Set 8:
Compton Scattering
Energy-Momentum Conservation
• Thomson scattering ei + γi → ef + γf where the frequencies
ωi = ωf (elastic scattering) cannot strictly be true
• Photons carry off E/c momentum and so to conserve momentum
the electron must recoil.
• The electron then carries away some of the available energy and so
the Thomson limit requires h̄ωi /mc2 1 in the electron rest
frame
• General case (arbitrary electron velocity)
photon
Pi
Pf
ni
nf
directional
vectors
electron
Qi
Qf
vi
vf
Energy-Momentum Conservation
• 4 momenta
Ef
Ei
(1, n̂f ) ,
Pi = (1, n̂i ) , Pf =
c
c
Qi = γi m(c, v̂i ) , Qf = γf m(c, v̂f ) ,
photon
electron
• To work out change in photon energy consider Lorentz invariants:
2
2
v
−
c
2 2
Qµ Qµ = γ 2 m2 (v 2 − c2 ) = m2
=
−m
c
2
2
1 − v /c
Pµ P µ = 0
Energy-Momentum Conservation
• Conservation
Pi + Qi = Pf + Qf
also
Pf · [Pi + Qi = Pf + Qf ]
Pf · Pi + Pf · Qi = Pf · Qf
• Some identities to express final state in terms of initial state
(Pi + Qi )µ (Pi + Qi )µ = (Pf + Qf )µ (Pf + Qf )µ
2Piµ Qµi − m2 c2 = 2Pf µ Qµf − m2 c2
Pi · Qi = Pf · Qf
• So
Pf · Pi + Pf · Qi = Pi · Qi
Energy-Momentum Conservation
• In three vector notation
Ef
Ei
Ei Ef
(n̂i · n̂f − 1) +
γi m(n̂f · vi − c) = γi m(n̂i · vi − c)
c c
c
c
• Introducing the scattering angle as n̂i · n̂f = cos θ and auxiliary
angles n̂f · vi = vi cos αf and n̂i · vi = vi cos αi
Ei Ef (cos θ − 1) + Ef γi mc2 (βi cos αf − 1) = Ei γi mc2 (βi cos αi − 1)
Ei γi mc2 (βi cos αi − 1)
Ef =
Ei (cos θ − 1) + γi mc2 (βi cos αi − 1)
• So the change in photon energy is given by
Ef
1 − βi cos αi
=
Ei
Ei
1 − βi cos αf + γmc
2 (1 − cos θ)
Recoil Effect
• Two ways of changing the energy: Doppler boost βi from
incoming electron velocity and Ei comparable to γmc2
• Take the incoming electron rest frame βi = 0 and γ = 1
1
Ef =
Ei
Ei rest 1 + mc
2 (1 − cos θ)
• Since −1 ≤ cos θ ≤ 1, Ef ≤ Ei , energy is lost from the recoil
except for purely forward scattering via comparing the incoming
photon energy Ei and the electron rest mass mc2
• The backwards scattering limit is easy to derive
2
Ei
1 2 1
2Ei
2Ei
|qf | = m|vf | = 2 ,
∆E = mvf = m
=
Ei
2
c
2
2
mc
mc
2Ei
Ei
Ef = Ei − ∆E = (1 −
)Ei ≈
2Ei
2
mc
1 + mc
2
Compton Wavelength
• Alternate phrasing in terms of length scales E = hν = hc/λ so
1
λi =
λf rest 1 + λ hc
2 (1 − cos θ)
i mc
λf
h
λc
− 1 =
(1 − cos θ) ≡ (1 − cos θ)
λi
mcλi
λi
rest
where the Compton wavelength is λc = h/mc ≈ 2.4 × 10−10 cm.
• Doppler effect: consider the limit of βi 1 then expand to first
order
Ef
Ei
= 1 − βi cos αi + βi cos αf −
(1 − cos θ)
2
Ei
mc
however averaging over angles the Doppler shifts don’t change the
energies
Second Order Doppler Shift
• To second order in the velocities, the Doppler shift transfers energy
from the electron to the photon in opposition to the recoil
Ei
Ef
2
2
= 1 − βi cos αi + βi cos αf + βi cos αf −
Ei
mc2
Ef
1 2
Ei
h i ≈ 1 + βi −
Ei
3
mc2
• For a thermal distribution of velocities
1
3kT
2
mhv i =
2
2
βi2
3kT
Ef
kT − Ei
≈
→h
− 1i ∼
2
mc
Ei
mc2
so that if Ei kT the photon gains energy and Ei kT it loses
energy → this is a thermalization process
Energy Transfer
• A more proper treatment of the radiative transfer equation (below)
for the distribution gives the Kompaneets equation and
4kT − Ei
Ef
−1=
Ei
mc2
where the coefficient reflects the fact that averaged over the Wien
tail hEi i = 3kT and hEi2 i = 12(kT )2 so that
Ef − Ei ∝ 4kT Ei − Ei2 = 0
in thermodynamic equilibrium. For multiple scattering events one
finds that the energy transfer goes as
k(Te − Tγ )
4
τ
2
mc
Inverse Compton Scattering
• Relativistic electrons boost energy of low energy photons by a
potentially enormous amount - a way of getting γ rays in
astrophysics
Ef
1 − βi cos αi
=
Ei
1 − βi cos αf + γiEmci 2 (1 − cos θ)
• Take βi → 1 and γ 1 but Ei < γi mc2 so that the scattering is
Thomson in the rest frame
• Qualitative: αi incoming angle wrt electron velocity can be
anything; BUT outgoing αf is strongly beamed in direction of
velocity with αf ∼ 1/γ 1 or cos αf ≈ 1 so
Ef
1 − βi cos αi
1
≈
= (1 − β)(1 + β) ≈ 2(1 − β)
2
Ei
1 − βi
γi
≈ (1 − βi cos αi )2γi2
Inverse Compton Scattering
• Since the incoming angle is random, a typical photon gains energy
by a factor of γi2 !
• From the electron’s perspective: see a bunch of (beamed) photons
coming head on boosted in energy by a factor of γi and scatter out
roughly isotropic preserving energy in the rest frame. Reverse the
boost and pick up another factor of γi in the lab frame beamed into
the direction of motion.
• Limit to energy transfer - total energy of the electron in lab frame
Ef − Ei ≈ Ef < γi mc2
γi2 Ei < γi mc2
γi Ei < mc2
Inverse Compton Scattering
• Maximum energy boost
Ef = γi (γi Ei ) < γi mc2 = γi (511keV)
so that Ef can be an enormous energy and the scattering is still
Thomson in the rest frame
• Is it consistent to neglect recoil compared with 1/γi2 in the lab
frame? The condition becomes
1
Ei
2
Recoil =
2
γi mc
γi
Ei γi mc2
Yes until the maximum energy is reached.
Inverse Compton Scattering
• With γi = 103
radio 1GHz = 109 Hz → 1015 Hz ≈ 300nm
optical
4 × 1014 Hz → 4 × 1020 Hz ≈ 1.6MeV
UV
gamma ray
• These energies are less than the maximal γi (512keV) = 512MeV
so still Thomson in rest frame.
Single Scattering
• Consider a distribution of photons (and electrons) in the lab frame.
Characterize the radiative transfer in the optically thin single
scattering regime
• Consider the energy density (A is a constant)
Z
Z
3
dp
3
Ef
=
A
f
p
dpdΩ
u=2
2
(2πh̄)
• In the electron rest frame the energy density transforms with the
aid of the Lorentz transformations
p0 = pγ(1 − βµ)
dΩ
dΩ = 2
γ (1 − βµ)2
f0 = f
0
2
→ p0 dΩ0 = p2 dΩ
Single Scattering
• Therefore energy density transforms as
Z
Z
3
u0 = A f 0 p0 dp0 dΩ0 = A f p0 p2 dp0 dΩ
Z
= A γ 2 (1 − βµ)2 f p3 dpdΩ
• Assume that f is isotropic (in lab frame) then the energy densities
are related as
Z
Z
u0 = dΩγ 2 (1 − βµ)2 A f p3 dp
Z
2
2 u
= dΩγ (1 − βµ)
4π
1 2
2
= γ (1 + β )u
3
Single Scattering
• In the rest frame the emitted power is given by the Thomson
scattering strength σT = P/S and power is a Lorentz invariant
dE 0
dErad
= 0 = cσT u0
dt
dt
1 2
2
= cσT γ [1 + β ]u
3
• The total power accounts for the “absorbed” energy of the
incoming photons
1 2
dE
2
= cσT [γ (1 + β ) − 1]u ,
dt
3
4
= σT cγ 2 β 2 u
3
γ 2 − 1 = γ 2β 2
Single Scattering
• If we assume a thermal distribution of electrons hβ 2 i = 3kT /mc2
we get
du
4kT
4kT
du
cσT ne u
→
u
=
=
2
2
dt
mc
dτ
mc
• Assume a power law distribution of electron energies dE = mc2 dγ
confined to a range γmin ≤ γ ≤ γmax
Z
ne = ne,γ dγ
ne,γ = Kγ −p
• Change in the radiation energy density given γ 1 (β ≈ 1)
Z
2
du
dE
dE
=
=
ne,γ dγ
dt
dtdV
dt
4
K
3−p
3−p
= σT cu
(γmax
− γmin
)
3
3−p
Single Scattering
• Energy spectrum in the power law case
2
dE
∝
dtdV
Z
γ 2−p dγ
and the scattered photon energy Ef ∝ γ 2 Ei
d2 E
dγ d2 E
1 2−p
(1−p)/2
1−p
=
∝ γ
=γ
∝ Ef
dtdV dEf
dEf dtdV dγ
γ
so the scattered spectrum is also a power law Ef−s but with a power
law index s = (p − 1)/2.
Radiative Transfer Equation
• For full radiative transfer, we must go beyond the single scattering,
optically thin limit. Generally (set h̄ = c = k = 1 and neglect
Pauli blocking and polarization)
Z 3
Z 3
Z
1
1
1
1
d pi
d qf
d3 q i
∂f
=
∂t
2E(pf )
(2π)3 2E(pi )
(2π)3 2E(qf )
(2π)3 2E(qi )
× (2π)4 δ(pf + qf − pi − qi )|M |2
× {fe (qi )f (pi )[1 + f (pf )] − fe (qf )f (pf )[1 + f (pi )]}
where the matrix element is calculated in field theory and is
Lorentz invariant. In terms of the rest frame α = e2 /h̄c (c.f. Klein
Nishina Cross Section)
E(pf )
2
2 2 E(pi )
|M | = 2(4π) α
+
− sin2 β
E(pf ) E(pi )
with β as the rest frame scattering angle
Kompaneets Equation
• The Kompaneets equation is the radiative transfer equation in the
limit that electrons are thermal
"
3/2 #
mTe
−(m−µ)/Te −q 2 /2mTe
−(m−µ)/Te
fe = e
e
ne = e
2π
3/2
2π
−q 2 /2mTe
=
ne e
mTe
and assume that the energy transfer is small (non-relativistic
electrons, Ei m
Ef − Ei
1
Ei
[O(Te /m, Ei /m)]
Kompaneets Equation
• Kompaneets equation (restoring h̄, c k)
kTe 1 ∂
∂f
∂f
4
= ne σ T c
x
+ f (1 + f )
2
2
∂t
mc x ∂x
∂x
x = h̄ω/kTe
• Equilibrium solution must be a Bose-Einstein distribution
∂f /∂t = 0
∂f
4
x
+ f (1 + f )
=K
∂x
∂f
K
+ f (1 + f ) = 4
∂x
x
Kompaneets Equation
Assume that as x → 0, f → 0 then K = 0 and
df
df
= −f (1 + f )
→
= dx
dx
f (1 + f )
f
f
ln
= −x + c
→
= e−x+c
1+f
1+f
e−x+c
1
f=
= x−c
−x+c
1−e
e
−1
Kompaneets Equation
• More generally, no evolution in the number density
Z
Z
nγ ∝ d3 pf ∝ dxx2 f
Z
∂f
∂nγ
2 1 ∂
4
∝ dxx 2
x
+ f (1 + f )
∂t
x ∂x
∂x
∞
4 ∂f
∝x
+ f (1 + f )
=0
∂x
0
• Energy evolution R ≡ ne σT c(kTe /mc2 )
Z
Z
Z 3
3
4
dp
p dpc
(kTe ) 1
3
u=2
f
=
x
dxf
Ef
=
2
≡
A
3
3
3
2
2
4
(2πh̄)
2π h̄
c h̄ π
Z
∂u
∂
∂f
4
= AR dxx
x
+ f (1 + f )
∂t
∂x
∂x
Kompaneets Equation
∂f
∂u
4
= −AR dxx
+ f (1 + f )
∂t
∂x
Z
Z
= AR dx4x3 f − AR dxx4 f (1 + f )
Z
kTe
= 4ne σT c 2 u − AR dxx4 f (1 + f )
mc
Z
Change in energy is difference between Doppler and recoil
• If f is a Bose-Einstein distribution at temperature Tγ
∂f
pc
= −f (1 + f )
xγ =
∂xγ
kTγ
Z
Z
Z
4
4 ∂f
3 dx
AR dxx f (1 + f ) = −AR dxx
= AR dx4x
f
∂xγ
dxγ
Kompaneets Equation
• Radiative transfer equation for energy density
Tγ
kTe
∂u
= 4ne σT c 2 1 −
u
∂t
mc
Te
1 ∂u
k(Te − Tγ )
= 4ne σT c
u ∂t
mc2
which is our original form from the energy-momentum
conservation argument
• The analogue to the optical depth for energy transfer is the
Compton y parameter
dτ = ne σT ds = ne σt cdt
k(Te − Tγ )
dy =
dτ
2
mc
Kompaneets Equation
• Example: hot X-ray cluster with kT ∼ keV and the CMB:
Te Tγ
• Inverse Compton scattering transfers energy to the photons while
conserving the photon number
• Optically thin conditions: low energy photons boosted to high
energy leaving a deficit in the number density in the RJ tail and an
enhancement in the Wien tail called a Compton-y distortion — see
problem set
• Compton scattering off high energy electrons can give low energy
photons a large boost in energy but cannot create the photons in
the first place
Kompaneets Equation
• Numerical solution of the Kompaneets equation going from a
Compton-y distortion to a chemical potential distortion of a
blackbody
∆T / Tinit
0.1
0
y–distortion
–0.1
µ-distortion
–0.2
10–3
10–2
10–1
p/Tinit
1
101
102
Compton Scattering Map
• WMAP measured the Compton “emission” of photons, i.e. the
Compton scattering of the CMB at z ∼ 1000. At 61 GHz:
• Red band at equator is galactic contamination from the next two
processes.
Compton Scattering Map
• Gamma ray bubble from center of galaxy thought to be inverse
Compton from synchrotron radiation seen at radio frequencies.
26
Fermi 1-5 GeV
Haslam 408 MHz
3.0
50
50
30
0
0
20
1.0
-50
-50
2.0
K
keV cm-2 s-1 sr-1
2.5
40
1.5
10
0.5
0
50
0
-50
50
Rosat Band 6 and 7
0
-50
WMAP 23 GHz haze
400
0.06
50
0.04
50
300
0.02
250
0
mK
10-6 counts s-1 arcmin-2
350
0
200
0.00
150
-50
-50
-0.02
100
50
-0.04
50
0
-50
50
0
-50
Fig. 18.— Comparison of the Fermi bubbles with features in other maps. Top left: Point-source subtracted 1 − 5 GeV Fermi-LAT 1.6
yr map, same as the lower left panel of Figure 3 with north and south bubble edges marked with green dashed line, and north arc in blue
dashed line. The approximate edge of the Loop I feature is plotted in red dotted line, and the “donut” in purple dot-dashed line. Top
right: The Haslam 408 MHz map overplotted with the same red dotted line as the top left panel. The red dotted line remarkably traces