Electric and Magnetic Fields due to Accelerated Charge

eb-8ef68b06-408f-4fca-9a0a-0f1c4a31a2ce
The Liénard-Wiechert potentials are given as:
Φ ( r → , t ) = 1 4 π ϵ 0 q c ( R c − R → ⋅ v → ) A → ( r → , t ) = v → c 2 Φ ( r → , t ) Φ ( r → , t ) = 1 4 π ϵ 0 q c ( R c − R → ⋅ v → ) A → ( r → , t ) = v → c 2 Φ ( r → , 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)=14πϵ0qc(Rc−R→⋅v→)A→(r→,t)=v→c2Φ(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 → = − ∇ → Φ − ∂ A → ∂ t B → = ∇ → × 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→∂tB→=∇→×A→
Calculation of E → E → vec(E)\mathbf{\vec{E}}E→ :
E → = − ∇ → Φ − ∂ A → ∂ t E → = − ∇ → Φ − ∂ A → ∂ 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 [ ∇ → ( 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 → ) 2 [ ∇ → ( R → ⋅ v → ) − 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=qc4πϵ0∇→[(Rc−R→⋅v→)−1]=qc4πϵ0(−1)(Rc−R→⋅v→)2∇→(Rc−R→⋅v→)⇒∇→Φ=qc4πϵ01(Rc−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=qc4πϵ01(Rc−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}tr=t−Rc⇒R=c(t−tr)
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∇→tr
Solving for 1st term in equation 1:\
Using the product rule:
∇ → ( X → ⋅ Y → ) = ( X → ⋅ ∇ → ) Y → + ( Y → ⋅ ∇ → ) X → + X → × ( ∇ → × 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→=(Rx∂∂x+Ry∂∂y+Rz∂∂z)v→(tr)=Rxdv→ dtr∂tr∂x+Rydv→ dtr∂tr∂y+Rzdv→ dtr∂tr∂z=a→(R→⋅∇→tr)
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→(tr)
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→=(vx∂∂x+vy∂∂y+vz∂∂z)(xi^+yj^+zk^)=vxi^+vyj^+vzk^=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→⋅∇→tr)
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→⋅∇→tr)
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→=(∂vz∂y−∂vy∂z)i^+(∂vx∂z−∂vz∂x)j^+(∂vy∂x−∂vx∂y)k^=(dvz dtr∂tr∂y−dvy dtr∂tr∂z)i^+⋯=−a→×∇→tr
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→×∇→tr)
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→×∇→tr]=v→×∇→tr
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→×∇→tr)
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→⋅∇→tr)+v→−v→(v→⋅∇→tr)−R→×(a→×∇→tr)+v→×(v→×∇→tr)
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→×∇→tr)=a→(R→⋅∇→tr)−∇→tr(R→⋅a→)v→×(v→×∇→tr)=v→(v→⋅∇→tr)−v2∇→tr
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→⋅∇→tr)+v→−v→(v→⋅∇→tr)−a→(R→⋅∇→tr)+∇→tr(R→⋅a→)+v→(v→⋅∇→tr)−v2∇→tr⇒∇→(R→⋅v→)=v→+(R→⋅a→−v2)∇→tr
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=qc4πϵ01(Rc−R→⋅v→)2[v→+(R→⋅a→−v2)∇→tr+c2∇→tr]
∇ → 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=qc4πϵ01(Rc−R→⋅v→)2[v→+(c2−v2+R→⋅a→)∇→tr],…(3)
To find ∇ → t r ∇ → t r vec(grad)t_(r)\vec{\nabla} t_r∇→tr, 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∇→tr=∇→R=∇→R→⋅R→=12R→⋅R→∇→(R→⋅R→)=1R[(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→⋅∇→tr)
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→×∇→tr\
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∇→tr=1R[R→−v→(R→⋅∇→tr)+R→×v→×∇→tr]
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∇→tr=1R[R→−(R→⋅v→)∇→tr]⇒∇→tr=−R→Rc−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}