March 28, 2011
HW 7 due Wed.Midterm #2 on Monday April 4Today: Relativistic Mechanics
Radiation in S.R.
Fields of a Uniformly Moving Charge
If we consider a charge q at rest in the K’ frame, the E and B fields are
3
3
3
r
zqE
r
yqE
r
xqE
z
y
x
0
0
0
z
y
x
B
B
B
where
€
r 3 = ′ x 2 + ′ y 2 + ′ z 2( )
3 / 2
Transform to frame K We’ll skip the derivation: see R&L p.130-132
The field of a moving charge is the expression we derived from the Lienard-Wiechart potential:
€
rE = q
r n −
r β ( ) 1−
r β 2( )
r κ 3R2
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
n1
We’ll consider some implications
Consider the following special case:
(1) charged particle at x=y=z=0 at t=0 v = (v,0,0) uniform velocity
(2) Observer at x = z = 0 and y = b sees
€
Ex = −qvγt
γ 2v2t 2 + b2( )
3 / 2
Ey =qvb
γ 2v2t 2 + b2( )
3 / 2
E z = 0
yz
yx
EB
BB
0
What do Ex(t), Ey(t) look like?
Ex(t) and Ey(t)
(1) max(Ey) >> max (Ex) particularly for gamma >>1
(2) Ey, Bz strong only for 2t0
(3)
v0 b
t
)max(
)max(
x
y
E
EAs particle goes faster, γ increases,E-field points in y-direction more
Ex(t) and Ey(t)
(4) The observer sees a pulse of radiation, of duration
(5) When gamma >>1, β~1 so
(6) To get the spectrum that the observer sees, take the fourier transform of E(t) E(w)
We can already guess the answer
v
2
b
yyz EEB
The spectrum will be 2
)(ˆ
EcdAd
dW
where )(ˆ E is the fourier transform
€
ˆ E (ω) =1
2πEy (t)e iωtdt∫
=qγb
2πγ 2v2t 2 + b2
( )−3 / 2
e−iωtdt−∞
∞
∫
The integral can be written in terms of the modified Bessel functionof order one, K1
vvv)(ˆ
1
b
Kb
b
qE
The spectrum cuts off for
b
v~
Rybicki & Lightman give expressions for d
dW
and some approximate analytic forms -- Eqns 4.74, 4.75
Bessel Functions see Numerical Recipes
Bessel functions are useful for solving differential equations for systems with cylindrical symmetry
Bessel Fn. of the First Kind Jn(z), n integerBessel Fn. of the Second Kind Yn(z), n integer
Jn, Yn are linearly independent solutions of
€
z2 d2y
dz2+ z
dy
dz+ z2 − n2y( ) = 0
Modified Bessel Functions:
tosolutions are )(),( zKzI nn
€
z2 d2y
dz2+ z
dy
dz− z2 − n2
( )y = 0
Relativistic Mechanics
4-momentum Ump 0
where m0 = rest mass, i.e. the mass in the inertial reference frame in which the particle is at rest.
For particles with mass:
p
cEp
/where v0V
mp
22420
2
220
pccmE
cmpp
For photons: kpE
,
k
cp
/
Can show that p and E transform in the following way: (v in x-direction)
x
zz
yy
xx
pEE
pp
pp
cpp
v
vE2
where
2/1
2
2v1
c
Relativistic Mass
2
2
0
cv
1
m
m
For what value of v/c will the relativistic mass exceed its rest massby a given fraction f?
0
0
m
mmf
0.001 0.0140.01 0.140.1 0.421.0 0.8710 0.994100 0.999
At high beta, m >> m0
The invariant quantity is
€
E02 = m0c
2
E 2 − cp( )2
= E02or
since 2222 pcEcpE
where
v0
20
mp
cmE
4-acceleration
d
dUa
Newton’s Second Law
4-Force F
F=ma
d
dp
amF
0
For the Lorentz force
The corresponding 4-vector is
€
rF = e
r E + 1
c
r v ×
r B [ ]
UF
c
eF
So the equation of motion for a charge is
UF
cm
ea
0
UF
c
e
d
dp
BEc
E
c
e
c
BBE
BBE
BBE
EEE
c
e
UFc
e
d
dp
z
y
x
xyz
xzy
xzx
zyx
v
v
v
v
v
0
0
0
0
UU
vv
1000
0100
0010
0001
cc
BEc
E
c
eUF
c
e
d
dp
v
v
Does this make sense?
repeat
0
Ec
ed
dW
cd
dp
v
10
W = particle energy
sov Ee
dt
dWOK!
t, not tau
3,2,1
Bv
1
cEe
d
dp
so for μ=1
€
dpx
dt= e Ex +
r v ×
r B ( )
x[ ]
similarly for y,z
Emission from Accelerated Relativistic Particles
Basic Idea:
(i) We know how to calculate the emission from an accelerating particle if <v> << c Larmor result
(ii) We can transform to an inertial frame, K’, in which the particle is instantaneously at rest calculate the radiation field transform back to lab frame, K
Consider a particle with change q in its instantaneous rest frame K’
Let tddW in time emittedenergy total'
0pd
momentum=0 since the particle is at rest
In K frame:
tddt
WddW
Ptd
Wd
dt
dWP
So power
so emitted energy / time =P is a Lorentz invariant
In rest frame, the Larmor result:
23
2
3
2u
c
qP velocityu
or2
3
2
3
2a
c
qP
onacceleratia
In terms of components of the acceleration parallel and perpendicularto the direction of motion between frames, recall that we derived:
aa
aa2
||3
||
provided the particle is instantaneously at rest in K’
€
P = P =2q2
3c 3
r ′ a
2
=2q2
3c 3′ a ||2 + ′ a ⊥
2( )
=2q2
3c 3γ 4 a⊥
2 + γ 2a||2
( )
Angular Distribution of Radiation:
In the rest frame K’, consider emission of energy dW’ into solid angle dΩ’
How do dW’ and dΩ’ transform?
dW’
Since energy and momentrum form a 4-vector, dW transforms like dt
xpdWddW v
For photons, c
W
c
hp
so if we let cos
WddW 1
d Let
ddd
cos
ddd
cos
aberration formula
1
22 1
d
dso
Since dd
22 1
d
d
so...
d
Wd
d
dW 33 1
Now of course we are interested in the power, i.e. the energy/time:
d
Pd
ddt
dW 1
'
'
so how does dt’ transform?
There are two possibilities for relating dt and dt’
(i) tddt This is the interval as seen in K frame emitted power Pe
so
€
dPe
dΩ= γ 2 1+ β ′ μ ( )
3 d ′ P
d ′ Ω
can also write
€
dPe
dΩ=
1
γ 4 1− βμ( )3
d ′ P
d ′ Ω
(ii) 1tddt c.f. Doppler formula
This is better – it’s what you would actually measure as a stationary observer in K
received power Pr
d
Pd
d
dPr 44 1
or
d
Pd
d
dPr44 1
1
We’ll adopt this
Suppose the acceleration of the particle is caused by a forcehaving components
FF and ||with respect to the particle’s velocity (see R&L problem 4.14)
||
||
2||
||||
v
Ftd
pd
xdc
td
Edc
pd
dt
dpF
since dE’ =dx’ =0
and
F
td
pd
xdc
td
pddt
dpF
1
1
€
P =2
3
q2
c 3γ 4 a⊥
2 + γ 2a||2
( )
P =2
3
q2
c 3
1
m2F||
2 + γ 2F⊥2
( )So F
||F
is more effective than
in producing radiation
Beaming As β 1, photons which are isotropic in the rest frame are “beamed” forward
In K:
€
=cosθ ≈1 −θ 2
2
β = 1−1
γ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2
=1−1
2γ 2
1
γ 4 1 − βμ( )4 ≈
2γ
1+ γ 2θ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
4
so
so
d
Pd
d
dP4
221
2
Strongly peaked at θ=0
1
~
What happens to a dipole?
Recall that is the rest frame of the emitting particle, Larmor’s result had a dipole angular distribution
2
2
22
sin4 c
aq
d
Pd
in K’
angle between the acceleration and thedirection of emission
writing aaa
||2
24
22||
2
3
2
sin14
aa
c
q
d
dP
Working out the result is messy; see R&L Eqn. 4.101 and 4.103
Qualitative Picture
Transformation of the Equation of Radiative Transfer
How does the specific intensity Iν transform?
Recall the radiation density
ddvolenergyu /// We showed that
c
Iu so volumeenergydd
c
I/
Now, the energy per volume is also
€
f hν p2dpdΩ where f = phase space density of photons
f = # photons / dV’ where dV’ = dx’dy’dz’ dpx’ dpy’ dpz’
How does the phase space volume transform?
zyxzyx pdpdpddpdpdp
zdydxddxdydz
1
VddV so phase space volume is an invariant
But since f = Number of photons/dV
f is a Lorentz invariant
OK:
€
Iνc
dν dΩ = f hν p2dpdΩ
ddc
hfdd
c
I 3
43
sincec
hp
Invariant Lorentz 3
I
What about the source function Sν ?
Recall SI
ds
dI
so S transforms like I
Invariant Lorentz 3
S
Optical depth
e gives fraction of photons absorbed
so Invariant Lorentz
Absorption coefficient
K
θ
v
l
yk
l
c
l
l
2
sin
1
sin
l =distance perp. to v
2
kc
Both l and ky are perpendicular to v transform the same way
yk
lis Lorentz invariant
So Invariant Lorentz
Emission coefficient jν
Sj
Invariant Lorentz is
Invariant Lorentz is
3
S
So Invariant Lorentz 2
j
So, to solve a radiative transfer problem:
• set up problem in emitter rest frame• solve the equation of radiative transfer• transform the result
etc. ,I