Feb. 28, 2011
Larmor Formula: radiation from nonrelativistic particles
Dipole Approximation
Thomson Scattering

The E, B field at point r and time t depends on
the retarded position r(ret) and retarded time t(ret) of the charge.
Let

(
u

r
)
0t
ret





u

r
(
t
)
0
ret

u


c
velocity
of
charged
particle
accelerati
on



1

n

r r
r r r
B(r,t)  n  E (r,t)


r
r2
r
r


rÝ
r r
(n  ) (1    ) q  n
 r r
E ( r ,t)  q
 
  3  ( n  )  
3 2
 R
c1 4 4R 442 4 4 4 43


1 4 44 2 4 4 43
"VELOCITY FIELD"
1
 2 Coulomb Law
R
Field of particle w/ constant velocity
"RADIATION FIELD"
1

R
Transverse field due to acceleration
Qualitative Picture:
transverse “radiation”
field propagates at velocity c
Radiation from Non-Relativistic Particles
For now, we consider non-relativistic particles, so
 u
 
1
c
Then E the RADIATION FIELD is
r
r
r r rÝ 
q  n
E rad   3  ( n   )    

c  R
simplifies to

 q r r rÝ
  2 n  ( n  u)
Rc


and
r
r r
Brad  n  E rad





In
u
,n
plane

B
into
of
paper
ra
dplane

Poynting
vector
in
n
directio
Magnitudes of E(rad) and B(rad):
  q

u
E

B

sin
ra
d
ra
d2
Rc

Poynting vector is in n direction with magnitude
22
c 2
S E
rad
4
 2
cq
u
S
 24sin

4
R
c
ASIDE:
r
 q r
r rÝ 
E rad   2 n  ( n  u) 
Rc

If
r
quÝ
E rad  2 sin
Rc
Show that

r
r r
rr r r
r r r
r
Need two identities: A  B  C  B
n n A
1 C  C A B
r r2 r2 r2
r r r r
 A  B A  B  2 A C A B cos
So…


 
 
  
r r rÝ r r rÝ rÝ r r
n  n  u  n n u  un n

Now
r rÝ r rÝ
n  u  n u cos
r r
n n  1


  


r
r rÝ r r rÝ rÝ r r
n  n  u  n n  u  un  n 
rÝ
rÝ rÝ
= u cos n  u
rÝ rÝ
= uÝcosn  u
r
r rÝ 2
rÝ rÝ2
n  n  u = uÝcos n  u
= uÝ2 cos2   uÝ2  2 uÝ2 cos2 
= uÝ2 1  cos2  
2
2
Ý
= u sin 
r
quÝ
E rad  2 sin
Rc

n
Energy flows out along direction
with energy dω emitted per time per solid angle dΩ
dW
SdA
dt
ergs/s/cm2
cm2
2 2
q uÝ
2
2

sin

R
d
1
2
3
2 3
4R c
dA
22
so

 2
dW
q
u
 3sin

dtd
4

c
Integrate over all dΩ to get total power
2 2
dW q uÝ
2
P

sin d
3 
dt 4c
2 2 1
q uÝ
2
 3  1  d
2c 1
2
2

2q u
P
3
3
c
LARMOR’S FORMULA
emission from a single
accelerated charge q
2
2

2q u
P
3
3
c
NOTES
1. Power ~ q2
2. Power ~ acceleration 2
3. Dipole pattern:
No radiation emitted along
the direction of acceleration.
dW 2

sin

dtd

Maximum radiation is emitted
perpendicular to acceleration.
4.

The direction of E
is determined by
ra d

u
If the particle is accelerated along a line, then the radiation is
100% linearly polarized in the plane of



uand
n
The Dipole Approximation
Generally, we will want to derive
with

Era d for a collection of particles

positions
r
i

velocities
u
i
charges
q
i
1
,2
,...
N
i

You could just add the Era d‘s given by the formulae derived
previously, but then you would have to keep track of all the
tretard(i) and Rretard(i)
One can treat, however, a system of size L with “time scale for changes”
tau where
L
 
c
so differences between tret(i) within the system are negligible
Note: since frequency of radiation
If
L
 
c
c
then


L
1

or

L
This will be true whenever the size of the system is small
compared to the wavelength of the radiation.
Can we use our non-relativistic expressions for
Let l = characteristic scale of particle orbit
u = typical velocity
tau ~ l/u
tau >> L/c  u/c << l/L
since l<L, u<c
 non-relativistic

Era d
? yes.
Using the non-relativistic expression for E(rad):



)

q
n

(
n

u
i
i
E


ra d
2
R
i c
i
If Ro = distance from field to system, then we can write






n
(n
d)
E
 2
rad
cR
o
where


d
qiri
i
Dipole Moment
L
Ro

n
dW
Recall

Poy
Vec
A
n
dt
Emitted Power


2
dP d
2
 3sin

d
4

c
power per solid angle
2
2d
P 3
3c
power
DIPOLE APPROXIMATION
FOR NON-RELATIVISTIC
PARTICLES
What is the spectrum for this Erad(t)?






n
(n
d)
E
 2
rad
cR
o


d
qiri
i
Simplify by assuming the dipole moment is always in same direction,
let

E
(t)E
then

d(t)d
sin



E
(t)
d
(t) 2
cR
o
Let fourier transform of

ˆ
d
(
t
)be
d
()

ˆ
fourier
ansfo
of
E(t)
be
E
(
)
then


i
t
ˆ
d
(
t)
e
d
)
d

 (



2

i

t


ˆ
d
(
t
)



d
(

)
ed




Then
sin 
ˆ
Ý(t)
E ( )  2  fourier transformdÝ
c Ro
sin 
 2
 2 dˆ ( )
c Ro
 


 2 sin ˆ
 2
d ( )
c Ro
Recall from the discussion of the Poynting vector:

energy
dW
c2
 E
(
t
)
time

area
dtdA
4
Integrate over time:


dW
c 2
 E
(
t)
dt
dA
4

(1)
Parseval’s Theorem for Fourier Transforms 

2





ˆ
E
(
t
)
dt

2

E

)d


(
2
(2)
Since E(t) is real

FT of E 
1
ˆ
E
(

) E
(t)eitdt
2

ˆ(
E
)
so
Thus (2) 
(See Lecture notes for Feb. 16)
ˆ
ˆ
E
(

)
E
(


)
2
2

2



0
ˆ
E
(
t
)
dt

4

E

)d


(
2

Substituting into (1) 
dW ˆ 2

c
E
(

)d


dA0
Thus, the energy per area per frequency 
substituting
2

sin

ˆ
ˆ
E
(

)
2 d
(

)
c
R
o
dW ˆ 2

c
E
(

)
dAd

and
dA

R
d





2
0
2
dW
1
4
2
ˆ

d
(
)
sin
3
d
d

c
integrate over solid angle
2
dW
8

4ˆ
3
d
(

)
d
3
c
2
dW
8

4ˆ
3
d
(

)
d
3
c
NOTE:
1. Spectrum ~ frequencies of oscillation of d (dipole moment)
2. This is for non-relativistic particles only.
Thomson Scattering
Rybicki & Lightman, Section 3.4
Thomson Scattering
EM wave scatters off a free charge. Assume non-relativistic: v<<c.




n
E field
e = charge
electron
Incoming E field
in direction 

Incoming wave: assume linearly polarized. Makes charge oscillate.
Wave exerts force:
 
F eEo s inot

ma


mr
r = position of charge
Dipole moment:
so
Integrate twice
wrt time, t
 
d  er
 
d  er
2

 e 
d
Eo sinot
m
2
 e


E
o
 2
d



sin

t
o
m


 o
So the wave induces an oscillating dipole
moment with amplitude
 e2E 
o
do  2 
m
o
What is the power?
Recall time averaged power / solid angle


2
dP d
2
 3sin

d 4
c
4 2
2
eE
 2
o sin
 2
sin
ot
3
m 4
c
 
1
t
2 1
sin
t
t
s
2
in
t
(see next slide)
o
t
1
4
2
o
2
o
So
1
2
42
o
23
eE 2
dP

sin

d
8

m
c
Aside:
Time Averages
The time average of the signal is denoted by angle brackets , i.e.,
1 T
x(t)  lim
x(t)dt

T  2T
T
If x(t) is periodic with period To, then

1
x(t) 
To
2sin2 x  1 cos2x
1
sin x
 1/2  cos2x
2
2
To sin 2 x 
t2
1
1
x  sin2x
2
4
t1

t o To
to
x(t)dt
The total power is obtained by integrating over all solid angle:
2
2d
P
3
3c
or
e4Eo2
P 2 3
3mc
What is the Thomson Cross-section?
Incident flux is given by the time-averaged Poynting Vector
c2 ergs
1
S
E
o
2
8
sec
cm

Define differential cross-section: dσ
scattering into solid angle dΩ
dP
d

S
d

d

Power per solid angle
erg /sec /ster
Time averaged
Poynting Vector
erg/sec /cm2
cross-section per
solid angle
cm2 /ster
Thus
so
since
dP
c 2 d

Eo
d 8
d
d dP 8 1

d d c Eo2
e 4 Eo2
dP
2

sin

2 3
d 8m c
we get
4
d
e
2
sin

24
d
m
c
d
r2sin
2

o
d

classical electron
radius
2
e
ro  2
mc
(polarized incident light)
Thomson cross section

13
r

2
.
82

10
cm
o
Integrate over dΩ to get TOTAL cross-section for Thomson scattering
d
T  d
d
2
ro2 sin
d
8 2
T  ro
3
Thomson Cross-section


0
.
6

1
6
cm
0
5

24
2
T
NOTES:
1. Thomson cross-section is independent of frequency.
Breaks down when hν >> mc2, can no longer ignore relativistic effects.
2. Scattered wave is linearly polarized in ε-n plane
Electron Scattering for un-polarized radiation
Unpolarized beam = superposition of 2 linearly polarized beams with
perpendicular axes




,
:
direc
of
line
pol
on
1
2

n

d
of
s
irecti
catte
wav

 

We
chose
1 to
be
in
n
kplane
k

direction
of
inciden
ave


 then

is
perpendicu
lar
n
to
2

angle
b1
etween
and
n



and
hence
the
angle
between
2and
n

angle
between
n
and
k
. Also,

is




2
2



Differential Cross-section
d 
1 d( ) 
 
 

d unpol 2 
 d pol
1 2
 ro 1 sin 2  
2

Thomson
cross-section
for unpolarized
light
d( 2 )  
 
 
 d pol 

Average for
2 components
d 1 2
2
 ro 1 cos  
d 2


angl
betwee
inci
eand
d

scattered
waves