& \vec{\nabla} \Phi=\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})^2}\left[\overrightarrow{\boldsymbol{v}}+\left(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}-v^2\right) \vec{\nabla} t_r+c^2 \vec{\nabla} t_r\right] \\
\end{aligned} ∇ → Φ = q c 4 π ϵ 0 1 ( R c − R → ⋅ v → ) 2 [ v → + ( R → ⋅ a → − v 2 ) ∇ → t r + c 2 ∇ → t r ]
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
v
→
+
(
c
2
−
v
2
+
R
→
⋅
a
→
)
∇
→
t
r
]
,
…
(
3
)
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
v
→
+
c
2
−
v
2
+
R
→
⋅
a
→
∇
→
t
r
,
…
3
vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc- vec(R)* vec(v))^(2))[ vec(v)+(c^(2)-v^(2)+ vec(R)* vec(a)) vec(grad)t_(r)],dots(3) \overrightarrow{\boldsymbol{\nabla}} \Phi=\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})^2}\left[\overrightarrow{\boldsymbol{v}}+\left(c^2-v^2+\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}\right) \overrightarrow{\boldsymbol{\nabla}} t_r\right],\ldots\left(\mathrm{3}\right)\\ ∇ → Φ = q c 4 π ϵ 0 1 ( R c − R → ⋅ v → ) 2 [ v → + ( c 2 − v 2 + R → ⋅ a → ) ∇ → t r ] , … ( 3 )
To find
∇
→
t
r
∇
→
t
r
vec(grad)t_(r) \vec{\nabla} t_r ∇ → t r , we know:
−
c
∇
→
t
r
=
∇
→
R
=
∇
→
R
→
⋅
R
→
=
1
2
R
→
⋅
R
→
∇
→
(
R
→
⋅
R
→
)
=
1
R
[
(
R
→
⋅
∇
→
)
R
→
+
R
→
×
(
∇
→
×
R
→
)
]
−
c
∇
→
t
r
=
∇
→
R
=
∇
→
R
→
⋅
R
→
=
1
2
R
→
⋅
R
→
∇
→
(
R
→
⋅
R
→
)
=
1
R
[
(
R
→
⋅
∇
→
)
R
→
+
R
→
×
(
∇
→
×
R
→
)
]
{:[-c vec(grad)t_(r)= vec(grad)R= vec(grad)sqrt( vec(R)* vec(R))=(1)/(2sqrt( vec(R)* vec(R))) vec(grad)( vec(R)* vec(R))],[=(1)/(R)[( vec(R)* vec(grad)) vec(R)+ vec(R)xx( vec(grad)xx vec(R))]]:} \begin{aligned}
-c \overrightarrow{\boldsymbol{\nabla}} t_r & =\overrightarrow{\boldsymbol{\nabla}} R=\overrightarrow{\boldsymbol{\nabla}} \sqrt{\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{R}}}=\frac{1}{2 \sqrt{\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{R}}}} \overrightarrow{\boldsymbol{\nabla}}(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{R}}) \\
& =\frac{1}{R}[(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla}) \overrightarrow{\boldsymbol{R}}+\overrightarrow{\boldsymbol{R}} \times(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{R}})]
\end{aligned} − c ∇ → t r = ∇ → R = ∇ → R → ⋅ R → = 1 2 R → ⋅ R → ∇ → ( R → ⋅ R → ) = 1 R [ ( R → ⋅ ∇ → ) R → + R → × ( ∇ → × R → ) ]
It can be shown that
(
R
→
⋅
∇
→
)
R
→
=
R
→
−
v
→
(
R
→
⋅
∇
→
t
r
)
(
R
→
⋅
∇
→
)
R
→
=
R
→
−
v
→
R
→
⋅
∇
→
t
r
( vec(R)* vec(grad)) vec(R)= vec(R)- vec(v)( vec(R)*( vec(grad))t_(r)) (\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla}) \overrightarrow{\boldsymbol{R}}=\overrightarrow{\boldsymbol{R}}-\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{R}} \cdot \vec{\nabla} t_r\right) ( R → ⋅ ∇ → ) R → = R → − v → ( R → ⋅ ∇ → t r )
and
∇
→
×
R
→
=
v
→
×
∇
→
t
r
∇
→
×
R
→
=
v
→
×
∇
→
t
r
vec(grad)xx vec(R)= vec(v)xx vec(grad)t_(r) \vec{\nabla} \times \overrightarrow{\boldsymbol{R}}=\overrightarrow{\boldsymbol{v}} \times \vec{\nabla} t_r ∇ → × R → = v → × ∇ → t r \
Therefore,
−
c
∇
→
t
r
=
1
R
[
R
→
−
v
→
(
R
→
⋅
∇
→
t
r
)
+
R
→
×
v
→
×
∇
→
t
r
]
−
c
∇
→
t
r
=
1
R
R
→
−
v
→
R
→
⋅
∇
→
t
r
+
R
→
×
v
→
×
∇
→
t
r
{:-c vec(grad)t_(r)=(1)/(R)[ vec(R)- vec(v)( vec(R)* vec(grad)t_(r))+ vec(R)xx vec(v)xx vec(grad)t_(r)]:} \begin{aligned}
-c \overrightarrow{\boldsymbol{\nabla}} t_r & =\frac{1}{R}\left[\overrightarrow{\boldsymbol{R}}-\overrightarrow{\boldsymbol{v}}\left(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{\nabla}} t_r\right)+\overrightarrow{\boldsymbol{R}} \times \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{\nabla}} t_r\right] \\
\end{aligned} − c ∇ → t r = 1 R [ R → − v → ( R → ⋅ ∇ → t r ) + R → × v → × ∇ → t r ]
Using BAC-CAB rule:
⇒
−
c
∇
→
t
r
=
1
R
[
R
→
−
(
R
→
⋅
v
→
)
∇
→
t
r
]
⇒
∇
→
t
r
=
−
R
→
R
c
−
R
→
⋅
v
→
,
…
(
4
)
⇒
−
c
∇
→
t
r
=
1
R
R
→
−
(
R
→
⋅
v
→
)
∇
→
t
r
⇒
∇
→
t
r
=
−
R
→
R
c
−
R
→
⋅
v
→
,
…
4
{:[=>-c vec(grad)t_(r)=(1)/(R)[ vec(R)-( vec(R)* vec(v))( vec(grad))t_(r)]],[=> vec(grad)t_(r)=(- vec(R))/(Rc- vec(R)* vec(v))”,”dots(4)]:} \begin{aligned}
\Rightarrow-c \overrightarrow{\boldsymbol{\nabla}} t_r & =\frac{1}{R}\left[\overrightarrow{\boldsymbol{R}}-(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}) \vec{\nabla} t_r\right] \\
\Rightarrow \overrightarrow{\boldsymbol{\nabla}} t_r & =\frac{-\overrightarrow{\boldsymbol{R}}}{R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}},\ldots\left(\mathrm{4}\right)\\
\end{aligned} ⇒ − c ∇ → t r = 1 R [ R → − ( R → ⋅ v → ) ∇ → t r ] ⇒ ∇ → t r = − R → R c − R → ⋅ v → , … ( 4 )
Inserting eq.(4) in eq.(3):
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
[
v
→
−
(
c
2
−
v
2
+
R
→
⋅
a
→
)
R
→
R
c
−
R
→
⋅
v
→
]
⇒
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
3
[
(
R
c
−
R
→
⋅
v
→
)
v
→
−
(
c
2
−
v
2
+
R
→
⋅
a
→
)
R
→
]
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
2
v
→
−
c
2
−
v
2
+
R
→
⋅
a
→
R
→
R
c
−
R
→
⋅
v
→
⇒
∇
→
Φ
=
q
c
4
π
ϵ
0
1
(
R
c
−
R
→
⋅
v
→
)
3
(
R
c
−
R
→
⋅
v
→
)
v
→
−
c
2
−
v
2
+
R
→
⋅
a
→
R
→
{:[ vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc- vec(R)* vec(v))^(2))[ vec(v)-((c^(2)-v^(2)+ vec(R)* vec(a)) vec(R))/(Rc- vec(R)* vec(v))]],[=> vec(grad)Phi=(qc)/(4piepsilon_(0))(1)/((Rc- vec(R)* vec(v))^(3))[(Rc- vec(R)* vec(v)) vec(v)-(c^(2)-v^(2)+ vec(R)* vec(a)) vec(R)]]:} \begin{aligned}
\vec{\nabla} \Phi & =\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})^2}\left[\overrightarrow{\boldsymbol{v}}-\frac{\left(c^2-v^2+\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}\right) \overrightarrow{\boldsymbol{R}}}{R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}\right] \\
\Rightarrow \vec{\nabla} \Phi & =\frac{q c}{4 \pi \epsilon_0} \frac{1}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})^3}\left[(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}) \overrightarrow{\boldsymbol{v}}-\left(c^2-v^2+\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}\right) \overrightarrow{\boldsymbol{R}}\right]
\end{aligned}