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(RcRv)A(r,t)=vc2Φ(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=ΦAtB=×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=ΦAt
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[(RcRv)1]=qc4πϵ0(1)(RcRv)2(RcRv)Φ=qc4πϵ01(RcRv)2[(Rv)cR]
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(RcRv)2[(Rv)cR],(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=tRcR=c(ttr)
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_rR=ctr
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}(XY)=(X)Y+(Y)X+X×(×Y)+Y×(×X)(Rv)=(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=(Rxx+Ryy+Rzz)v(tr)=Rxdv dtrtrx+Rydv dtrtry+Rzdv dtrtrz=a(Rtr)
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=rw(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=(vxx+vyy+vzz)(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(vtr)
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=vv(vtr)
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=(vzyvyz)i^+(vxzvzx)j^+(vyxvxy)k^=(dvz dtrtrydvy dtrtrz)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}(Rv)=a(Rtr)+vv(vtr)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(AC)C(AB)R×(a×tr)=a(Rtr)tr(Ra)v×(v×tr)=v(vtr)v2tr
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}(Rv)=a(Rtr)+vv(vtr)a(Rtr)+tr(Ra)+v(vtr)v2tr(Rv)=v+(Rav2)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(RcRv)2[v+(Rav2)tr+c2tr]
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(RcRv)2[v+(c2v2+Ra)tr],(3)
To find t r t r vec(grad)t_(r)\vec{\nabla} t_rtr, 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}ctr=R=RR=12RR(RR)=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=Rv(Rtr)
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}ctr=1R[Rv(Rtr)+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}