Feb. 28, 2011

Larmor Formula: radiation from non-relativistic particles

Dipole ApproximationThomson Scattering

The E, B field at point r and time t depends onthe retarded position r(ret) and retarded time t(ret) of the charge.

Let

nc

u

tru

tru

ret

ret

1

onaccelerati )(

particle charged of velocity )(

0

0

€

rB (

r r , t) =

r n ×

r E (

r r , t)[ ]

€

rE (

r r , t) = q

(r n −

r β )(1−

r β 2)

κ 3R2

⎡

⎣ ⎢

⎤

⎦ ⎥

"VELOCITY FIELD"

∝1

R 2 Coulomb Law

1 2 4 4 4 3 4 4 4 +

q

c

r n

κ 3R× (

r n −

r β ) ×

r β ⎧ ⎨ ⎩

⎫ ⎬ ⎭

⎡ ⎣ ⎢

⎤ ⎦ ⎥

"RADIATION FIELD"

∝1

R

1 2 4 4 4 4 3 4 4 4 4

Field of particle w/ constant velocity 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 1c

u

Then E the RADIATION FIELD is

€

rE rad =

q

c

r n

κ 3R× (

r n −

r β ) ×

r β { } ⎡ ⎣ ⎢

⎤ ⎦ ⎥

simplifies to

=q

Rc2

r n × (

r n ×

r u ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥ and

€

rB rad =

r n ×

r E rad[ ]

plane ,In nu

direction in vector Poynting

paper of plane into

n

Brad

Magnitudes of E(rad) and B(rad):

sin2Rc

uqBE radrad

Poynting vector is in n direction with magnitude2

4 radEc

S

242

22

sin4 cR

uqcS

€

rE rad =

q

Rc2

r n × (

r n ×

r u ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥ASIDE: 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

€

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

Energy flows out along direction n

with energy dω emitted per time per solid angle dΩ

dASdt

dW

ergs/s/cm2 cm2

€

=q2 ˙ u 2

4πR2c 3 sin2 θ R2dΩdA

1 2 3

23

22

sin4 c

uq

dtd

dW

so

Integrate over all dΩ to get total power

€

P =dW

dt=

q2 ˙ u 2

4πc 3sin2 θdΩ∫

=q2 ˙ u 2

2c 31− μ 2( )dμ

−1

1

∫

3

22

3

2

c

uqP

LARMOR’S FORMULA

emission from a singleaccelerated charge q

3

22

3

2

c

uqP

NOTES

1. Power ~ q2

2. Power ~ acceleration 2

3. Dipole pattern: 2sindtd

dW No radiation emitted alongthe direction of acceleration.

Maximum radiation is emittedperpendicular to acceleration.

4. The direction of is determined by

If the particle is accelerated along a line, then the radiation is100% linearly polarized in the plane of

radE

u

nu and

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

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.

Can we use our non-relativistic expressions for radE

? yes.

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

Using the non-relativistic expression for E(rad):

i

i

i

irad R

unn

c

qE

)(2

If Ro = distance from field to system, then we can write

orad Rc

dnnE

2

)(

where i

iirqd

Dipole Moment

L

n

oR

Emitted PowerArea Vector x Poynting Recall

dt

dW

23

2

sin4 c

d

d

dP

power per solid angle

3

2

3

2

c

dP

power

DIPOLE APPROXIMATIONFOR NON-RELATIVISTICPARTICLES

What is the spectrum for this Erad(t)?

orad Rc

dnnE

2

)(

i

iirqd

Simplify by assuming the dipole moment is always in same direction,let

EtE

)( dtd

)(

then

oRctdtE

2

sin)()(

Let fourier transform of )(ˆ be )( dtd

)(E be E(t) of ansformfourier tr

then

ddetd ti )(ˆ)(

dedtd ti)(ˆ)( 2

€

ˆ E (ω) =sinθ

c 2Ro

× fourier transform ˙ d (t)( )

=sinθ

c 2Ro

−ω 2 ˆ d (ω)( )

= −ω 2 sinθ

c 2Ro

ˆ d (ω)

Then

Recall from the discussion of the Poynting vector:

)(4

2 tEc

dtdA

dW

areatime

energy

Integrate over time:

dttEc

dA

dW)(

42

(1)

Parseval’s Theorem for Fourier Transforms

dEdttE2

2 )(ˆ2)( (2)

Since E(t) is real

FT of E

)(ˆ

)(2

1)(ˆ

E

dtetEE ti

so22

)(ˆ)(ˆ EE

Thus (2)

0

22 )(ˆ4)( dEdttE

(See Lecture notes for Feb. 16)

Substituting into (1)

0

2

)(ˆ dEcdA

dW

Thus, the energy per area per frequency 2

)(ˆ

EcdAd

dW

substituting )(ˆsin)(ˆ

2

2

dRc

Eo

and dRdA 20

22

43

sin)(ˆ1d

cdd

dW

integrate over solid angle 24

3)(ˆ

3

8

dcd

dW

24

3)(ˆ

3

8

dcd

dW

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

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

Dipole moment:

tEm

ed

red

red

oo sin

2

tm

Eed o

o

o

sin2

2

2

2

o

oo m

Eed

so

Integrate twicewrt time, t

So the wave induces an oscillating dipolemoment with amplitude

What is the power?

Recall time averaged power / solid angle

tcm

Ee

c

d

d

dP

oo

23

2

2

24

23

2

sin4

sin

sin4

2

12sin

4

1sin 2

1212 t

too

o ttt

232

24

sin8 cm

Ee

d

dP o

So

(see next slide)

Aside:Time Averages

The time average of the signal is denoted by angle brackets , i.e.,

If x(t) is periodic with period To, then

€

x(t) = limT →∞

1

2Tx(t)dt

−T

+T

∫

€

x(t) =1

To

x(t)dtto

to +To∫

€

2sin2 x =1− cos2x

sin2 x =1/2 −1

2cos2x

To sin2 x =1

2x −

1

4sin2x

t1

t2

The total power is obtained by integrating over all solid angle:

3

2

3

2

c

dP

or

32

24

3 cm

EeP o

What is the Thomson Cross-section?

22 1

sec

8 cm

ergsE

cS o

Incident flux is given by the time-averaged Poynting Vector

Define differential cross-section: dσ scattering into solid angle dΩ

d

dS

d

dP

Power per solid angleerg /sec /ster Time averaged

Poynting Vectorerg/sec /cm2

cross-section persolid anglecm2 /ster

Thus

232

24

2

2

sin8

188

cm

Ee

d

dP

Ecd

dP

d

dd

dE

c

d

dP

o

o

o

so

since

we get 242

4

sincm

e

d

d

(polarized incident light)

22 sinord

d

2

2

mc

ero cmro

131082.2 classical electronradius

Thomson cross section

Integrate over dΩ to get TOTAL cross-section for Thomson scattering

dr

dd

d

o

T

22 sin

Thomson Cross-section2

3

8oT r

22410665.0 cmT

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

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