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}∇→Φ=qc4πϵ01(Rc−R→⋅v→)2[v→−(c2−v2+R→⋅a→)R→Rc−R→⋅v→]⇒∇→Φ=qc4πϵ01(Rc−R→⋅v→)3[(Rc−R→⋅v→)v→−(c2−v2+R→⋅a→)R→]
∇ → 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 → ] , … ( 5 ) ∇ → 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 → , … 5 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)],dots(5)\overrightarrow{\boldsymbol{\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],\ldots\left(\mathrm{5}\right)\\∇→Φ=qc4πϵ01(Rc−R→⋅v→)3[(Rc−R→⋅v→)v→−(c2−v2+R→⋅a→)R→],…(5)
From the Liénard-Wiechert potential
A → ( r → , t ) = v → c 2 Φ ( r → , t ) A → ( r → , t ) = v → c 2 Φ ( r → , t ) vec(A)( vec(r),t)=( vec(v))/(c^(2))Phi( vec(r),t)\overrightarrow{\boldsymbol{A}}(\overrightarrow{\boldsymbol{r}}, t)=\frac{\overrightarrow{\boldsymbol{v}}}{c^2} \Phi(\overrightarrow{\boldsymbol{r}}, t)A→(r→,t)=v→c2Φ(r→,t)
and
Φ ( r → , t ) = 1 4 π ϵ 0 q c ( R c − R → ⋅ v → ) Φ ( r → , t ) = 1 4 π ϵ 0 q c ( R c − R → ⋅ v → ) Phi( vec(r),t)=(1)/(4piepsilon_(0))(qc)/((Rc- vec(R)* vec(v)))\Phi(\overrightarrow{\boldsymbol{r}}, t)=\frac{1}{4 \pi \epsilon_0} \frac{q c}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})}Φ(r→,t)=14πϵ0qc(Rc−R→⋅v→)
Hence,
A → ( r → , t ) = 1 4 π ϵ 0 c q v → ( R c − R → ⋅ v → ) A → ( r → , t ) = 1 4 π ϵ 0 c q v → ( R c − R → ⋅ v → ) vec(A)( vec(r),t)=(1)/(4piepsilon_(0)c)(q vec(v))/((Rc- vec(R)* vec(v)))\overrightarrow{\boldsymbol{A}}(\overrightarrow{\boldsymbol{r}}, t)= \frac{1}{4 \pi \epsilon_0 c} \frac{q \overrightarrow{\boldsymbol{v}}}{(R c-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}})}A→(r→,t)=14πϵ0cqv→(Rc−R→⋅v→)
Therefore,
∂ A → ∂ t = q 4 π ϵ 0 c [ ∂ v → ∂ t ( 1 R c − R → ⋅ v → ) + v → ∂ ∂ t ( 1 R c − R → ⋅ v → ) ] , … ( 6 ) ∂ A → ∂ t = q 4 π ϵ 0 c ∂ v → ∂ t 1 R c − R → ⋅ v → + v → ∂ ∂ t 1 R c − R → ⋅ v → , … 6 {:(del vec(A))/(del t)=(q)/(4piepsilon_(0)c)[(del vec(v))/(del t)((1)/(Rc- vec(R)* vec(v)))+ vec(v)(del)/(del t)((1)/(Rc- vec(R)* vec(v)))]”,”dots(6):}\begin{aligned} \frac{\partial \overrightarrow{\boldsymbol{A}}}{\partial t}= \frac{q}{4 \pi \epsilon_0 c}\left[\frac{\partial\overrightarrow{{\boldsymbol{v}}}}{\partial t}\left(\frac{1}{Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}\right)+\overrightarrow{\boldsymbol{v}}\frac{\partial}{\partial t}\left(\frac{1}{Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}\right)\right], \ldots\left(\mathrm{6}\right)\\ \end{aligned}∂A→∂t=q4πϵ0c[∂v→∂t(1Rc−R→⋅v→)+v→∂∂t(1Rc−R→⋅v→)],…(6)
Now, solving for 1st term in eq. (6),\
∂ v → ∂ t = d v ( t r ) d t r ∂ t r ∂ t = a → ( t r ) ∂ t r ∂ t , … ( 7 ) ∂ v → ∂ t = d v t r d t r ∂ t r ∂ t = a → ( t r ) ∂ t r ∂ t , … 7 {:(del vec(v))/(del t)=(d(v(t_(r))))/(dt_(r))(delt_(r))/(del t)= vec(a)(t_(r))(delt_(r))/(del t)”,”dots(7):}\begin{aligned} \frac{\partial\overrightarrow{{\boldsymbol{v}}}}{\partial t}=\frac{d{\boldsymbol{v}\left(t_r\right)}}{dt_r}\frac{\partial t_r}{\partial t}=\overrightarrow{\boldsymbol{a}}(t_r)\frac{\partial t_r}{\partial t},\ldots\left(\mathrm{7}\right)\\ \end{aligned}∂v→∂t=dv(tr)dtr∂tr∂t=a→(tr)∂tr∂t,…(7)
Calculation of ∂ t r ∂ t ∂ t r ∂ t (delt_(r))/(del t)\mathbf{\frac{\partial t_r}{\partial t}}∂tr∂t:\
Since,\
R = c ( t − t r ) = c Δ t r , R → = r → − w → ( t r ) ⇒ R → ⋅ R → = c 2 ( t − t r ) 2 ⇒ ∂ ∂ t c 2 ( t − t r ) 2 = ∂ ∂ t ( R → ⋅ R → ) ⇒ c . c ( t − t r ) . ( 1 − ∂ t r ∂ t ) = R → ∂ ∂ t [ r → − w → ( t r ) ] ⇒ c R ( 1 − ∂ t r ∂ t ) = R → [ ∂ ∂ t r → − ∂ ∂ t w → ( t r ) ] R = c ( t − t r ) = c Δ t r , R → = r → − w → ( t r ) ⇒ R → ⋅ R → = c 2 ( t − t r ) 2 ⇒ ∂ ∂ t c 2 ( t − t r ) 2 = ∂ ∂ t R → ⋅ R → ⇒ c . c ( t − t r ) . ( 1 − ∂ t r ∂ t ) = R → ∂ ∂ t [ r → − w → ( t r ) ] ⇒ c R ( 1 − ∂ t r ∂ t ) = R → [ ∂ ∂ t r → − ∂ ∂ t w → ( t r ) ] {:[R=c(t-t_(r))=c Deltat_(r)”,” vec(R)= vec(r)- vec(w)(t_(r))],[=> vec(R)* vec(R)=c^(2)(t-t_(r))^(2)],[=>(del)/(del t)c^(2)(t-t_(r))^(2)=(del)/(del t)( vec(R)* vec(R))],[=>c.c(t-t_(r)).(1-(delt_(r))/(del t))= vec(R)(del)/(del t)[ vec(r)- vec(w)(t_(r))]],[=>cR(1-(delt_(r))/(del t))= vec(R)[(del)/(del t) vec(r)-(del)/(del t) vec(w)(t_(r))]]:}\begin{gathered} R=c(t-t_r)= c \Delta t_r, \overrightarrow{\boldsymbol{R}}= \overrightarrow{\boldsymbol{r}}- \overrightarrow{\boldsymbol{w}}(t_r)\\ \Rightarrow\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{R}} = c^2 (t- t_r)^2\\ \Rightarrow\frac{\partial}{\partial t} c^2 (t-t_r)^2= \frac{\partial}{\partial t}\left(\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{R}}\right)\\ \Rightarrow c.c (t-t_r).(1-\frac{\partial t_r}{\partial t})= \overrightarrow{\boldsymbol{R}} \frac{\partial}{\partial t}[\overrightarrow{\boldsymbol{r}}-\overrightarrow{\boldsymbol{w}}(t_r)]\\ \Rightarrow c R (1-\frac{\partial t_r}{\partial t}) = \overrightarrow{\boldsymbol{R}} [\frac{\partial}{\partial t}\overrightarrow{\boldsymbol{r}}-\frac{\partial}{\partial t}\overrightarrow{\boldsymbol{w}}(t_r)]\\ \end{gathered}R=c(t−tr)=cΔtr,R→=r→−w→(tr)⇒R→⋅R→=c2(t−tr)2⇒∂∂tc2(t−tr)2=∂∂t(R→⋅R→)⇒c.c(t−tr).(1−∂tr∂t)=R→∂∂t[r→−w→(tr)]⇒cR(1−∂tr∂t)=R→[∂∂tr→−∂∂tw→(tr)]
Since,
∂ ∂ t r → = 0 , ∂ w → ( t r ) ∂ t = ∂ w → ( t r ) ∂ t r ∂ t r ∂ t = v → ( t r ) ∂ t r ∂ t ∂ ∂ t r → = 0 , ∂ w → ( t r ) ∂ t = ∂ w → ( t r ) ∂ t r ∂ t r ∂ t = v → ( t r ) ∂ t r ∂ t {:(del)/(del t) vec(r)=0″,”(del vec(w)(t_(r)))/(del t)=(del vec(w)(t_(r)))/(delt_(r))(delt_(r))/(del t)= vec(v)(t_(r))(delt_(r))/(del t):}\begin{aligned} \frac{\partial}{\partial t}\overrightarrow{\boldsymbol{r}}=0,\frac{\partial \overrightarrow{\boldsymbol{w}}(t_r)}{\partial t}= \frac{\partial\overrightarrow{\boldsymbol{w}}(t_r)}{\partial t_r} \frac{\partial t_r}{\partial t}= \overrightarrow{\boldsymbol{v}}(t_r) \frac{\partial t_r}{\partial t}\\ \end{aligned}∂∂tr→=0,∂w→(tr)∂t=∂w→(tr)∂tr∂tr∂t=v→(tr)∂tr∂t
Hence,
∂ R → ∂ t = − v → ( t r ) ∂ t r ∂ t , ∂ R ∂ t = c ( 1 − ∂ t r ∂ t ) … ( 8 ) ∂ R → ∂ t = − v → ( t r ) ∂ t r ∂ t , ∂ R ∂ t = c 1 − ∂ t r ∂ t … 8 {:(del vec(R))/(del t)=- vec(v)(t_(r))(delt_(r))/(del t)”,”(del R)/(del t)=c(1-(delt_(r))/(del t))dots(8):}\begin{aligned} \frac{\partial \overrightarrow{\boldsymbol{R}}}{\partial t} = -\overrightarrow{\boldsymbol{v}}(t_r) \frac{\partial t_r}{\partial t}, \frac{\partial R}{\partial t} = c\left(1- \frac{\partial t_r}{\partial t}\right)\ldots\left(\mathrm{8}\right)\\ \end{aligned}∂R→∂t=−v→(tr)∂tr∂t,∂R∂t=c(1−∂tr∂t)…(8)
Therefore,
c R ( 1 − ∂ t r ∂ t ) = − R → ⋅ v → ( t r ) ∂ t r ∂ t c R ( 1 − ∂ t r ∂ t ) = − R → ⋅ v → ( t r ) ∂ t r ∂ t {:cR(1-(delt_(r))/(del t))=- vec(R)* vec(v)(t_(r))(delt_(r))/(del t):}\begin{aligned} c R (1-\frac{\partial t_r}{\partial t}) = -\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}(t_r) \frac{\partial t_r}{\partial t}\\ \end{aligned}cR(1−∂tr∂t)=−R→⋅v→(tr)∂tr∂t
Rearranging the terms, we get,
∂ t r ∂ t = R c R c − R → ⋅ v → ( t r ) , … ( 9 ) ∂ t r ∂ t = R c R c − R → ⋅ v → ( t r ) , … 9 {:(delt_(r))/(del t)=(Rc)/(Rc- vec(R)* vec(v)(t_(r)))”,”dots(9):}\begin{aligned} \frac{\partial t_r}{\partial t}= \frac{Rc}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}(t_r)},\ldots\left(\mathrm{9}\right)\\ \end{aligned}∂tr∂t=RcRc−R→⋅v→(tr),…(9)
Using eq. (9) in eq. (7), we get;
∂ v → ∂ t = a → ( t r ) R c R c − R → ⋅ v → ( t r ) , … ( 10 ) ∂ v → ∂ t = a → ( t r ) R c R c − R → ⋅ v → ( t r ) , … 10 {:(del vec(v))/(del t)=( vec(a)(t_(r))Rc)/(Rc- vec(R)* vec(v)(t_(r)))”,”dots(10):}\begin{aligned} \frac{\partial\overrightarrow{{\boldsymbol{v}}}}{\partial t}=\frac{\overrightarrow{\boldsymbol{a}}(t_r) Rc}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}(t_r)},\ldots\left(\mathrm{10}\right)\\ \end{aligned}∂v→∂t=a→(tr)RcRc−R→⋅v→(tr),…(10)
Now, solving for 2nd term in eq. (6),\
∂ ∂ t [ 1 R c − R → ⋅ v → ( t r ) ] = − 1 [ R c − R → ⋅ v → ] 2 ∂ ∂ t [ R c − R → ⋅ v → ] = − 1 [ R c − R → ⋅ v → ] 2 [ c ∂ R ∂ t − R → ⋅ ∂ v → ∂ t − v → ⋅ ∂ R → ∂ t ] , … ( 11 ) ∂ ∂ t 1 R c − R → ⋅ v → ( t r ) = − 1 [ R c − R → ⋅ v → ] 2 ∂ ∂ t [ R c − R → ⋅ v → ] = − 1 [ R c − R → ⋅ v → ] 2 c ∂ R ∂ t − R → ⋅ ∂ v → ∂ t − v → ⋅ ∂ R → ∂ t , … 11 {:[(del)/(del t)[(1)/(Rc- vec(R)* vec(v)(t_(r)))]=(-1)/([Rc- vec(R)* vec(v)]^(2))(del)/(del t)[Rc- vec(R)* vec(v)]],[=(-1)/([Rc- vec(R)* vec(v)]^(2))[c(del R)/(del t)- vec(R)*(del vec(v))/(del t)- vec(v)*(del vec(R))/(del t)]”,”dots(11)]:}\begin{aligned} \frac{\partial}{\partial t}\left[\frac{1}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}(t_r)}\right] = \frac{-1}{[Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}]^2} \frac{\partial}{\partial t} [Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}]\\ = \frac{-1}{[Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}]^2} \left[c\frac{\partial R}{\partial t} – \overrightarrow{\boldsymbol{R}} \cdot \frac{\partial \overrightarrow{\boldsymbol{v}}}{\partial t}-\overrightarrow{\boldsymbol{v}} \cdot \frac{\partial \overrightarrow{\boldsymbol{R}}}{\partial t}\right] ,\ldots\left(\mathrm{11}\right)\\ \end{aligned}∂∂t[1Rc−R→⋅v→(tr)]=−1[Rc−R→⋅v→]2∂∂t[Rc−R→⋅v→]=−1[Rc−R→⋅v→]2[c∂R∂t−R→⋅∂v→∂t−v→⋅∂R→∂t],…(11)
Solving each term in eq. (11) using eq. (8-10);
c ∂ R ∂ t = c 2 ( 1 − ∂ t r ∂ t ) = c 2 [ − R → ⋅ v → R c − R → ⋅ v → ] R → ⋅ ∂ v → ∂ t = R → ⋅ a → ( t r ) R c R c − R → ⋅ v → v → ⋅ ∂ R → ∂ t = − v 2 ∂ t r ∂ t = − v 2 R c R c − R → ⋅ v → c ∂ R ∂ t = c 2 1 − ∂ t r ∂ t = c 2 − R → ⋅ v → R c − R → ⋅ v → R → ⋅ ∂ v → ∂ t = R → ⋅ a → ( t r ) R c R c − R → ⋅ v → v → ⋅ ∂ R → ∂ t = − v 2 ∂ t r ∂ t = − v 2 R c R c − R → ⋅ v → {:[c(del R)/(del t)=c^(2)(1-(delt_(r))/(del t))=c^(2)[(- vec(R)* vec(v))/(Rc- vec(R)* vec(v))]],[ vec(R)*(del vec(v))/(del t)=( vec(R)* vec(a)(t_(r))Rc)/(Rc- vec(R)* vec(v))],[ vec(v)*(del vec(R))/(del t)=-v^(2)(delt_(r))/(del t)=(-v^(2)Rc)/(Rc- vec(R)* vec(v))]:}\begin{gathered} c\frac{\partial R}{\partial t}= c^2\left(1- \frac{\partial t_r}{\partial t}\right) = c^2 \left[\frac{-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}}\right]\\ \overrightarrow{\boldsymbol{R}} \cdot \frac{\partial \overrightarrow{\boldsymbol{v}}}{\partial t}= \frac{\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{a}}(t_r) Rc}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}}\\ \overrightarrow{\boldsymbol{v}} \cdot \frac{\partial \overrightarrow{\boldsymbol{R}}}{\partial t}= -v^2 \frac{\partial t_r}{\partial t}= \frac{-v^2Rc}{Rc-\overrightarrow{\boldsymbol{R}} \cdot\overrightarrow{\boldsymbol{v}}} \\ \end{gathered}c∂R∂t=c2(1−∂tr∂t)=c2[−R→⋅v→Rc−R→⋅v→]R→⋅∂v→∂t=R→⋅a→(tr)RcRc−R→⋅v→v→⋅∂R→∂t=−v2∂tr∂t=−v2RcRc−R→⋅v→
Using the above calculated terms in eq. (11) and using eq. (10) and eq. (11) in eq. (6) we get:\
∂ A → ∂ t = q 4 π ϵ 0 c [ R c a → ( R c − R → ⋅ v → ) 2 − v → ( R c − R → ⋅ v → ) 2 ( c 2 − R c 3 R c − R → ⋅ v → − R → ⋅ a → R c R c − R → ⋅ v → + R c v 2 R c − R → ⋅ v → ) ] , ⇒ ∂ A → ∂ t = q 4 π ϵ 0 c 1 ( R c − R → ⋅ v → ) 3 [ R c a → ( R c − R → ⋅ v → ) − v → c 2 ( R c − R → ⋅ v → ) + R c 3 v → + R → ⋅ a → R c − v 2 R c v → ] ⇒ ∂ A → ∂ t = q c 4 π ϵ 0 1 ( R c − R → ⋅ v → ) 3 [ ( R c − R → ⋅ v → ) ( R c a → − v → ) + ( c 2 − v 2 + R → ⋅ a → ) R c v → ] , … ( 12 ) ∂ A → ∂ t = q 4 π ϵ 0 c R c a → R c − R → ⋅ v → 2 − v → R c − R → ⋅ v → 2 c 2 − R c 3 R c − R → ⋅ v → − R → ⋅ a → R c R c − R → ⋅ v → + R c v 2 R c − R → ⋅ v → , ⇒ ∂ A → ∂ t = q 4 π ϵ 0 c 1 R c − R → ⋅ v → 3 R c a → ( R c − R → ⋅ v → ) − v → c 2 ( R c − R → ⋅ v → ) + R c 3 v → + R → ⋅ a → R c − v 2 R c v → ⇒ ∂ A → ∂ t = q c 4 π ϵ 0 1 R c − R → ⋅ v → 3 ( R c − R → ⋅ v → ) ( R c a → − v → ) + ( c 2 − v 2 + R → ⋅ a → ) R c v → , … 12 {:[(del vec(A))/(del t)=(q)/(4piepsilon_(0)c)[(Rc vec(a))/((Rc- vec(R)* vec(v))^(2))-( vec(v))/((Rc- vec(R)* vec(v))^(2))(c^(2)-(Rc^(3))/(Rc- vec(R)* vec(v))-( vec(R)* vec(a)Rc)/(Rc- vec(R)* vec(v))+(Rcv^(2))/(Rc- vec(R)* vec(v)))]”,”],[=>(del vec(A))/(del t)=(q)/(4piepsilon_(0)c)(1)/((Rc- vec(R)* vec(v))^(3))[Rc vec(a)(Rc- vec(R)* vec(v))- vec(v)c^(2)(Rc- vec(R)* vec(v))+Rc^(3) vec(v)+ vec(R)* vec(a)Rc-v^(2)Rc vec(v)]],[=>(del vec(A))/(del t)=(qc)/(4piepsilon_(0))(1)/((Rc- vec(R)* vec(v))^(3))[(Rc- vec(R)* vec(v))((R)/(c) vec(a)- vec(v))+(c^(2)-v^(2)+ vec(R)* vec(a))(R)/(c) vec(v)]”,”dots(12)],[]:}\begin{aligned} &\frac{\partial \overrightarrow{\boldsymbol{A}}}{\partial t}= \frac{q}{4 \pi \epsilon_0 c}\left[\frac{Rc\overrightarrow{\boldsymbol{a}}}{\left(Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}\right)^2}-\frac{\overrightarrow{\boldsymbol{v}}}{\left(Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}\right)^2}\left(c^2- \frac{Rc^3}{Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}-\frac{\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{a}}Rc}{Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}+\frac{Rcv^2}{Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}}\right)\right],& \\ &\Rightarrow\frac{\partial \overrightarrow{\boldsymbol{A}}}{\partial t}=\frac{q}{4 \pi \epsilon_0 c}\frac{1}{\left(Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}\right)^3}\left[Rc\overrightarrow{\boldsymbol{a}}(Rc-\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{v}})-\overrightarrow{\boldsymbol{v}}c^2(Rc-\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{v}})+Rc^3\overrightarrow{\boldsymbol{v}}+\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{a}}Rc-v^2Rc\overrightarrow{\boldsymbol{v}}\right]&\\ &\Rightarrow\frac{\partial \overrightarrow{\boldsymbol{A}}}{\partial t}=\frac{qc}{4 \pi \epsilon_0}\frac{1}{\left(Rc-\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{v}}\right)^3}\left[(Rc-\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{v}})(\frac{R}{c}\overrightarrow{\boldsymbol{a}}-\overrightarrow{\boldsymbol{v}})+(c^2-v^2+\overrightarrow{\boldsymbol{R}}\cdot\overrightarrow{\boldsymbol{a}})\frac{R}{c}\overrightarrow{\boldsymbol{v}}\right], \ldots\left(\mathrm{12}\right)\\&\\ \end{aligned}∂A→∂t=q4πϵ0c[Rca→(Rc−R→⋅v→)2−v→(Rc−R→⋅v→)2(c2−Rc3Rc−R→⋅v→−R→⋅a→RcRc−R→⋅v→+Rcv2Rc−R→⋅v→)],⇒∂A→∂t=q4πϵ0c1(Rc−R→⋅v→)3[Rca→(Rc−R→⋅v→)−v→c2(Rc−R→⋅v→)+Rc3v→+R→⋅a→Rc−v2Rcv→]⇒∂A→∂t=qc4πϵ01(Rc−R→⋅v→)3[(Rc−R→⋅v→)(Rca→−v→)+(c2−v2+R→⋅a→)Rcv→],…(12)
Now introduce a new vector given as:
u → ≡ c R ^ − v → u → ≡ c R ^ − v → vec(u)-=c hat(R)- vec(v)\overrightarrow{\boldsymbol{u}} \equiv c \hat{\boldsymbol{R}}-\overrightarrow{\boldsymbol{v}}u→≡cR^−v→
Then, eq. (5) and eq. (12) may be combined to obtain the electric field as;
⇒ E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 [ ( c 2 − v 2 ) u → + R → × ( u → × a → ) ] , … ( 13 ) ⇒ E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 c 2 − v 2 u → + R → × ( u → × a → ) , … 13 {:=> vec(E)( vec(r)”,”t)=(q)/(4piepsilon_(0))(R)/(( vec(R)* vec(u))^(3))[(c^(2)-v^(2)) vec(u)+ vec(R)xx( vec(u)xx vec(a))]”,”dots(13):}\begin{aligned} \Rightarrow \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, t) & =\frac{q}{4 \pi \epsilon_0} \frac{R}{(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{u}})^3}\left[\left(c^2-v^2\right) \overrightarrow{\boldsymbol{u}}+\overrightarrow{\boldsymbol{R}} \times(\overrightarrow{\boldsymbol{u}} \times \overrightarrow{\boldsymbol{a}})\right], \ldots\left(\mathrm{13}\right)\\ \end{aligned}⇒E→(r→,t)=q4πϵ0R(R→⋅u→)3[(c2−v2)u→+R→×(u→×a→)],…(13)
Calculation of B → B → vec(B)\mathbf{\vec{B}}B→ :
B → = ∇ → × A → A → = v → c 2 Φ B → = ∇ → × A → A → = v → c 2 Φ {:[ vec(B)= vec(grad)xx vec(A)],[ vec(A)=(( vec(v)))/(c^(2))Phi]:}\begin{aligned} & \overrightarrow{\boldsymbol{B}}=\vec{\nabla} \times \vec{A} \\ & \overrightarrow{\boldsymbol{A}}=\frac{\vec{v}}{c^2} \Phi \end{aligned}B→=∇→×A→A→=v→c2Φ
Therefore, we have,
∇ → × A → = 1 c 2 ∇ → × ( Φ v → ) = 1 c 2 [ Φ ( ∇ → × v → ) − v → × ( ∇ → Φ ) ] ∇ → × A → = 1 c 2 ∇ → × ( Φ v → ) = 1 c 2 [ Φ ( ∇ → × v → ) − v → × ( ∇ → Φ ) ] {:[ vec(grad)xx vec(A)=(1)/(c^(2)) vec(grad)xx(Phi vec(v))],[=(1)/(c^(2))[Phi( vec(grad)xx vec(v))- vec(v)xx( vec(grad)Phi)]]:}\begin{aligned} \overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{A}} & =\frac{1}{c^2} \overrightarrow{\boldsymbol{\nabla}} \times(\Phi \overrightarrow{\boldsymbol{v}}) \\ & =\frac{1}{c^2}[\Phi(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{v}})-\overrightarrow{\boldsymbol{v}} \times(\overrightarrow{\boldsymbol{\nabla}} \Phi)] \end{aligned}∇→×A→=1c2∇→×(Φv→)=1c2[Φ(∇→×v→)−v→×(∇→Φ)]
We have already obtained Φ , ∇ → Φ Φ , ∇ → Φ Phi, vec(grad)Phi\Phi, \vec{\nabla} \PhiΦ,∇→Φ and ∇ → × v → = − a → × ∇ → t r ∇ → × v → = − a → × ∇ → t r vec(grad)xx vec(v)=- vec(a)xx vec(grad)t_(r)\vec{\nabla} \times \overrightarrow{\boldsymbol{v}}=-\overrightarrow{\boldsymbol{a}} \times \vec{\nabla} t_r∇→×v→=−a→×∇→tr. Inserting these and with u → u → vec(u)\vec{u}u→, we get
B → ( r → , t ) = ∇ → × A → = − 1 c q 4 π ϵ 0 1 ( u → ⋅ R → ) 3 R → × [ ( c 2 − v 2 ) v → + ( R → ⋅ a → ) v → + ( R → ⋅ u → ) a → ] , … ( 14 ) B → ( r → , t ) = ∇ → × A → = − 1 c q 4 π ϵ 0 1 ( u → ⋅ R → ) 3 R → × c 2 − v 2 v → + ( R → ⋅ a → ) v → + ( R → ⋅ u → ) a → , … 14 vec(B)( vec(r),t)= vec(grad)xx vec(A)=-(1)/(c)(q)/(4piepsilon_(0))(1)/(( vec(u)* vec(R))^(3)) vec(R)xx[(c^(2)-v^(2)) vec(v)+( vec(R)* vec(a)) vec(v)+( vec(R)* vec(u)) vec(a)],dots(14)\overrightarrow{\boldsymbol{B}}(\overrightarrow{\boldsymbol{r}}, t) =\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{A}} = -\frac{1}{c} \frac{q}{4 \pi \epsilon_0} \frac{1}{(\overrightarrow{\boldsymbol{u}} \cdot \overrightarrow{\boldsymbol{R}})^3} \overrightarrow{\boldsymbol{R}} \times\left[\left(c^2- v^2\right) \overrightarrow{\boldsymbol{v}}+(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{a}}) \overrightarrow{\boldsymbol{v}}+(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{u}}) \overrightarrow{\boldsymbol{a}}\right] ,\ldots\left(\mathrm{14}\right)\\B→(r→,t)=∇→×A→=−1cq4πϵ01(u→⋅R→)3R→×[(c2−v2)v→+(R→⋅a→)v→+(R→⋅u→)a→],…(14)
E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 [ ( c 2 − v 2 ) u → + R → × ( u → × a → ) ] , … ( 13 ) E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 c 2 − v 2 u → + R → × ( u → × a → ) , … 13 {: vec(E)( vec(r)”,”t)=(q)/(4piepsilon_(0))(R)/(( vec(R)* vec(u))^(3))[(c^(2)-v^(2)) vec(u)+ vec(R)xx( vec(u)xx vec(a))]”,”dots(13):}\begin{aligned} \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, t) & =\frac{q}{4 \pi \epsilon_0} \frac{R}{(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{u}})^3}\left[\left(c^2-v^2\right) \overrightarrow{\boldsymbol{u}}+\overrightarrow{\boldsymbol{R}} \times(\overrightarrow{\boldsymbol{u}} \times \overrightarrow{\boldsymbol{a}})\right], \ldots\left(\mathrm{13}\right)\\ \end{aligned}E→(r→,t)=q4πϵ0R(R→⋅u→)3[(c2−v2)u→+R→×(u→×a→)],…(13)
From the above, it can be shown that
B → ( r → , t ) = 1 c R ^ × E → ( r → , t ) B → ( r → , t ) = 1 c R ^ × E → ( r → , t ) vec(B)( vec(r),t)=(1)/(c) hat(R)xx vec(E)( vec(r),t)\overrightarrow{\boldsymbol{B}}(\overrightarrow{\boldsymbol{r}}, t)=\frac{1}{c} \hat{\boldsymbol{R}} \times \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, t)B→(r→,t)=1cR^×E→(r→,t)
Therefore, the ( E → , B → ) ( E → , B → ) ( vec(E), vec(B))(\overrightarrow{\boldsymbol{E}}, \overrightarrow{\boldsymbol{B}})(E→,B→) due to an accelerated point charge are:
E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 [ ( c 2 − v 2 ) u → + R → × ( u → × a → ) ] B → ( r → , t ) = 1 c R ^ × E → ( r → , t ) E → ( r → , t ) = q 4 π ϵ 0 R ( R → ⋅ u → ) 3 c 2 − v 2 u → + R → × ( u → × a → ) B → ( r → , t ) = 1 c R ^ × E → ( r → , t ) {:[ vec(E)( vec(r)”,”t)=(q)/(4piepsilon_(0))(R)/(( vec(R)* vec(u))^(3))[(c^(2)-v^(2)) vec(u)+ vec(R)xx( vec(u)xx vec(a))]],[ vec(B)( vec(r)”,”t)=(1)/(c) hat(R)xx vec(E)( vec(r)”,”t)]:}\begin{gathered} \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, t)=\frac{q}{4 \pi \epsilon_0} \frac{R}{(\overrightarrow{\boldsymbol{R}} \cdot \overrightarrow{\boldsymbol{u}})^3}\left[\left(c^2-v^2\right) \overrightarrow{\boldsymbol{u}}+\overrightarrow{\boldsymbol{R}} \times(\overrightarrow{\boldsymbol{u}} \times \overrightarrow{\boldsymbol{a}})\right] \\ \overrightarrow{\boldsymbol{B}}(\overrightarrow{\boldsymbol{r}}, t)=\frac{1}{c} \hat{\boldsymbol{R}} \times \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, t) \\ \end{gathered}E→(r→,t)=q4πϵ0R(R→⋅u→)3[(c2−v2)u→+R→×(u→×a→)]B→(r→,t)=1cR^×E→(r→,t)
Observations:
  • • B → B → vec(B)\overrightarrow{\boldsymbol{B}}B→ due to a point charge is always perpendicular to E → E → vec(E)\overrightarrow{\boldsymbol{E}}E→\
  • •1st term in eq.(7) varies as 1 / R 2 1 / R 2 1//R^(2)1 / R^21/R2 : Velocity field or generalized Coulomb field. If v → = 0 v → = 0 vec(v)=0\overrightarrow{\boldsymbol{v}}=0v→=0 and a → = 0 a → = 0 vec(a)=0\overrightarrow{\boldsymbol{a}}=0a→=0, then E → E → vec(E)\overrightarrow{\boldsymbol{E}}E→ reduces to
E → = 1 4 π ϵ 0 q R 2 R ^ E → = 1 4 π ϵ 0 q R 2 R ^ vec(E)=(1)/(4piepsilon_(0))(q)/(R^(2)) hat(R)\overrightarrow{\boldsymbol{E}}=\frac{1}{4 \pi \epsilon_0} \frac{q}{R^2} \hat{\boldsymbol{R}}E→=14πϵ0qR2R^
  • •2nd term in eq.(7) varies as 1 / R 1 / R 1//R1 / R1/R : Acceleration field or Radiation field
Scroll to Top