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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 -ray energies. In this chapter, we consider the total energy radiated by relativistic 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 electromagnetic radiation imphotons pinges on an electron, the electron oscillates and radiates electroIncident photons Scattered magnetic radiation (photons) in all directions. Oscillating electron

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We concentrate on the number flux of photons. Let dN inc Incident no of photons per unit time -------------- = dtdA per unit area dN scat No of photons per unit time per steradian --------------- = dtd scattered by the electron The differential number of scattered photons is defined in terms of the cross-section by: dN inc d T dN scat --------------- = -------------- --------dtdA d dtd

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The differential cross section for Thomson scattering is: d T 12 --------- = -- r 0 ( 1 + cos2 ) 2 d r 0 = Classical electron radius
­ 15 e2 = ------------------------- = 2.818 â10 m 4 0 m e c 2

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The classical electron radius, r 0 , 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. e2 ----------------- = m e c 2 4 0 r 0 Units Note the units for the equation describing Thomson scattering Number per unit time Area per unit Number per unit time = â per unit area solid angle per unit solid angle

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The total cross-section for scattering into all solid angles is given by dN inc dN scat --------------- = T -------------dtdA dt
2 d T 2r0 2 ) d = 8 r 2 ----- 0 T = --------- d = ----------- ( 1 + cos d 20 3 4

= 6.65 â10

­ 29

m2

The quantity T is the Thomson cross-section.
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2.2 Quantum-mechanical particle approach The above can 1 = h 1 be derived by approximating photons classi = h cally as an electromagnetic wave. It is also E Electron useful to treat the scattering from a particle point of view. To do so, we consider the collision between a photon and an electron in the rest frame of the electron, as described in the figure.
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The parameters describing the collision are: = h = Initial photon energy 1 = h 1 = Final photon energy m e c 2 = Initial electron energy E = Final electron energy = Angle of deflection of photon

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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 derived in the following way. In relativity, the conservation of energy and momentum becomes the conservation of four momentum. As we described in the chapter on relativistic effects, the 4­momentum of a particle with energy E = mc 2 moving in the direction of the unit vector n is described by:

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EE P = mc m v = --- --- v -c cc E = --- [ 1, n ] c n = Unit vector in the direction of motion Limiting case of photon In the limit where the mass goes to zero and 1 but the energy remains finite, we have a photon with 4­momentum P = -- [ 1, n ] c
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Let P i = Initial 4-momentum of the photon = -- 1 n i c 1 P f = Final 4 momentum of the photon = ---- 1 n f c P ei = Initial 4-momentum of the electron = m e c 0 E P ef = Final 4-momentum of the electron = --- 1 n ef c Conservation of 4-momentum, tells us that P i + P ei = P f + P ef
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We can rearrange this equation in such a way that the electron momentum drops out. First put P ef = P i + P ei ­ P f then take the 4-dimensional modulus of this equation. Remember that 1. The modulus of a vector A µ is given by A 2 = µ A µ A = ­ ( A 0 ) 2 + ( A 1 ) 2 + ( A 2 ) 2 + ( A 3 ) 2 2. The scalar product of A and B is A B = µ A µ B = ­ A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3
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3. The magnitude of the 4-momentum of a particle is given by: P 2 = ­m 2 c 2 4. Magnitude of 4-momentum of a photon is: P2 = 0

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Now take the modulus of the equation for P ef : P ef 2 = P i + P ei ­ P f 2
2 ­ m e c 2 = P i 2 + P ei 2 + P f 2 + 2 P i P ei ­ 2 P i P f ­ 2 ( P ei P f )

2 c 2 = 0 ­ m 2 c 2 + 0 + 2 ­ -- m c ­me e c e 1 1 1 ­ 2 ­ ------- + ------- n n f ­ 2 ­ m e c ---- c c2 i c2
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This simplifies to: me ­ 1 1 ------- ­ ------- cos ­ m e 1 = 0 c2 c2

1 [ m e c 2 + ( 1 ­ cos ) ] = 1 = ----------------------------------------------------- 1 + ------------ ( 1 ­ cos ) m e c 2

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One can see immediately from this equation that: « me c 2 1 2.3 Energy and wavelength change The wavelength of a photon is given by: hc hc = ----- = ---- The above equation for the energy can be expressed as:

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1 ­ = c ( 1 ­ cos ) h c = --------- = Compton wavelength of electron 0.0246 A me c The wavelength change after scattering is of order the Compton wavelength. For long wavelengths, » c , the change in wavelength is small compared to the initial wavelength. Equivalently, when « m e c 2 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.
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2.4 The Klein-Nishina cross-section When m e c 2 as well as the relativistic effects implied by conservation of energy and momentum, quantum mechanical effects also change the electron cross-section from the classical value. The differential cross-section is given by the KleinNishina formula:
22 r 0 1 d KN -------------- = ---- ----2 2 d



1 ---- + ---- ­ sin2 1

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As 1
2 d KN r0 d T 12 -------------- ---- ( 2 ­ sin2 ) = -- r 0 ( 1 + cos2 ) = --------d 2 2 d

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Integrating the above expression over solid angle gives the following expression for the total Klein-Nishina cross-section: 3 1 + x 2 x(1 + x) KN = -- ----------- ---------------------- ­ ln ( 1 + 2 x ) 4 x3 1 + 2 x 1 (1 + 3x) + ----- ln ( 1 + 2 x ) ­ ---------------------2x ( 1 + 2 x )2 h x = -----------me c 2

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Limits Nonrelativistic regime ( x « 1 ) : 26 x 2 = T 1 ­ 2 x + ---------- 5 Extreme relativistic regime ( x » 1 ) : 3 ­ 1 ln 2 x + 1 = -- T x - 8 2 0 as x That is, electrons are less efficient scatterers of high energy photons.
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3 Scattering from electrons in motion The above applies to an electron at rest. For most applications, 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.

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3.1 Rest frame and electron frame Here we use a device that it used a lot in High Energy Astrophysics, 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 ) h « m e c 2 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:
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Notation: (lab frame)
1 = h 1
1

S (rest frame) 1 1

= h



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.
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= Initial photon energy in lab frame Angle between initial photon direction = and electron velocity in lab frame = Initial photon energy in electron frame Angle between initial photon direction = and electron velocity in rest frame 1 = Scattered photon energy in lab frame 1 = Scattered photon angle in lab frame 1 = Scattered photon energy in rest frame 1 = Scattered photon angle in rest frame
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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 S the electron has velocity v and Lorentz factor v 2 ­ 1 / 2 = ( 1 ­ 2 ) ­1 / 2 = 1 ­ --- c 2 v = -c

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Recall the transformation between energies between two relatively moving frames. For massive particles: vp E = E 1 ­ ----- cos c where E is the energy of the particle and v p is its velocity. For photons (where we denote energy by ): where = ( 1 ­ cos ) = ( 1 ­ µ ) µ = cos

The Lorentz factor = ( 1 ­ 2 ) ­ 1 / 2 .

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The reverse and forward transformations are: = ( 1 ­ cos ) = ( 1 ­ µ ) 1 = 1 ( 1 + cos 1 ) = 1 ( 1 + µ 1 ) Hence, except for values of near 0 ( µ 1 ), the photon picks up a factor of when we transform to the rest frame and except for values of 1 near ( µ 1 ­ 1 ), we pick up a further factor of when we transform the energy of the scattered photon back to the lab frame.

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Energy gain from Thomson scattering Assuming that Thomson scattering applies in the rest frame, (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: : 2 Hence, in being scattered by an electron, a photon increases in energy by a factor of order 2 . Obviously for relativistic electrons, this can be substantial.

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Condition for Thomson scattering in the rest frame Since, the energy of the photon in the rest frame is of order , then the condition for Thomson scattering to apply in the rest frame is: « m e c 2 h « m e c 2 me c 2 h « -----------

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Example Consider scattering of radio emitting photons by electrons with a Lorentz factor of order 10 4 . First, assuming that the Thomson limit applies in the rest frame, the typical photon frequency produced is 2 10 8 â 10 9 Hz 10 17 Hz i.e. X-ray frequencies.

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Is the condition for the Thomson limit satisfied? We require the initial soft photon energy to satisfy:
­ 31 82 me c 2 9.11 â10 â ( 3 â10 ) « ------------ = --------------------------------------------------------- = 10 16 ­ 34 h 6.6 â10 â 10 4 and this is easily satisfied for radio photons.

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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 photons. When this is the case, the process is known as inverse Compton scattering.

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4.1 Single electron power Isotropic distribution of photons Photons impinging on a population of electrons have a distribution of directions. For simplicity and with physical applications in mind, we consider a distribution of photons which is isotropic in the lab frame. Let f ( p ) d 3 p = No density of photons within range d 3 p n ( ) d = No density of photons within d In the lab frame, where the photons are assumed isotropic: 4 p 2 f ( p ) dp = n ( ) d
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The calculation of the single electron power follows the following scheme. Isotropic distribution of photons in lab frame Anisotropic distribution of photons in rest frame of electron

Transform back to lab frame and calculate power in that frame

Compute power in rest frame, using conservation of energy in rest frame
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Electron rest frame
S 1 1

We consider the geometry at the left. Consider photons incident at an angle to the x axis which are scattered into a range of angles indicated by 1 . Consider the incident flux on the electron due to photons within a region d 3 p



of momentum space, where d 3 p = p 2 dp sin d

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The number density of photons in this region of momentum space is: n = f ( p)d 3 p Therefore: Incident photon flux per unit area = n â c = cf ( p ) d 3 p per unit time No of photons scattered per unit time = T â cf ( p ) d 3 p

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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: P = T â â cf ( p ) d 3 p Integrating over all momenta in the rest frame: P = c T f ( p ) d 3 p Transformation back to lab frame We know that the distribution function is invariant under Lorentz transformations: f( p) = f ( p)
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We also need to determine the transformation law for and d 3 p as a result of the Lorentz transformations between S and S . Determining the transformation of d 3 p involves determining the Jacobean of the transformation from S to S .Using the transformations for photons derived from the 4-momentum, we have for the spatial momentum components:

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p ­ -- = ( p ­ p ) px = x x c pz = pz py = py px px p ---------- = 1 ­ -------- = 1 ­ ----- px p px px ---------py = ( 1 ­ cos ) = ( 1 ­ µ ) py pz px ---------- = ­ ---= ­ ----p p pz

py pz ---------- = --------- = 1 py pz
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For the energy px = ( ­ cp x ) = 1 ­ ----- p = ( 1 ­ cos ) = ( 1 ­ µ ) Jacobean of the transformation between lab momentum space and rest frame momentum space pz py ( 1 ­ µ ) ­ ----- ­ ---p p 0 1 0 0 0 1

J=

= ( 1 ­ µ )

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Hence, d 3 p = ( 1 ­ cos ) d 3 p f ( p ) d 3 p = 2 ( 1 ­ cos ) 2 f ( p ) d 3 p The power radiated in the rest frame is: P = c T f ( p ) d 3 p = c T 2 ( 1 ­ cos ) 2 f ( p ) p 2 d p d c T 2 = --------------- ( 1 ­ cos ) 2 n ( ) d sin d d 4

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The integral is over variables in the lab frame where the photon distribution is isotropic. We obtain for the power: 2 P = c T 2 1 + ----- n ( ) d 3 0 The quantity
U ph = n ( ) d 0

is the photon energy density in the lab frame.

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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 dE 1 P = ----------dt dE 1 P = --------dt

where dE 1 and dE 1 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:

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dt + dx- = dt dt = --------- c since in the rest frame of the electron, dx = 0 . Consider a bunch of photons emitted with energy dE 1 and x -momentum d p x = ( dE 1 / c ) cos 1 at the angle 1 . The Lorentz transformed energy of this bunch of photons in the lab frame is dE 1 = ( dE 1 + cd p x ) = ( dE 1 ) ( 1 + cos 1 )

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However, scatterings with cos 1 > 0 and cos 1 < 0 are equally likely, because of the symmetry of the Thomson cross-section, so that the averaged contribution to dE 1 is dE 1 = dE 1 Hence the power in the lab frame is: dE 1 dE 1 dE 1 --------------- = -------------- = ----------dt dt dt That is, P = P
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Hence the scattered power in the lab frame is 2 P = c T 2 1 + ----- U ph 3 4.2 Number of scatterings per unit time In the electron rest frame: dN -------- = Number of scatterings per unit time dt = c T f ( p ) d 3 p

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Again we transform this equation to the lab frame: dN -------- = c T ( 1 ­ cos ) f ( p ) p 2 d p sin d d dt 2 c T = ------------------- ( 1 ­ cos ) n ( ) d sin d 4 = c T N ph where the photon number density is N ph = n ( ) d

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In transforming dN / dt to the lab frame, we note that dN is a number and so is Lorentz invariant, so that dN ­ 1 dN- = c N ------ = -------T ph dt dt

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

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Hence, the number of scatterings per unit time per unit photon energy by a single electron is c T n ( ) . Hence the energy removed from photons within d is c T n ( ) d . Hence, the energy removed from the photon field is given by: dE 1 --------- = ­ c T n ( ) d = ­ c T U ph dt

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Therefore, the nett energy radiated is: dE rad 2 1 + 1 2 ­ 1 -------------- = P compt = c T U ph 3 dt 4 = -- T c 2 2 U ph 3 where we have used 22 = 2 ­ 1

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4.4 Comparison of synchrotron and inverse Compton power We already know that the synchrotron power of an electron is given by 4 P synch = -- T c 2 2 U B 3 B2 U B = -------2 µ0 Hence, U ph P compton ---------------------- = --------P synch UB
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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 comparable to the magnetic energy density.
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Consider the energy density of blackbody radiation:
­ 16 85k 4 4 BB = aT a = ----------------- = 7.57 â10 J m ­ 3 K ­ 4 15 h 3 c 3 This is equal to the energy density of a magnetic field when

B2 -------- = aT 4 B = 2 µ0 Therefore, when

2 = 4.4 â10­ 11 T 2 2 µ 0 aT

For the microwave background, T = 2.7 K .
­ 10

B < B crit = 3.2 â10

T = 3.2 µ G
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the radiation from inverse Compton emission is more important 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 distribution, then, 2 1 3 kT v 2 2 = ---------- = -------- mc 2 c2
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Hence, the total Compton emission from a single electron is given by: 4 kT --------- c U P Compton = mc 2 T ph The volume emissivity from the plasma with electron number density n e is: 4 kT --------- c n U j Compton = mc 2 T e ph

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Compton scattering of "soft" photons by hot thermal electrons in the coronae of accretion disks is thought to be responsible 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.

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4 -- T c 2 2 U ph 3 1 = -----------------------------------c T N ph 422 4 2 2 U ph = -- --------- = -- 3 3 N ph

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Inverse Compton emission from relativistic electrons Thus for photons scattering off relativistic electrons, the mean amplification of energy per scattering is 4 2 / 3 , supporting the order of magnitude estimate of 2 for this parameter. Clearly, for relativistic electrons, the energy gain is substantial, underlining the importance of inverse Compton emission as an astrophysical process.

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