Moving charge (Homework 7-323)

1. Recall the point charge moving at a constant velocity. One interesting result is that the Liénard-Wiechert potentials can be written in terms of the “present” coordinates of the charge rather than its “retarded” coordinates. Demonstrate this is so.

Lienard Wiechert

Assume first the moving point charge q is somewhat extended over a (small) volume $V’$. The potential at point P can be written as
$V(r,t)=1/(4πϵ_0 )*∫ρ(r’,t_r )/R dV’$
When the volume V’ shrinks the above is
$V(r,t)=1/(4πϵ_0 )*1/R*∫ ρ(r’,t_r )dV’=1/(4πϵ_0 )*q/R(1-R ̂*v ⃗/c)$
$ρ(r’,t_r )→q$  and $dV’→dV/(1-R ̂*v ⃗/c)$    (see the proof in the book at page 451)   with $dV→0$
In the same way since
$J ⃗=qv ⃗$   one has $A ⃗(r ⃗,t)=μ_0/4π*(qv ⃗)/R(1-R ̂*v ⃗/c)$
Written as
$V(r ⃗,t)=1/(4πϵ_0 )*q/((R-R ⃗*v ⃗/c))$    and $A ⃗(r ⃗,t)=μ_0/4π*(qv ⃗)/((R-R ⃗*v ⃗/c))$
One can demonstrate that $V$ and $A ⃗$  depend just on r ⃗  and t (coordinates at the present)
One has
$R ⃗=r ⃗(t)-(r’ ) ⃗(t_r )$    and also $R=c(t-t_r )$
And assume for simplicity (remember $v ⃗$  is constant)
$(r’ ) ⃗(t_r )=v ⃗*t_r$    which means $(r’ ) ⃗(0)=0$
$R ⃗=r ⃗(t)-v ⃗t_r$   and $R=c(t-t_r)$
Square the above expression and equate them
$r^2-2r ⃗v ⃗*t_r+v^2 t_r^2=c^2 t^2-2c^2*t*t_r+c^2*t_r^2$
$(c^2-v^2 ) t_r^2+2(r ⃗v ⃗-c^2 t) t_r+(c^2 t^2-r^2 )=0$
$t_r=((c^2 t^2-r ⃗v ⃗ )-√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2)))/(c^2-v^2 )$
$(c^2 t^2-r ⃗v ⃗)-(c^2-v^2 ) t_r=√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2))$
The sign is – (minus) since
$t_r=t-r/c$    is retarded
We try to write the denominator from the blue equations differently
$(R-R ⃗*v ⃗/c)=c(t-t_r )-(r ⃗-v ⃗t_r )*v ⃗/c=c(t-t_r )-(r ⃗v ⃗)/c+(v^2 t_r)/c=$
$=1/c [(c^2 t-r ⃗v ⃗)-(c^2 t_r-v^2 t_r)]$
So that
$(R-R ⃗*v ⃗/c)=1/c*√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2))$
$V(r ⃗,t)=1/(4πϵ_0 )*qc/√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2 ))$    and
$A ⃗(r ⃗,t)=μ_0/4π*(qcv ⃗)/√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2 ) )=v ⃗/c^2 *V(r ⃗,t)$

DEPEND JUST ON $r ⃗$  and $t$ coordinates at present


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