feb. 28, 2011 larmor formula: radiation from non- relativistic particles dipole approximation...
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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
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[ ]
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
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
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
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