March 2, 2011
Fill in derivation from last lecture
Polarization of Thomson Scattering
No class Friday, March 11



In
u
,n
plane

B
into
of
paper
ra
dplane

Poynting
vector
in
n
directio
r
 q r
r rÝ 
E rad   2 n  ( n  u) 
Rc

If
r
quÝ
E rad  2 sin
Rc
Show that
Need two identities:

So…

r
r r
r r r
r r r
r r
n n A
1 C  C A B
A BC B
r r2 r2 r2
r r r r
A  B A  B  2 A C A B cos

 
 
  
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
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
c 2
S E
rad
4
22
 2
cq
u
S
 24sin

4
R
c
The Dipole Approximation
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)
The Dipole Approximation
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
or
1


L
This will be true whenever the size of the system is small
compared to the wavelength of the radiation.
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
Energy per second, Cross-sections:
time averaged power / solid angle
42
o
23
eE 2
dP

sin

d
8

m
c
The total power is obtained by integrating over all solid angle:
Cross section / solid angle,
Polarized incoming light
Total cross-section,
Integrated over solid angle
e4Eo2
P 2 3
3mc
d
r2sin
2

o
d



0
.
6

1
6
cm
0
5

24
2
T
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
NOTES:
d
d
is independent of frequency of
incoming wave
Total cross-section is same as unpolarized case.
Forward-Backward symmetry
d

12
2
r

cos

o1
d

2

8 2
 r0
3
r

class
ele
ra
0

d

12
2
r

cos

o1
d

2


incident
d 1 2
 ro 1 cos2  
d 2


d
 12
r
2
)
o(
d
2

Polarization of scattered radiation
The scattered radiation is polarized, even if the incident radiation is unpolarized.


scatters
into
new
compon
ipl
nof
an
pa
1
1

r
0
2


scatters
into
new
compo
ipl
nof
an
pa
1

r
cos

2
2
2
0

2
T
new
wa
hav
he
wi
int
ra
1
:
c
Incident light linear
polarized in direction #1
Incident light linear
polarized in direction #2
No light along dipole axis
Unpolarized
incident light
Degree of linear polarization:
Degree of
linear polarization

I pol
I
polarized intensity
total intensity
2
1

cos




0
2
1

cos

generally, get net
polarization
Π=100%
complete polarization
incident
wave,
unpolarized
Π=0 No net polarization in forward
direction
Summary of Thomson Scattering
(1)
(2)
42
e
E
1
o
P
 23 2
3
m
c m
dP
2

sin

d

(3) σ(pol) = σ(unpolar)
So electrons are much more effective
scatterers than protons
 by factor of 4x10^6
forward and backward scattering
identical
in our classical treatment of the electron,
it has no preferred direction (i.e. no spin)
(4) Scattered E-field is polarized
2
1

cos



2
1

cos

= percent polarization
(5) Thomson scattering σT is independent of frequency (“grey”)
R&L Problem 3.2: Cyclotron Radiation
A particle of mass m, charge e, moves in a circle of radius a
at speed V┴ << c.
Define x-y-z coordinate system
such that n is in the y-z plane
(a) What is the power emitted per unit solid angle in a direction n,
at angle θ to the axis of the circle?
(a) What is the power emitted per unit solid angle in a direction n,
at angle θ to the axis of the circle?
Consider a point at distance r from the origin
magnitude of
poynting vector
dP
S 
dA
c

Erad
4
2
dA

r
d

energy
1
area
sec
2
so power/solid angle
dP
c
22
 E
r
rad
d
4

Er a d radiation part of
E field
 

e 

E


n

u
ra
d2 n
c
r
What is Erad?
ˆ1
ˆ
a
x
Define unit vectors
normal to the plane
containing n, i.e. y-z plane
ˆ
a
in
y
z
plane
2
Particle has speed V┴ in circle of radius a, so

V
a

angular velocity of particle

position of particle
ˆ
ˆ
r

a
x
cos
t

a
y
sin
t
at time t
 

u
r
ˆsin
ˆcos


a

x

t
a

y

t
ˆsin
ˆcos


V
x

t
V
y

t


velocity of particle




acceleration of particle
ˆ
ˆ


u


V
x
cos
t

y
sin
t

at time t
Now unit vectors are related by
So
ˆ
ˆ
ˆ
n
y
sin


z
cos

ˆ
ˆ
ˆ
a


y
cos


z
sin

2
 
 


ˆ
ˆ
ˆ


n

u


V
a
cos
sin
t

a
cos
t

1
2


ˆ
ˆ
ˆ
ˆ


n

(
n

u
)


V
a
cos
t

a
cos
sin
t
1
 
2

e  

E

n

n

u
rad 2
cr
e
ˆ
ˆ
a

2 V

cos

t
a
cos

sin

t

1
2
cr
Power / steradian
dP
c
2 2

E rad r
d 4
e 2V2 2
2
2
2

cos

t

cos

sin
t 
3 
4c
as a function of t
The time-average
power/steradian

dP
e 2V2 2
2

1

cos

3
d
8c

since

1
cos t  sin t 
2
2
2
(b) What is the polarization of the radiation as a function of θ ?

e
Erad   2 V aˆ1 cos t  aˆ 2 cos  sin t 
c r
Recall the discussion of Stokes parameters: we write E in terms of
x- and y- components
Ex  1 cos(t  1 )
E y   2 cos(t  2 )
Identify x-component with
then
aˆ1
y-component with
aˆ2
eV
1   2
rc
eV
 2   2 cos
rc
1  0
2 

2
and the Stokes parameters are


 A1  cos  
I   12   22  A 1  cos2 
Q   12   22
2
U  2 1 2 cos1  2   0
V  2 1 2 sin 1  2   2 A cos
Where we have let
So
I 2  U 2  Q2  V 2
(1) the radiation is 100% elliptically polarized
(2) principle axes of polarization ellipse are
(2) At θ=0 left-hand circular polarization
θ=π/2 linear polarization along
aˆ
θ=π right-hand circular polarization 1
 eV 
A 2 
 rc 
aˆ1 and aˆ2
2
(c) What is the spectrum of the radiation?

e
Erad   2 V aˆ1 cos t  aˆ 2 cos  sin t 
c r
Only cos ωt and sin ωt terms 
spectrum is monochromatic at frequency ω
(d) Suppose the particle is moving in a constant magnetic field, B
What is ω, and total power P?
B
F
 e  
F  V B
c
Lorentz force

e
F  V B
c
e
 rB
c
Lorentz force is balanced by centripetal force
So
erB
 m 2 r
c
eB

mc
 m 2 r
gyro frequency of particle in B field
P 
eV 

8c 2
2
2

dP
d
d
2 2

2
d

(
1

cos
 ) sin d
 
0
0
2 e 2V2 2

3 c2
2 2 2 2
P  ro c  B
3
where
e2
ro 
m c2
V
 
c
from part (a)
(e) What is the differential and total cross-section for Thomson scattering
of circularly polarized radiation?
Equate the electric part of the Lorentz force = eE
with centripetal force = mrω2
and use our expression for <dP/dΩ> for a circularly moving charge
m r 2  eE
eE
r 
m
eE
V 
m
Then
dP
e 2V2 2
2

1

cos

2
d
8c


r02cE 2

1  cos2 
4
Recall


dP
d
 S
d
d
So, differential cross section
Total cross-section
cE 2
S 
4
d 1 2
 r0 1  cos 2 
d 2

8ro2
d
 
d 
d
3

Thomson
cross-section
Rybicki & Lightman Problem 3.4
Consider an optically thin cloud surrounding a luminous source.
The cloud consists of ionized gas.
Assume that Thomson scattering is the only important source of
optical depth, and that the luminous source emits unpolarized radiation.
(a) If the cloud is unresolved, what polarization is observed?
If the angular size of the cloud is smaller
than the angular resolution of the detector,
the polarization of the different parts of the
cloud cancel
 no net polarization
R=1pc
(b) If the cloud is resolved, what is the direction of the polarization
as a function of position on the sky?
Assume only “single-scattering” – i.e. each photon scatters only once.
θ
At each θ, the incident, unpolarized wave
can be decomposed into 2 linearly polarized
waves: one in the plane of the paper, one normal
to the plane of the paper.
These scatter into new waves in ratio
cos2θ : 1
Thus, the component normal to the paper dominates
the other
Integrating over every θ along a
given line of sight results in net
polarization which is normal to
radial line:
Net result:
(c) If the central object is clearly seen, what is an upper bound for the
electron density of the cloud, assuming that the cloud is homogeneous?
To see the central object,
 1
  ne T R
ne  electrondensity
 T  T homsoncross- section 0.66510-24 cm2
R  1 pc  3x1018 cm
1
5
3
  1  ne 
 5 10 cm
R T