1. The document discusses Maxwell's equations and their solutions, which are travelling waves in free space determined by charges and currents that change over time. 2. It then considers a simple case of sources being a very small volume element, much smaller than the wavelength, where the solution takes a spherical wave form. 3. The document derives an expression for the retarded potentials generated by a moving point charge, known as the Lienard-Wiechert potentials. This involves finding the position and time at which the charge emitted the radiation intercepted at a given observation point and time.

1. 1
The (GENERAL) solution of
Maxwell’s equations
Then for very small r, outside the charge region but near it, in the sense of
“small” defined above, we can approximate in the solution “t-r/c” with “t”.
We have so far studied and understood the form of the Maxwell’s
equations in free space. We have understood that the solutions are
travelling waves in free space. That the form of the waves is
determined by what happens at the sources, i.e. the charges and the
currents changing with time. By extrapolating from simple cases we
have understood the relative directions of the wave propagation and
of the electric and magnetic fields. It is time that we understand how
the sources determine the travelling waves.
Let us now consider a simple case of sources: very small volume
element. Very small is a relative statement: I must specify very
small--- relative to what? Which is the scale I am considering? Well,
let us take as example harmonic waves of frequency ν:
the unit of length will be most naturally a wavelength, that is: λ = cτ
= c/ν. Well then, I consider a system of charges and currents
contained in a volume of linear dimensions much smaller than the
wavelength of the charge and current changes inside – which are of
course known. Then I may expect that the solution is some sort of a
spherical wave ψ.
Which will be the solution of the inhomogeneous wave equation
πρ
ψ
ψ 4
1
2
2
2
2
−=
∂
∂
−∇
tc
r
tf
r
crtf
tr
)()/(
),( ≈
−
=ψ
)/(
r
1
t)z,y,(x, crtf −=ψ

2. 2
Our job is to find the solution for the potentials once we know the
sources: distributions of charge and of current. We could give the
formula for the solution and just say that it is a solution, but we
would rather like to understand that it is a solution and how can we
find it. So now instead of starting from a s (source) and trying to
find f(t-r/c) we will do the vice versa, i.e. try to find which s can
generate a given f.
Well, we already know that the function ψ(r,t)=f(t)/r satisfies the
Poisson equation
)()(4]/)([22
rtfrtf δπψ ⋅−=∇=∇
The situation is similar to the Poisson equation generated by a nearly
pointlike source… a charge distribution over a tiny volume, such that
r
dV
V
∫
=≅Φ
−=Φ∇
ρ
πρ
r
Q
solutionwith42
Where s is a source of the time-variable potentials, it can be the
charge distribution or one of the components of the current
distribution. Carrying through the analogy, we have that
then our equation for ψ(t)=f(t)/r (for r small).
is a solution for r small of the equation
)()(42
rtf δπψ ⋅−=∇
),(42
trsπψ −=∇
dVtrstf
V
∫= ),()(
dVQ
V
∫= ρ
So, if such a charge distribution satisfies the Poisson eq. with ρ

3. 3
The solution for small r is then
∫
∫
==
V
V
dVtrs
r
dVtrs
tr '),'(f(t)where,
'),'(
),(ψ
We have found a solution valid for a small volume (in it the retardation
term r/c is small compared to the typical “period” of the wave) of the
time- varying equations. It turns out that it is a solution of the equation in
which we have neglected the term
2
2
2
1
tc ∂
∂
−
ψ
We also know that f(t) is an approximation of f(t-r/c) for r/c small.
It is at this point easy to generalize and write the general solution, also
because we have learnt previously that the effect of the term:
is to introduce the dependence on -r/c beside that on t
∫
∫
−
=
−
=Φ
2
2
2
12
12
2
12
12
)/,(
),(
)/,(
),(
V
V
dV
cr
crt
t
dV
r
crt
t
2
1
2
1
rJ
rA
r
r
ρ
2
2
2
1
tc ∂
∂
−
ψ
In these equations, Φ (A) are the values of the potentials at the point
r1 at the instant t. They have been generated by the sources at point r2
at the earlier time t – r12/c, and are called
Retarded Potentials

4. 4
They are called “Retarded Potentials” because the potential
at the point where I want to know it and when I want to know it is
retarded with respect to the time when it was generated.
How is a retarded potential calculated?
An essential condition for calculating a retarded potential is to
know beforehand the motion of all charges – at least before the
time at which we want to know the potential.
Suppose we want to know the potential at the point O (that we
choose as origin of the system of reference) and at the time t0.
We have to find the time each charge generated the potential.
We start sending light around backward in time at time t0 at the
point O and take frames of the positions of charges at each
spherical shell of thickness cδt: those which are within the shell
of thickness cδt will contribute; we keep moving the radius of
this anomalous light ray and keep adding up all the charge
contributions that we find.
I expect to find, in this process, a dependence on the velocity of
the charges.

5. 5
Retarded potentials of a moving charge
or Lienard-Wiechert potentials
Let β(r,t) J(r,t) be due to the motion in space of a charged
particle.
vrrJ ),(),( tt ρ=
Where v is the velocity of the charge, which we assume to have a
finite extension: a cube of side “a”.
What we have to do now is to use the recipe given before for
calculating the retarded potentials, i.e. have the spherical light shell
moving backwards in time and intercepting ONCE the trajectory
of the charge.

6. 6
The charge has to be spotted for times earlier and earlier as we move
away from the Origin O that we have located where we wanted to
compute the potentials. My light shell moving backwards in time in
frames of thickness δt will eventually get to intercept the moving
charge at time t0. Since the charge has a finite size, it will take some
finite time, however small, for the light to exit (at instant t1) from the
other side of the charge volume. And… during this time the charge will
have moved a bit.
This “bit” is a distance, which we can call D, to indicate that it is a
vector,
)()( 01 ttt −⋅−= vd
So… the charge has radiated of course all the time. BUT, the
radiated potential which has arrived to O at time t has been radiated
between times t1 and t0, which is a different time than a/c!

7. 7
The distance over which what we have called the light pulse travelling
back in time will have seen the moving charge (of side “a”) is L, with
)(
)()(
01
0101
ttcL
ttatt
r
aL
−⋅=
−⋅•+=−•+= rεv
r
v
where the 2 equations are respectively the side of the charge cube plus
the distance moved by the charge, and the distance moved by the light
pulse during their transient crossing.→
rε is the unitary vector that from the position of the charge points
towards the origin (i.e. where we are calculating the potentials).
The calculation of the potentials includes an integration over the charge
volume. But, as far as the integration of the 1/r term in the integral is
concerned, the distance being anything – and usually much larger than
the charge size- we can take 1/r out of the integral and consider it
constant.
a
qL
cRtRcRtR
aL
dVcrt
c
RtR
dV
r
tO
V
V
)/(
1
)/(
)/,(
)
(
1
r/c)-t,(
),(
2
−
=
−
⋅⋅
=
=⋅−
−
≅
==Φ
∫
∫
ρ
ρ
ρ
r
r
We find with the last formula
that the electrostatic
potential q/R is modified by a
factor L/a. To calculate it, we
use the two formulas at the
top of this page, to find that:
r
r
c
a
c
L
aL
ε
v
εv
•−
=⋅•+=
1
And we have found the kinematical
factor L/a which, as expected,
depends on the velocity of the
charge.

8. 8
We are now ready to calculate the retarded potential:
)/(
)/(
1)/(
1
),(
crt
rc
cRt
q
cRtR
tO
−•
−
−−
=Φ
εv
If we want to calculate the potential in a point r1; and indicating with
the sign ‘ “prime” the quantities computed at the time
ctt /||' 21 rr −−=
We obtain the equations:
Don’t forget this one! It will come back over
and over again. And it has a physical
meaning!
)'1('
'
)'1('
ε'β
β
A
ε'β
•−
⋅
=
•−
=Φ
r
q
r
q
These are the Potentials of Lienard and Wiechert:
to compute the fields, it is “sufficient” to compute their
derivatives wrt time and the three space coordinates.
Unfortunately, that “sufficient” which is a small problem from the
general theoretical point of view, is not a simple problem at all in
practice.

9. 9
Differentiating the Lienard-Wiechert potentials:
the Lienard-Wiechert fields
Writing the formulas for the fields once we have the expressions for
the potentials is an easy business:
)(
][][
')'('
1|)'('|
)'('
)
')'('
1(|)'('|
)'('1
)
')'('
1(|)'('|
c
t
tc
tq
c
t
tc
tq
dt
d
c
c
t
tc
q
εv
rr
v
B
εv
rr
v
εv
rr
E
•
−−
×∇=
•
−−
−
•
−−
−∇=
Now, how do we calculate gradients and curls and time derivatives??
Simple, gradients and curls are also derivatives wrt space coordinates.
Which space coordinates? We have r and r’ ...well, if we move r by
δx there will be some “direct” derivatives, i.e. derivatives of r wrt x
(y, z) but t’ and r’(t’) will also change. So, if I calculate the derivative
wrt t of, say , v’ , I will have to take into account the fact that
moving “t” I will also move the point r’ and time t’ where the charge
irradiates the potentials I am after. And I have to take into account
that indirect variation as easily as possible, That will be done by
writing:
dt
dt
dt
d
dt
d '
'
[...][...]
=
… and a similar treatment for gradient and curl.

10. 10
We know that
c
tt
||
'
r'r −
−= From this formula we easily calculate
''
'
|}'{|
εv
rr
•−=
−
dt
d
dt
dt'
is a bit more difficult to get, and we do it starting from the
same formula for t’ in this page:
c
dt
dt
cdt
dt
dt
dt
cdt
dt
tcdt
dt
ε'v
ε'v
ε'v
rr
•
−
=⇒=•−⋅
⋅•+=
∂
−∂
−= ⋅
'
1
1'
1'
1
1
'
'
'
1
1
'
'
|)'(|1
1
'
}{
}{
But… wait a minute! This is just the kinematical factor dependent
on the charge velocity that we obtained before with an euristic
calculation! It is actually dt’/ dt ! It is coherent with the way we
calculated it before: it is “how long does the charge radiate for
an interval dt at the observer to differentiate the potential”.
Now we must go on and differentiate t’ over space, i.e. compute
the gradient of t’. That will be done in precisely the same way as
we just did the ∂t’/ ∂t. We only have to derive the same formula
for t’ wrt x (y,z) instead of t as we just did.

11. 11
)(
}{
'
1
'
'
'
|)'(|11
'
c
c
t
t
tcc
t
ε'v
ε'
rr
ε'
•
−
−
=∇
∇
∂
−∂
−−=∇
⇒
⋅
In the formulas for the potentials, which we must derive, there
are the following primed variables – which we have to derive wrt
x,y,z and t: r’, v’,ε’. For them all we have to calculate the
gradient and the derivative wrt to time t. Let us go on with ε’.
|'|
)''(
|'|
'')''(
|'|
'
|'|
)')(''(
|'|
'
''
}{
rr
vεε'
rr
vεεv
rr
v
rr
rrεv
rr
rrε'
2
−
××
=
−
−•
=
=
−
−
−
−•
=
−
−
∂
∂
=
∂
∂
tt
We have now all the elements to calculate the most awkward
element in the formulas of the potentials that we have to derivate:
( )
)
'
'(')'(
|'|
'''
'''
|'|)
''
1('')
''
1(|'|
'
][][
cc
cccct
v
εvε'a
rr
vεε
vεa
rr
εv
εv
εv
rr
−−•⋅
−
−
×ו−
•
⋅−−
•
−⋅•−
•
−⋅−
∂
∂
=
==
In this formula we notice the appearance (for the first time!) of
the charge acceleration!!! We now have all the elements we need
to write down in detail the fields: E and B.