PHYS 4011 Ð HEA Lec. 5 Lecture 5: Radiation from Moving Charges

In the previous lectures, we found that particles interacting with scattering centres (i.e. other particles or fields) can be accelerated to very high energies. When these accelerated particles are charges, they produce electromagnetic waves. Radiation is an irreversible flow of electromagnetic energy from the source (charges) to infinity. This is possible only because the 2 electromagnetic fields associated with accelerating charges fall off as 1/r instead of 1/r as is the case for charges at rest or moving uniformly. So the total energy flux obtained from the Poynting flux is finite at infinity.

Note: The material in this lecture is largely a review of some of the content presented in the Advanced Electromagentic Theory courses in 3rd and 4th years. As such, it is non-examinable for this course.

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PHYS 4011 Ð HEA Lec. 5 5.1 Overview of the Radiation Field of Single Moving Charges

Consider a radiating charge moving along a trajectory r0(t). Suppose we wish to measure the radiation field at a point P at a time t. Let the location of this field point be r(t). At time t, the charge is at point S, located at r0(t). But the radiation measured at P was actually emitted by ! ! the particle when it was at point S at an earlier time t . This is because an EM wave has a ! finite travel time |r(t) − r0(t )|/c before arriving at point P . Thus, the radiation field at P ! needs to be specified in terms of the time of emission t , referred to as the retarded time:

! ! |r(t) − r0(t )| t = t − (1) c

Information from the charged particle’s trajectory arriving at field point P has propagated a finite dis- tance and taken a finite time to reach there at time t. At most, only one source point on the trajectory is in communication with P at time t.

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The radiation field at P at time t is calculated from the retarded scalar and vector potentials V and A using E = −∇V − ∂A/∂t and B = ∇ × A. But Maxwell’s equations define these in terms of the continuous charge and current densities ρ and J. So to evaluate V and A for a point charge, it is necessary to integrate over the volume distribution at one instant in time taking the limit as the size of the volume goes to zero. Formally, this can be done by taking

ρ(r, t) = qδ(r − r0(t)) and J(r, t) = qv(t)δ(r − r0(t)), where v(t) = r˙ 0(t) is the velocity of the charge. Similarly, another delta function is introduced to single out only the source point at the retarded time that we are interested in. After integrating over the volume, we have ! r r ! r q δ(t − t + | (t) − 0(t )|/c) ! V ( , t) = ! dt (2) 4π%0 |r(t) − r0(t )| " ! r r ! A r µ0q v ! δ(t − t + | (t) − 0(t )|/c) ! ( , t) = (t ) ! dt 4π |r(t) − r0(t )| " After introducing the simplifying notation ! ! ! ! ! R(t ) R(t ) = r(t) − r0(t ) ,R(t ) = |r(t) − r0(t )| , Rˆ = R(t!)

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the integral can be solved with a change of variables giving ! µ0 qv(t ) v A(r, t) = = V (r, t) Lienard–Wiechart´ potentials (3) 4π (1 − Rˆ · v(t!)/c)R c2

These are the famous retarded potentials for a moving point charge.

Points to note: ! 1. The factor (1 − Rˆ · v(t )/c) implies geometrical beaming. It means that the potentials ! are strongest at field points lying ahead of the source point S and closely aligned with the particle’s trajectory. The effect is enhanced when the particle speed becomes relativistic. 2. Retardation is what makes it possible for a charged particle to radiate. To see why, note that the potentials fall off as 1/r. Differentiation to retrieve the fields would yield a 1/r2 fall-off if there were no other r-dependence in the potentials. This does not give rise to a net electromagnetic energy flux as r → ∞ and hence, no radiation field. (Recall that the rate of change of EM energy goes as S · dA). However, there is an implicit r-dependence in the retarded time that leads! to a 1/r-dependence in the fields, upon differentiation of the potentials. This does result in a net flow of EM energy towards infinity.

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The radiation field

The differentiation of the Lienard–Weichart´ potentials to obtain the radiation field of a single moving charge is lengthly, but straightforward (see Jackson for details). Writing the charged ! particle’s velocity at the retarded time as βc = r˙ 0(t ) and its corresponding acceleration as ! β˙ c = ¨r0(t ), the fields are

2 q (Rˆ − β)(1 − β ) Rˆ 1 ˙ E(r, t) = + × (Rˆ − β) × β (4) 4π%0 # (1 − Rˆ · β)3R2 (1 − Rˆ · β)3R c & $ % 1 B(r, t) = Rˆ × E(r, t) (5) c

The first term on the RHS of the E-field is the velocity field. It falls off as 1/R2 and is just the generalisation of Coulomb’s Law to uniformly moving charges. The second term is the acceleration field contribution. It falls off as 1/R, is proportional to the particle’s acceleration and is perpendicular to Rˆ .

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This electric field, along with the corresponding magnetic field, constitute the radiation field of a moving charge:

q Rˆ 1 ˙ Erad(r, t) = × (Rˆ − β) × β 4π%0 #(1 − Rˆ · β)3R c & $ % 1 Brad(r, t) = Rˆ × Erad(r, t) (6) c

Note that Erad, Brad and Rˆ are mutually perpendicular.

45 PHYS 4011 Ð HEA Lec. 5 5.2 Radiation from Nonrelativistic Charged Particles

The Larmor formula

When β & 1, the radiation fields simplify to Rˆ E r q 1 Rˆ ˙ rad( , t) = × 2 × v (7) 4π%0 ' R c * ( )

and Brad(r, t) follows from (6). Note that Erad lies in the plane containing Rˆ and v˙ (i.e. the plane of polarisation) and Brad is perpendicular to this plane. If we let θ be the angle between Rˆ and v˙ , then E B q v˙ | rad| = c| rad| = 2 sin θ 4π%0 Rc The Poynting vector is in the direction of Rˆ and has the magnitude

2 2 1 2 µ0 q v˙ 2 S = Erad = 2 2 sin θ (8) µ0c 16π c R

We now want to express this as an emission coefficient.

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Since S is the EM energy dW emitted per unit time dt per unit area dA (i.e. S = dW/(dtdA)), we can write dA = R2dΩ , where dΩ is the solid angle about the direction Rˆ of S. So the power emitted per solid angle is

dW µ0 = SR2 = q2v˙2 sin2 θ dtdΩ 16π2c 2 Note the characteristic dipole pattern ∝ sin θ: there is no emission in the direction of acceleration and the maximum radiation is emitted perpendicular to the acceleration. The total electromagnetic power emitted into all angles is obtained by integrating this:

2π π +1 dW µ0 2 2 2 µ0 2 2 2 P = = 2 q v˙ sin θ sin θ dθ dφ = q r¨0 (1 − µ ) dµ dt 16π c 8πc − "0 "0 " 1 The integral gives a factor of 4/3, whence we arrive at the Larmor formula for the electromagnetic power emitted by an accelerating charge:

2 2 µ0q r¨0 P = Larmor formula (9) 6πc

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The dipole approximation

To calculate the radiation field from a system of many moving charges, we must keep track of the phase relations between the radiating sources, because the retarded times will differ for each charge. In some situations, however, it is possible to neglect this complication and use the principle of superposition to determine the properties of the radiation field at large r.

Suppose we have a collection of particles with positions ri, velocities vi and charges qi. Suppose further that these particles are confined to a region of size L and that the typical timescale over which the system changes is τ . If τ is much longer than the light travel time across the system (i.e. if τ ( L/c) then the differences in retarded time across the source are negligible. The timescale τ is also the characteristic timescale over which Erad varies, the above condition is equivalent to λ ( L, where λ ∼ cτ is the characteristic wavelength of the emitted radiation. The timescale τ also represents the characteristic time a particle takes to change its velocity substantially. Then τ ( L/c implies v & c, so we can use the nonrelativistic limits of the radiation fields derived above.

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Applying the superposition rule, we have Rˆ E qi i 1 Rˆ r rad = × 2 i × ¨i 4π%0 R c i ' i * + ( ) where Ri is the distance between each source point ri and the field point r. But the

differences between the Ri are negligible, particularly as r → ∞. So we can just keep ! R = |r − r0(t )| as a characteristic distance and use the definition for the dipole moment, viz.

d = qiri (10) i + to get Rˆ E 1 1 Rˆ d¨ rad = × 2 × (11) 4π%0 ' R c * ( ) Then, as before, we find dW µ0 = d¨2 sin2 θ dtdΩ 16π2c

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and the total power is 2 µ0d¨ P = (12) 6πc which is directly analogous to Larmor’s formula (9) for an individual charge. As with the case for a single charge, the instantaneous polarisation of E lies in the plane of d¨and R and the 2 radiation pattern is ∝ sin θ. So there is zero radiation along the axis of the dipole and a maximum perpendicular to the axis. Since there is no azimuthal dependence, the 3D intensity profile looks like a doughnut (N.B. This is also true for the single charge case).

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PHYS 4011 Ð HEA Lec. 5 5.3 The Radiation Spectrum

For astrophysical applications, we wish to specify a spectrum of radiation. This specifies how the power is distributed over frequency. First we introduce the Fourier transform of the acceleration of a particle through the Fourier transform pair: +∞ v 1 v ˙ (t) = 1 2 ˙ (ω) exp(−iωt) dω (2π) / −∞ " +∞ v 1 v ˙ (ω) = 1 2 ˙ (t) exp(iωt) dt (13) (2π) / −∞ " Then we use Parseval’s theorem, which relates this as follows: +∞ +∞ 2 2 |v˙ (ω)| dω = |v˙ (t)| dt (14) ∞ ∞ " " We can also use the relation ∞ 0 2 2 |v˙ (ω)| dω = |v˙ (ω)| dω = (15) −∞ "0 " which is valid provided v˙ (t) is real. Applying these realtions to the Larmor formula (9), we can

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determine the total energy radiated by a single charged particle with an acceleration history v˙ (t): +∞ +∞ 2 2 ∞ µ0q 2 µ0q 2 P dt = |v˙ (t)| dt = |v˙ (ω)| dω (16) −∞ −∞ 6πc 3πc 0 " " ∞ " Since the total emitted energy must also equal 0 (dW/dω) dω, then the energy per unit bandwidth is just ! 2 dW µ0q 2 = |v˙ (ω)| (17) dω 3πc Clearly, this energy is just emitted during the period that the particle is experiencing acceleration. In the dipole approximation for multiple particles, this expression becomes

dW µ0 4 2 = ω |d(ω)| (18) dω 3πc This is because differentiating wrt t twice introduces a factor ω2 in the Fourier transform – c.f. ∞ −1/2 + d(t) = (2π) −∞ d(ω) exp(−iωt) dω =⇒ ∞ ¨ −1/2 + 2 v 2d d(t) = −(2π) ! −∞ ω d(ω) exp(−iωt) dω, so e ˙ (ω) = ω (ω). ! 52 PHYS 4011 Ð HEA Lec. 6

Lecture 6: Bremsstrahllung Radiiatiion

When a high speed electron encounters the Coulomb field of another charge, it emits bremsstrahlung radiation, also known as free-free emission. The word bremsstrahlung means braking radiation because the electron rapidly decelerates when the other charge is a massive ion. The derivation can be done classically using the dipole approximation for nonrelativistic particles, with quantum corrections added as “Gaunt factors” to the classical formulas. The quantum corrections become important when photon energies become comparable to energies of the emitting particles. We only need to consider electron-ion bremsstrahlung because for collisions between like charges (e.g. electron-electron), the dipole approximation predicts zero radiation and a higher order calculation is required. This also means that less radiation is emitted for collisions between like particles. In electron-ion bremsstrahlung, the electrons are the primary emitters because their accleration is m /m times greater. ∼ p e

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PHYS 4011 Ð HEA Lec. 6 6..1 Emiissiion ffrom Siinglle Speed Ellecttrons Consider an electron moving with velocity v past an ion of charge Ze with impact parameter b. We will assume small-angle scattering so there is negligible deviation in the electron’s trajectory from a straight line (see figure).

Let R be the position vector of the electron from the ion. Then the dipole moment is d = eR and its second time derivative is d¨ = ev˙ . We want an emission spectrum for the − − bremsstrahlung radiation using eqn. (17) in Lec. 5, viz. 2 dW µ0e = v˙ (ω) 2 (1) dω 3πc | | and using ∞ 1 + v˙ (ω) = v˙ (t) exp(iωt) dt (2) (2π)1/2 !−∞ 55 PHYS 4011 Ð HEA Lec. 6

Consider first that the electron is interacting with the ion only over a finite collision time τ b/v. When ωτ 1 the exponential in the integrand oscillates rapidly and the resulting # $ integral is small. When ωτ 1, on the other hand, the exponential term is approximately % unity and the resulting integral is just v˙ = dv/dt ∆v over the time interval dt τ . So we ≈ ≈ have 1 1/2 ∆v ωτ 1 v˙ (ω)  (2π) % (3) | | #  0 ωτ 1 $ So our radiation spectrum goes as 

2 µ0e 2 dW 2 ∆v b v/ω  6π c | | % (4) dω #  0 b v/ω $ 

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Now we can work out ∆v by noting that the total Coloumb force on the electron is Ze2/(4π$ R2) in the Rˆ direction. The perpendicular component of acceleration is the 0 − strongest and will thus make the dominant contribution to the radiation spectrum, so we can write 1 Ze2 b dt 1 Ze2 2 (5) ∆v 2 2 2 3/2 = # 4π$0 me ! (b + v t ) 4π$0 me bv (the integral turns out to be elementary). Substituting this into the expression for the radiation spectrum gives 2 6 8Z e dW 3 3 2 2 2 b v/ω  3πc (4π"0) me b v % (6) dω #  0 b v/ω $ This is the spectrum for small angle scatterings by a single electron off a single ion. Next we want to generalise this to the case of a realistic plasma in which we have many electrons interacting with many ions.

57 PHYS 4011 Ð HEA Lec. 6 Radiation spectrum for an electron-ion plasma

Let the ion and electron number densities in the plasma be ni and ne. Then the flux of electrons incident on an ion is nev for a fixed electron speed v. The element of area about an ion over which an electron encounter occurs is approximately 2πb db. So the emission per unit time per unit volume per unit frequency range is ∞ dW dW (b) = neni2πv b db (7) dωdV dt !bmin dω

where bmin is a minimum impact parameter to be chosen. Now it is difficult to see how the solution for dW/dω obtained above in the asymptotic limits b v/ω and b v/ω can be % $ used to solve this integral over a full range of impact parameters. However, it turns out that the solution can be well approximated by using the non-zero asymptotic solution for dW/dω because the integral is just logarithmic in b: 2 6 dW 16neniZ e bmax (8) 3 3 2 ln dωdV dt # 3c (4π$0) me v % bmin & where b v/ω is some value beyond which the b v/ω limit no longer applies and the max ∼ % contribution to the integral becomes negligible. We can set bmax = v/ω, even though it is

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uncertain because it is inside the logarithm. An appropriate value of bmin can be chosen to correspond to the break down of the small-angle scattering approximation. So when ∆v v, ∼ (5) implies b 2Ze2/(4π$ m v2). min # 0 e The exact expression for the radiation spectrum can be obtained with a full quantum treatment. For convenience, a quantum correction is added to the classical formula. This correction term is known as the Gaunt factor,

√3 bmax Gff (v, ω) = ln free-free Gaunt factor (9) π % bmin & giving 2 6 dW 16πneniZ e (10) 3/2 3 3 2 Gff (v, ω) dωdV dt # 3 c (4π$0) me v This is now the bremsstrahlung radiation emitted per unit time per unit volume per unit frequency by single-speed electrons interacting with many ions. Next we compute the volume emissivity for a thermal distribution of electron speeds.

59 PHYS 4011 Ð HEA Lec. 6 6..2 Emiissiion ffrom a Thermall Diisttriibuttiion off Ellecttrons

In a thermal plasma, the velocity distribution of the electrons (and ions) is Maxwellian, which is isotropic. The number of thermal particles with velocity v in the range d3v is mv2 dn(v) = f(v)d3v = f(v)4πv2 dv exp v2 dv ∝ %− 2kT &

We need to average the single-speed radiation spectrum over this distribution function for all 1 2 > electron speeds satisfying 2 mev ∼ hω/2π, i.e. ∞ dW (v,ω) 2 2 dW (T, ω) vmin dωdV dt v exp( mv /2kT ) dv = ∞ − (11) dωdV dt ' v2 exp( mv2/2kT ) dv 0 − ' Using dω = 2π dν, the final result is the expression for the free-free volume emissivity:

6 1/2 ff dW 1 32πe 2π 2 hν ¯ jν = Z neni exp Gff (12) 3 3/2 3 ≡ dV dtdν (4π$0) 3me c %3kT & %−kT &

where G¯ff is now the velocity averaged Gaunt factor. Its value is typically of order unity.

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Points to note:

1. The only frequency dependence is in the exponential term exp( hν/kT ). So the − spectrum declines exponentially at frequencies hν kT , but is approximately flat for $ hν kT . % − 2. The emissivity is also proportional to T 1/2. So for hν kT , the spectrum is lower for % higher T . But for hν kT , the exponential cutoff extends to higher frequencies for $ higher T , so there is more high-energy emission. − − 3. The units of emissivity are W m 3 Hz 1. So to calculate the total radiative power (i.e. luminosity) in bremsstrahlung emission from a real astrophysical source, we simply ff integrate jν over an appropriate source volume and over the frequency bandwidth that we are interested in.

61 PHYS 4011 Ð HEA Lec. 7 Lecture 7: Radiiatiion from Moviing Charges IIII Emission from Relativistic Particles

In Lec. 5 (Sec. 5.1, eqn. 6), we found the expressions for the radiation field resulting from a nonuniformly moving charge, viz.

q Rˆ 1 Erad r Rˆ β β˙ (1) ( , t) = 3 × ( − ) × 4π"0 !(1 − Rˆ · β) R c $ " # Here, R is the displacement from the retarded source point to the field point and βc is the particle’s velocity evaluated at the retarded time. We used this equation in the nonrelativistic limit (β # 1) to calculate the Poynting flux and then derive the Larmor formula (c.f. eqn. 9 in Lec. 5) for the total power emitted by a nonrelativistic particle. How do we calculate the total power emitted by a relativistic particle? We could follow the same procedure, keeping terms involving β, but the derivation is complicated by the fact that the Poynting flux at the field point where we observe the radiation is not the same as the rate at which the energy left the source point, because the charge is moving. There is an easier way which takes advantage of the Lorentz invariance of the total power. We then need to consider how the radiation is emitted in the particle’s rest frame and how it is received in an observer’s rest frame. 62

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Total power emitted

Step 1. Consider an instantaneous rest frame K! such that a particle has zero velocity at a certain time and moves nonrelativistically for infinitessimally neighbouring times. We can then use the Larmor formula to calculate the total power P ! emitted in K!. We then need to how to transform back to the rest frame K of an observer who measures P . The two references frames have a relative speed βc. First note that P ! = dW !/dt!, where dW ! is the energy emitted in time dt! in frame K!. These quantities transform as

dW = γdW ! , dt = γdt!

2 1 2 where γ = (1 − β ) / is the Lorentz factor. Note that the Lorentz factors cancel out, i.e. dW γdW ! dW ! P = = = = P ! (2) dt γdt! dt! so the total emitted power is Lorentz invariant.

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2 ! 2 ! µ0q |a | ! Step 2. Now we have the identity P = P = 6πc . In this expression, a is the 3-acceleration in K! and it is possible to relate this to the 4-acceleration, thereby allowing us to express P in the more general covariant form (i.e. obeying special relativity). The 4-acceleration can be defined as µ 2 µ µ dv d x 2 µ 0 2 a ≡ = = , a = a a = −a a0 + |a| (3) dτ dτ 2 µ where vµ = γ(c, v) is the 4-velocity. Note that the 4-velocity and 4-acceleration are µ µ 1 µ 1 2 ! orthogonal: a vµ = (dv /dτ)vµ = 2 d(v vµ)/dτ = 2 d(−c )/dτ = 0. Now in K , we ! 2 !2 !0 ! !µ can write |a | = a + a a0. But in the particle’s rest frame, v = (c, 0) and since !µ ! !0 a vµ = 0, we must have a = 0. Thus, we can write

! 2 !µ ! µ |a | = a aµ = a aµ

and 2 µ0q P = aµa (4) 6πc µ which is now in a manifestly covariant form (i.e. can be evaluated in any frame).

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µ !2 Step 3. Since a aµ = |a| , we can keep P in terms of the 3-acceleration and we can write the acceleration in terms of components parallel and perpendicular to the particle’s velocity, i.e. ! ! a" and a⊥. The transformation properties of these components are

! 3 ! 2 a" = γ a" , a⊥ = γ a⊥

Thus, we have 2 2 2 µ0q 2 µ0q 2 2 µ0q 4 2 2 2 P = |a!| = (a! + a! ) = γ (γ a + a ) (5) 6πc 6πc " ⊥ 6πc " ⊥ Clearly, the power emitted increases drastically as a particle becomes relativistic. This expression can be written another way:

2 µ0q 6 2 2 P = γ a − |β × a| relativistic Larmor formula (6) 6πc % & We will use this later to derive an expression for power emitted in synchrotron radiation by relativsitic electrons in a magnetic field.

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Angular distribution of radiation

Whilst the total radiation power emitted by a relativistic particle is Lorentz invariant, its angular distribution is not. Consider again the rest frame of the particle K!. We want to find how the emitted energy per solid angle transforms from K! to an observer rest frame K. Suppose the relative velocity v between these two frames is along the x-axis. Then the change in energy transforms as ! ! dW = γ(dW + vdpx) µ µ from the transformation properties of the 4-momentum p = (E/c, p) = m0v . For photons, we have |p| = E/c, so if we define an angle ϑ measured w.r.t. the x-axis, then

px = (E/c) cos ϑ, so dW = γ(1 + β cos ϑ!)dW ! (7)

Now we want to find how much energy is radiated in the solid angle dΩ! = sin ϑ!dϑ!dφ! = d cos ϑ!dφ! about ϑ! and we want to know how dW !/dΩ! transforms to give dW/dΩ, where dΩ = d cos ϑdφ.

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! Geometry for dipole emission from a particle instantaneously at rest in frame K .

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The transformation of cos ϑ! is given by the aberration of light formula: cos ϑ! + β cos ϑ = (8) 1 + β cos ϑ! Differentiating gives d cos ϑ! d cos ϑ = γ2(1 + β cos ϑ!)2 The azimuthal angle is invariant, so dφ = dφ!. Thus, d cos ϑ!dφ! dΩ! dΩ = d cos ϑdφ = = γ2(1 + β cos ϑ!)2 γ2(1 + β cos ϑ!)2 and so we have ! dW 3 3 dW = γ (1 + β cos ϑ!) (9) dΩ dΩ! To get the angular distribution of the power, we divide through by the time interval dt = γ(1 − β cos ϑ)dt!, which includes a Doppler correction resulting from the motion of the source (particle). This then gives us an expression for the received power per solid angle in frame K: ! ! dP 4 4 dP 1 dP = γ (1 + β cos ϑ!) = (10) dΩ dΩ! γ4(1 − β cos ϑ)4 dΩ! 68

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If the emission is isotropic in the particle’s rest frame K!, then in K, it will be peaked forward in the direction of motion (i.e. ϑ → 0). We know that in K!, we can use the dipole approximation , which gives (c.f. Lec. 5)

! 2 dP µ0q 2 2 = a! sin θ! (11) dΩ! 16π2c where θ is the angle between the acceleration and the direction of emission. So in the particle’s rest frame, the emission drops to zero in the direction of acceleration and peaks in ! ! ! the direction perpendicular to it (i.e. the dipole torus pattern). Writing a = a" + a⊥ and using the transformations defined earlier yields 2 2 2 2 2 ! dP µ0q (γ a" + a⊥) 2 3P sin θ = sin θ! = (12) dΩ 16π2c (1 − β cos ϑ)4 8πγ4 (1 − β cos ϑ)4 Note the strong dependence on the factor 1 − β cos ϑ in the denominator. This term dominates when ϑ → 0 and β → 1. In other words, the radiation is observed to be strong in the forward direction with respect to the particle’s motion. This is referred to as relativistic beaming. We still have to transform θ! back to angles in K and this is difficult to do for the general case. We can do it for some special cases.

69 PHYS 4011 Ð HEA Lec. 7 Case 1: acceleration parallel to velocity ! ! Here, a⊥ = 0 and θ = ϑ and the transformation can be done using the aberration formula (8) to give 2 2 sin θ sin θ! = (13) γ2(1 − β cos ϑ)2 Substituting into (12) yields

2 2 dP µ0q 2 sin θ = a (14) dΩ 16π2c " (1 − β cos ϑ)6

which peaks in the forward direction at ϑ ∼ 1/γ.

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Case 2: acceleration perpendicular to velocity ! ! ! We put a" = 0 and use cos θ = sin ϑ cos φ . The transformation is

2 2 2 sin ϑ cos φ sin θ! = 1 − (15) γ2(1 − β cos ϑ)2

giving 2 2 2 dP µ0q 2 1 sin ϑ cos φ = a 1 − (16) dΩ 16π2c ⊥ (1 − β cos ϑ)4 γ2(1 − β cos ϑ)2 ' ( Again, this peaks in the forward direction near ϑ ∼ 1/γ, with a smaller peak at larger ϑ due to the azimuthal dependence.

forward beaming (observer rest frame) dipole emission (particle rest frame)

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Case 3: ultrarelativistic limit When γ ' 1, the term 1 − β cos ϑ becomes extremely small and since this appears in the denominator in dP/dΩ, this term dominates and the emission pattern becomes strongly peaked in the forward direction (i.e. direction of motion). In fact, we can write 1 + γ2ϑ2 1 − β cos ϑ ∼ (17) 2γ2

Substituting this limit back into (12) we get for the parallel and perpendicular cases

2 2 2 2 dP" 4µ0q a" 10 γ ϑ ∼ γ (18) dΩ π2c (1 + γ2ϑ2)6 2 2 2 2 4 4 dP⊥ µ0q a 8 1 − 2γ ϑ cos(2φ) + γ θ ∼ ⊥ γ (19) dΩ π2c (1 + γ2ϑ2)6

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Lecture 8: Synchrotron Radiiatiion II

Charged particles in a magnetic field radiate because they experience an acceleratio perpendicular to the field. If the particles are nonrelativistic, the radiation is referred to as cyclotron emission and the frequency of emission is directly related to the particle gyration fre- quency. This results in a discrete emission spectrum which usually does not extend beyond optical/UV fre- quencies in most astrophysical situations. When the particles are relativistic, however, the radiation is re- ferred to as synchrotron emission and results in a continuum spectrum because the frequency of emis- sion extends over many higher order harmonics of the gyration frequency. Synchrotron emission by relativis- tic particles in a magnetic field is a prevalent radia- tion process in astrophysics. The emissivity is broad- band and extends all the way from radiofrequencies to X-ray and γ-ray energies. Many real high-energy sources in astrophysics are also sources of strong ra- dio emission due to synchrotron radiation (e.g. the radio galaxy 3C223, right, shown with X-ray colour contours and radio line contours overlaid).

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8..1 Tottall Emiitttted Synchrottron Power

Consider a relativistic particle of 3-momentum p = γmv in a steady magnetic field B. The equation of motion of the particle is dp d = (γmv) = qv × B (1) dt dt We now make the assumption that γ ≈ const. This is a valid assumption provided the emitted radiation field does not have a back reaction on the particle’s motion (i.e. provided d(γmc2)/dt = qv · E ≈ 0). So γ ≈ const gives mγdv/dt = qv × B and we can separate the velocity into components parallel and perpendicular to B:

dv! dv⊥ q = 0 , = v × B (2) dt dt γm ⊥

Thus, v! = const and since the total |v| = const (because γ = const), then |v⊥| = const also. Thus, there is uniform circular motion in the plane normal to B and the acceleration is perpendicular to the velocity in this plane.

74 PHYS 4011 Ð HEA Lec. 8 The combination of circular motion ⊥ to B and uniform motion $ to B results in helical motion (see figure). The magnitude of the acceleration is

|q|v⊥B a = = Ωv (3) ⊥ γm ⊥

where Ω = |q|B/γm is the gyrofrequency (or cyclotron frequency). We can substitute this into the expression we found for total power emitted by a relativistic particle with Helical motion of an electron in a magnetic field acceleration perpendicular to velocity, eqn. B results from the combination of uniform motion (5) in Lec. 7, viz. along B and circular motion perpendicular to B.

2 2 2 2 4 µ0q µ0q q B 1 q 4 2 4 2 2 2 2 2 (4) P = γ a⊥ = γ 2 2 v⊥ =⇒ P = 2 γ β B sin α 6πc 6πc γ m 6π#0c m where α is the pitch angle between v and B. Note the dependence on m: synchrotron emission is much less efficient for protons than for electrons.

75

PHYS 4011 Ð HEA Lec. 8 For an isotropic distribution of velocities, it is necessary to average over all pitch angles for a given speed β: 2 +1 2 β 2 1 2 2 2 2 &β⊥' = sin α dΩ = β (1 − cos α) d cos α = β (5) 4π ! 2 !−1 3 So the total power emitted by an electron, averaged over all pitch angles is 2 1 q4 2 2 2 (6) P = 2 γ β B 3 6π#0c m This is also sometimes expressed in terms of the Thomson cross-section, 4 8 2 e −29 2 T m (7) σ = πr0 = 2 2 4 = 6.65 × 10 3 6π#0me c 2 −15 where r0 = e /(4π#0mec) = 2.82 × 10 m is the classical electron radius, obtained by 2 2 equating the electrostatic potential energy, e /(4π#0r0), with the rest mass energy mec . So

2 c 2 2 2 4 2 2 P = σTγ β B = σTcγ β UB total synchrotron power (8) 3 µ0 3

2 where UB = B /2µ0 is the magnetic energy density (i.e. energy per unit volume).

76 PHYS 4011 Ð HEA Lec. 8

8..2 Synchrottron Specttrum – Qualliittattiive Treattmentt Because of beaming effects, the radiation emit- ted by a relativistic particle appears to an ob- server as being concentrated in a narrow range of directions about the particle’s velocity. Since the acceleration of an electron in a magnetic field is perpendicular to its velocity, the radiation pattern is like the one shown in the figure.

In the ultrarelativistic limit, the electron velocity is close to the speed of light, so the electron appears as though it is trying to catch up to the photons it produces.

An observer at rest will see a pulse of electromagnetic radiation E(t) confined to a time

interval δt ∼ 1/f + T , where T = 2π/Ω = 2πγme/(eB) is the gyration period. Thus, the spectrum will be spread over δω , Ω.

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Consider the diagram here in which an observer’s line-of-sight intercepts the emission cones of a relativistic electron. The observer will thus detect pulses of radiation from points 1 and 2

along the electron’s helical path. The times t1 and t2 at which the electron passes points 1

and 2 satisfy v(t2 − t1) = ∆s, where ∆s is the distance travelled by the electron along its path. If a is the radius of curvature, then ∆s = a∆θ, where ∆θ in this diagram is just equal to 2/γ, from the geometry, so ∆s = 2a/γ. But from the equation of motion, we also have ∆v evB sin α ) ∆t γm

78 PHYS 4011 Ð HEA Lec. 8

and since ∆v ) v∆θ and ∆t ) ∆s/v, we have ∆θ evB sin α Ω ) = sin α ∆s γmv v and hence, a ) ∆s/∆v ) v/(Ω sin α) and 2v ∆s ) γΩ sin α So the time interval between the emitted pulses is ∆s 2 ∆t = t − t ) ) (9) 2 1 v γΩ sin α a a a The time interval between the arrival times ∆t = t2 − t1 at the observer will be less than ∆t by an amount ∆s/c = (v/c)∆t, so we have

a a a ∆s ∆s v 2 v ∆t = t2 − t1 ) ∆t − = 1 − ) 1 − (10) c v " c # γΩ sin α " c # For γ , 1, we have v 1 1 − ∼ c 2γ2

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and so ∆ta ∼ (γ3Ω sin α)−1 (11)

So the time interval between pulses and the width of the individual pulses, E(t), are smaller than the gyration period by a factor ∼ γ3. When we take the Fourier transform of the pulses, we expect to get a broad spectrum up to some cutoff frequency ω ∼ 1/∆ta. In fact, in the exact treatment to follow, we will use the following definition

3 3 3 eB 2 ωc ≡ γ Ω sin α = γ sin α critical frequency (12) 2 2 me

The spectrum should fall off sharply at frequencies above ωc. We can estimate that the power

per unit frequency emitted per electron is P (ω) ∼ P/ωcF (ω/ωc), where F (ω/ωc) is a

dimensionless function that describes the correct behaviour of the spectrum near ωc. Using

eqn. (4) for P and ωc defined above, we have (for β ) 1) 1 e3B ω P (ω) ∼ F sin α (13) 9π#0c me $ωc %

80 PHYS 4011 Ð HEA Lec. 8 8..3 Specttrall IIndex ffor a Power-Law Ellecttron Diisttriibuttiion Note that in the above expression for P (ω), there is no explicit dependence on γ, only an

implicit dependence in the as yet undefined function F (ω/ωc). This means that when we want to calculate the spectrum for a distribution of electron energies, we only need to integrate over that function. We can define the number density of electrons with energies between ε and ε + dε (or γ and γ + dγ) as

N(ε)dε ∝ ε−pdε , N(γ)dγ ∝ γ−pdγ power law distribution (14)

< < < < over the range ε1 ∼ ε ∼ ε2 (or γ1 ∼ γ ∼ γ2). Then the total power emitted per unit volume per unit frequency is just Ptot(ω) = P (ω)N(γ)dγ. Thus,

& γ2 ω −p Ptot(ω) ∝ F γ dγ (15) !γ1 $ωc % 2 If we change integration variables to x = ω/ωc, noting that ωc ∝ γ , we have

x2 −(p−1)/2 (p−3)/2 Ptot(ω) ∝ ω F (x)x dx (16) !x1

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If the energy limits are sufficiently wide, then we can take x1 ) 0 and x2 ) ∞ and the resulting integral is approximately constant. In that case, we have

−(p−1)/2 Ptot(ω) ∝ ω (17)

This is a power law spectrum and the spectral index s of the emitted radiation spectrum is directly related to the particle distribution index p:

1 s = (p − 1) spectral index (18) 2

Although this relation has been derived qualitatively here, the result is the same for the exact treatment.

82 PHYS 4011 Ð HEA Lec. 9 Lecture 9: Synchrotron Radiiatiion IIII

9..1 Specttrum off Synchrottron Radiiattiion

The procedure for deriving the spectrum of synchrotron radiation is lengthy. In brief, we use the Poynting flux and take the Fourier transform of the retarded radiation field. This field can be decomposed into components parallel and perpendicular to the projection of the magnetic field on the plane of propagation. These components are the polarisation modes.

Step 1 – set-up. The Poynting flux gives the power emitted per unit area (c.f. Lec. 5):

dW 1 2 = Erad(t) (1) dtdA µ0c| | and the radiation field for a moving electron is that given by eqn. (6) in Lec. 5, viz

e Rˆ 1 ˙ Erad(r, t) = (Rˆ β) β (2) 4π"0 !(1 Rˆ β)3R × c − × $ − · " # where Rˆ is the propagation direction of the radiation to the observer. Note that the quantities on the RHS are evaluated at the retarded time t!. 83

PHYS 4011 Ð HEA Lec. 9

The energy emitted per unit area is thus

dW +∞ = c" E (t) 2 dt dA 0 | rad | %−∞ and Parseval’s theorem says

+∞ +∞ E(t) 2 dt = E(ω) 2 dω (3) | | | | %−∞ %−∞ −1/2 +∞ E for E(ω) = (2π) −∞ E(t) exp(iωt)dt. Since rad(t) is real and E(ω) is +∞ ∞ | | symmetric, we have E(ω) 2 dω = 2 E(ω) 2 dω and so the energy emitted per −∞& | | 0 | | unit area is & & dW ∞ = 2c" E(ω) 2 dω (4) dA 0 | | %0 This implies that the energy emitted per unit area per unit frequency bandwidth is dW = 2c" E(ω) 2 (5) dAdω 0| |

84 PHYS 4011 Ð HEA Lec. 9

Substituting dA = R2dΩ and inverting the Fourier transform gives

+∞ 2 dW c"0 = 2c" R2 E(ω) 2 = RE (t) exp(iωt) dt (6) dωdΩ 0 | | π rad )%−∞ ) ) ) ) ) Step 2 – Fourier transform of the retarded field. W) e need to solve the integral on )the RHS of 1 e +∞ Rˆ [(Rˆ β) β˙] RE(ω) = × − × eiωt dt (7) 1/2 3 (2π) 4πc"0 −∞ (1 Rˆ β) % − · We first change the integration variable from t to t!, using the definition of retarded time t! = t R(t!)/c and R = r r to give − | − 0| ! dt ˆ ! dt = = (1 R β) dt (8) (∂t!/∂t) − · So now we have 1 e +∞ Rˆ [(Rˆ β) β˙] RE(ω) = × − × eiωt dt! (9) 1/2 2 (2π) 4πc"0 −∞ (1 Rˆ β) % − · Next we express eiωt = exp[iω(t! + R/c)] in terms of t! only. We assume the radiation is ! being observed far enough away from the source that r(t) r0(t ), so that R r. 85 $ #

PHYS 4011 Ð HEA Lec. 9

Then we expand r to first order in r , which gives R(t!) r Rˆ r . Now we have 0 # | | − · 0 +∞ Rˆ Rˆ ˙ 1 e [( β) β] ! ! ! RE(ω) = × − × exp iω t Rˆ r0(t )/c dt 1/2 2 (2π) 4πc"0 −∞ (1 Rˆ β) − · % − · " ' ((#10) Then we use the identity

Rˆ [(Rˆ β) β˙] d Rˆ (Rˆ β) × − × = × × (1 Rˆ β)2 dt! (1 Rˆ β) − · − · and integrate by parts. Substituting all the above into (6) gives the expression for the energy emitted per unit frequency bandwidth per solid angle:

2 dW e2ω2 +∞ = Rˆ (Rˆ β) exp iω t! Rˆ r (t!)/c dt! (11) dωdΩ 16π3" c × × − · 0 0 )%−∞ ) ) " ' (# ) This is now in a form that)can be integrated. ) ) )

86 PHYS 4011 Ð HEA Lec. 9 Step 3 – evaluation of integral. To simplify the integration, we need to simplify the triple cross product. Consider the diagram below. An electron moves along an orbital trajectory with radius of curvature a. The coordinate system is set up such that the electron is trav- elling in the x y plane and passes through − the origin at retarded time t! = 0 with an in- stantaneous velocity in the x-direction. The

unit vector e⊥ is along the y axis and e% = Rˆ e . Thus, e and e define a plane × ⊥ % ⊥ perpendicular to an observer’s line of sight de- fined by the direction Rˆ . This is the plane of propagation, defined by the triple cross prod- uct Rˆ (Rˆ β) in the integral. × ˆ × The magnetic field B must also be in the plane containing R and β, so e% and e⊥ define directions parallel and perpendicular to the projection of the magnetic field on the plane of propagation.

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At any arbitrary retarded time t!, for β 1, we have | | # vt! vt! Rˆ (Rˆ β) = e cos sin ϑ e sin (12) × × % a − ⊥ a * + * + The exponential term in the integral in (11) is simplified using a small angle expansion:

! ! 2 2 !3 Rˆ r0(t ) a vt 1 c γ t t! · t! cos ϑ sin (1 + γ2θ2)t! + (13) − c # − c a # 2γ2 3a2 , - where 1 v/c 1/2γ2 and v c has been used elsewhere. Now we can calculate the − # # spectrum in the two polarisation states with

dW dW% dW⊥ = + (14) dωdΩ dωdΩ dωdΩ 2 2 2 and where, defining ϑγ = 1 + γ ϑ , we have dW e2ω2ϑ2 iω c2γ2t!3 2 % = exp ϑ2 t! + dt! dωdΩ 16π3" c 2γ2 γ 3a2 0 )% , * +- ) 2 2 ) ! 2 2 !3 ) 2 dW⊥ e ω ) ct iω c γ t ) = ) exp ϑ2 t! + )dt! (15) dωdΩ 16π3" c a 2γ2 γ 3a2 0 )% , * +- ) ) ) ) 88 ) ) ) PHYS 4011 Ð HEA Lec. 9 Now we make a change of variables:

! 3 ct ωaϑγ y γ , η = 3 ≡ aϑγ ≡ 3cγ which gives

2 dW e2ω2ϑ2 aϑ 2 +∞ 3 1 % = γ exp iη y + y3 dy dωdΩ 16π3" c γc 2 3 0 * + )%−∞ , * +- ) 2 ) ) 2 2 2 2 +∞ 2 dW⊥ e ω ϑ aϑγ ) 3 1 ) = ) y exp iη y + y3 d)y (16) dωdΩ 16π3" c γ2c 2 3 0 . / )%−∞ , * +- ) ) ) ) ) The integrals can be expressed in terms of th)e modified Bessel functions of 1/3 and)2/3 order: dW e2ω2ϑ2 aϑ 2 % = γ K2 (η) dωdΩ 16π3" c γc 1/3 0 * + 2 2 2 2 2 dW⊥ e ω ϑ aϑγ 2 (17) = 3 2 K2/3(η) dωdΩ 16π "0c . γ c /

89

PHYS 4011 Ð HEA Lec. 9 We next integrate over solid angle to give the energy per frequency emitted by an electron per orbit in the plane of propagation. During one orbit, the emission is almost completely confined to within an angle 1/γ around a cone of half-angle α (the pitch angle). So we use dΩ 2π sin αdϑ. Thus, # dW e2ω2a2 sin α +∞ % ϑ2 ϑ2K2 (η)dϑ dω # 6π2" c3γ2 γ 1/3 0 %−∞ 2 2 2 +∞ dW⊥ e ω a sin α = ϑ4 K2 (η)dϑ (18) dω 6π2" c3γ4 γ 2/3 0 %−∞ These integrals were first solved by Westfold (1959). They give

dW √3e2γ sin α % [F (x) G(x)] dω # 8π"0c − 2 dW⊥ √3e γ sin α = [F (x) + G(x)] (19) dω 8π"0c where ∞ F (x) x K (ζ)dζ , G(x) xK (x) (20) ≡ 5/3 ≡ 2/3 %x 90 PHYS 4011 Ð HEA Lec. 9

and , where 3 eB 2 is the critical frequency (see Lec. 8). To convert x = ω/ωc ωc = 2 me γ sin α this to power per unit frequency, we divide by the orbital period T = 2π/Ω = 2πγme/eB, which gives

√3e3B sin α P%(ω) = 2 [F (x) G(x)] 16π "0mec − √3e3B sin α (21) P⊥(ω) = 2 [F (x) + G(x)] 16π "0mec These are the components of the single-electron synchrotron power per unit frequency corresponding to polarisation modes parallel and perpendicular to B. The total synchrotron power per unit frequency is

√3e3B sin α (22) P (ω) = P%(ω) + P⊥(ω) = 2 F (x) 8π "0mec single electron synchrotron power spectrum

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The functions F (x) and G(x) that appear in the synchrotron power spectrum are plotted below in linear and logarithmic scales.

Both functions have similar shapes and reach similar asympototic values at large x, but G(x) < F (x) for x < 1. The asympototic behaviour of the functions goes as

F (x), G(x) x1/2e−x , x 1 F (x), G(x) x1/3 , x 1 (23) ∼ $ ∼ ( Implications for the synchrotron spectrum: 1. emission is broadband (∆ω/ω 1) ∼ 2. spectrum is power-law at small x = ω/ωc < 1 3. emission peaks near x 0.3 # 92 PHYS 4011 Ð HEA Lec. 9 9..2 Emiissiiviitty ffor a Power-llaw Ellecttron Diisttriibuttiion The expression P (ω) for the synchrotron power of a single electron given by (23) is valid for a single energy γ. To obtain a volume emissivity for a distribution of electron energies, we need to integrate P (ω) over the energy distribution. A nonthermal (power law) distribution is appropriate for ultrarelativistic electrons and so we use

−p < < Ne(γ) = Keγ , γ1 ∼ γ ∼ γ2 (24)

2 for the number of electrons with energy γmec . The total number of electrons per unit volume γ2 is . Note that the electron pitch angles will in general also be spread Ne = γ1 Ne(γ) dγ α k Rˆ B around a &direction α defined by a characteristic angle α0 between and . We assume the electron distribution is isotropic in pitch angle, so that Ne(γ) is independent of kα. The synchrotron volume emissivity for a nonthermal distribution of relativistic electrons thus has the following form:

γ2 γ2 jsyn = 2πK P (ω) γ−p dγ F (x)γ−p dγ (25) ν e ∝ %γ1 %γ1 where the factor 2π enters because 93

PHYS 4011 Ð HEA Lec. 9

P (ω) = dW/(dωdt) = dW/(2πdνdt) = jν dV/2π. We now change the integration variable from γ to x = ω/ωc using the relation&s 3 eBx sin α −1/2 1 3 eB sin α −1/2 γ = , dγ = x−3/2 dx (26) 2 ωm −2 2 ωm * e + * e + So we now have an integral of the form

x1 jsyn x(p−3)/2F (x) dx (27) ν ∝ %x2 where x1,2 are related to γ1,2 via the relation (26).

To evaluate this integral analytically, the following approximation is made. Consider the relations ω ω x1 = , x2 = ωc(γ1) ωc(γ2) If γ γ , then ω (γ ) ω (γ ) and we can have a wide range of frequencies satisfying 1 ( 2 c 1 ( c 2 ω ω (γ ) so that x . Similarly, we can have ω ω (γ ), so that x 0. Thus, $ c 1 1 → ∞ ( c 2 2 →

94 PHYS 4011 Ð HEA Lec. 9 we have an integral of the form ∞ x(p−3)/2F (x) dx %0 ∞ (p−3)/2 as well as an analogous one of the form 0 x G(x)dx if we are interested in separating the polarisation modes. These& integrals have solutions involving gamma functions Γ(ξ), where ξ is related to p.

The final result for the synchrotron volume emissivity is p 19 p 1 jsyn = 3p/22−(p+7)/2π−(p+3)/2Γ( + )Γ( )(p + 1)−1 (28) ν 4 12 4 − 12 e2 eB (p+1)/2 K sin α ν−(p−1)/2 synchrotron emissivity " c e m 0 0 * e + This is clearly a power-law spectrum, with spectral index 1 α = (p 1) (29) 2 −

If the magnetic field B does not have a fixed direction (e.g. randomly oriented), then we need

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to perform a further integration over directions α0. The required integration is 1 2π π sin(p+1)/2 α = dφ sin(p+1)/2 α sin α dα , 0- 4π 0 0 0 %0 %0 1 π = sin(p+3)/2 α dα 2 0 0 %0 √π Γ( 5+p ) = 4 (30) 2 7+p Γ( 4 )

96 PHYS 4011 Ð HEA Lec. 9

The final and most often used expression for the synchrotron emissivity is the following for the randomly oriented magnetic field case:

5+p p 19 p 1 syn p/2 −(p+13)/2 −(p+2)/2 Γ( 4 )Γ( 4 + 12 )Γ( 4 12 ) jν = 3 2 π 7+p − Γ( 4 ) e2 eB (p+1)/2 K ν−(p−1)/2 (31) " c e m 0 * e + synchrotron emissivity (pitch angle averaged)

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PHYS 4011 Ð HEA Lec. 9 9..3 Pollariisattiion off Synchrottron Radiiattiion Radiation from a single electron will be in general elliptically polarised, i.e. the tip of the vector

Erad(t) will sweep out an ellipse in the plane of propagation perpendicular to the observer’s line of sight. However, the left-hand and right-hand components will tend to cancel out for a distribution of emitting electrons that varies smoothly with pitch angle, so only linear

components of Erad(t) remain. Thus, synchrotron radiation is partially linearly polarised. The degree of linear polarisation for particles with a single energy γ is defined by

P⊥(ω) P%(ω) G(x) Π(ω) = − = (32) P⊥(ω) + P%(ω) F (x) The frequency integrated value is Π 75 %, which is quite high. For a power-law distribution # of electron energies, the degree of linear polarisation is p + 1 Π = 7 (33) p + 3 This is 70 %, which is still quite high. Observationally, such high polarisations are never ∼ seen and this is usually interpreted as being due to propagation effects, which can reduce Π to values down to a few percent.

98 PHYS 4011 Ð HEA Lec. 10

Lecture 10: Compton Scatteriing

Compton scattering is the scattering of photons off electrons. For low photon energies, it reduces to the classical case of Thomson scattering. For relativistic electrons, lower energy photons can be efficiently upscattered to energies reaching X-ray and γ-ray wavelengths. The photon upscattering process is referred to as Comptonisation. When referring to cooling of the electrons, the radiation process is called inverse Compton scattering. The emission (scattered) spectrum can be calculated analytically for single scatterings only. For multiple scatterings, numerical simulations are usually necessary.

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10..1 Energy Transffer and Scatttteriing Cross Secttiion

In the classical (Thomson) limit, an electron will oscillate and emit radiation in response to incident electromagnetic waves. Quantum effects, however, modify the kinematics and interaction cross-section. Because a photon possesses momentum as well as energy, the recoil of the electron must be taken into account so the scattering cannot be elastic. It is easiest to treat the momentum transfer from a particle approach. Consider a photon of energy ε = hν incident upon an electron initially at rest. The photon scatters through an angle Θ

w.r.t. its initial propagation direction kˆi. The energy of the photon and electron after the

scattering event are ε1 = hν1 and E, respectively (see figure).

Initial and final 4-momenta of photons: ε ε P = (1, kˆ ) , P = 1 (1, kˆ ) γi c i γf c f Initial and final 4-momenta of electron:

Pei = (mec, 0) , Pef = (E/c, p)

100 PHYS 4011 Ð HEA Lec. 10 Conservation of 4-momentum requires

Pei + Pγi = Pef + Pγf (1)

Rearranging and squaring gives

2 2 |Pef | = |Pei + Pγi − Pγf | (2) 2 2 2 = |Pγi| + |Pei| + |Pγf | + 2PγiPei − 2PγiPγf − 2Pγf Pei

Note that the modulus of a 4-vector Aµ is defined as 2 µ 0 2 1 2 2 2 3 2 A = A Aµ = −(A ) + (A ) + (A ) + (A ) . This implies that the magnitudes of 2 2 2 2 the 4-momenta for a photon and an electron are Pγ = 0 and Pe = −me c , respectively. So 2 2 2 2 2 2 in the above expression, we have |Pef | = −me c = |Pei| and |Pγi| = 0 = |Pγf | , which leaves behind only those terms with a factor of 2 in front. These terms are: εε1 εε1 ˆ ˆ PγiPei = −εme, PγiPγf = − c2 + c2 ki · kf , and Pγf Pei = −ε1me. Substituting these in, rearranging and using kˆi · kˆf = cos Θ gives the following expression for the energy of the scattered photon: ε ε1 = ε (3) 2 1 + mec (1 − cos Θ)

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In terms of wavelength, λ = hc/ε, we have a change

∆λ = λ1 − λ = λC(1 − cos Θ) (4)

where h λC ≡ # 0.0243 A˚ Compton wavelength (5) mec

Thus, the wavelength change is of order λC. For long wavelengths (λ $ λC) or equivalently, 2 ε % mec , the scattering is approximately elastic (i.e. ε1 # ε). This is the Thomson regime.

102 PHYS 4011 Ð HEA Lec. 10

The Klein-Nishina cross section

In addition to the effects of photon momentum, quantum corrections also modify the cross section for Compton scattering. The exact expression for the differential cross section for Compton scattering is derived from quantum electrodynamics and is known as the Klein-Nishina formula: 2 dσKN 1 ε ε ε1 2 1 2 (6) = r0 2 + − sin Θ dΩ 2 ε #ε1 ε $

2 −15 where r0 = e /(4π'0mec) = 2.82 × 10 m is the classical electron radius (defined in

Lec. 8). This reduces to the classical differential Thomson cross section in the limit ε1 ∼ ε, 1 2 2 viz. dσT/dΩ = 2 r0 (1 + cos Θ). The total cross section is obtained by integrating over +1 solid angle, σKN = 2π −1 (dσKN/dΩ) d cos Θ: % 3 1 + x 2x(1 + x) 1 1 + 3x σKN = σT − ln(1 + 2x) + ln(1 + 2x) − 4 & x3 ' 1 + 2x ( 2x (1 + 2x)3 ) (7) 2 where x ≡ hν/mec .

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The overall effect of σKN is to reduce the scattering cross section relative to σT at high photon energies. Thus, Compton scattering becomes less efficient at high energies. The decline is shown in the plot below.

3 −1 1 For x $ 1, the asymptotic solution is σKN ∼ 8 σTx ln 2x + 2 ! " 104 PHYS 4011 Ð HEA Lec. 10

Scattering from electrons in motion

In general, electrons will not be at rest, but will be moving, sometimes with relativistic velocities. Whenever a moving electron has energy greater than that of an incident photon, the energy transfer is from electron to photon. This is inverse Compton scattering. The results for scattering by a stationary electron are extended to a moving electron using a Lorentz " transformation. Let K be the observer’s frame and K be the rest frame of an electron. The − relative velocity βc defines the Lorentz factor γ = (1 − β2) 1/2. A scattering event as seen in each frame is shown in the figure below. In K, the electron’s velocity is in the x direction and all angles in both frames are measured from this axis.

! observer’s frame K electron rest frame K

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" In K , all the previous formulas for scattering from stationary electrons are valid. Transforming " the photon’s initial energy into K :

" ε = εγ(1 − β cos Θ) (8)

and transforming the scattered photon energy back into K:

" " ε1 = ε1γ(1 + β cos Θ1) (9)

Thus, in transforming to the electron rest frame, the photon picks up a factor γ and in transforming back to the lab frame, it picks up an additional factor γ. Hence, the photon energy can increase by a factor γ2 in the lab frame, implying that Compton scattering by relativistic electrons can be quite efficient.

A maximum gain of ∼ γ2 is only possible for scatterings that are in the Thomson regime in the " " " > rest frame (i.e. ε1 # ε ) and which have Θ, Θ1 ∼ π/2. The condition for Thomson scattering in the rest frame is " 2 2 ε % mec =⇒ γhν % mec (10)

106 PHYS 4011 Ð HEA Lec. 10 10..2 Siinglle Scatttteriing Power

We want to obtain an expression for the average inverse Compton power due to an isotropic distribution of photons scattering off electrons. As before, the procedure is to derive all quantities in the electron rest frame, calculate the scattering in the Thomson limit in the rest " " frame (i.e. let ε1 # ε ) and then transform everything back into the lab (observer) frame. Let n(ε)dε be the number density of photons having energy in the range ε + dε. The total power emitted (i.e. scattered) in the electron’s rest frame is given by " dE1 " " " " = cσT ε n (ε ) dε (11) dt" * 1 Now we know that the emitted power is an invariant. Another invariant is the quantity n(ε)dε/ε. So dE1 " " " "2 ndε = cσT ε n dε = cσT ε (12) dt * * ε " Now we substitute ε = εγ(1 − β cos Θ) from eqn. (8) to get

dE1 2 2 = cσTγ (1 − β cos Θ) εn dε (13) dt *

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PHYS 4011 Ð HEA Lec. 10 which now only contains quantities in frame K. For an isotropic distribution of photons, we have 1 )(1 − β cos Θ)2* = 1 + β2 3 giving

dE1 2 1 2 = cσTγ 1 + β U (14) dt # 3 $ γ

where Uγ = εndε is the initial photon energy density. Now dE1/dt is the rate at which the electron loses%energy. The nett power converted into increased radiation is this minus the rate

at which the initial photon energy distribution decreases, dε/dt = σTcUγ . So

dErad dE1 dε 2 1 2 = − = cσTU γ 1 + β − 1 (15) dt dt dt ph ' # 3 $ (

which gives the following for the inverse Compton power for a single electron:

4 2 2 P = σTcγ β U inverse Compton power (16) ic 3 γ

This has used γ2 − 1 = γ2β2.

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Emitted power for a distribution of electrons

−p For a nonthermal power law distribution of electrons N(γ) = Keγ , we can obtain the total power per unit volume from γ2 Pic,tot = PicN(γ)dγ *γ1 This gives, for β # 1,

4 −1 3−p 3−p P = σTcU K (3 − p) (γ − γ ) nonthermal power-law electrons (17) ic,tot 3 γ e 2 1

2 2 For a thermal distribution of electrons, γ = 1 and )β * = 3kTe/mec . The total power needs to be derived from the single electron power in the more general case where energy transfer in the electron rest frame is not neglected. The result is

4kTe T thermal electrons (18) Pic,tot = σ cUγNe 2 mec

where Ne is the total electron number density.

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10..3 Siinglle Scatttteriing Specttra

The spectrum resulting from single scattering events between a distribution of photons and a distribution of relativistic electrons depends on both the specified distributions. The spectrum can be calculated for a scattering event with a single photon energy and single electron energy and the nett spectrum is obtained by averaging over the electron and incident photon distributions. The derivation for the spectrum due to inverse Compton scattering is different from the derivations for true emission processes. The treatment deals with intensity based on photon number and the full details are omitted. The relevant expression is that for the total scattered power per unit volume per energy due to a nonthermal power law distribution of electrons (i.e. volume emissivity per unit energy rather than frequency):

γ2 3 n(ε) −(p+2) ε1 T (19) jε1 = σ cε1Ke dε dγ γ f 2 16π * ε *γ1 #4γ ε$

where the function f is defined by f(x) = 2x ln x + x + 1 − 2x2.

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For sufficiently large limits on the γ integral, we have

2 ic 3 p−2 p + 4p + 11 −(p−1)/2 (p−1)/2 j = 2 σTcε Ke ε n(ε) dε (20) ε1 π (p + 1)(p + 3)2(p + 5) 1 *

Thus, inverse Compton scattering also predicts a power law spectrum with a spectral index 1 α = (p − 1) (21) 2 identical to the case of synchrotron emission. The power law spectrum is independent of the incident photon distribution.

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Scattering of blackbody photons

The above derivation implies that if the incident photon distribution is a blackbody spectrum, the resulting spectrum after a single scattering by nonthermal electrons should be a power law. For a blackbody, we have 8π ε −1 2 (22) n(ε) = 3 ε exp − 1 (hc) + ,kT - .

Inserting this into the expression jε1 above, and solving the integrals gives

σT −(p−1)/2 jic,bb = f (p)(kT )(p+5)/2K ε (23) ε1 h3c2 bb e 1 where 2p+1(p2 + 4p + 11) p + 5 p + 5 f (p) = 3 γ ζ bb (p + 1)(p + 3)2(p + 5) # 2 $ # 2 $ where ζ is the Riemann zeta function.

112 PHYS 4011 Ð HEA Lec. 10 Synchrotron Self-Comptonisation

A particularly interesting case of inverse Compton scattering is that in which the seed photons are synchrotron photons emitted by the scattering electrons. In this case, the incident photon spectrum is the synchrotron power law spectrum, which can be written as

−(p−1)/2 Uγ (ε0) ε < n(ε) = , εmin ∼ εmax (24) ε0 #ε0 $

where ε0 is some fiducial seed photon energy. The solution for the synchrotron self-Compton volume emissivity is

−(p−1)/2 ssc εmax ν1 jν1 = f(p)σTcKeUγν0 ln (25) # εmin $ #ν0 $

where the the relation jν1 = hjε1 has been used and where

2 3 − p + 4p + 11 f(p) = 2p 2 π (p + 1)(p + 3)2(p + 5)

The term ln(εmax/εmin) is known as the Compton logarithm.

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PHYS 4011 Ð HEA Lec. 10 10..4 Mullttiiplle Scatttteriings:: tthe Comptton y Parametter The spectrum resulting from repeated scatterings is usually calculated numerically using Monte Carlo techniques. Qualitatively, however, we can expect that the more scatterings that occur, the more the seed photon distribution becomes distorted. A useful parameter that measures the importance of scattering in a medium is the Compton y parameter:

y ≡ fractional energy change × mean no. of scatterings (26)

The mean number of scatterings is determined by the optical depth, τ = σNer, where r is the size of the scattering region. A value of τ ∼ 1 means that on average, a photon will scatter once before escaping the region. Specifically, we have

mean no. of scatterings # max(τ, τ 2) (27)

The scattering regimes are defined in terms of the y parameter as follows:

y % 1 negligible spectral changes

y ∼< 1 power law spectrum, with exponential cut-off (28) y $ 1 saturated Comptonisation

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For y ∼< 1, it is possible to obtain a power law scattered spectrum, even if the scattering electrons have a thermal distribution. For a thermal electron distribution, the Compton y parameter is defined as

4kTe 4kTe 2 max T (29) y = 2 1 + 2 (τ , τT ) mec # mec $ y parameter for thermal Comptonisation (30)

where τT = σTNer is the Thomson optical depth of the scattering region of size r. In the saturated Comptonisation limit (y $ 1), the incident photon spectrum is completely distorted beyond recognition. The resulting spectrum approaches a similar distribution to the scattering electrons, implying that the photons and electrons come into thermal equilibrium. An incident nonthermal (i.e. power law) photon spectrum, for example, will become thermalised by the scatterings and approach a blackbody spectrum at the temperature of the scattering electrons,

so the spectrum will peak at hν1 # 2.8kTe.

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Some example spectra of multiple Compton scatterings calculated from Monte Carlo simulations (see e.g. Sunyaev & Titarchuk, 1980, Astron. Astrophys., 86, 121.):

Emergent spectra from a spherical region with varying optical depths containing electrons with kTe = 2 0.7mec . The incident seed photons are injected at the centre with a black- body spectrum at kT ! kTe. The Compton y parameter thus ranges from y " 0.5 for the τT = 0.05 3 spectrum, to y " 10 for the τT = 0.05 spectrum.

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