Compton Scattering I

1 Introduction

Compton scattering is the process whereby photons gain or lose energy from collisions with electrons. It is an important source of radiation at high energies, particularly at X-ray to γ-ray energies. In this chapter, we consider the total energy radiated by rela- tivistic electrons as a result of scattering of soft photons.

High Energy Astrophysics © Geoffrey V. Bicknell

2 Scattering from electrons at rest 2.1 Classical approach (Thomson scattering) When a flux of electro- magnetic radiation im- pinges on an electron, the electron oscillates Incident photons Scattered photons and radiates electro- Θ magnetic radiation (photons) in all direc- tions. Oscillating electron

High Energy Astrophysics: Compton Scattering I 2/59

We concentrate on the number flux of photons. Let

dNinc Incident no of photons per unit time ------= dtdA per unit area

dNscat No of photons per unit time per steradian ------= dtdΩ scattered by the electron The differential number of scattered photons is defined in terms of the cross-section by: σ dNscat dNincd T ------= ------dtdΩ dtdA dΩ

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The differential cross section for Thomson scattering is: σ d T 1 ------= ---r2()1 + cos2Θ dΩ 2 0

r0 = Classical electron radius e2 Ð15 ==------2.818×10 m πε 2 4 0mec

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The classical electron radius, r0 , is the “radius” derived by treating the electron as a classical particle and assuming that the its rest-mass is equal to its electrostatic potential, i.e. 2 e 2 ------πε - = mec 4 0r0 Units Note the units for the equation describing Thomson scattering Number per unit time Number per unit time Area per unit = × per unit solid angle per unit area solid angle

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The total cross-section for scattering into all solid angles is given by

dNscat dNinc ------= σ ------dt T dtdA σ π 2 d T 2 r0 π 8π σ ==∫ ------dΩ ------∫ ()Θ1 + cos2Θ d =------r2 T dΩ 2 0 3 0 4π Ð29 = 6.65×10 m2

σ The quantity T is the Thomson cross-section.

High Energy Astrophysics: Compton Scattering I 6/59 2.2 Quantum-mechanical particle approach ε ν The above can 1 = h 1 be derived by approximating photons classi- ε ν = h Θ cally as an elec- tromagnetic wave. It is also Electron E useful to treat the scattering from a particle point of view. To do so, we con- sider the collision between a photon and an electron in the rest frame of the electron, as described in the figure.

High Energy Astrophysics: Compton Scattering I 7/59 The parameters describing the collision are: ε ==hν Initial photon energy ε ν 1 ==h 1 Final photon energy 2 mec = Initial electron energy E = Final electron energy Θ = Angle of deflection of photon

High Energy Astrophysics: Compton Scattering I 8/59 Treatment of collision using 4-vectors The conservation of momentum and energy gives the final photon energy in terms of the initial energy. This can be de- rived in the following way. In relativity, the conservation of energy and momentum be- comes the conservation of four momentum. As we described in the chapter on relativistic effects, the 4Ðmomentum of a particle with energy E = γmc2 moving in the direction of the unit vector n is described by:

High Energy Astrophysics: Compton Scattering I 9/59 E Ev P ==γmc γmv ------c c c E = --- []1, βn c n = Unit vector in the direction of motion Limiting case of photon In the limit where the mass goes to zero and β → 1 but the en- ergy ε remains finite, we have a photon with 4Ðmomentum ε P = -- []1, n c

High Energy Astrophysics: Compton Scattering I 10/59 Let ε P ==Initial 4-momentum of the photon -- 1 n γi c γi ε 1 P ==Final 4 momentum of the photon ----- 1 n γf c γf

Pei ==Initial 4-momentum of the electron mec 0 E P ==Final 4-momentum of the electron --- 1 n ef c ef Conservation of 4-momentum, tells us that

Pγi + Pei = Pγf + Pef High Energy Astrophysics: Compton Scattering I 11/59 We can rearrange this equation in such a way that the electron momentum drops out. First put

Pef = Pγi + Pei Ð Pγf then take the 4-dimensional modulus of this equation. Remember that

1. The modulus of a vector Aµ is given by 2 µ ν 0 2 1 2 2 2 3 2 A ==ηµνA A Ð ()A +++()A ()A ()A 2. The scalar product of AB and is µ ν 0 0 1 1 2 2 3 3 AB⋅η==µνA B ÐA B +++A B A B A B

High Energy Astrophysics: Compton Scattering I 12/59 3. The magnitude of the 4-momentum of a particle is given by: P2 = Ðm2c2 4. Magnitude of 4-momentum of a photon is: P2 = 0

High Energy Astrophysics: Compton Scattering I 13/59 Now take the modulus of the equation for Pef : 2 2 Pef = Pγi + Pei Ð Pγf ⇒ 2 2 2 2 2 Ðme c = Pγi ++Pei Pγf ⋅ ⋅ ()⋅ + 2Pγi Pei Ð 2Pγi Pγf Ð 2 Pei Pγf ε Ð0m2c2 = Ð02m2c2 ++ Ð-- m c e e c e εε εε ε 1 1 ⋅ 1 Ð22 Ð ------+ ------n n f Ð Ðmec----- c2 c2 i c

High Energy Astrophysics: Compton Scattering I 14/59 This simplifies to: εε εε 1 1 εm Ð ------Ð ------cosΘ Ð0m ε = e c2 c2 e 1 ε []ε2 ε()Θ 1 mec + 1 Ð cos = ε ε = ------1 ε 1 + ------()1 Ð cosΘ 2 mec

High Energy Astrophysics: Compton Scattering I 15/59 One can see immediately from this equation that: ε 2 ⇒ ε ≈ ε Ç mec 1

2.3 Energy and wavelength change The wavelength of a photon is given by:

ε hc ⇒ λ hc = -----λ- = -----ε- The above equation for the energy can be expressed as:

High Energy Astrophysics: Compton Scattering I 16/59 λ λ λ ()θ 1 Ð = c 1 Ð cos λ h ≈ c ==------Compton wavelength of electron 0.0246 A mec The wavelength change after scattering is of order the Comp- λλ ton wavelength. For long wavelengths, È c , the change in wavelength is small compared to the initial wavelength. ε 2 ()ε ε Equivalently, when Ç mec energy is conserved 1 = to a good approximation. The above treatment follows very closely the treatment given in the text book Special Relativity by W. Rindler.

High Energy Astrophysics: Compton Scattering I 17/59 2.4 The Klein-Nishina cross-section ε ∼ 2 When mec as well as the relativistic effects implied by conservation of energy and momentum, quantum mechanical effects also change the electron cross-section from the classi- cal value. The differential cross-section is given by the Klein- Nishina formula: σ 2ε2 ε d KN r0 1 1 ε ------= ----------- + ----- Ð sin2Θ Ω 2 ε ε d 2 ε 1

High Energy Astrophysics: Compton Scattering I 18/59 As εε→ 1 σ 2 σ d KN r0 1 d T ------→ ----- ()2 Ð sin2Θ ==---r2()1 + cos2Θ ------dΩ 2 2 0 dΩ

High Energy Astrophysics: Compton Scattering I 19/59 Integrating the above expression over solid angle gives the following expression for the total Klein-Nishina cross-sec- tion:

3 1 + x2x()1 + x σ = ------------Ð ln() 1+ 2 x KN 4 x3 12+ x 1 ()13+ x +12------ln()+ x Ð ------2x ()12+ x 2 hν x = ------2 mec

High Energy Astrophysics: Compton Scattering I 20/59 Limits Nonrelativistic regime ()x Ç 1 : 26x2 σσ= 12Ð x + ------T 5 Extreme relativistic regime ()x È 1 : 3 1 σ = ---σ xÐ1 ln2x + --- 8 T 2 → 0 as x → ∞ That is, electrons are less efficient scatterers of high energy photons.

High Energy Astrophysics: Compton Scattering I 21/59 3 Scattering from electrons in motion

The above applies to an electron at rest. For most applica- tions, the electrons are moving, sometimes with relativistic velocities so that we need to consider the details of electron scattering in this case. We do so by extending the results for scattering by a stationary electron to moving electrons using a change of frame defined by the Lorentz transformation.

High Energy Astrophysics: Compton Scattering I 22/59 3.1 Rest frame and electron frame Here we use a device that it used a lot in High Energy Astro- physics, we determine the energy transfer in an arbitrary frame by Lorentz transforming to and from a frame in which the electron is at rest. In this application “something” is the electron. Assume that in the rest frame ()S′ ν′ 2 h Ç mec so that the energy change in the rest frame can be neglected. The photon-electron collision in the two frames is as depicted in the following diagram:

High Energy Astrophysics: Compton Scattering I 23/59 Notation: (lab frame) S′ (rest frame) ε ′ ε ν 1 1 = h 1 θ θ θ′ θ ′ 1 1

ε = hν ε′ Electron scattering in laboratory frame and rest frame of electron

Note that all angles are measured clockwise from the positive -axis defined by the electron velocity.

High Energy Astrophysics: Compton Scattering I 24/59 ε = Initial photon energy in lab frame Angle between initial photon direction θ = and electron velocity in lab frame ε′ = Initial photon energy in electron frame Angle between initial photon direction θ′ = and electron velocity in rest frame ε 1 = Scattered photon energy in lab frame θ 1 = Scattered photon angle in lab frame ε ′ 1 = Scattered photon energy in rest frame θ ′ 1 = Scattered photon angle in rest frame High Energy Astrophysics: Compton Scattering I 25/59 3.2 Transformation between frames The frame S′ is the frame in which the electron is at rest. The various angles refer to the angle between the direction of the photon (pre- or post-collision) and the x -axis. In the lab frame Sv the electron has velocity and Lorentz factor 2 Ð12/ γ v ()β2 Ð12/ ==1 Ð ----- 1 Ð c2

v β = -- c

High Energy Astrophysics: Compton Scattering I 26/59 Recall the transformation between energies between two rel- atively moving frames. For massive particles:

vp E′γ= E 1 Ð β------cosθ c where Ev is the energy of the particle and p is its velocity. For photons (where we denote energy by ε ): ε′== γε()γε1 Ð βθcos ()1 Ð βµ where µθ= cos The Lorentz factor γ = ()1 Ð β2 Ð12/ .

High Energy Astrophysics: Compton Scattering I 27/59 The reverse and forward transformations are: ε′== γε()γε1 Ð βθcos ()1 Ð βµ ε γε ′()γεβθ ′()βµ 1 ==1 1 + cos 1 1 1 + 1 Hence, except for values of θ near 0 (µ ≈ 1 ), the photon picks up a factor of γ when we transform to the rest frame and ex- θ πµ≈ cept for values of 1 near (1 Ð1 ), we pick up a further factor of γ when we transform the energy of the scattered photon back to the lab frame.

High Energy Astrophysics: Compton Scattering I 28/59 Energy gain from Thomson scattering Assuming that Thomson scattering applies in the rest frame, ε ′ε′ (i.e. 1 = ) then the ratios of energies in going from lab frame to rest frame and then back to lab frame are of order 1:γ :γ 2 Hence, in being scattered by an electron, a photon increases in energy by a factor of order γ 2 . Obviously for relativistic electrons, this can be substantial.

High Energy Astrophysics: Compton Scattering I 29/59 Condition for Thomson scattering in the rest frame Since, the energy of the photon in the rest frame is of order γε, then the condition for Thomson scattering to apply in the rest frame is: γε 2 ⇒ γ ν 2 Ç mec h Ç mec m c2 ν e h Ç ------γ

High Energy Astrophysics: Compton Scattering I 30/59 Example Consider scattering of radio emitting photons by electrons with a Lorentz factor of order 104 . First, assuming that the Thomson limit applies in the rest frame, the typical photon frequency produced is γ 2ν ∼∼108 × 109 Hz 1017 Hz i.e. X-ray frequencies.

High Energy Astrophysics: Compton Scattering I 31/59 Is the condition for the Thomson limit satisfied? We require the initial soft photon energy to satisfy: 2 Ð31 8 2 mec 9.11×10× () 3×10 ν Ç ------==------1 0 16 hγ Ð34 6.6×10× 104 and this is easily satisfied for radio photons.

High Energy Astrophysics: Compton Scattering I 32/59 4 Emitted power resulting from inverse Compton scattering

The scattering of photons by energetic electrons, frequently results in a transfer of energy from the electrons to the pho- tons. When this is the case, the process is known as inverse Compton scattering.

High Energy Astrophysics: Compton Scattering I 33/59 4.1 Single electron power Isotropic distribution of photons Photons impinging on a population of electrons have a distri- bution of directions. For simplicity and with physical appli- cations in mind, we consider a distribution of photons which is isotropic in the lab frame. Let f ()p d3 p = No density of photons within range d3 p n()ε dε = No density of photons within dε

In the lab frame, where the photons are assumed isotropic: 4π p2 fp()dp= n()ε dε High Energy Astrophysics: Compton Scattering I 34/59 The calculation of the single electron power follows the fol- lowing scheme.

Isotropic distribu- Anisotropic distribu- tion of photons in tion of photons in lab frame rest frame of electron

Compute power in rest Transform back to lab frame, using conserva- frame and calculate tion of energy in rest power in that frame frame

High Energy Astrophysics: Compton Scattering I 35/59 We consider the geometry at Electron rest frame the left. Consider photons in- ε ′ S′′′ 1 cident at an angle θ′ to the x - axis which are scattered into a θ′ θ ′ 1 range of angles indicated by θ ′ 1 . Consider the incident ε′ flux on the electron due to photons within a region d3 p′ of momentum space, where d3 p′ = p′2dp′θ′sin dφ′

High Energy Astrophysics: Compton Scattering I 36/59 The number density of photons in this region of momentum space is: δnf= ()p′ d3 p′ Therefore: Incident photon flux per unit area ==δnc× cf ()p′ d3 p′ per unit time

σ × ()′ 3 ′ No of photons scattered per unit time = T cf p d p

High Energy Astrophysics: Compton Scattering I 37/59 In the rest frame the energy of the scattered photons remains the same. Hence, the energy per unit time, i.e. the power, of the scattered radiation contributed by a single electron, is: δ ′σ× ε′ × ′()′ 3 ′ P = T cf p d p Integrating over all momenta in the rest frame: ′ σ ε′ ′()′ 3 ′ P = c T ∫ f p d p Transformation back to lab frame We know that the distribution function is invariant under Lorentz transformations: f′()p′ = f ()p

High Energy Astrophysics: Compton Scattering I 38/59 We also need to determine the transformation law for ε and d3 p as a result of the Lorentz transformations between S and S′. Determining the transformation of d3 p involves deter- mining the Jacobean of the transformation from S′ to S .Using the transformations for photons derived from the 4-momen- tum, we have for the spatial momentum components:

High Energy Astrophysics: Compton Scattering I 39/59 ε p ′γ==p Ð -- β γ ()p Ð pβ x x c x ′ ′ py = py pz = pz ∂p ′ ∂ p x γ p β γ xβ ------∂ - ==1 Ð ∂------1 Ð ------px px p = γ ()1 Ð βθcos = γ ()1 Ð βµ ∂p ′ p ∂p ′ p x γβ y x γβ z ------∂ - = Ð ------∂ - = Ð ----- py p pz p ∂ ′ ∂ ′ py pz ------∂ - ==------∂ 1 py pz High Energy Astrophysics: Compton Scattering I 40/59 For the energy

px ε′== γ()γε εÐ βcp 1 Ð β------x p = γε()γε1 Ð βθcos = ()1 Ð βµ Jacobean of the transformation between lab mo- mentum space and rest frame momentum space

py pz γ ()γβ1 Ð βµ Ð ------Ðγβ----- J ==p p γ ()1 Ð βµ 010 001

High Energy Astrophysics: Compton Scattering I 41/59 Hence, d3 p′γ= ()1 Ð βθcos d3 p

ε′f ()p′ d3 p′γ= 2()1 Ð βθcos 2εfp()d3 p The power radiated in the rest frame is: ′ σ ε ()′ 3 ′ σ γ 2()βθ2ε ()2 Ω P ==c T ∫ f p d p c T ∫ 1 Ð cos fpp dpd σ γ 2 c T = ------∫()1 Ð βθcos 2εn()εε d sin θdθdφ 4π

High Energy Astrophysics: Compton Scattering I 42/59 The integral is over variables in the lab frame where the pho- ton distribution is isotropic. We obtain for the power: β2 ∞ P′ = cσ γ 2 1 + ------∫ εn()εε d T 3 0 The quantity ∞ U = ∫ εn()εε d ph 0 is the photon energy density in the lab frame.

High Energy Astrophysics: Compton Scattering I 43/59 Emitted power in the lab frame How does the power emitted in the rest frame relate to the power emitted in the lab frame? We can write these powers as ′ dE1 dE1 P′ = ------P = ------dt′ dt ′ where dE1 and dE1 are the energies in bundles of radiation emitted in time intervals dt and dt′ respectively. From the Lorentz transformation between the electron rest frame and the lab frame:

High Energy Astrophysics: Compton Scattering I 44/59 βdx′ dt ==γ dt′ + ------γdt′ c since in the rest frame of the electron, dx′ = 0 . ′ Consider a bunch of photons emitted with energy dE1 and ′ ()θ′ ⁄ θ ′ xdp-momentum x = dE1 c cos 1 at the angle 1 . The Lorentz transformed energy of this bunch of photons in the lab frame is γ ()γ′β ′ ()′ ()βθ′ dE1 ==dE1 + cdpx dE1 1 + cos 1

High Energy Astrophysics: Compton Scattering I 45/59 θ ′ >θ′ < However, scatterings with cos1 0 and cos1 0 are equally likely, because of the symmetry of the Thomson cross-section, so that the averaged contribution to dE1 is 〈〉γ′ dE1 = dE1 Hence the power in the lab frame is: 〈〉γ ′ ′ dE1 dE1 dE1 ------==------dt γdt′ dt′ That is, PP= ′

High Energy Astrophysics: Compton Scattering I 46/59 Hence the scattered power in the lab frame is β2 Pc= σ γ 2 1 + ------U T 3 ph

4.2 Number of scatterings per unit time In the electron rest frame: dN′ ------= Number of scatterings per unit time dt′ σ ()′ 3 ′ = c T ∫ f p d p

High Energy Astrophysics: Compton Scattering I 47/59 Again we transform this equation to the lab frame: dN′ ------= cσ ∫γ ()1 Ð βθcos fp()p2dpsinθdθdφ dt′ T πγ σ 2 c T = ------∫()1 Ð βθcos n()εε d sin θdθ 4π γ σ = c T Nph where the photon number density is ()εε Nph = ∫n d

High Energy Astrophysics: Compton Scattering I 48/59 In transforming dN′ ⁄ dt′ to the lab frame, we note that dN′ is a number and so is Lorentz invariant, so that dN dN′ ------==γ Ð1------cσ N dt dt′ T ph

4.3 Nett energy radiated We write the above equation as dN ------==cσ N cσ ∫n()εε d dt T ph T

High Energy Astrophysics: Compton Scattering I 49/59 Hence, the number of scatterings per unit time per unit pho- σ ()ε ton energy by a single electron is c T n . Hence the energy ε σ ε ()ε ε removed from photons within d is c T n d . Hence, the energy removed from the photon field is given by:

dE1 ------==Ðcσ ∫εn()εε d Ðcσ U dt T T ph

High Energy Astrophysics: Compton Scattering I 50/59 Therefore, the nett energy radiated is:

dErad 1 ------==P cσ U γ 2 1 + ---β2 Ð 1 dt compt T ph 3 4 = ---σ cγ 2β2U 3 T ph where we have used γ 2β2 = γ 2 Ð 1

High Energy Astrophysics: Compton Scattering I 51/59 4.4 Comparison of synchrotron and inverse Comp- ton power We already know that the synchrotron power of an electron is given by 4 P = ---σ cγ 2β2U synch 3 T B B2 UB = ------µ 2 0 Hence,

Pcompton Uph ------= ------Psynch UB High Energy Astrophysics: Compton Scattering I 52/59 From this expression, we can see that the inverse Compton power can be comparable to the synchrotron power, when the photon energy density is comparable to the magnetic energy density. This is often the case at the bases of jets, so that these regions are often strong X-ray emitters.

5 Inverse Compton emission from the microwave background

A regime in which inverse Compton emission is important is in the extended regions of radio galaxies, where the energy density of the microwave background radiation may be com- parable to the magnetic energy density.

High Energy Astrophysics: Compton Scattering I 53/59 Consider the energy density of blackbody radiation:

8π5k4 Ð16 ε = aT4 a ==------7.57×10 J mÐ3 KÐ4 BB 15h3c3 This is equal to the energy density of a magnetic field when 2 B 4 ⇒ µ 2 × Ð11 2 ------µ = aT B ==2 0aT 4.4 10 T 2 0 For the microwave background, T = 2.7 K . Therefore, when < × Ð10 µ BBcrit ==3.2 10 T 3.2 G

High Energy Astrophysics: Compton Scattering I 54/59 the radiation from inverse Compton emission is more impor- tant than synchrotron radiation as an energy loss mechanism for the electrons.

6 Inverse Compton emission from a thermal plasma

The above expressions are derived without any restriction on the energies of the electrons. If we consider a thermal distri- bution, then, 〈〉v2 3kT γ 2 ≈β1 〈〉2 ==------c2 mc2

High Energy Astrophysics: Compton Scattering I 55/59 Hence, the total Compton emission from a single electron is given by:

4kT σ P = ------c U Compton mc2 T ph The volume emissivity from the plasma with electron number density ne is:

4kT σ j = ------c n U Compton mc2 T e ph

High Energy Astrophysics: Compton Scattering I 56/59 Compton scattering of “soft” photons by hot thermal elec- trons in the coronae of accretion disks is thought to be respon- sible for the X-ray emission from AGN and for the X-ray emissivity of galactic black hole candidates.

7 Mean energy of scattered photons

By dividing the radiated power by the number of scattered photons per unit time, we can calculate the mean energy per scattered photon.

High Energy Astrophysics: Compton Scattering I 57/59 4σ γ 2β2 --- T c Uph 〈〉ε 3 1 = ------σ - c T Nph

4 Uph 4 ==---γ 2β2------γ 2β2 〈〉ε 3 Nph 3

High Energy Astrophysics: Compton Scattering I 58/59 Inverse Compton emission from relativistic elec- trons Thus for photons scattering off relativistic electrons, the mean amplification of energy per scattering is 4γ 2 ⁄ 3 , sup- porting the order of magnitude estimate of γ 2 for this param- eter.

Clearly, for relativistic electrons, the energy gain is substan- tial, underlining the importance of inverse Compton emission as an astrophysical process.

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