Set 8: Compton - Conservation

Thomson scattering ei + γi ef + γf where the frequencies • → ωi = ωf () cannot strictly be true carry off E/c momentum and so to conserve momentum • the must recoil. The electron then carries away some of the available energy and so • 2 the Thomson limit requires hω¯ i/mc 1 in the electron rest  frame General case (arbitrary electron velocity) • P P i f ni nf directional vectors electron Qi Qf vi vf Energy-Momentum Conservation 4 momenta •

E E P = i (1, nˆ ) ,P = f (1, nˆ ) , photon i c i f c f Qi = γim(c, vˆi) ,Qf = γf m(c, vˆf ) , electron To work out change in photon energy consider Lorentz invariants: • 2 2 µ 2 2 2 2 2 v c 2 2 QµQ = γ m (v c ) = m − = m c − 1 v2/c2 − µ − PµP = 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 • µ µ (Pi + Qi)µ(Pi + Qi) = (Pf + Qf )µ(Pf + Qf ) µ 2 2 µ 2 2 2PiµQ m c = 2PfµQ m c i − f − Pi Qi = Pf Qf · · So •

Pf Pi + Pf Qi = Pi Qi · · · Energy-Momentum Conservation In three vector notation • Ei Ef Ef Ei (nˆi nˆf 1) + γim(nˆf vi c) = γim(nˆi vi c) c c · − c · − c · −

Introducing the scattering angle as nˆi nˆf = cos θ and auxiliary • · angles nˆf vi = vi cos αf and nˆi vi = vi cos αi · · 2 2 EiEf (cos θ 1) + Ef γimc (βi cos αf 1) = Eiγimc (βi cos αi 1) − − − E γ mc2(β cos α 1) E = i i i i f 2 − Ei(cos θ 1) + γimc (βi cos αi 1) − − So the change in photon energy is given by •

Ef 1 βi cos αi = − Ei Ei 1 βi cos αf + 2 (1 cos θ) − γmc − Recoil Effect

Two ways of changing the energy: Doppler boost βi from • 2 incoming electron velocity and Ei comparable to γmc

Take the incoming electron rest frame βi = 0 and γ = 1 • E 1 f = Ei Ei rest 1 + (1 cos θ) mc2 − Since 1 cos θ 1, Ef Ei, energy is lost from the recoil • − ≤ ≤ ≤ except for purely forward scattering via comparing the incoming 2 photon energy Ei and the mc The backwards scattering limit is easy to derive •  2 Ei 1 2 1 2Ei 2Ei qf = m vf = 2 , ∆E = mv = m = Ei | | | | c 2 f 2 mc mc2 2Ei Ei Ef = Ei ∆E = (1 )Ei mc2 2Ei − − ≈ 1 + mc2 Compton Alternate phrasing in terms of length scales E = hν = hc/λ so • λ 1 i = hc λf rest 1 + 2 (1 cos θ) λimc − λ h λ f 1 = (1 cos θ) c (1 cos θ) λi − rest mcλi − ≡ λi − 10 where the is λc = h/mc 2.4 10− cm. ≈ × Doppler effect: consider the limit of βi 1 then expand to first •  order E E f = 1 β cos α + β cos α i (1 cos θ) i i i f 2 Ei − − mc − however averaging over angles the Doppler shifts don’t change the 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

E E f = 1 β cos α + β cos α + β2 cos2 α i i i i f i f 2 Ei − − mc E 1 E f 1 + β2 i i 2 h Ei i ≈ 3 − mc For a thermal distribution of velocities • 1 3kT 3kT E kT E m v2 = β2 f 1 i i 2 −2 2 h i 2 ≈ mc → h Ei − i ∼ mc

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

E 4kT E f 1 = i −2 Ei − mc where the coefficient reflects the fact that averaged over the Wien 2 2 tail Ei = 3kT and E = 12(kT ) so that h i h i i 2 Ef Ei 4kT Ei E = 0 − ∝ − i in thermodynamic equilibrium. For multiple scattering events one finds that the energy transfer goes as

k(Te Tγ) 4 − τ mc2 Inverse Compton Scattering Relativistic boost energy of low energy photons by a • potentially enormous amount - a way of getting γ rays in

Ef 1 βi cos αi = − Ei Ei 1 βi cos αf + 2 (1 cos θ) − γimc − 2 Take βi 1 and γ 1 but Ei < γimc 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 − 2 = (1 β)(1 + β) 2(1 β) Ei ≈ 1 βi γi − ≈ − − 2 (1 βi cos αi)2γ ≈ − i Inverse Compton Scattering Since the incoming angle is random, a typical photon gains energy • 2 by a factor of γi ! 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 • 2 Ef Ei Ef < γimc − ≈ 2 2 γi Ei < γimc 2 γiEi < mc Inverse Compton Scattering Maximum energy boost • 2 Ef = γi(γiEi) < γimc = 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/γ2 in the lab • i frame? The condition becomes E 1 Recoil = i 2 2 γimc  γi 2 Eiγi mc  Yes until the maximum energy is reached. Inverse Compton Scattering 3 With γi = 10 • radio 1GHz = 109Hz 1015Hz 300nm UV → ≈ optical 4 1014Hz 4 1020Hz 1.6MeV × → × ≈

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 d3p Z u = 2 Ef = A fp3dpdΩ (2πh¯)2 In the electron rest frame the energy density transforms with the • aid of the Lorentz transformations

p0 = pγ(1 βµ) − dΩ 2 2 dΩ0 = p0 dΩ0 = p dΩ γ2(1 βµ)2 → − f 0 = f Single Scattering Therefore energy density transforms as • Z Z 3 2 u0 = A f 0p0 dp0dΩ0 = A fp0p dp0dΩ Z = A γ2(1 βµ)2fp3dpdΩ − Assume that f is isotropic (in lab frame) then the energy densities • are related as Z Z 2 2 3 u0 = dΩγ (1 βµ) A fp dp − Z u = dΩγ2(1 βµ)2 − 4π 1 = γ2(1 + β2)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

dErad dE0 = = cσT u0 dt dt0 1 = cσ γ2[1 + β2]u T 3 The total power accounts for the “absorbed” energy of the • incoming photons

dE 2 1 2 2 2 2 = cσT [γ (1 + β ) 1]u , γ 1 = γ β dt 3 − − 4 = σ cγ2β2u 3 T Single Scattering If we assume a thermal distribution of electrons β2 = 3kT/mc2 • h i we get du 4kT du 4kT = cσT neu = u dt mc2 → dτ mc2 Assume a power law distribution of electron energies dE = mc2dγ • confined to a range γ γ γ min ≤ ≤ max Z p ne = ne,γdγ ne,γ = Kγ−

Change in the radiation energy density given γ 1 (β 1) •  ≈ du d2E Z dE = = n dγ dt dtdV dt e,γ

4 K 3 p 3 p = σT cu (γ − γ − ) 3 3 p max − min − Single Scattering Energy spectrum in the power law case • 2 Z d E 2 p γ − dγ dtdV ∝ 2 and the scattered photon energy Ef γ Ei ∝ 2 2 d E dγ d E 1 2 p 1 p (1 p)/2 = γ − = γ − Ef − dtdV dEf dEf dtdV dγ ∝ γ ∝

s so the scattered spectrum is also a power law Ef− 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 h¯ = c = k = 1 and neglect Pauli blocking and polarization) ∂f 1 Z d3p 1 Z d3q 1 Z d3q 1 = i f i 3 3 3 ∂t 2E(pf ) (2π) 2E(pi) (2π) 2E(qf ) (2π) 2E(qi) 4 2 (2π) δ(pf + qf pi qi) M × − − | | 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/hc¯ (c.f. Klein Nishina Cross Section)  E(p ) E(p )  M 2 = 2(4π)2α2 i + f 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# 2 mT f = e (m µ)/Te e q /2mTe n = e (m µ)/Te e e − − − e − − 2π  3/2 2 2π q /2mTe = nee− mTe and assume that the energy transfer is small (non-relativistic

electrons, Ei m 

Ef Ei − 1 [ (Te/m, Ei/m)] Ei  O Kompaneets Equation Kompaneets equation (restoring h¯, c k) • ∂f  kT  1 ∂  ∂f  = n σ c e x4 + f(1 + f) x =hω/kT ¯ ∂t e T mc2 x2 ∂x ∂x e Equilibrium solution must be a Bose-Einstein distribution • ∂f/∂t = 0

 ∂f  x4 + f(1 + f) = K ∂x ∂f K + f(1 + f) = ∂x x4 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 x+c ln = x + c = e− 1 + f − → 1 + f e x+c 1 f = − = 1 e x+c ex c 1 − − − − Kompaneets Equation More generally, no evolution in the number density • Z Z 3 2 nγ d pf dxx f ∝ ∝ ∂n Z 1 ∂  ∂f  γ dxx2 x4 + f(1 + f) ∂t ∝ x2 ∂x ∂x ∂f ∞ x4 + f(1 + f) = 0 ∝ ∂x 0 2 Energy evolution R neσT c(kTe/mc ) • ≡ Z d3p Z p3dpc (kT )4 1  Z u = 2 Ef = 2 f = e A x3dxf (2πh¯)3 2π2h¯3 c4h¯3 π2 ≡ ∂u Z ∂  ∂f  = AR dxx x4 + f(1 + f) ∂t ∂x ∂x Kompaneets Equation

∂u Z ∂f  = AR dxx4 + f(1 + f) ∂t − ∂x Z Z = AR dx4x3f AR dxx4f(1 + f) − Z kTe 4 = 4neσT c u AR dxx f(1 + f) mc2 − 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 ∂f Z dx AR dxx4f(1 + f) = AR dxx4 = AR dx4x3 f − ∂xγ dxγ Kompaneets Equation Radiative transfer equation for energy density •

∂u kT  T  = 4n σ c e 1 γ u e T 2 ∂t mc − Te

1 ∂u k(Te Tγ) = 4neσT 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 = neσT ds = neσtcdt

k(Te Tγ) dy = − dτ mc2 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

0.1

0 y–distortion i n t T

/

∆ T –0.1 µ-distortion

–0.2

10–3 10–2 10–1 1 101 102 p/Tinit 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 40

2.5 50 50 30 -1 2.0 sr -1 s K -2 0 0 20 1.5 keV cm 1.0 10 -50 -50

0.5 0 50 0 -50 50 0 -50

Rosat Band 6 and 7 WMAP 23 GHz haze 400 0.06

350 50 50 0.04 -2 300

0.02 arcmin 250 mK -1

0 0 200 0.00 counts s

-6 150 10 -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 the edge of the bright Loop I feature in the Haslam soft synchrotron map. Bottom left: the ROSAT 1.5 keV X-ray map is shown together with the same color lines marking the prominent Fermi bubble features. Bottom right: WMAP haze at K-band 23 GHz overplotted with Fermi bubble edges. The ROSAT X-ray features and the WMAP haze trace the Fermi bubbles well, suggesting a common origin for these features. shown in Figure 16, the Loop I correlated emission also with no evidence of spatial variation across the bub- has a softer spectrum than the Fermi bubble emission. bles. As shown in Figure 23, an electron population 2 2.5 The Loop I feature in the ROSAT map similarly has a with dN /dE E− − is required to produce these e ∼ softer spectrum than the limb-brightened X-ray bubble gamma rays by IC scattering: this is comparable to the edges: as shown in Figure 20, when a low-energy map spectrum of electrons accelerated by supernova shocks or is subtracted from a higher-energy map in such a way polar cap acceleration (Biermann et al. 2010). However, that Loop I vanishes, the bubble edges remain bright. diffusive propagation and cooling would be expected to We also see additional shell structures which follow the soften the spectrum, making it difficult to explain the Fermi bubble edges and the northern arc in the Haslam Fermi bubbles by IC scattering from a steady-state pop- 408 MHz map (top row of Figure 26). ulation of these electrons (a single brief injection of elec- 2 The Fermi bubbles are morphologically and spectrally trons with dN/dE E− could generate a sufficiently ∼ distinct from both the π0 emission and the IC and hard spectrum for the bubbles if there was a mechanism emission from the disk electrons. As we to transport them throughout the bubble without sig- have shown in Figure 12 to Figure 17, the Fermi bub- nificant cooling). The facts strongly suggest that a dis- 2 bles have a distinctly hard spectrum, dN /dE E− , tinct electron component with a harder spectrum than γ ∼