March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11
Dec 22, 2015
March 2, 2011
Fill in derivation from last lecturePolarization of Thomson Scattering
No class Friday, March 11
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rE rad =
q
Rc2
r n × (
r n ×
r u ) ⎡ ⎣ ⎢
⎤ ⎦ ⎥
If
Show that
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rE rad =
q ˙ u
Rc 2 sinθ
Need two identities:
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rA ×
r B ×
r C ( ) =
r B
r A ⋅
r C ( ) −
r C
r A ⋅
r B ( )
r A −
r B
2=
r A
2+
r B
2− 2
r A ⋅
r C
r A ⋅
r B cosθ
So…
€
rn ×
r n ×
r u ( ) =r n
r n ⋅
r u ( ) −r u
r n ⋅
r n ( )
€
rn ⋅
r n =1
€
rn ⋅
r u =r n
r u cosθr n ⋅
r n =1
Now
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rn ×
r n ×
r u ( ) =r n
r n ⋅
r u ( ) −r u
r n ⋅
r n ( )
=r u cosθ
r n −r u
= ˙ u cosθr n −
r u
€
rn ×
r n ×
r u ( )2
= ˙ u cosθ r n −
r u 2
= ˙ u 2 cos2 θ + ˙ u 2 − 2 ˙ u 2 cos2 θ
= ˙ u 2 1 − cos2 θ( )
= ˙ u 2sin2θ
€
rE rad =
q ˙ u
Rc 2 sinθ
Magnitudes of E(rad) and B(rad):
sin2Rc
uqBE radrad
Poynting vector is in n direction with magnitude
2
4 radEc
S
242
22
sin4 cR
uqcS
The Dipole Approximation
radE
Niq
u
r
i
i
i
,...2,1 charges
velocities
positions
Generally, we will want to derive for a collection of particles
with
You could just add the ‘s given by the formulae derived previously, but then you would have to keep track of all thetretard(i) and Rretard(i)
radE
The Dipole Approximation
One can treat, however, a system of size L with “time scale for changes”tau where
c
L
so differences between tret(i) within the system are negligible
Note: since frequency of radiation
1
If c
L then L
c
or L
This will be true whenever the size of the system is smallcompared to the wavelength of the radiation.
The Dipole Approximation
Thomson Scattering
n
EM wave scatters off a free charge. Assume non-relativistic: v<<c.
E field
electron
e = charge
Incoming E fieldin direction
Incoming wave: assume linearly polarized. Makes charge oscillate.Wave exerts force:
rm
am
tEeF oo
sin
r = position of charge
Energy per second, Cross-sections:
time averaged power / solid angle
232
24
sin8 cm
Ee
d
dP o
32
24
3 cm
EeP oThe total power is obtained by integrating over all solid angle:
22 sinord
d
Cross section / solid angle,Polarized incoming light
22410665.0 cmTTotal cross-section,
Integrated over solid angle
Electron Scattering for un-polarized radiation
on wavespolarizatilinear of directions :, 21
wavescattered ofdirection n
aveincident w ofdirection k
Unpolarized beam = superposition of 2 linearly polarized beams with perpendicular axes
n
and between angle 1 kn
and between angle
22
2
2
1
Also, . is
and between angle thehence and
lar toperpendicu is then
plane in be to chose We
n
n
kn
Differential Cross-section
€
dσ
dΩ
⎛
⎝ ⎜
⎞
⎠ ⎟unpol
=1
2
dσ (φ)
dΩ
⎛
⎝ ⎜
⎞
⎠ ⎟pol
+dσ (π
2 )
dΩ
⎛
⎝ ⎜
⎞
⎠ ⎟pol
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=1
2ro
2 1+ sin2 φ( )
Average for2 components
€
dσ
dΩ=
1
2ro
2 1+ cos2 θ( )Thomson cross-sectionfor unpolarizedlight
wavesscattered
andincident between angle
€
dσ
dΩ=
1
2ro
2 1+ cos2 θ( )
NOTES:
d
d
203
8r
radiuselectron classical0 r
22 cos12
1
ord
d
)2(2
1 2ord
d
22 cos12
1
ord
d
is independent of frequency ofincoming wave
Total cross-section is same as unpolarized case.
Forward-Backward symmetry
incident
Polarization of scattered radiation
0
1
2
1
paper of planein component new into scatters
r
20
2
cos2
1
paper of planein component new into scatters
r
The scattered radiation is polarized, even if the incident radiation is unpolarized.
2cos:1 ratiointensity have will wavenew The
Incident light linearpolarized in direction #1
Incident light linearpolarized in direction #2
Unpolarizedincident light
No light along dipole axis
Degree of linear polarization:
I
I pol
0 cos1
cos12
2
Degree oflinear polarization
polarized intensity
total intensity
generally, get netpolarization
incidentwave,unpolarized
Π=100%complete polarization
Π=0 No net polarization in forward direction
Summary of Thomson Scattering
232
24 1
3 mcm
EeP o (1) So electrons are much more effective
scatterers than protons by factor of 4x10^6
2sind
dP(2) forward and backward scattering
identical
(3) σ(pol) = σ(unpolar) in our classical treatment of the electron, it has no preferred direction (i.e. no spin)
(4) Scattered E-field is polarized
cos1
cos12
2
= 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 systemsuch 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
2
4 radE
cdA
dPS
sec
1
area
energy
radE radiation part ofE field
drdA 2 so power/solid angle
22
4rE
c
d
dPrad
What is Erad? unnrc
eErad
2
Define unit vectors xa ˆˆ1
plane z-yin ˆ2a
normal to the planecontaining n, i.e. y-z plane
Particle has speed V┴ in circle of radius a, so
a
V
tyatxar sinˆcosˆ
tyVtxV
tyatxa
ru
cosˆsinˆ
cosˆsinˆ
angular velocity of particle
position of particleat time t
velocity of particle
tytxVu sinˆcosˆ acceleration of particle
at time t
Now unit vectors are related by
sinˆcosˆˆ
cosˆsinˆˆ
2 zya
zyn
So tataVunn
tataVun
sincosˆcosˆ)ˆ(ˆ
cosˆsincosˆˆ
21
21
tataVrc
e
unnrc
eErad
sincosˆcosˆ 212
2
€
dP
dΩ=
c
4πE rad
2r2
=e2V⊥
2ω 2
4πc 3 cos2 ωt + cos2 θ sin2 ωt( )
Power / steradian
as a function of t
The time-average power/steradian
23
222
cos18
c
Ve
d
dP
2
1sincos 22 tt since
(b) What is the polarization of the radiation as a function of θ ?
tataVrc
eErad sincosˆcosˆ 212
Recall the discussion of Stokes parameters: we write E in terms ofx- and y- components
)cos(
)cos(
22
11
tE
tE
y
x
Identify x-component with y-component with1a 2a
then
2 0
cos
21
2221
rc
eV
rc
eV
and the Stokes parameters are
cos2sin2
0cos2
cos1
cos1
2121
2121
222
21
222
21
AV
U
AQ
AI
2
2
rc
eVA
Where we have let
So 2222 VQUI (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 θ=π right-hand circular polarization
1a
1a
and2a
(c) What is the spectrum of the radiation?
tataVrc
eErad sincosˆcosˆ 212
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?
F
B
BVc
eF
rBc
e
BVc
eF
Lorentz force is balanced by centripetal force
Lorentz force
rm 2
So rmc
rBe 2
mc
eB
gyro frequency of particle in B field
(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
eEV
m
eEr
eEmr
2
Then
222
0
22
222
cos14
cos18
cEr
c
Ve
d
dP
Recall
d
dS
d
dP 4
2cES
So, differential cross section 220 cos1
2
1
r
d
d
3
8 2ord
d
d
Total cross-section Thomsoncross-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 ofoptical 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 wavecan be decomposed into 2 linearly polarizedwaves: one in the plane of the paper, one normalto the plane of the paper.These scatter into new waves in ratio cos2θ : 1Thus, the component normal to the paper dominatesthe other
Integrating over every θ along a given line of sight results in netpolarization which is normal to radial line: