Top Banner
March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11
38

March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

Dec 22, 2015

Download

Documents

Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

March 2, 2011

Fill in derivation from last lecturePolarization of Thomson Scattering

No class Friday, March 11

Page 2: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

plane ,In nu

direction in vector Poynting

paper of plane into

n

Brad

Page 3: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

rE rad =

q

Rc2

r n × (

r n ×

r u ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥

If

Show that

rE rad =

q ˙ u

Rc 2 sinθ

Need two identities:

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

Page 4: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 5: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 6: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 7: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 8: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

Thomson Scattering

Rybicki & Lightman, Section 3.4

Page 9: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 10: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 11: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 12: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.
Page 13: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

Differential Cross-section

⎝ ⎜

⎠ ⎟unpol

=1

2

dσ (φ)

⎝ ⎜

⎠ ⎟pol

+dσ (π

2 )

⎝ ⎜

⎠ ⎟pol

⎣ ⎢ ⎢

⎦ ⎥ ⎥

=1

2ro

2 1+ sin2 φ( )

Average for2 components

dΩ=

1

2ro

2 1+ cos2 θ( )Thomson cross-sectionfor unpolarizedlight

wavesscattered

andincident between angle

Page 14: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 15: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 16: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.
Page 17: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.
Page 18: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

Incident light linearpolarized in direction #1

Incident light linearpolarized in direction #2

Unpolarizedincident light

No light along dipole axis

Page 19: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 20: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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)

Page 21: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(4) Scattered E-field is polarized

cos1

cos12

2

= percent polarization

(5) Thomson scattering σT is independent of frequency (“grey”)

Page 22: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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?

Page 23: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Page 24: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 25: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 26: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 27: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Page 28: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 29: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.
Page 30: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Page 31: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Page 32: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

2

222

0

22

02

222

3

2

sin)cos1(8

c

Ve

ddc

Ve

dd

dPP

222

3

2BcrP o

from part (a)

where

c

Vmc

ero

2

2

Page 33: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Page 34: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 35: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

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

Page 36: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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:

Page 37: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

Net result:

Page 38: March 2, 2011 Fill in derivation from last lecture Polarization of Thomson Scattering No class Friday, March 11.

(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

Rn Te

cmpcR

cm

n

T

e

18

224-

3x10 1

10 0.665 section -crossThomson

densityelectron

351051

1 cmR

nT

e