Transcript Mar4

March 4, 2011
Turn in HW 5; Pick up HW 6
Today: 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 system
such that n is in the y-z plane
 
dP
c
22
 E
r
rad
d
4


e  

E

n

n

u
rad 2
cr
e
ˆ
ˆ
a

2 V

cos

t
a
cos

sin

t

1
2
cr
b. Polarization
(c) What is the spectrum of the radiation?


 e
ˆ
ˆ


E


V
a
cos
t

a
cos
sin
t
rad

1
2
2
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 2

m
r
c
eB

mc
2

m

r
gyro frequency of particle in B field
P 
dP
d
d
2 2
eV 

8c2
2 2


2
d

(
1

cos
)sin
d
 
0
0
2eV 

3 c
2 2 2

2
22 2 2
P r

B
oc

3
where
e2
ro 
mc 2
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
mr 
2
 eE
eE
r 
m
eE
V 
m
Then
2 2
dP e2V
2


1

cos

2
d 8
c




2
r02cE
2

1cos

4

Recall
dP
d

S
d

d

So, differential cross section
Total cross-section
2
cE
S
4
d

12
2
r
1

cos

0
d

2

2
8

r
d

o


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
“Geometrical dilution”
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

n

R
e
T
n

electron
density
e


Thomson
cross
section

0.665

10
cm
24
2
T
18
R

1
pc

3x10
cm

1
5 
3

1

n
 
5

10
cm
e
R
T

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”
Power radiated = energy lost

Frad
 Prad
r
r
 F

u
rad
123
work done
by field on
charge

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
mu2
T~
Prad
2 2

2eu
P
~ 3
rad
3c
in the dipole approximation
so
2
3
mc
u
T~ 2  

2
e u
3
u
~
orbit
scale
for
particle

time
t
P

u
2
2
3
r
2
e
2
1
e
2
3mc 1
o
where



Let

3
2
3
m
3
c
c
m
3
c
c
2e2 
r

clas
ele
ra
o

So


23
~
electro
light
crossi
~
time
10
s

F
can
be
ignored
if
T

t
rad
P
For the electron,


23
t


10
s
P
So on time scales longer than 10-23 s, the radiation reaction is
a small perturbation

What
is
F
?
rad
we wrote
 

F

u

P
rad
rad

2
2

 2
e
u

F

u
 3
P
rad
rad
3
c

This can’t be right because P(rad) depends on u
only, and in the above equation, F(rad) depends
 but also 
not only on 
u
u
Instead, consider the work done by F(rad) in time interval (t1,t2)
  t2 2e2  
u

dt
F

u


dt
u
rad
t
t 3c3
1
1
t2
2 t2
2



t2
2e
2
e



 3 dtuu 3 uu
t1
3c t1
3c
Integrate by parts
If the motion is periodic, then

 
 so


u

u
(
t
)

u

u
(
t
)
2
1
Collecting terms:
=0
2




2
e



F

u
u
dt

0
ra
d
3 


3
c


t
1
t
2
so
2


2e 

F

u
rad
3c3



m
u
2 e2
 3
3mc
Abraham-Lorentz
Force
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: Relativistic
Abraham-Lorentz-Dirac-Langevin Force: Relativistic, quantum mechanical scales
The equation of motion for the charged particle,
with some external force (i.e. the EM wave) applied is then
r
rÝ
rÝ
m u  m uÝ  Fext

F=ma

radiation
force
Abraham-Lorentz
Eqn. of Motion
external force

The radiation force is a damping term which will damp out
the oscillations of the particle induced by the external force
Fext
Radiation from Harmonically Bound Particles
Up to now, we’ve been talking about the motion of free electrons
in an oscillation E, B field
Now we will consider a particle which is harmonically bound to
a center of force, i.e. where

 

2
F


k
r


m
r
o
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
 






x

x

x

0
2
o
restoring force
Radiation force

Frad  m u
2e2

3mc3
F=ma

F


m

r
2
o
But

is very small, so
 x
is small
The harmonic solution in first order is (ignoring
at t=0,
x) and requiring that
x(0)  x o
xÝo (0)  0
So that we have we keep the cosine term and
x(t)  cos o t 

Further, this implies
 x oe
2




x


ox
 o 2t / 2
cos( o t)
so the equation of motion becomes


2
2



x

x

x

0
o
o
Let the solution

t
x
(
t)
e
Then the equation for α is







0
2 2
o
2
o
       0
2
So we need to solve
2
o
2
o


1


 


4

22


From the quadratic formula:
for
2
o

4
2
o
2
o
4
2
2
But
since
is
sma
ver
com
4
to

o
o


1


 
4

2
o
2 2
2
o
2


o

i
o
2
Let

2
o
so

 i
o
2
then the solution for x(t) can be written


 
t


i



o


t


i



o
2


x
o 2 
x
(
t
)

e

e

2




where x0 = arbitrary constant
= position at t=0
The spectrum will be related to the Fourier Transform of x(t)

1
i
t
ˆ(
x

) x
(
t)e
dt
2
0


x
1
1
o
 




4
2
i(



) 2
i(



)
o
o
ˆ()
x
becomes large at



and 

0
0
We are only interested in the physically valid ω>0 so
x
1
0
ˆ
x
(

)


4


i
(



)
2
0
Power per frequency
dW 8 ˆ 2
 3 d()
d 3c
4
8 2
2
ˆ()
 3 ex
3c
4
dW 8 4 e 2 x 02
1

d
3c 3 4 2    0 2   22
or


dW
2
2
1
 2 kx0 
2
2

d
   0    2
Initial potential energy


Line shape (Lorentz profile)
2
w
k

m
here

spri
con
0


dW
2
2
1
 2 kx0 
2
2

d
   0    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:









m
x


m
x

m
x

eE
e
2
0
F=ma
Harmonic
oscillator
restoring
force
i
t
0
Radiation
damping
force
Note: ω0 = frequency of harmonic oscillator
ω = frequency of incoming EM wave
External driving
force:
sinusoidally
varying E field
(1)
To find the solution, we let
x

Re
y
i

t
0
y
(
t
)
y
e
y  iy
To derive y0, substitute (2) into (1)
y   2 y
y  i 2 y
(1) becomes

3E 0 y
 y   y  i y 
m y0
2
2
0
3
eE
1
0
y

 2 2 3
0
m

i
0


(2)
So


y
(
t
)

y
e
i
(t

)
0
where
3






arctan
2
2





0

phase shift between
driving force and
particle oscillation
Re(y)  x(t)
y 0 i(t  ) i(t  )

e
e


2
So


x
(
t
)

x
cos(
t

)
0
eE 0
x0  
m
1

2

    
2 2
0
3
2
3






arctan
2
2



0

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
(3) “Resonance” at ω = ±ω0
Particle leads Fext
Particle lags Fext
Time averaged, total power in dipole approximation:
dipole oscillating with
frequency ω,
amplitude
e x0 
P 
3c 3
e 4 E 02
4

3m 2c 3  2   02 2   3 2
2
2
4
x0
Divide by time-averaged Poynting vector to get the cross-section
c 2
S E
8
 0

 ( )   T

where  T  T homson cross-section
4
2

    
2 2
0
3
2
8e 4
2

cm
3m 2c 4
 ( )
4

2
2
2
2
3
T
   0     
Note:
(1) ω>>ω0

(

)

 electron becomes unbound
T
(2)

0









Resonance
2


0


(

)(

)

(

)
2
0
0
0
0

T

2
(
)
2
2
 (
(2)2
o)

2
2e2
Looks like free-oscillation
2

2
Lorentzian
mc(
(2)2
o)
2
2
0
 ( )
4

2
2
2
2
3
T
   0     
(3) ω<<ω0

 
(
)

T


 0
4
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