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Transcript
We have now calculated all the intermediate derivatives which we need
for calculating the fields E and B. We will now write down the
procedure to follow and then the results. We start by making clear
HOW we will carry all the derivatives through.

   ( ) t 'cos t . 
t '
t '
1 A
1 A t '


c t
c t ' t
A
  A  (  A) t 'cos t  ( )  t '
t '
Note that some derivatives at constant t’ are not zero:
q  (ε' v' / c)
 ()t 'const 
v'ε' 2
2
| r  r ' | (1 
)
c
ε'  r  r '
and also
 [
] 0
x x | r  r ' |
Inserting all these various derivatives, gradients and curls in the
equations for the fields, we obtain at the end the analytical
expressions for E and
B, which turn out to be especially simple:
Advanced EM - Master
in Physics 2011-2012
1
The FIELDS OF Lienard-Wiechert
 q[a'(a'ε' )ε' ]  q  ε'(a'β' ) q  (ε'β' )(1   '2 )
E(r, t ) 

2
3
c | r  r ' | (1  β'ε' )
| r  r ' |2 (1  β'ε' )3
q(β'ε' )(1   ' )
 qa' q[ε'(a'β' )]
B(r, t ) 

ε'

| r  r ' |2 (1  β'ε' )3
c 2 | r  r ' | (1  β'ε' )3
2
Well,… after all, they do not look so simple. We should look at them
carefully and in detail. The first remark is that B is just the vector
product of
B  ε'E
ε’ and E.
This simple move already halves the complication.
Next step, we shall look only at
E, confident that whatever we find
will also apply to B. And, what we find is that E is the sum of two
terms, which exhibit significant differences.
The first term has a dependence on distance as
1/r:
 It is the radiation!!
And, second remark, which is also not a casual fact: it is proportional
to the charge acceleration
a’ (“retarded”, of course).
The other term has the usual 1/r2 dependence, and it is the static
field attached to the charge.
Advanced EM - Master
in Physics 2011-2012
2
Another expression (Actually, more than two formulations exist…)
for the fields exist, and is due to Feynman (FeynmanII 21-1;
Lovitch-Rosati pp 399-400). It can be obtained by leaving
unchanged the derivatives wrt t ( I mean, not resolving their
indirect dependence on the “retarded” parameters), but
calculating instead the gradient (derivatives wrt x, y, z).
[
]
ε' r ' d ε'
1 d 2ε'
E  q 2 
 2 2
2
r'
c dt r '
c dt
B  ε'E
( )
In this formulation, the acceleration of the charge is implicitly stated
inside the term
2
1 d ε'
c 2 dt 2
ε'
Since
is a unitary vector, its derivatives can only be orthogonal to its
direction. But its direction is the direction towards the observer.
Therefore, the electric field can only be orthogonal to the direction of
propagation of the wave. This property is also evident from the first
formulas given here: the two parts of the radiation term are both
orthogonal to the direction of sight. That is where the transversity of
the electromagnetic waves comes from.
 q[a'(a'ε' )ε' ]  q  ε'(a'β' ) q  (ε'β' )(1   '2 )
E(r, t ) 

2
3
c | r  r' | (1  β'ε' )
| r  r' |2 (1  β'ε' )3
Radiation term: acceleration fields
Advanced EM - Master
in Physics 2011-2012
Velocity fields
3
 q[a'(a'ε' )ε' ]  q  ε'(a'β' )
q  (ε'β' )(1   '2 )
E(r, t ) 

2
3
c | r  r' | (1  β'ε' )
| r  r' |2 (1  β'ε' )3
On which basis was decided that one term was the radiation and the
other an electrostatic type of field? It was decided on the basis of
the dependence from the distance “r”: this is 1/r in one case and 1/r2
in the other. Note moreover that B being equal to the vector product
of ε’ and E must be orthogonal to both.
Another remark: E and B are orthogonal….. The wave parts of E and B,
of course! For the other parts, customarily called the velocity fields,
they can take other directions! (Just think of a light ray passing near
a magnet….).
In the derivation of the formulas we can also see the reason why,
having the potentials a 1/r dependence, their derivatives had also a
part with 1/r dependence instead of only 1/r2. The reason can be
found in the form of the Lienard – Wiechert scalar potential:
q


r ' (1  β 'ε' )
In the case of an electrostatic potential, taking a derivative of this
potential to obtain the fields only gives an 1/r2 dependence. But in
this case taking derivatives wrt space gives two terms, one is the
derivative of 1/r (which is 1/r2) and the other 1/r times the
derivative of
1 / 1  β'ε'
This part is dependent on the movement of the charge, its
velocities and accelerations. Just imagine p.ex. a rotating charge
which, seen from a side, has position, velocity and acceleration
with sinusoidal behaviour.
Advanced EM - Master
in Physics 2011-2012
4
Main forms of the Lienard -Wiechert fields
We have already seen the expressions by Schwartz and Feynman
for the Lienard-Wiechert radiation electric field. They are
repeated here, together with the form used by the book of
Jackson, which is basically a simplification of Schwartz’s.
Schwartz:
 q[a'(a'ε' )ε' ]  q  ε'(a'β' ) q  (ε'β' )(1   '2 )
E(r, t ) 

2
3
c | r  r' | (1  β'ε' )
| r  r' |2 (1  β'ε' )3
Jackson:
E

q
ε'β'
q  '{( ε'β' )  β'}


r '2  '2 (1  β'ε' )3 cr ' (1  β'ε' )3
Feynman
[
]
ε' r ' d ε'
1 d 2ε'
E  q 2 
 2 2
2
r'
c dt r '
c dt
( )
Advanced EM - Master
in Physics 2011-2012
5
Velocity fields
Velocity fields are the fields generated by a charge in motion
without acceleration, i.e. a charge in uniform motion.
I choose my reference
system with the x axis
coincident with the
trajectory of the charge,
and put the origin where
the charge is at time 0.
Then, x = vt is the
equation of motion of my
charge.
I want to calculate the potentials and the fields generated –in point P and
at time t- by the motion of that charge. Let’s call R the retarded point
(i.e. the point where the charge was when it generated the fields which
are seen in P at time t), and A the actual point, i.e. the point where the
particle is at the time the potential and fields in P are calculated. I have
the usual 2 equations for the retarded time (speed of light of the spheric
light shell and position of the charge at the retarded time).
r'
c
r '2  ( x  vt' ) 2  y 2  z 2  c 2 (t  t ' ) 2
t'  t 
Solving this last equation in t’, we obtain…
(1   2 )t '  t 
x 1
c

c
( x  vt) 2  (1   2 )  ( y 2  z 2 )
r '  c(t  t ' )
Advanced EM - Master
in Physics 2011-2012
6
We need to calculate the potential
q

r ' (1  β'ε' )
q

r 'β'r'

For that, we start to compute
'
v2
r 'β'r'  c(t  t ' )   ' ( x  vt' )  c[t 
x  (1  2 )t ' ]
c
c
And then, we would like to replace t’ with its value as a function of t.
But we remember that we had just done that the previous page:
(1   2 )t '  t 
x 1
c

c
( x  vt) 2  (1   2 )  ( y 2  z 2 )
So, we replace this in that, and arrive to:
r'β'r'
1  2 
(
x  vt
1  2
)2  y
2
 z2
Which we can immediately use for the expression of the potentials as
functions of only the coordinates of the measuring point P, i.e. x, y, z
and t; and NOT of the retarded time or position.
 ( x, y , z , t ) 
A  β
1  2 
(
q
x  vt
1  2
)2  y
Advanced EM - Master
in Physics 2011-2012
2
 z2
7
In the derivation of this formula we have used only the EofM, but it is
obvious that the equations we end up with do smell of Lorentz
transformation, even if we have not studied it in detail: The
(
x  vt
1  2
)
expression really invites us to study the problem
with the Lorentz transformation.
In conclusion, we have calculated the Lienard-Wiechert potentials of a
charge in uniform linear motion. If the motion is directed along the “x”
axis, then
A=(βΦ,0,0), and it is easy to compute the gradient, curl and
time derivative of the potentials in order to find the fields, because in the
final formulae the potentials are given as a function of x,y,z,and t, not as
functions of x’, t’ and all that.
The formulas for the electric field components are:
Ex  
 1 Ax
q ( x  vt)



2
2
2
2
3
x c t {  ( x  vt)  y  z }
q
 ( x  vt)
 2

 ( x  vt) 2  y 2  z 2  2 ( x  vt) 2  y 2  z 2
Ey  


qy


2
2
2
2
3
y {  ( x  vt)  y  z }
q
y

 2 ( x  vt) 2  y 2  z 2  2 ( x  vt) 2  y 2  z 2
Ez  

qz

z {  2 ( x  vt) 2  y 2  z 2 }3
Advanced EM - Master
in Physics 2011-2012
8
Now, first thing to notice is that in the formulas the coordinates of
P enter as (x-vt), y and z. And, these are the coordinates of P in a
reference frame obtained by the previous one by displacement of
the origin along the x axis to bring the origin to the point “A”.
Second thing, the direction of the electric fields always point away
from “A”: in fact
E x x  vt

;
Ey
y
Ey
Ez

y
z
Therefore, if we make a drawing of the electric field, it seems to
stem out of the “actual” position of the charge. But in reality, as we
have seen, the fields were “generated” when the charge was in the
position “R”, well before “A”. So the fields seem to be flowing out of
the point “A”, and they move with the charge, being always
centered on it. And here comes the third observation: the electric
field – compared to the electrostatic one – has the component along
the direction of motion a factor γ2 smaller than in the
electrostatic case, while in the other two directions the fields are a
factor of γ larger than in the electrostatic case. The field status
for a (fast) particle is shown in the figures.
Advanced EM - Master
in Physics 2011-2012
9
Advanced EM - Master
in Physics 2011-2012
10
Advanced EM - Master
in Physics 2011-2012
11