Lecture 5: Radiation from Moving Charges - Uniuddeangeli/fismod/Radiation.pdf · 2012-11-07 ·...

PHYS 4011 – HEA Lec. 5

Lecture 5: Radiation from Moving ChargesLecture 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

electromagnetic fields associated with accelerating charges fall off as 1/r instead of 1/r2 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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5.1 Overview of the Radiation Field of Single Moving Charges5.1 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:

t′ = t −|r(t) − r0(t′)|

c(1)

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 timet. At most, only one source point on the trajectory

is in communication with P at time t.

41


The radiation field at P at time t is calculated from the retarded scalar and vector potentials V

andA usingE = −∇V − ∂A/∂t andB = ∇× A. But Maxwell’s equations define these

in terms of the continuous charge and current densities ρ and J. So to evaluate V andA 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) = r0(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

V (r, t) =q

4πε0

∫δ(t′ − t + |r(t) − r0(t′)|/c)

|r(t) − r0(t′)|dt′ (2)

A(r, t) =µ0q

4π

∫v(t′)

δ(t′ − t + |r(t) − r0(t′)|/c)

|r(t) − r0(t′)|dt′

After introducing the simplifying notation

R(t′) = r(t) − r0(t′) , R(t′) = |r(t) − r0(t

′)| , R =R(t′)

R(t′)

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the integral can be solved with a change of variables giving

A(r, t) =µ0

4π

qv(t′)

(1 − R · v(t′)/c)R=

v

c2V (r, t) Lienard–Wiechart potentials (3)

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

Points to note:

1. The factor (1 − R · 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 = r0(t′) and its corresponding acceleration as

βc = r0(t′), the fields are

E(r, t) =q

4πε0

(R − β)(1 − β2)

(1 − R · β)3R2+

R

(1 − R · β)3R×

1

c

[(R − β) × β

](4)

B(r, t) =1

cR × E(r, t) (5)

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 R.

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

moving charge:

Erad(r, t) =q

4πε0

R

(1 − R · β)3R×

1

c

[(R − β) × β

]

Brad(r, t) =1

cR × Erad(r, t) (6)

Note thatErad,Brad and R are mutually perpendicular.

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5.2 Radiation from Nonrelativistic Charged Particles5.2 Radiation from Nonrelativistic Charged Particles

The Larmor formula

When β & 1, the radiation fields simplify to

Erad(r, t) =q

4πε0

[R

R×

1

c2

(R × v

)](7)

andBrad(r, t) follows from (6). Note that Erad lies in the plane containing R and v (i.e. the

plane of polarisation) andBrad is perpendicular to this plane. If we let θ be the angle between

R and v, then

|Erad| = c|Brad| =q

4πε0

v

Rc2sin θ

The Poynting vector is in the direction of R and has the magnitude

S =1

µ0cE2

rad =µ0

16π2c

q2v2

R2sin2 θ (8)

We now want to express this as an emission coefficient.

46


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 R of S. So the power emitted per solid angle is

dW

dtdΩ= S R2 =

µ0

16π2cq2v2 sin2 θ

Note the characteristic dipole pattern∝ sin2 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:

P =dW

dt=

µ0

16π2cq2v2

∫ 2π

0

∫ π

0

sin2 θ sin θ dθ dφ =µ0

8πcq2r2

0

∫ +1

−1

(1 − µ2) dµ

The integral gives a factor of 4/3, whence we arrive at the Larmor formula for the

electromagnetic power emitted by an accelerating charge:

P =µ0q2r2

0

6πcLarmor formula (9)

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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

Erad =∑

i

qi

4πε0

[Ri

Ri×

1

c2

(Ri × ri

)]

whereRi is the distance between each source point ri and the field point r. But the

differences between theRi 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 =∑

i

qiri (10)

to get

Erad =1

4πε0

[R

R×

1

c2

(R × d

)](11)

Then, as before, we find

dW

dtdΩ=

µ0

16π2cd2 sin2 θ

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and the total power is

P =µ0d2

6πc(12)

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 andR and the

radiation pattern is ∝ sin2 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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5.3 The Radiation Spectrum5.3 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(t) =1

(2π)1/2

∫ +∞

−∞

v(ω) exp(−iωt) dω

v(ω) =1

(2π)1/2

∫ +∞

−∞

v(t) exp(iωt) dt (13)

Then we use Parseval’s theorem, which relates this as follows:∫ +∞

∞

|v(ω)|2 dω =

∫ +∞

∞

|v(t)|2 dt (14)

We can also use the relation∫∞

0

|v(ω)|2 dω =

∫ 0

−∞

|v(ω)|2 dω = (15)

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): ∫ +∞

−∞

P dt =

∫ +∞

−∞

µ0q2

6πc|v(t)|2 dt =

µ0q2

3πc

∫∞

0

|v(ω)|2 dω (16)

Since the total emitted energy must also equal∫∞

0(dW/dω) dω, then the energy per unit

bandwidth is just

dW

dω=

µ0q2

3πc|v(ω)|2 (17)

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

dω=

µ0

3πcω4|d(ω)|2 (18)

This is because differentiating wrt t twice introduces a factor ω2 in the Fourier transform – c.f.

d(t) = (2π)−1/2∫ +∞

−∞d(ω) exp(−iωt) dω =⇒

d(t) = −(2π)−1/2∫ +∞

−∞ω2d(ω) exp(−iωt) dω, so ev(ω) = ω2d(ω).

52


Lecture 6: Bremsstrahlung RadiationLecture 6: Bremsstrahlung Radiation

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 quantumcorrections 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∼ mp/me times greater.

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6.1 Emission from Single Speed Electrons6.1 Emission from Single Speed Electrons

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).

LetR 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.

dW

dω=

µ0e2

3πc|v(ω)|2 (1)

and using

v(ω) =1

(2π)1/2

∫ +∞

−∞

v(t) exp(iωt) dt (2)

55


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

|v(ω)| #

1(2π)1/2 ∆v ωτ % 1

0 ωτ $ 1(3)

So our radiation spectrum goes as

dW

dω#

µ0e2

6π2c |∆v|2 b % v/ω

0 b $ v/ω(4)

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Now we can work out∆v by noting that the total Coloumb force on the electron is

Ze2/(4πε0R2) in the−R direction. The perpendicular component of acceleration is the

strongest and will thus make the dominant contribution to the radiation spectrum, so we can

write

∆v # 1

4πε0

Ze2

me

∫b dt

(b2 + v2t2)3/2=

1

4πε0

Ze2

me

2

bv(5)

(the integral turns out to be elementary). Substituting this into the expression for the radiation

spectrum gives

dW

dω#

8Z2e6

3πc3(4πε0)3m2eb2v2 b % v/ω

0 b $ v/ω(6)

This is the spectrum for small angle 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.

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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

dωdV dt= neni2πv

∫∞

bmin

dW (b)

dωb db (7)

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:

dW

dωdV dt# 16neniZ2e6

3c3(4πε0)3m2ev

ln

(bmax

bmin

)(8)

where bmax ∼ v/ω is some value beyond which the b % v/ω limit no longer applies and the

contribution to the integral becomes negligible. We can set bmax = v/ω, even though it is

58


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 bmin # 2Ze2/(4πε0mev2).

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,

Gff(v, ω) =

√3

πln

(bmax

bmin

)free-free Gaunt factor (9)

giving

dW

dωdV dt# 16πneniZ2e6

33/2c3(4πε0)3m2ev

Gff(v, ω) (10)

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


6.2 Emission from a Thermal Distribution of Electrons6.2 Emission from a Thermal Distribution of Electrons

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

dn(v) = f(v)d3v = f(v)4πv2 dv ∝ exp

(−mv2

2kT

)v2 dv

We need to average the single-speed radiation spectrum over this distribution function for all

electron speeds satisfying12mev2 >

∼hω/2π, i.e.

dW (T, ω)

dωdV dt=

∫∞

vmin

dW (v,ω)dωdV dt v2 exp(−mv2/2kT ) dv∫∞

0 v2 exp(−mv2/2kT ) dv(11)

Using dω = 2π dν, the final result is the expression for the free-free volume emissivity:

jffν ≡ dW

dV dtdν=

1

(4πε0)332πe6

3m3/2e c3

(2π

3kT

)1/2

Z2neni exp

(− hν

kT

)Gff (12)

where Gff 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 areW m−3 Hz−1. So to calculate the total radiative power (i.e.

luminosity) in bremsstrahlung emission from a real astrophysical source, we simply

integrate jffν over an appropriate source volume and over the frequency bandwidth that we

are interested in.

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Lecture 7: Radiation from Moving Charges IILecture 7: Radiation from Moving Charges II

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.

Erad(r, t) =q

4πε0

R

(1 − R · β)3R×

1

c

[(R− β) × β

](1)

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.

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Total power emitted

Step 1. Consider an instantaneous rest frameK′ 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 powerP ′ emitted in K′. We then need to how to

transform back to the rest frameK 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 frameK′. These quantities transform as

dW = γdW ′ , dt = γdt′

where γ = (1 − β2)1/2 is the Lorentz factor. Note that the Lorentz factors cancel out, i.e.

P =dW

dt=

γdW ′

γdt′=

dW ′

dt′= P ′

(2)

so the total emitted power is Lorentz invariant.

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Step 2. Now we have the identity P = P ′ = µ0q2|a′|2

6πc . In this expression, a′ is the

3-acceleration inK′ 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

aµ ≡dvµ

dτ=

d2xµ

dτ2= , a2 = aµaµ = −a0a0 + |a|2 (3)

where vµ = γ(c,v) is the 4-velocity. Note that the 4-velocity and 4-acceleration are

orthogonal: aµvµ = (dvµ/dτ)vµ = 1

2d(vµvµ)/dτ = 1

2d(−c2)/dτ = 0. Now inK′, we

can write |a′|2 = a′2 + a′0a′0. But in the particle’s rest frame, v

′µ = (c,0) and since

a′µv′µ = 0, we must have a′0 = 0. Thus, we can write

|a′|2 = a′µa′µ = aµaµ

and

P =µ0q2

6πcaµaµ (4)

which is now in a manifestly covariant form (i.e. can be evaluated in any frame).

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Step 3. Since aµaµ = |a|′2, 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

a′‖ = γ3a‖ , a′

⊥ = γ2a⊥

Thus, we have

P =µ0q2

6πc|a′|2 =

µ0q2

6πc(a′2

‖ + a′2⊥) =

µ0q2

6πcγ4(γ2a2

‖ + a2⊥) (5)

Clearly, the power emitted increases drastically as a particle becomes relativistic. This

expression can be written another way:

P =µ0q2

6πcγ6

(a2 − |β × a|2

)relativistic Larmor formula (6)

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 particleK′. We want to find how the

emitted energy per solid angle transforms fromK′ to an observer rest frameK . Suppose the

relative velocity v between these two frames is along the x-axis. Then the change in energy

transforms as

dW = γ(dW ′ + vdp′x)

from the transformation properties of the 4-momentum pµ = (E/c,p) = m0vµ. 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) cosϑ, so

dW = γ(1 + β cosϑ′)dW ′(7)

Now we want to find how much energy is radiated in the solid angle

dΩ′ = sinϑ′dϑ′dφ′ = d cosϑ′dφ′ about ϑ′ and we want to know how dW ′/dΩ′

transforms to give dW/dΩ, where dΩ = d cosϑdφ.

66


Geometry for dipole emission from a particle instantaneously at rest in frameK′.

67


The transformation of cosϑ′ is given by the aberration of light formula:

cosϑ =cosϑ′ + β

1 + β cosϑ′(8)

Differentiating gives

d cosϑ =d cosϑ′

γ2(1 + β cosϑ′)2

The azimuthal angle is invariant, so dφ = dφ′. Thus,

dΩ = d cosϑdφ =d cosϑ′dφ′

γ2(1 + β cosϑ′)2=

dΩ′

γ2(1 + β cosϑ′)2

and so we havedW

dΩ= γ3(1 + β cosϑ′)3

dW ′

dΩ′(9)

To get the angular distribution of the power, we divide through by the time interval

dt = γ(1 − β cosϑ)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

frameK :dP

dΩ= γ4(1 + β cos ϑ′)4

dP ′

dΩ′=

1

γ4(1 − β cosϑ)4dP ′

dΩ′(10)

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If the emission is isotropic in the particle’s rest frameK′, then inK , it will be peaked forward

in the direction of motion (i.e. ϑ → 0). We know that inK′, we can use the dipole

approximation , which gives (c.f. Lec. 5)

dP ′

dΩ′=

µ0q2

16π2ca′2 sin2 θ′ (11)

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

dP

dΩ=

µ0q2

16π2c

(γ2a2

‖ + a2⊥)

(1 − β cosϑ)4sin2 θ′ =

3P

8πγ4

sin2 θ′

(1 − β cosϑ)4(12)

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.

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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

sin2 θ′ =sin2 θ

γ2(1 − β cosϑ)2(13)

Substituting into (12) yields

dP

dΩ=

µ0q2

16π2ca2

‖

sin2 θ

(1 − β cosϑ)6(14)

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′ = sinϑ′ cosφ′. The transformation is

sin2 θ′ = 1 −sin2 ϑ cos2 φ

γ2(1 − β cosϑ)2(15)

giving

dP

dΩ=

µ0q2

16π2ca2⊥

1

(1 − β cosϑ)4

[1 −

sin2 ϑ cos2 φ

γ2(1 − β cos ϑ)2

](16)

Again, this peaks in the forward direction near ϑ ∼ 1/γ, with a smaller peak at larger ϑ due to

the azimuthal dependence.

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

71


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 − β cosϑ ∼1 + γ2ϑ2

2γ2(17)

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

dP‖

dΩ∼

4µ0q2a2

‖

π2cγ10 γ2ϑ2

(1 + γ2ϑ2)6(18)

dP⊥

dΩ∼

µ0q2a2⊥

π2cγ8 1 − 2γ2ϑ2 cos(2φ) + γ4θ4

(1 + γ2ϑ2)6(19)

72


Lecture 8: Synchrotron Radiation ILecture 8: Synchrotron Radiation I

Charged particles in a magnetic field radiate becausethey experience an acceleratio perpendicular to thefield. If the particles are nonrelativistic, the radiation isreferred to as cyclotron emission and the frequency ofemission is directly related to the particle gyration fre-quency. This results in a discrete emission spectrumwhich usually does not extend beyond optical/UV fre-quencies in most astrophysical situations. When theparticles are relativistic, however, the radiation is re-ferred to as synchrotron emission and results in acontinuum spectrum because the frequency of emis-sion extends over many higher order harmonics of thegyration 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 radiofrequenciesto X-ray and γ-ray energies. Many real high-energysources in astrophysics are also sources of strong ra-dio emission due to synchrotron radiation (e.g. theradio galaxy 3C223, right, shown with X-ray colourcontours and radio line contours overlaid).

73


8.1 Total Emitted Synchrotron Power8.1 Total Emitted Synchrotron Power

Consider a relativistic particle of 3-momentum p = γmv in a steady magnetic fieldB. The

equation of motion of the particle is

dp

dt=

d

dt(γmv) = qv × B (1)

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 givesmγdv/dt = qv × B and we can

separate the velocity into components parallel and perpendicular toB:

dv‖

dt= 0 ,

dv⊥

dt=

q

γmv⊥ × B (2)

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 toB and the

acceleration is perpendicular to the velocity in this plane.

74


The combination of circular motion ⊥ to B

and uniform motion ‖ toB results in helical

motion (see figure). The magnitude of the

acceleration is

a⊥ =|q|v⊥B

γm= Ωv⊥ (3)

whereΩ = |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

acceleration perpendicular to velocity, eqn.

(5) in Lec. 7, viz.

Helical motion of an electron in a magnetic field

B results from the combination of uniform motionalongB and circular motion perpendicular toB.

P =µ0q2

6πcγ4a2

⊥ =µ0q2

6πcγ4 q2B2

γ2m2v2⊥ =⇒ P =

1

6πε0c

q4

m2γ2β2B2 sin2 α (4)

where α is the pitch angle between v andB. Note the dependence onm: synchrotron

emission is much less efficient for protons than for electrons.

75


For an isotropic distribution of velocities, it is necessary to average over all pitch angles for a

given speed β:

〈β2⊥〉 =

β2

4π

∫sin2 α dΩ =

1

2β2

∫ +1

−1(1 − cos2 α) d cosα =

2

3β2

(5)

So the total power emitted by an electron, averaged over all pitch angles is

P =2

3

1

6πε0c

q4

m2γ2β2B2

(6)

This is also sometimes expressed in terms of the Thomson cross-section,

σT =8

3πr2

0=

e4

6πε20m2ec

4= 6.65 × 10−29

m2

(7)

where r0 = e2/(4πε0mec) = 2.82 × 10−15 m is the classical electron radius, obtained by

equating the electrostatic potential energy, e2/(4πε0r0), with the rest mass energymec2. So

P =2

3

c

µ0σTγ2β2B2 =

4

3σTcγ2β2UB total synchrotron power (8)

where UB = B2/2µ0 is the magnetic energy density (i.e. energy per unit volume).

76


8.2 Synchrotron Spectrum – Qualitative Treatment8.2 Synchrotron Spectrum – Qualitative Treatment

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 radiationE(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.

77


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

∆t)

evB sinα

γm

78


and since∆v ) v∆θ and∆t ) ∆s/v, we have

∆θ

∆s)

evB sinα

γmv=

Ω

vsinα

and hence, a ) ∆s/∆v ) v/(Ω sin α) and

∆s )2v

γΩ sinα

So the time interval between the emitted pulses is

∆t = t2 − t1 )∆s

v)

2

γΩ sinα(9)

The time interval between the arrival times∆ta = ta2 − ta1 at the observer will be less than

∆t by an amount∆s/c = (v/c)∆t, so we have

∆ta = ta2 − ta1 ) ∆t −∆s

c=

∆s

v

(1 −

v

c

))

2

γΩ sinα

(1 −

v

c

)(10)

For γ , 1, we have

1 −v

c∼

1

2γ2

79


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

ωc ≡3

2γ3Ω sinα =

3

2

eB

meγ2 sinα critical frequency (12)

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)

P (ω) ∼1

9πε0c

e3B

meF

(ω

ωc

)sinα (13)

80


8.3 Spectral Index for a Power-Law Electron Distribution8.3 Spectral Index for a Power-Law Electron Distribution

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,

Ptot(ω) ∝

∫ γ2

γ1

F

(ω

ωc

)γ−pdγ (15)

If we change integration variables to x = ω/ωc, noting that ωc ∝ γ2, we have

Ptot(ω) ∝ ω−(p−1)/2

∫ x2

x1

F (x)x(p−3)/2dx (16)

81


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

Ptot(ω) ∝ ω−(p−1)/2(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:

s =1

2(p − 1) spectral index (18)

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

treatment.

82


Lecture 9: Synchrotron Radiation IILecture 9: Synchrotron Radiation II

9.1 Spectrum of Synchrotron Radiation9.1 Spectrum of Synchrotron Radiation

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

dtdA=

1

µ0c|Erad(t)|2 (1)

and the radiation field for a moving electron is that given by eqn. (6) in Lec. 5, viz

Erad(r, t) =e

4πε0

R

(1 − R · β)3R× 1

c

[(R− β) × β

](2)

where R 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


The energy emitted per unit area is thus

dW

dA= cε0

∫ +∞

−∞|Erad(t)|2 dt

and Parseval’s theorem says∫ +∞

−∞|E(t)|2 dt =

∫ +∞

−∞|E(ω)|2 dω (3)

for E(ω) = (2π)−1/2∫ +∞−∞ E(t) exp(iωt)dt. Since |Erad(t)| is real and E(ω) is

symmetric, we have∫ +∞−∞ |E(ω)|2 dω = 2

∫ ∞0 |E(ω)|2 dω and so the energy emitted per

unit area is

dW

dA= 2cε0

∫ ∞

0|E(ω)|2 dω (4)

This implies that the energy emitted per unit area per unit frequency bandwidth is

dW

dAdω= 2cε0|E(ω)|2 (5)

84


Substituting dA = R2dΩ and inverting the Fourier transform gives

dW

dωdΩ= 2cε0R

2|E(ω)|2 =cε0π

∣∣∣∣∫ +∞

−∞RErad(t) exp(iωt) dt

∣∣∣∣2

(6)

Step 2 – Fourier transform of the retarded field. We need to solve the integral on the RHS of

RE(ω) =1

(2π)1/2

e

4πcε0

∫ +∞

−∞

R × [(R − β) × β]

(1 − R · β)3eiωt dt (7)

We first change the integration variable from t to t′, using the definition of retarded time

t′ = t − R(t′)/c andR = |r − r0| to give

dt =dt′

(∂t′/∂t)= (1 − R · β) dt′ (8)

So now we have

RE(ω) =1

(2π)1/2

e

4πcε0

∫ +∞

−∞

R × [(R − β) × β]

(1 − R · β)2eiωt dt′ (9)

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 thatR # r.85


Then we expand r to first order in r0, which givesR(t′) # |r|− R · r0. Now we have

RE(ω) =1

(2π)1/2

e

4πcε0

∫ +∞

−∞

R × [(R − β) × β]

(1 − R · β)2exp

[iω

(t′ − R · r0(t

′)/c)]

dt′

(10)

Then we use the identity

R × [(R − β) × β]

(1 − R · β)2=

d

dt′R × (R × β)

(1 − R · 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:

dW

dωdΩ=

e2ω2

16π3ε0c

∣∣∣∣∫ +∞

−∞R × (R × β) exp

[iω

(t′ − R · r0(t

′)/c)]

dt′∣∣∣∣2

(11)

This is now in a form that can be integrated.

86


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‖ =

R × e⊥. Thus, e‖ and e⊥ define a plane

perpendicular to an observer’s line of sight de-

fined by the direction R. This is the plane of

propagation, defined by the triple cross prod-

uct R× (R× β) in the integral.

The magnetic fieldB 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.

87


At any arbitrary retarded time t′, for |β| # 1, we have

R × (R× β) = e‖ cos

(vt′

a

)sinϑ − e⊥ sin

(vt′

a

)(12)

The exponential term in the integral in (11) is simplified using a small angle expansion:

t′ − R · r0(t′)

c# t′ − a

ccos ϑ sin

vt′

a# 1

2γ2

[(1 + γ2θ2)t′ +

c2γ2t′3

3a2

](13)

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

dωdΩ=

dW‖

dωdΩ+

dW⊥

dωdΩ(14)

and where, defining ϑ2γ = 1 + γ2ϑ2, we have

dW‖

dωdΩ=

e2ω2ϑ2

16π3ε0c

∣∣∣∣∫

exp

[iω

2γ2

(ϑ2

γt′ +c2γ2t′3

3a2

)]dt′

∣∣∣∣2

dW⊥

dωdΩ=

e2ω2

16π3ε0c

∣∣∣∣∫

ct′

aexp

[iω

2γ2

(ϑ2

γt′ +c2γ2t′3

3a2

)]dt′

∣∣∣∣2

(15)

88


Now we make a change of variables:

y ≡ γct′

aϑγ, η =≡ ωaϑ3

γ

3cγ3

which gives

dW‖

dωdΩ=

e2ω2ϑ2

16π3ε0c

(aϑγ

γc

)2 ∣∣∣∣∫ +∞

−∞exp

[3

2iη

(y +

1

3y3

)]dy

∣∣∣∣2

dW⊥

dωdΩ=

e2ω2ϑ2

16π3ε0c

(aϑ2

γ

γ2c

)2 ∣∣∣∣∫ +∞

−∞y exp

[3

2iη

(y +

1

3y3

)]dy

∣∣∣∣2

(16)

The integrals can be expressed in terms of the modified Bessel functions of 1/3 and 2/3

order:

dW‖

dωdΩ=

e2ω2ϑ2

16π3ε0c

(aϑγ

γc

)2

K21/3(η)

dW⊥

dωdΩ=

e2ω2ϑ2

16π3ε0c

(aϑ2

γ

γ2c

)2

K22/3(η) (17)

89


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‖

dω# e2ω2a2 sinα

6π2ε0c3γ2

∫ +∞

−∞ϑ2

γϑ2K21/3(η)dϑ

dW⊥

dω=

e2ω2a2 sinα

6π2ε0c3γ4

∫ +∞

−∞ϑ4

γK22/3(η)dϑ (18)

These integrals were first solved by Westfold (1959). They give

dW‖

dω#

√3e2γ sinα

8πε0c[F (x) − G(x)]

dW⊥

dω=

√3e2γ sinα

8πε0c[F (x) + G(x)] (19)

where

F (x) ≡ x

∫ ∞

xK5/3(ζ)dζ , G(x) ≡ xK2/3(x) (20)

90


and x = ω/ωc, where ωc = 32

eBme

γ2 sinα is the critical frequency (see Lec. 8). To convert

this to power per unit frequency, we divide by the orbital period T = 2π/Ω = 2πγme/eB,

which gives

P‖(ω) =

√3e3B sinα

16π2ε0mec[F (x) − G(x)]

P⊥(ω) =

√3e3B sinα

16π2ε0mec[F (x) + G(x)] (21)

These are the components of the single-electron synchrotron power per unit frequency

corresponding to polarisation modes parallel and perpendicular toB. The total synchrotron

power per unit frequency is

P (ω) = P‖(ω) + P⊥(ω) =

√3e3B sinα

8π2ε0mecF (x) (22)

single electron synchrotron power spectrum

91


The functions F (x) andG(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 # 0.392


9.2 Emissivity for a Power-law Electron Distribution9.2 Emissivity for a Power-law Electron Distribution

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

Ne(γ) = Keγ−p , γ1

<∼ γ <

∼ γ2 (24)

for the number of electrons with energy γmec2. The total number of electrons per unit volume

isNe =∫ γ2

γ1Ne(γ) dγ. Note that the electron pitch angles α will in general also be spread

around a direction kα defined by a characteristic angle α0 between R andB. We assume

the electron distribution is isotropic in pitch angle, so thatNe(γ) is independent of kα.

The synchrotron volume emissivity for a nonthermal distribution of relativistic electrons thus

has the following form:

jsynν = 2πKe

∫ γ2

γ1

P (ω) γ−p dγ ∝∫ γ2

γ1

F (x)γ−p dγ (25)

where the factor 2π enters because

93


P (ω) = dW/(dωdt) = dW/(2πdνdt) =∫

jνdV/2π. We now change the integration

variable from γ to x = ω/ωc using the relations

γ =

(3

2

eBx sinα

ωme

)−1/2

, dγ = −1

2

(3

2

eB sinα

ωme

)−1/2

x−3/2 dx (26)

So we now have an integral of the form

jsynν ∝

∫ x1

x2

x(p−3)/2F (x) dx (27)

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

x1 =ω

ωc(γ1), x2 =

ω

ωc(γ2)

If γ1 ( γ2, then ωc(γ1) ( ωc(γ2) and we can have a wide range of frequencies satisfying

ω $ ωc(γ1) so that x1 → ∞. Similarly, we can have ω ( ωc(γ2), so that x2 → 0. Thus,

94


we have an integral of the form ∫ ∞

0x(p−3)/2F (x) dx

as well as an analogous one of the form∫ ∞0 x(p−3)/2G(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

jsynν = 3p/22−(p+7)/2π−(p+3)/2Γ(

p

4+

19

12)Γ(

p

4− 1

12)(p + 1)−1

(28)

e2

ε0cKe

(eB

mesinα0

)(p+1)/2

ν−(p−1)/2synchrotron emissivity

This is clearly a power-law spectrum, with spectral index

α =1

2(p − 1) (29)

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

95


to perform a further integration over directions α0. The required integration is

〈sin(p+1)/2 α0〉 =1

4π

∫ 2π

0dφ

∫ π

0sin(p+1)/2 α0 sinα0dα0

=1

2

∫ π

0sin(p+3)/2 α0dα0

=

√π

2

Γ( 5+p4 )

Γ( 7+p4 )

(30)

96


The final and most often used expression for the synchrotron emissivity is the following for the

randomly oriented magnetic field case:

jsynν = 3p/22−(p+13)/2π−(p+2)/2 Γ( 5+p

4 )Γ(p4 + 19

12 )Γ(p4 − 1

12 )

Γ( 7+p4 )

e2

ε0cKe

(eB

me

)(p+1)/2

ν−(p−1)/2(31)

synchrotron emissivity (pitch angle averaged)

97


9.3 Polarisation of Synchrotron Radiation9.3 Polarisation of Synchrotron Radiation

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 ofErad(t) remain. Thus, synchrotron radiation is partially linearly polarised. The

degree of linear polarisation for particles with a single energy γ is defined by

Π(ω) =P⊥(ω) − P‖(ω)

P⊥(ω) + P‖(ω)=

G(x)

F (x)(32)

The frequency integrated value isΠ # 75%, which is quite high. For a power-law distribution

of electron energies, the degree of linear polarisation is

Π =p + 1

p + 73

(33)

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 reduceΠ to

values down to a few percent.

98


Lecture 10: Compton ScatteringLecture 10: Compton Scattering

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.

99


10.1 Energy Transfer and Scattering Cross Section10.1 Energy Transfer and Scattering Cross Section

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 angleΘ

w.r.t. its initial propagation direction ki. 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:

Pγi =ε

c(1, ki) , Pγf =

ε1

c(1, kf )

Initial and final 4-momenta of electron:

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

100


Conservation of 4-momentum requires

Pei + Pγi = Pef + Pγf (1)

Rearranging and squaring gives

|Pef |2 = |Pei + Pγi − Pγf |

2 (2)

= |Pγi|2 + |Pei|

2 + |Pγf |2 + 2PγiPei − 2PγiPγf − 2PγfPei

Note that the modulus of a 4-vectorAµ is defined as

A2 = AµAµ = −(A0)2 + (A1)2 + (A2)2 + (A3)2. This implies that the magnitudes of

the 4-momenta for a photon and an electron are P 2γ = 0 and P 2

e = −m2ec

2, respectively. So

in the above expression, we have |Pef |2 = −m2ec

2 = |Pei|2 and |Pγi|2 = 0 = |Pγf |2,

which leaves behind only those terms with a factor of 2 in front. These terms are:

PγiPei = −εme, PγiPγf = − εε1

c2 + εε1

c2 ki · kf , and PγfPei = −ε1me. Substituting

these in, rearranging and using ki · kf = cosΘ gives the following expression for the energy

of the scattered photon:

ε1 =ε

1 + εmec2 (1 − cosΘ)

(3)

101


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

∆λ = λ1 − λ = λC(1 − cosΘ) (4)

where

λC ≡h

mec# 0.0243 A Compton wavelength (5)

Thus, the wavelength change is of order λC. For long wavelengths (λ $ λC) or equivalently,

ε % mec2, the scattering is approximately elastic (i.e. ε1 # ε). This is the Thomson regime.

102


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:

dσKN

dΩ=

1

2r20

ε21

ε2

(ε

ε1+

ε1

ε− sin2 Θ

)(6)

where r0 = e2/(4πε0mec) = 2.82 × 10−15 m is the classical electron radius (defined in

Lec. 8). This reduces to the classical differential Thomson cross section in the limit ε1 ∼ ε,

viz. dσT/dΩ = 12r2

0(1 + cos2 Θ). The total cross section is obtained by integrating over

solid angle, σKN = 2π∫ +1−1 (dσKN/dΩ) d cosΘ:

σKN = σT

3

4

1 + x

x3

[2x(1 + x)

1 + 2x− ln(1 + 2x)

]+

1

2xln(1 + 2x) −

1 + 3x

(1 + 2x)3

(7)

where x ≡ hν/mec2.

103


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.

For x $ 1, the asymptotic solution is σKN ∼ 38σTx−1

(ln 2x + 1

2

)104


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. LetK be the observer’s frame andK′ 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. InK , the electron’s velocity is in the x direction

and all angles in both frames are measured from this axis.

observer’s frameK electron rest frameK′

105


InK′, all the previous formulas for scattering from stationary electrons are valid. Transforming

the photon’s initial energy intoK′:

ε′ = εγ(1 − β cos Θ) (8)

and transforming the scattered photon energy back intoK :

ε1 = ε′1γ(1 + β cosΘ′

1) (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. ε′1 # ε′) and which haveΘ, Θ′

1>∼π/2. The condition for Thomson scattering

in the rest frame is

ε′ % mec2 =⇒ γhν % mec

2 (10)

106


10.2 Single Scattering Power10.2 Single Scattering 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 ε′1 # 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

dE′

1

dt′= cσT

∫ε′1n

′(ε′) dε′ (11)

Now we know that the emitted power is an invariant. Another invariant is the quantity

n(ε)dε/ε. SodE1

dt= cσT

∫ε′n′dε′ = cσT

∫ε′2

ndε

ε(12)

Now we substitute ε′ = εγ(1 − β cosΘ) from eqn. (8) to get

dE1

dt= cσTγ2

∫(1 − β cosΘ)2εndε (13)

107


which now only contains quantities in frameK . For an isotropic distribution of photons, we

have

〈(1 − β cosΘ)2〉 = 1 +1

3β2

giving

dE1

dt= cσTγ2

(1 +

1

3β2

)Uγ (14)

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

dt=

dE1

dt−

dε

dt= cσTUph

[γ2

(1 +

1

3β2

)− 1

](15)

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

Pic =4

3σTcγ2β2Uγ inverse Compton power (16)

This has used γ2 − 1 = γ2β2.

108


Emitted power for a distribution of electrons

For a nonthermal power law distribution of electronsN(γ) = Keγ−p, we can obtain the total

power per unit volume from

Pic,tot =

∫ γ2

γ1

PicN(γ)dγ

This gives, for β # 1,

Pic,tot =4

3σTcUγKe(3 − p)−1(γ3−p

2 − γ3−p1 ) nonthermal power-law electrons (17)

For a thermal distribution of electrons, γ = 1 and 〈β2〉 = 3kTe/mec2. 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

Pic,tot = σTcUγNe4kTe

mec2thermal electrons (18)

whereNe is the total electron number density.

109


10.3 Single Scattering Spectra10.3 Single Scattering Spectra

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):

jε1=

3

16πσTcε1Ke

∫dε

n(ε)

ε

∫ γ2

γ1

dγ γ−(p+2)f

(ε1

4γ2ε

)(19)

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

110


For sufficiently large limits on the γ integral, we have

jicε1

=3

π2p−2 p2 + 4p + 11

(p + 1)(p + 3)2(p + 5)σTcε−(p−1)/2

1 Ke

∫ε(p−1)/2n(ε) dε (20)

Thus, inverse Compton scattering also predicts a power law spectrum with a spectral index

α =1

2(p − 1) (21)

identical to the case of synchrotron emission. The power law spectrum is independent of the

incident photon distribution.

111


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

n(ε) =8π

(hc)3ε2

[exp

( ε

kT

)− 1

]−1(22)

Inserting this into the expression jε1above, and solving the integrals gives

jic,bbε1

=σT

h3c2fbb(p)(kT )(p+5)/2Keε

−(p−1)/21 (23)

where

fbb(p) = 32p+1(p2 + 4p + 11)

(p + 1)(p + 3)2(p + 5)γ

(p + 5

2

)ζ

(p + 5

2

)

where ζ is the Riemann zeta function.

112


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

n(ε) =Uγ(ε0)

ε0

(ε

ε0

)−(p−1)/2

, εmin<∼ εmax (24)

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

volume emissivity is

jsscν1

= f(p)σTcKeUγν0ln

(εmax

εmin

)(ν1

ν0

)−(p−1)/2

(25)

where the the relation jν1= hjε1

has been used and where

f(p) =3

π2p−2 p2 + 4p + 11

(p + 1)(p + 3)2(p + 5)

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

113


10.4 Multiple Scatterings: the Compton y Parameter10.4 Multiple Scatterings: the Compton y Parameter

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, τ2) (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

114


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

y =4kTe

mec2

(1 +

4kTe

mec2

)max(τT, τ2

T) (29)

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.

115


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 sphericalregion with varying optical depths

containing electrons with kTe =

0.7mec2. The incident seed photons

are injected at the centre with a black-

body spectrum at kT ! kTe. TheCompton y parameter thus rangesfrom y " 0.5 for the τT = 0.05

spectrum, to y " 103for the τT =

0.05 spectrum.

116

Lecture 5: Radiation from Moving Charges - Uniuddeangeli/fismod/Radiation.pdf · 2012-11-07 ·...

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Transcript of Lecture 5: Radiation from Moving Charges - Uniuddeangeli/fismod/Radiation.pdf · 2012-11-07 ·...