March 4, 2011

Turn in HW 5; Pick up HW 6Today: Finish R&L Chapter 3

Next: Special Relativity

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

22

4rE

c

d

dPrad

tataVrc

e

unnrc

eErad

sincosˆcosˆ 212

2

b. Polarization

(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

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

(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“Geometrical dilution”

(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:

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

Rn Te

cmpcR

cm

n

T

e

18

224-

3x10 1

10 0.665 section -crossThomson

densityelectron

351051

1 cmR

nT

e

Radiation Reaction

Rybicki & Lightman Sections 3.5 & 3.6

When an accelerating charged particle radiates, it is losing energy. there must be a force acting on the particle

This is the “radiation reaction force”radF

Power radiated = energy lost

€

≡Prad

= −r F rad ⋅

r u

work doneby field oncharge

1 2 3 mass cm^2/sec^3 ergs/sec

When is this important? When the particle motion is significantly modified on the time-scale of interest

Define T=time scale of change in kinetic energy due to Frad

radP

muT

2

~

3

22

3

2~

c

uePrad

in the dipole approximation so

2

2

3

2

3~

u

u

e

mcT

Ptparticlefor scale-orbit time ~ u

u

Let1

2

32

3

e

mcc

r

mc

e

cmc

e o

3

21

3

2

3

22

2

3

2

where

radiuselectron classicalor

So s2310~ timecrossinglight electron ~

Prad tF T if ignored becan

For the electron, stP2310

So on time scales longer than 10-23 s, the radiation reaction is a small perturbation

we wrote

? isWhat radF

radrad PuF

radrad Pc

ueuF

3

22

3

2

This can’t be right because P(rad) depends ononly, and in the above equation, F(rad) depends

not only on but also

u

u

u

Instead, consider the work done by F(rad) in time interval (t1,t2)

2

1

2

1

2

1

2

1

3

2

3

2

3

2

3

2

3

2

3

2

t

t

t

t

t

t

t

t

rad

uuc

euudt

c

e

uuc

edtuFdt

Integrate by parts

If the motion is periodic, then

)()( 12 tuutuu so = 0

Collecting terms:

03

22

1

3

2

dtuu

c

eF

t

t

rad

so

3

2

3

2

3

2

3

2

mc

eum

uc

eFrad

This is the radiation force in a time averaged sense for periodic motion,so one needs to be careful how one uses this expression.

Valid for “classical” non-relativistic electron

Abraham-Lorentz-Dirac Force: RelativisticAbraham-Lorentz-Dirac-Langevin Force: Relativistic, quantum mechanical scales

Abraham-LorentzForce

The equation of motion for the charged particle, with some external force (i.e. the EM wave) applied is then

€

m r ̇ u − τm

r ˙ ̇ u ( ) =r F ext

Abraham-LorentzEqn. of Motion

F=ma radiation force

external force

The radiation force is a damping term which will damp outthe oscillations of the particle induced by the external forceFext

Radiation from Harmonically Bound Particles

Up to now, we’ve been talking about the motion of free electronsin an oscillation E, B field

Now we will consider a particle which is harmonically bound to a center of force, i.e. where

rmrkF o

2

so that it oscillates sinusodially around the origin

The important applications of this situation include:• Rayleigh Scattering• Classical treatment of spectral line transitions in atoms

First, we consider

Undriven, harmonically bound particles damped by radiation force

then

Driven, harmonically bound particles, where forced oscillations occur due to incident radiation

Undriven, harmonically bound particle

For oscillations along the x-axis, the equation of motion is

0 2 xxx o

Radiation forceF=ma

restoring force

3

2

3

2

mc

e

umFrad

rmF o

2

But is very small, so x is small

The harmonic solution in first order is (ignoring ) and requiring that at t=0,

x

€

x(t) ∝ cos ωot( )

= xoe−ω o

2τt / 2 cos(ωot)Further, this implies

xx o 2

€

x(0) = xo

˙ x o(0) ≈ 0

So that we have we keep the cosine term and

so the equation of motion becomes

0 22 xxx oo

Let the solution tetx )(

Then the equation for α is

0222 oo

So we need to solve

€

2 + ωo2τα + ωo

2 = 0 for α

From the quadratic formula:224

2

42

1

2 ooo

2o

24 4 tocompared small very is sinceBut o

22

42

1

2 oo

2

2 ooi

Let 2o so oi

2

then the solution for x(t) can be written

oit

o

eex

txit

o

2

2

2)(

where x0 = arbitrary constant = position at t=0

The spectrum will be related to the Fourier Transform of x(t)

)(

1

)(

1

4

)(2

1)(ˆ

22

0

oo

o

ti

ii

x

dtetxx

)(ˆ x becomes large at 0 and 0

We are only interested in the physically valid ω>0 so

)(

1

4)(ˆ

02

0

i

xx

Power per frequency

223

4

2

3

4

)(ˆ 3

8

)(ˆ3

8

xec

dcd

dW

€

dW

dω=

8πω4

3c 3

e2x02

4π( )2

1

ω −ω0( )2

+ Γ2( )

2

or

€

dW

dω= 1

2 kx02

( )Γ

2

ω −ω0( )2

+ Γ2( )

2

Initial potential energy Line shape (Lorentz profile)

constant spring 20 mkwhere

€

dW

dω= 1

2 kx02

( )Γ

2

ω −ω0( )2

+ Γ2( )

2

The shape of the emitted spectrum is the Lorentz Profile

Γ = FWHM

FREE OSCILLATIONS OF A HARMONIC OSCILLATOR

Driven Harmonically Bound Particles

Now consider the forced oscillations of a harmonic oscillator, where the forcing is due to an incident beam of radiation:

tieeExmxmxm 020

F=ma Harmonicoscillatorrestoring force

Radiationdampingforce

External driving force: sinusoidally varying E field

Note: ω0 = frequency of harmonic oscillator ω = frequency of incoming EM wave

(1)

To find the solution, we let

yx Re tieyty 0)( (2)

To derive y0, substitute (2) into (1)

yiy

yy

yiy

2

2

(1) becomes

€

− 2y = −ω02y − iω3τy +

3E0

m

y

y0

320

20

0

1

im

eEy

So)(

0)( tieyty

where

20

2

3

arctan

phase shift between driving force andparticle oscillation

€

Re(y) = x(t)

=y0

2e i(ωt +δ ) + e−i(ωt +δ )

[ ]

So )cos()( 0 txtx

€

x0 = −eE0

m

1

ω2 −ω02

( )2

+ ω3τ( )2

20

2

3

arctan

Note:

(1) x(t) is an oscillating dipole, for charge q and frequency ω

(2) Phase shift, δ, is not symmetric about ±ω0

0

0

0

0

Particle leads Fext

Particle lags Fext

(3) “Resonance” at ω = ±ω0

Time averaged, total power in dipole approximation:

€

P =e2 x0

2ω4

3c 3

=e4 E0

2

3m2c 3

ω4

ω2 −ω02

( )2

+ ω3τ( )2

dipole oscillating withfrequency ω, amplitude

0x

Divide by time-averaged Poynting vector to get the cross-section

208

Ec

S

€

(ω) = σ T

ω4

ω2 −ω02

( )2

+ ω3τ( )2

€

where σ T = Thomson cross - section

= 8πe4

3m2c 4 cm2

€

(ω)

σ T

=ω4

ω2 −ω02

( )2

+ ω3τ( )2

Note:(1) ω>>ω0

T )( electron becomes unbound

(2) 0 Resonance 2

0

000020

2 2)())((

22

22

22

22

22

)()(

2

)()(2)(

o

o

T

mc

e

Looks like free-oscillationLorentzian

€

(ω)

σ T

=ω4

ω2 −ω02

( )2

+ ω3τ( )2

(3) ω<<ω04

0

)(

T

Rayleigh Scattering

Since ω<<ω0 the E-field appears static, and the driving force is effectively constant the electron responds directly to the incident field

ω4 dependance blue sky, red sun at sunrise and sunset

blue light scattered

Color at zenith at sunrise - sunset

Red (6500 A) 0.96 0.21

Green (5200 A) 0.90 0.024

Violet (4100 A) 0.76 0.000065

Jackson E&M