Time Derivatives of the Lienard-Wiechert Potentials
📂ElectrodynamicsTime Derivatives of the Lienard-Wiechert Potentials
Overview

The time derivative of the Liénard-Wiechert potential is as follows.
∂t∂V∂t∂A=4πϵ0qc(
c−
⋅v)21(c2−c2∂tr∂t−v2+
⋅a)∂t∂tr=4πϵ0qc(
c−
⋅v)31[(
c+
⋅v)(
a/c−v)+c
v(c2−v2+
⋅a)]
Lemma
The time derivative of the retarded time is as follows.
∂t∂tr=
c−
⋅v
c=
⋅u
c
At this time, it is u=c
−v.
Proof
According to the definition of retarded time,
c(t−tr)=
Therefore, squaring both sides,
c2(t−tr)2=
2=
⋅
Differentiating both sides by t,
2c2(t−tr)(1−∂t∂tr)=2∂t∂
⋅
(1)
At this time, since it is
=r−w(tr),
∂t∂
=−∂t∂w=−∂tr∂w∂t∂tr=−v∂t∂tr
It is. Substituting this into (1) and rearranging,
⟹⟹c
(1−∂t∂tr)c
−c
∂t∂tr∂t∂tr=−
⋅v∂t∂tr=−
⋅v∂t∂tr=c
−
⋅vc
=
⋅uc
■
Proof
Liénard-Wiechert potential
Retarded time The potential for a point charge moving at speed tr from v to q is as follows.
V(r,t)A(r,t)=4πϵ01(
c−
⋅v)qc=4πμ0(
c−
⋅v)qcv=c2vV(r,t)
At this time,
=r−w(tr) is the vector from the retarded position to the observation point, and w(tr) is the retarded position, which is the position of the point charge at the retarded time.
Differentiating the scalar potential V by t,
∂t∂V=4πϵ0qc∂t∂(
c−
⋅v1)=4πϵ0qc∂tr∂(
c−
⋅v1)∂t∂tr
At this time, since it is dxd(f(x)1)=df(x)d(f(x)1)dxdf(x)=(f(x))2−1f′(x),
∂t∂V=4πϵ0qc(
c−
⋅v)2−1∂tr∂(
c−
⋅v)∂t∂tr=4πϵ0qc(
c−
⋅v)2−1(c∂tr∂
−∂tr∂
⋅v−
⋅∂tr∂v)∂t∂tr
Since it is
=r−w(tr),
∂tr∂
=−∂tr∂w(tr)=−v(tr)
Since it is ∣
∣=
=c(t−tr),
∂tr∂
=c∂tr∂t−c
Therefore,
∂t∂V=4πϵ0qc(
c−
⋅v)2−1(−c2+c2∂tr∂t+v2−
⋅a)∂t∂tr=4πϵ0qc(
c−
⋅v)21(c2−c2∂tr∂t−v2+
⋅a)∂t∂tr
Using the lemma and (a), it can be easily obtained.
∂t∂A=∂t∂(c2vV(r,t))=c21(∂t∂vV+v∂t∂V)=c21[∂tr∂v∂t∂trV+v4πϵ0qc(
c−
⋅v)21(c2−c2∂tr∂t−v2+
⋅a)∂t∂tr]=c21∂t∂tr[a4πϵ0qc
c−
⋅v1+v4πϵ0qc(
c−
⋅v)21(c2−c2∂tr∂t−v2+
⋅a)]=c21∂t∂tr4πϵ0qc(
c−
⋅v)21[a(
c+
⋅v)+v(c2−c2∂tr∂t−v2+
⋅a)]
According to the lemma,
∂t∂A=c21
c−
⋅v
c4πϵ0qc(
c−
⋅v)21[a(
c+
⋅v)+v(c2−c2
c
c−
⋅v−v2+
⋅a)]=c
4πϵ0qc(
c−
⋅v)31[a(
c+
⋅v)+c2v−c2v(
c
c−
⋅v)−v2v+(
⋅a)v]=c
4πϵ0qc(
c−
⋅v)31[(
c+
⋅v)(a−
cc2v)+v(c2−v2+
⋅a)]=4πϵ0qc(
c−
⋅v)31[(
c+
⋅v)(
a/c−v)+c
v(c2−v2+
⋅a)]
■