Chapter 11 Radiation
11.1 Dipole Radiation
11.2 Point Charges
11.1 Dipole Radiation
11.1.1 What is Radiation?
11.1.2 Electric Dipole Radiation
11.1.1 What is Radiation?
The source of EM waves is from the radiation of accelerating
charges and changing currents.
The signature of radiation is the irreversible flow of energy
away from the source.
Assume that the source is localized
near the origin.
The total power passing out through
the surface at radius r is
1
P(r )   S  da 
 ( E  B)  da
0
The power radiated is Prad  lim P(r )
r 
For static fields, E  1N , B  1N ,
r
Prad  lim [
r 

r 2N
r

N  2,   da  4r 2 ,
 r2 ]  0
 Static field does not radiate.
11.1.1 (2)
But, Jefimenko’s equations:
 


 
( r , t r ) ˆ  ( r , t r ) ˆ J( r , t r )
1
E( r, t ) 
[
R
R  2 ]d

2
4 0
cR
R
c R
 
 
 

 0 J( r , t r ) J( r , t r ) ˆ
B( r, t ) 
[

]  Rd

2
4
cR
R



In the and J terms, E  1 , B  1 ,
r
r
 
1

  ( E  B)  da  O( 2  4r 2 )  O(1)
r 
r 
r
These terms are responsible for EM radiation
11.1.2 Electric Dipole Radiation
q( t )  q 0 cos(t )

dq
I ( t )  zˆ  q 0  sin( t )zˆ
dt

P( t )  P0 cos(t )zˆ P0  q 0 d
R
R
R   r 2  rd cos   (d / 2)2

V( r, t ) 
1 q 0 cos[( t  R  / c)] q 0 cos[( t  R  / c)]
{

}
4 0
R
R
approximation 1:
R   r(1 
d
cos )
2r
d  r
1 1
d
 (1  cos )
R r
2r
cos[( t  R  / c)]  cos[( t  r / c) 
d
cos ]
2c
d
d
 cos[( t  r / c)] cos( cos )  sin[ ( t  r / c)] sin( cos )
2c
2c
11.1.2 (2)
approximation 2:
c
d 

[i.e., d  ]
cos[ (t  R / c)]  cos[ (t  r / c)]
d
2c
cos  sin[ (t  r / c)]
P0 cos  
1
V( r, , t ) 
{ sin[ ( t  r / c)]  cos[( t  r / c)]}
4 0 r
c
r
[static li mit ,   0,
V
P0 cos 
]
2
4 0 r
c
[ d] [i.e., r    d]

radiation zone
approximation 3: r 
P0  cos 
V( r, , t )  
(
) sin[ ( t  r / c)]
4 0 c r
11.1.2 (3)

  0 J( r, t r )
A
d

4
R
 
0 d  q0 sin[ (t  R / c)] zˆ
A(r , t ) 
dz
2d

4 2
R
order of d
Keep only zeroth order of R ; i.e., R  r
[ approximation 1: d  r, approximation 2: d  c /  ]

 P
A( r, , t )   0 0 sin[ ( t  r / c)]zˆ
4r
11.1.2 (4)
V
1 V ˆ
rˆ 

r
r 
P
1
r

r
sin 
r
  0 {cos (  2 sin[ ( t  )]  cos[( t  )]) rˆ  2 sin[ ( t  )]ˆ }
4 0 c
c
rc
c
c
r
r
P0 2 cos 
r

(
) cos[( t  )]rˆ [  approximation 3]
2
r
c
4 0 c
V 
zˆ  cos  rˆ  sin  ˆ

 0 P0 
A
sin[ ( t  r / c)](cos rˆ  sin  ˆ )
4r

 0 P0 
A

cos[( t  r / c)](cos rˆ  sin  ˆ )
t
4


 0 P0 2 sin 
A
 E   V 

(
) cos[( t  r / c)]ˆ
t
4
r
11.1.2 (5)
 1 
A r ˆ
  A  [ ( rA  ) 
]
r r

1  0 P0 

sin[ ( t  r / c)] 
 (
){(  sin ) sin[ ( t  r / c)] 
cos }ˆ
r
4
r
r

 0 P0  
sin 
 (
){ sin  cos[( t  r / c)] 
sin[ ( t  r / c)]} ˆ
4r c
r
approximat ion 3
0 P0 2 sin 
B   A  
(
) cos[ (t  r / c)]ˆ
4 c
r
B
E
c


B
[Recall   E   ]
t
11.1.2 (6)
The energy radiated by an oscillating electric dipole is
determined by the Poynting vector:
 1  
 0 P0 2 sin 
S  ( E  B) 
{
(
) cos[( t  r / c)]}2 rˆ
0
c 4
r
2

 0 P0 4 sin2 
 S  (
) 2 rˆ
2
32  c
r
The total power radiated is
2 4
2
2

 0 P0 4
  0 P0  sin  2
 P    S  da 
r sin dd 

2
2
12 c
32  c
r
11.1.2 (7)
Example 11.1
(a) Why the sky is blue during the day ?
Answer :
•Sun’s radiation is white-light.
•The absorption of the white-light
by the atmosphere atoms is more
on higher frequency.

•The reradiation S  4 sin2 
so more on the blue than the red.
 The sky is blue .
11.1.2 (8)
(b)Why, during the sunset, the sky is more red ?
Longer, more scattering, especially for
Blue lights.
 more red-lights reach.
11.2 Point Charges
The fields of a point charge q in arbitrary motion
 
E( r, t ) 


  
q
R
2
2 
  3 (c  v )u  R  ( u  a )
4 0 ( R  u )
Velocity field
 
1ˆ  
B( r, t )  R  E( r, t )
c


u  cRˆ  v
Acceleration field
Radiation field
The Poynting vector

 1  

 
1 
1
2ˆ
ˆ
ˆ

S  ( E  B) 
E  ( R  E ) 
E R  ( R  E )E
0
0c
0c

11.2 (2)
The radiation energy is the stuff that detaches itself from the
charge and propagates off to infinity.

 
Srad  lim  S  da
A  A
4R 2

1
 Only terms of S  2 remain
R
1 1
(the terms  3 , 4  0 )
R R

That is, only the terms of E  1 survive.
R
11.2 (3)
 


E( r, t )  E vel.  Eacc.

q
R
1
2
2 
E vel. 
  3 (c  v )u  2
4 0 ( R  u )
R

  
q
R
1


E acc . 
R

(
u

a
)


4 0 ( R  u )3
R

 E rad.

E rad .  Rˆ

1 2 ˆ
Srad . 
E rad .R
0c


11.2 (4)
if


v( t  t r )  0, u  cRˆ

0q ˆ  ˆ 
q

ˆ
ˆ



E rad. 
R  (R  a ) 
( R  a )R  a 
2
4R
4 0 c R

1 0q 2 2
 2 ˆ
ˆ
Srad. 
(
) a  (R  a ) R
 0 c 4R


 0 q 2 a 2 sin2  ˆ

( 2 )R
2
16  c R
The total power radiated
2 2
2

  0 q a sin  2
P   Srad  da 
R sin  d d

2
2
16  c
R
 0q 2a 2
P
6c
Larmor formula

R
11.2 (5)
if

v0
 

ˆ
Rv
Ru
N g  N t (1 
)  Nt
c
Rc
Rate of gun shoot Rate of strike on the target
R
 t  tr 
c
 

R  robserver  rcharge
 
R  RR



rc 
t
1 R 1 1 1  rc  
Rˆ  v
1 

  (  )  R  R  (  )  
t r
c t r c 2 R  R  t r
t r 
c

t
Rˆ  v
1
t r
c
 

ˆ
Δt
Rv
Ru
N

N
(
)

N
(
1

)

N
N g  Δ t r  1  N tΔ t
g
t
t
t
Δ tr
c
Rc
11.2 (6)
The power radiated by the particle
 
 
Ru 
 Ru 1 2 2
P
S

d
a

E rad .R sin  d d
rad


Rc
Rc  0 c
dΩ
The power radiated by the particle into R 2dΩ per solid angle dΩ
 

  
q
R
dP R  u 1 2 2

E rad R
Erad 
  3 R  (u  a )
dΩ
Rc  0 c
4 0 ( R  u )
   2
2
2
2
R  (u  a )
  R
   2
q
q
R


 (R  u )
R  (u  a ) 
ˆ

2
5
2
2   6
u

c
R

v
16   0 ( Rˆ  u )
 0 c ( 4 0 ) ( R  u )
The total power radiated is
dP
P
sin  d d
dΩ
 2
0q 2  6 2 v  a
P
(a 
)
6c
c

1
v2
1 2
c


Rˆ

a
Lienard`s generalization
of the Larmor formula
11.2 (7)
Example 11.3
q

Rˆ
 
v, a
Solution: 

u  cRˆ  v
dP
?
dΩ
P?

 
u  a  cRˆ  a
 2
  2
2
2
ˆ
ˆ
ˆ
R  (u  a )
dP
q
q c R  (R  a )




2
5
d 16  0 ( Rˆ  u )
16 2  0 (c  Rˆ  v)5
2
ˆ  (R
ˆ  a )  (R
ˆ  a )R
ˆ  a
R
 2



ˆ
ˆ
R  ( R  a )  a 2  ( Rˆ  a )2  2( Rˆ  a )Rˆ  a

 a 2  ( Rˆ  a )2  a 2 (1  cos 2 )  a 2 sin 2 

c  Rˆ  v  c  v cos   c(1   cos )

v
c
11.2 (8)
0q 2a 2
dP
q 2c 2
a 2 sin2 
sin2 


2
5
5
d 16  0 c (1   cos )
162 c (1   cos )5
dP 0 q 2 a 2 2
 0

sin 
2
d 16 c
d dP
 1
( )
0
d d max
2 sin  m cos  m (1   cos  m )  sin 2  m ( 5) (  sin  m )  0
5
cos m   cos2 m   sin2 m  0
2
3
cos m  (1  sin2 m )  0
2

m 
0
2

m 
 1
m  0 (but m  0)
 
EB
11.2 (9)
The total power radiated
0q 2a 2
dP
sin2 
P   dR 
sin  d d

2
5
d
16 c (1   cos )
x  cos   0 q 2 a 2 1 1  x 2
dx

5

1
8c
(1  x )
0q 2a 2  6
P
6c
a＞0
a＜0
4
(1  2 )3
3

1
2 2
(1   )
accelerati ng
decelerati ng
Same radiation result.
11.2 (10)
producting radiation (or photon)
(bremsstrahlung)
e
e
e
(annilation)
Prob. 11.16

v

a
synchrotron radiation