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.

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

$V(r,t)=1/(4πϵ_0 )*1/R*∫ ρ(r’,t_r )dV’=1/(4πϵ_0 )*q/R(1-R ̂*v ⃗/c)$

Because

$ρ(r’,t_r )→q$ 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 $(r’ ) ⃗(0)=0$

Therefore

$R ⃗=r ⃗(t)-v ⃗t_r$ 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

$(R-R ⃗*v ⃗/c)=1/c*√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2))$

AND

$V(r ⃗,t)=1/(4πϵ_0 )*qc/√((c^2 t^2-r ⃗v ⃗ )-(c^2-v^2 )(c^2 t^2-r^2 ))$ 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