PHYS 4011 – 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
electromagnetic fields associated with accelerating charges fall off as 1/r instead of 1/r 2 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 – 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−
c
!
(1)
Information from the charged particle’s trajectory
arriving at field point P has propagated a finite distance 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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PHYS 4011 – HEA
Lec. 5
The radiation field at P at time t is calculated from the retarded scalar and vector potentials V
= −∇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
and A using E
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) = ṙ0 (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
"
q
δ(t! − t + |r(t) − r0 (t! )|/c) !
V (r, t) =
dt
4π%0
|r(t) − r0 (t! )|
"
µ0 q
δ(t! − t + |r(t) − r0 (t! )|/c) !
A(r, t) =
v(t! )
dt
4π
|r(t) − r0 (t! )|
(2)
After introducing the simplifying notation
!
!
R(t ) = r(t) − r0 (t )
,
!
!
R(t ) = |r(t) − r0 (t )|
,
R(t! )
R̂ =
R(t! )
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PHYS 4011 – HEA
Lec. 5
the integral can be solved with a change of variables giving
A(r, t) =
µ0
v
qv(t! )
= 2 V (r, t)
4π (1 − R̂ · v(t! )/c)R
c
Liénard–Wiechart potentials
(3)
These are the famous retarded potentials for a moving point charge.
Points to note:
1. The factor (1 − R̂ · 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/r 2 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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PHYS 4011 – HEA
Lec. 5
The radiation field
The differentiation of the Liénard–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
= ṙ0 (t! ) and its corresponding acceleration as
β̇c = r̈0 (t! ), the fields are
&
#
%
q
1$
(R̂ − β)(1 − β 2 )
R̂
(R̂ − β) × β̇
E(r, t) =
+
(4)
×
4π%0
c
(1 − R̂ · β)3 R2
(1 − R̂ · β)3 R
B(r, t) =
1
R̂ × E(r, t)
c
(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 R̂.
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PHYS 4011 – HEA
Lec. 5
This electric field, along with the corresponding magnetic field, constitute the radiation field of a
moving charge:
#
Erad (r, t) =
q
4π%0
Brad (r, t) =
1
R̂ × Erad (r, t)
c
%
1$
(R̂ − β) × β̇
×
c
(1 − R̂ · β)3 R
R̂
Note that Erad , Brad and R̂ are mutually perpendicular.
45
&
(6)
PHYS 4011 – HEA
Lec. 5
5.2 Radiation from Nonrelativistic Charged Particles
The Larmor formula
When β
& 1, the radiation fields simplify to
Erad (r, t) =
q
4π%0
'
)
R̂
1 (
× 2 R̂ × v̇
R
c
*
(7)
and Brad (r, t) follows from (6). Note that Erad lies in the plane containing R̂ and v̇ (i.e. the
plane of polarisation) and Brad is perpendicular to this plane. If we let θ be the angle between
R̂ and v̇, then
|Erad | = c|Brad | =
q
v̇
sin θ
4π%0 Rc2
The Poynting vector is in the direction of R̂ and has the magnitude
S=
1 2
µ0 q 2 v̇ 2
sin2 θ
Erad =
µ0 c
16π 2 c R2
(8)
We now want to express this as an emission coefficient.
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PHYS 4011 – HEA
Lec. 5
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 = R2 dΩ , where dΩ is the solid angle about the
direction R̂ of S. So the power emitted per solid angle is
dW
µ0 2 2 2
= S R2 =
q v̇ sin θ
dtdΩ
16π 2 c
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:
dW
µ0 2 2
P =
=
q v̇
dt
16π 2 c
"
2π
0
"
π
µ0 2 2
sin θ sin θ dθ dφ =
q r̈0
8πc
2
0
"
+1
(1 − µ2 ) dµ
−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:
P =
µ0 q 2 r̈02
6πc
Larmor formula
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PHYS 4011 – HEA
Lec. 5
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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PHYS 4011 – HEA
Lec. 5
Applying the superposition rule, we have
Erad
+ qi
=
4π%0
i
'
*
)
R̂i
1 (
× 2 R̂i × r¨i
Ri
c
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=
+
qi r i
(10)
i
to get
Erad
Then, as before, we find
1
=
4π%0
'
)
R̂
1 (
× 2 R̂ × d̈
R
c
dW
µ0 ¨2 2
=
d sin θ
dtdΩ
16π 2 c
49
*
(11)
PHYS 4011 – HEA
Lec. 5
and the total power is
µ0 d¨2
P =
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¨ and R 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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PHYS 4011 – 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:
v̇(t) =
=
v̇(ω)
1
(2π)1/2
1
(2π)1/2
+∞
"
v̇(ω) exp(−iωt) dω
−∞
" +∞
v̇(t) exp(iωt) dt
(13)
−∞
Then we use Parseval’s theorem, which relates this as follows:
"
+∞
2
|v̇(ω)| dω =
∞
"
+∞
|v̇(t)|2 dt
(14)
∞
We can also use the relation
"
∞
2
|v̇(ω)| dω =
0
"
0
|v̇(ω)|2 dω =
(15)
−∞
which is valid provided v̇(t) is real. Applying these realtions to the Larmor formula (9), we can
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PHYS 4011 – HEA
Lec. 5
determine the total energy radiated by a single charged particle with an acceleration history
v̇(t):
+∞
+∞
"
µ0 q 2
µ0 q 2 ∞
2
P dt =
|v̇(ω)|2 dω
(16)
|v̇(t)| dt =
3πc 0
−∞
−∞ 6πc
!∞
Since the total emitted energy must also equal 0 (dW/dω) dω , then the energy per unit
"
"
bandwidth is just
µ0 q 2
dW
=
|v̇(ω)|2
dω
3πc
(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
µ0 4
=
ω |d(ω)|2
dω
3πc
(18)
This is because differentiating wrt t twice introduces a factor ω 2 in the Fourier transform – c.f.
! +∞
d(t) = (2π)−1/2 −∞ d(ω) exp(−iωt) dω =⇒
!
¨ = −(2π)−1/2 +∞ ω 2 d(ω) exp(−iωt) dω , so ev̇(ω) = ω 2 d(ω).
d(t)
−∞
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PHYS 4011 – HEA
Lec. 6
Lecture 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 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 ∼
mp /me times greater.
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PHYS 4011 – HEA
Lec. 6
6.1 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).
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 d̈ = −ev̇. We want an emission spectrum for the
bremsstrahlung radiation using eqn. (17) in Lec. 5, viz.
µ0 e2
dW
=
|v̇(ω)|2
dω
3πc
(1)
and using
1
v̇(ω) =
(2π)1/2
!
+∞
v̇(t) exp(iωt) dt
−∞
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(2)
PHYS 4011 – 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 v̇ = dv/dt ≈ ∆v over the time interval dt ≈ τ . So we
have
|v̇(ω)| #
So our radiation spectrum goes as


1
∆v
(2π)1/2
ωτ % 1
 0


(3)
ωτ $ 1
µ e2
2
0
dW
6π 2 c |∆v|
#
 0
dω
b % v/ω
(4)
b $ v/ω
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PHYS 4011 – HEA
Lec. 6
Now we can work out ∆v by noting that the total Coloumb force on the electron is
Ze2 /(4π$0 R2 ) in the −R̂ direction. The perpendicular component of acceleration is the
strongest and will thus make the dominant contribution to the radiation spectrum, so we can
write
1 Ze2
∆v #
4π$0 me
!
b dt
1 Ze2 2
=
4π$0 me bv
(b2 + v 2 t2 )3/2
(5)
(the integral turns out to be elementary). Substituting this into the expression for the radiation
spectrum gives


2 6
8Z e
dW
3πc3 (4π"0 )3 m2e b2 v 2
#
 0
dω
b % v/ω
(6)
b $ v/ω
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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PHYS 4011 – 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 ne v 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
= ne ni 2πv
dωdV dt
!
∞
bmin
dW (b)
b db
dω
(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:
16ne ni Z 2 e6
dW
# 3
ln
dωdV dt
3c (4π$0 )3 m2e v
%
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
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PHYS 4011 – HEA
Lec. 6
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
(5) implies bmin
# 2Ze2 /(4π$0 me v 2 ).
∼ v,
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
Gff (v, ω) =
ln
π
%
bmax
bmin
&
free-free Gaunt factor
(9)
giving
dW
16πne ni Z 2 e6
Gff (v, ω)
# 3/2 3
dωdV dt
3 c (4π$0 )3 m2e v
(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.
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PHYS 4011 – HEA
Lec. 6
6.2 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 d3 v is
&
%
mv 2
v 2 dv
dn(v) = f (v)d v = f (v)4πv dv ∝ exp −
2kT
3
2
We need to average the single-speed radiation spectrum over this distribution function for all
electron speeds satisfying 12 me v 2 >
∼ hω/2π , i.e.
dW (T, ω)
=
dωdV dt
Using dω
jνff
dW (v,ω) 2
v exp(−mv 2 /2kT ) dv
vmin dωdV dt
'∞
v 2 exp(−mv 2 /2kT ) dv
0
'∞
(11)
= 2π dν , the final result is the expression for the free-free volume emissivity:
32πe6
dW
1
≡
=
dV dtdν
(4π$0 )3 3me3/2 c3
%
2π
3kT
&1/2
&
%
hν
Z ne ni exp −
G¯ff
kT
2
(12)
where G¯ff is now the velocity averaged Gaunt factor. Its value is typically of order unity.
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PHYS 4011 – HEA
Lec. 6
Points to note:
1. The only frequency dependence is in the exponential term exp(−hν/kT ). So the
spectrum declines exponentially at frequencies hν
hν % kT .
$ kT , but is approximately flat for
2. The emissivity is also proportional to T −1/2 . So for hν
higher T . But for hν
% kT , the spectrum is lower for
$ 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
integrate jνff over an appropriate source volume and over the frequency bandwidth that we
are interested in.
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PHYS 4011 – HEA
Lec. 7
Lecture 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
4π"0
!
R̂
(1 − R̂ · β)3 R
×
1"
c
(R̂ − β) × β̇
$
#
(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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PHYS 4011 – HEA
Lec. 7
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 !
where γ
,
dt = γdt!
= (1 − β 2 )1/2 is the Lorentz factor. Note that the Lorentz factors cancel out, i.e.
P =
dW
dW !
γdW !
=
= P!
=
dt
γdt!
dt!
so the total emitted power is Lorentz invariant.
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PHYS 4011 – HEA
Lec. 7
Step 2. Now we have the identity P
= P! =
µ0 q 2 |a! |2
. In this expression, a! is the
6πc
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
aµ ≡
dv µ
d2 xµ
=
=
dτ
dτ 2
a2 = aµ aµ = −a0 a0 + |a|2
,
(3)
where v µ
= γ(c, v) is the 4-velocity. Note that the 4-velocity and 4-acceleration are
orthogonal: aµ vµ = (dv µ /dτ )vµ = 12 d(v µ vµ )/dτ = 12 d(−c2 )/dτ = 0. Now in K ! , we
can write |a! |2 = a!2 + a!0 a!0 . But in the particle’s rest frame, v !µ = (c, 0) and since
a!µ vµ! = 0, we must have a!0 = 0. Thus, we can write
|a! |2 = a!µ a!µ = aµ aµ
and
P =
µ0 q 2 µ
a aµ
6πc
(4)
which is now in a manifestly covariant form (i.e. can be evaluated in any frame).
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PHYS 4011 – HEA
Lec. 7
Step 3. Since aµ aµ
= |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!" = γ 3 a"
,
a!⊥ = γ 2 a⊥
Thus, we have
P =
µ0 q 2 ! 2
µ0 q 2 !2
µ0 q 2 4 2 2
)
=
|a | =
(a" + a!2
γ (γ a" + a2⊥ )
⊥
6πc
6πc
6πc
(5)
Clearly, the power emitted increases drastically as a particle becomes relativistic. This
expression can be written another way:
P =
&
µ0 q 2 6 % 2
γ a − |β × a|2
6πc
relativistic Larmor formula
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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PHYS 4011 – HEA
Lec. 7
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 ! + vdp!x )
from the transformation properties of the 4-momentum pµ
= (E/c, p) = m0 v µ . 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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PHYS 4011 – HEA
Lec. 7
!
Geometry for dipole emission from a particle instantaneously at rest in frame K .
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PHYS 4011 – HEA
Lec. 7
The transformation of cos ϑ! is given by the aberration of light formula:
cos ϑ =
cos ϑ! + β
1 + β cos ϑ!
(8)
Differentiating gives
d cos ϑ!
γ 2 (1 + β cos ϑ! )2
The azimuthal angle is invariant, so dφ = dφ! . Thus,
d cos ϑ =
dΩ = d cos ϑdφ =
and so we have
d cos ϑ! dφ!
dΩ!
=
γ 2 (1 + β cos ϑ! )2
γ 2 (1 + β cos ϑ! )2
dW
dW !
= γ 3 (1 + β cos ϑ! )3
dΩ
dΩ!
(9)
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
dP !
dP !
1
= γ 4 (1 + β cos ϑ! )4 ! = 4
dΩ
dΩ
γ (1 − β cos ϑ)4 dΩ!
(10)
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PHYS 4011 – HEA
Lec. 7
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)
dP !
µ0 q 2 ! 2 2 !
=
a sin θ
dΩ!
16π 2 c
(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
2 2
2
3P
µ0 q 2 (γ a" + a⊥ )
dP
sin2 θ !
2 !
sin
θ
=
=
dΩ
16π 2 c (1 − β cos ϑ)4
8πγ 4 (1 − β cos ϑ)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.
69
PHYS 4011 – 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
sin2 θ ! =
sin2 θ
γ 2 (1 − β cos ϑ)2
(13)
Substituting into (12) yields
µ0 q 2 2
dP
sin2 θ
=
a
dΩ
16π 2 c " (1 − β cos ϑ)6
which peaks in the forward direction at ϑ
(14)
∼ 1/γ .
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PHYS 4011 – HEA
Lec. 7
Case 2: acceleration perpendicular to velocity
We put a"
= 0 and use cos θ ! = sin ϑ! cos φ! . The transformation is
sin2 ϑ cos2 φ
sin θ = 1 − 2
γ (1 − β cos ϑ)2
(15)
'
(
sin2 ϑ cos2 φ
µ0 q 2 2
dP
1
1− 2
=
a
dΩ
16π 2 c ⊥ (1 − β cos ϑ)4
γ (1 − β cos ϑ)2
(16)
2
!
giving
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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PHYS 4011 – HEA
Lec. 7
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 cos ϑ ∼
1 + γ 2 ϑ2
2γ 2
(17)
Substituting this limit back into (12) we get for the parallel and perpendicular cases
dP"
dΩ
∼
dP⊥
dΩ
∼
4µ0 q 2 a2"
γ 2 ϑ2
(1 + γ 2 ϑ2 )6
(18)
µ0 q 2 a2⊥ 8 1 − 2γ 2 ϑ2 cos(2φ) + γ 4 θ 4
γ
π2 c
(1 + γ 2 ϑ2 )6
(19)
π2 c
γ 10
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PHYS 4011 – HEA
Lec. 8
Lecture 8: Synchrotron Radiation I
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 frequency. This results in a discrete emission spectrum
which usually does not extend beyond optical/UV frequencies in most astrophysical situations. When the
particles are relativistic, however, the radiation is referred to as synchrotron emission and results in a
continuum spectrum because the frequency of emission extends over many higher order harmonics of the
gyration frequency. Synchrotron emission by relativistic particles in a magnetic field is a prevalent radiation process in astrophysics. The emissivity is broadband 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 radio 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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PHYS 4011 – HEA
Lec. 8
8.1 Total Emitted Synchrotron 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
dt
dt
We now make the assumption that γ
(1)
≈ 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!
=0
dt
,
dv⊥
q
=
v⊥ × B
dt
γm
(2)
= 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
Thus, v!
acceleration is perpendicular to the velocity in this plane.
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PHYS 4011 – 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
a⊥ =
where Ω
|q|v⊥ B
= Ωv⊥
γm
(3)
= |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
B results from the combination of uniform motion
along B and circular motion perpendicular to B.
acceleration perpendicular to velocity, eqn.
(5) in Lec. 7, viz.
P =
µ0 q 2 4 2
µ0 q 2 4 q 2 B 2 2
v =⇒
γ a⊥ =
γ
6πc
6πc γ 2 m2 ⊥
P =
1 q4 2 2 2 2
γ β B sin α
6π#0 c m2
(4)
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.
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PHYS 4011 – HEA
Lec. 8
For an isotropic distribution of velocities, it is necessary to average over all pitch angles for a
given speed β :
2
&β⊥
'
β2
=
4π
!
1
sin α dΩ = β 2
2
2
!
+1
(1 − cos2 α) d cos α =
−1
2 2
β
3
(5)
So the total power emitted by an electron, averaged over all pitch angles is
P =
2 1 q4 2 2 2
γ β B
3 6π#0 c m2
(6)
This is also sometimes expressed in terms of the Thomson cross-section,
σT =
8 2
e4
= 6.65 × 10−29 m2
πr0 =
3
6π#20 m2e c4
(7)
= e2 /(4π#0 me c) = 2.82 × 10−15 m is the classical electron radius, obtained by
equating the electrostatic potential energy, e2 /(4π#0 r0 ), with the rest mass energy me c2 . So
where r0
P =
where UB
2 c
4
σT γ 2 β 2 B 2 = σT cγ 2 β 2 UB
3 µ0
3
total synchrotron power
= B 2 /2µ0 is the magnetic energy density (i.e. energy per unit volume).
76
(8)
PHYS 4011 – HEA
Lec. 8
8.2 Synchrotron Spectrum – Qualitative Treatment
Because of beaming effects, the radiation emitted by a relativistic particle appears to an observer 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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PHYS 4011 – HEA
Lec. 8
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
evB sin α
∆v
)
∆t
γm
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PHYS 4011 – HEA
and since ∆v
Lec. 8
) v∆θ and ∆t ) ∆s/v , we have
evB sin α
Ω
∆θ
)
= sin α
∆s
γmv
v
and hence, a
) ∆s/∆v ) v/(Ω sin α) and
∆s )
2v
γΩ sin α
So the time interval between the emitted pulses is
∆t = t2 − t1 )
∆s
2
)
v
γΩ sin α
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
"
v#
2
v#
∆s
∆s "
a
a
a
1−
)
1−
∆t = t2 − t1 ) ∆t −
=
c
v
c
γΩ sin α
c
For γ
, 1, we have
1−
(9)
(10)
v
1
∼ 2
c
2γ
79
PHYS 4011 – HEA
Lec. 8
and so
∆ta ∼ (γ 3 Ω sin α)−1
(11)
So the time interval between pulses and the width of the individual pulses, E(t), are smaller
γ 3 . 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
than the gyration period by a factor ∼
exact treatment to follow, we will use the following definition
ωc ≡
3 3
3 eB 2
γ sin α
γ Ω sin α =
2
2 me
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/ωc F (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 e3 B
ω
sin α
P (ω) ∼
F
(13)
9π#0 c me
ωc
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PHYS 4011 – HEA
Lec. 8
8.3 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−p dε
N (γ)dγ ∝ γ −p dγ
,
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
γ −p dγ
Ptot (ω) ∝
F
ωc
γ1
= ω/ωc , noting that ωc ∝ γ 2 , we have
! x2
−(p−1)/2
Ptot (ω) ∝ ω
F (x)x(p−3)/2 dx
(15)
If we change integration variables to x
(16)
x1
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PHYS 4011 – HEA
Lec. 8
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
(p − 1)
2
spectral index
(18)
Although this relation has been derived qualitatively here, the result is the same for the exact
treatment.
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PHYS 4011 – HEA
Lec. 9
Lecture 9: Synchrotron Radiation II
9.1 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
1
=
|Erad (t)|2
dtdA
µ0 c
(1)
and the radiation field for a moving electron is that given by eqn. (6) in Lec. 5, viz
e
Erad (r, t) =
4π"0
!
$
#
1"
(R̂ − β) × β̇
×
c
(1 − R̂ · β)3 R
R̂
(2)
where R̂ 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 – HEA
Lec. 9
The energy emitted per unit area is thus
dW
= c"0
dA
%
+∞
|Erad (t)|2 dt
−∞
and Parseval’s theorem says
%
+∞
%
2
−∞
|E(t)| dt =
+∞
−∞
|E(ω)|2 dω
(3)
& +∞
= (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
for E(ω)
unit area is
dW
= 2c"0
dA
%
∞
0
|E(ω)|2 dω
(4)
This implies that the energy emitted per unit area per unit frequency bandwidth is
dW
= 2c"0 |E(ω)|2
dAdω
84
(5)
PHYS 4011 – HEA
Lec. 9
= R2 dΩ and inverting the Fourier transform gives
)2
)%
)
dW
c"0 )) +∞
2
2
RErad (t) exp(iωt) dt))
= 2c"0 R |E(ω)| =
)
dωdΩ
π
−∞
Substituting dA
(6)
Step 2 – Fourier transform of the retarded field. We need to solve the integral on the RHS of
1
e
RE(ω) =
(2π)1/2 4πc"0
%
+∞
R̂ × [(R̂ − β) × β̇]
(1 − R̂ ·
−∞
β)3
eiωt dt
(7)
We first change the integration variable from t to t! , using the definition of retarded time
t! = t − R(t! )/c and R = |r − r0 | to give
dt =
dt!
= (1 − R̂ · β) dt!
(∂t! /∂t)
(8)
So now we have
1
e
RE(ω) =
(2π)1/2 4πc"0
%
+∞
R̂ × [(R̂ − β) × β̇]
(1 − R̂ ·
−∞
Next we express eiωt
β)2
eiωt dt!
(9)
= 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 – HEA
Lec. 9
Then we expand r to first order in r0 , which gives R(t! )
1
e
RE(ω) =
1/2
4πc"0
(2π)
%
+∞
# |r| − R̂ · r0 . Now we have
R̂ × [(R̂ − β) × β̇]
−∞
(1 − R̂ · β)2
" '
(#
exp iω t! − R̂ · r0 (t! )/c dt!
(10)
Then we use the identity
R̂ × [(R̂ − β) × β̇]
(1 − R̂ ·
β)2
=
d R̂ × (R̂ × β)
dt! (1 − R̂ · 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
" '
e2 ω 2 )) +∞
dW
=
R̂ × (R̂ × β) exp iω t! − R̂ · r0 (t! )/c dt! ))
dωdΩ
16π 3 "0 c ) −∞
This is now in a form that can be integrated.
86
(11)
PHYS 4011 – 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 travelling in the x − y plane and passes through
the origin at retarded time t!
= 0 with an instantaneous velocity in the x-direction. The
unit vector e⊥ is along the y axis and e% =
R̂ × e⊥ . Thus, e% and e⊥ define a plane
perpendicular to an observer’s line of sight defined by the direction R̂. This is the plane of
propagation, defined by the triple cross prod-
uct R̂ × (R̂ × β) in the integral.
The magnetic field B must also be in the plane containing R̂ and β , so e% and e⊥ define
directions parallel and perpendicular to the projection of the magnetic field on the plane of
propagation.
87
PHYS 4011 – HEA
Lec. 9
At any arbitrary retarded time t! , for |β|
# 1, we have
* !+
* !+
vt
vt
sin ϑ − e⊥ sin
R̂ × (R̂ × β) = e% cos
a
a
(12)
The exponential term in the integral in (11) is simplified using a small angle expansion:
,
R̂ · r0 (t! )
a
c2 γ 2 t!3
vt!
1
!
2 2 !
t −
# t − cos ϑ sin
# 2 (1 + γ θ )t +
c
c
a
2γ
3a2
!
where 1 − v/c
(13)
# 1/2γ 2 and v # c has been used elsewhere. Now we can calculate the
spectrum in the two polarisation states with
dW%
dW⊥
dW
=
+
dωdΩ
dωdΩ dωdΩ
and where, defining ϑ2γ
dW%
dωdΩ
dW⊥
dωdΩ
= 1 + γ 2 ϑ2 , we have
)%
*
+- )2
,
)
c2 γ 2 t!3
e2 ω 2 ϑ2 ))
iω
2 !
!)
ϑ
dt
t
+
=
exp
γ
)
16π 3 "0 c )
2γ 2
3a2
)%
*
+- )2
,
)
c2 γ 2 t!3
e2 ω 2 )) ct!
iω
2 !
!)
ϑ
dt
t
+
=
exp
γ
)
16π 3 "0 c )
a
2γ 2
3a2
(14)
88
(15)
PHYS 4011 – HEA
Lec. 9
Now we make a change of variables:
ct!
y≡γ
aϑγ
ωaϑ3γ
η =≡
3cγ 3
,
which gives
dW%
dωdΩ
=
dW⊥
dωdΩ
=
+2 )% +∞
+- )2
,
*
)
)
1
3
3
)
)
dy
exp
iη
y
+
y
)
)
2
3
−∞
.
/2 )%
+- )2
,
*
)
) +∞
3
e2 ω 2 ϑ2 aϑ2γ
1
3
)
)
dy
y
exp
iη
y
+
y
)
)
16π 3 "0 c γ 2 c
2
3
−∞
e2 ω 2 ϑ2
16π 3 "0 c
*
aϑγ
γc
(16)
The integrals can be expressed in terms of the modified Bessel functions of 1/3 and 2/3
order:
dW%
dωdΩ
dW⊥
dωdΩ
=
e2 ω 2 ϑ2
16π 3 "0 c
*
aϑγ
γc
=
e2 ω 2 ϑ2
16π 3 "0 c
.
aϑ2γ
γ2c
+2
/2
2
K1/3
(η)
2
K2/3
(η)
(17)
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PHYS 4011 – 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%
dω
dW⊥
dω
e2 ω 2 a2 sin α
6π 2 "0 c3 γ 2
#
2
2 2
e ω a sin α
6π 2 "0 c3 γ 4
=
+∞
%
−∞
+∞
%
−∞
2
ϑ2γ ϑ2 K1/3
(η)dϑ
2
ϑ4γ K2/3
(η)dϑ
(18)
These integrals were first solved by Westfold (1959). They give
dW%
dω
#
dW⊥
dω
=
where
F (x) ≡ x
%
√
√
3e2 γ sin α
[F (x) − G(x)]
8π"0 c
3e2 γ sin α
[F (x) + G(x)]
8π"0 c
(19)
∞
K5/3 (ζ)dζ
,
x
90
G(x) ≡ xK2/3 (x)
(20)
PHYS 4011 – HEA
and x
Lec. 9
= ω/ωc , where ωc =
3 eB 2
2 me γ
sin α 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
√ 3
3e B sin α
P% (ω) =
[F (x) − G(x)]
16π 2 "0 me c
√ 3
3e B sin α
[F (x) + G(x)]
P⊥ (ω) =
16π 2 "0 me c
(21)
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
√
3e3 B sin α
P (ω) = P% (ω) + P⊥ (ω) =
F (x)
8π 2 "0 me c
(22)
single electron synchrotron power spectrum
91
PHYS 4011 – HEA
Lec. 9
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) ∼ x1/2 e−x , x $ 1
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
F (x), G(x) ∼ x1/3 , x ( 1
(23)
PHYS 4011 – HEA
Lec. 9
9.2 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) = Ke γ −p
,
<
γ1 <
∼ γ ∼ γ2
(24)
for the number of electrons with energy γme c2 . The total number of electrons per unit volume
is Ne
=
& γ2
γ1
Ne (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 R̂ and B. 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:
jνsyn
= 2πKe
%
γ2
P (ω) γ
−p
γ1
dγ ∝
%
γ2
F (x)γ −p dγ
(25)
γ1
where the factor 2π enters because
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PHYS 4011 – HEA
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&
P (ω) = dW/(dωdt) = dW/(2πdνdt) = jν dV /2π . We now change the integration
variable from γ to x = ω/ωc using the relations
*
+−1/2
+−1/2
*
1 3 eB sin α
3 eBx sin α
γ=
,
dγ = −
x−3/2 dx (26)
2 ωme
2 2 ωme
So we now have an integral of the form
jνsyn
∝
%
x1
x(p−3)/2 F (x) dx
(27)
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
x1 =
ω
ωc (γ1 )
,
If γ1
x2 =
ω
ωc (γ2 )
( 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,
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PHYS 4011 – HEA
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we have an integral of the form
%
∞
x(p−3)/2 F (x) dx
0
&∞
as well as an analogous one of the form 0 x(p−3)/2 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
jνsyn = 3p/2 2−(p+7)/2 π −(p+3)/2 Γ( + )Γ( − )(p + 1)−1
4 12
4 12
*
+
(p+1)/2
e2
eB
sin α0
ν −(p−1)/2
synchrotron emissivity
Ke
"0 c
me
(28)
This is clearly a power-law spectrum, with spectral index
α=
1
(p − 1)
2
(29)
If the magnetic field B does not have a fixed direction (e.g. randomly oriented), then we need
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PHYS 4011 – HEA
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to perform a further integration over directions α0 . The required integration is
,sin
(p+1)/2
α0 -
=
=
=
% π
% 2π
1
dφ
sin(p+1)/2 α0 sin α0 dα0
4π 0
0
% π
1
sin(p+3)/2 α0 dα0
2 0
√
5+p
π Γ( 4 )
2 Γ( 7+p
4 )
96
(30)
PHYS 4011 – HEA
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The final and most often used expression for the synchrotron emissivity is the following for the
randomly oriented magnetic field case:
jνsyn = 3p/2 2−(p+13)/2 π −(p+2)/2
p
Γ( 5+p
4 )Γ( 4 +
e2
Ke
"0 c
p
19
1
12 )Γ( 4 − 12 )
Γ( 7+p
4 )
*
+(p+1)/2
eB
−(p−1)/2
me
ν
(31)
synchrotron emissivity (pitch angle averaged)
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9.3 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 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
Π(ω) =
The frequency integrated value is Π
P⊥ (ω) − P% (ω)
G(x)
=
P⊥ (ω) + P% (ω)
F (x)
(32)
# 75 %, which is quite high. For a power-law distribution
of electron energies, the degree of linear polarisation is
Π=
p+1
p + 73
This is ∼
(33)
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.
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Lecture 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.
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10.1 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 angle Θ
w.r.t. its initial propagation direction k̂i . 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
Pγi = (1, k̂i )
c
,
Pγf =
ε1
(1, k̂f )
c
Initial and final 4-momenta of electron:
Pei = (me c, 0)
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,
Pef = (E/c, p)
PHYS 4011 – HEA
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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 + 2Pγi Pei − 2Pγi Pγf − 2Pγf Pei
Note that the modulus of a 4-vector Aµ is defined as
A2 = Aµ Aµ = −(A0 )2 + (A1 )2 + (A2 )2 + (A3 )2 . This implies that the magnitudes of
the 4-momenta for a photon and an electron are Pγ2 = 0 and Pe2 = −m2e c2 , respectively. So
in the above expression, we have |Pef |2 = −m2e c2 = |Pei |2 and |Pγi |2 = 0 = |Pγf |2 ,
which leaves behind only those terms with a factor of 2 in front. These terms are:
εε1
1
Pγi Pei = −εme , Pγi Pγf = − εε
c2 + c2 k̂i · k̂f , and Pγf Pei = −ε1 me . Substituting
these in, rearranging and using k̂i · k̂f = cos Θ gives the following expression for the energy
of the scattered photon:
ε1 =
ε
1+
ε
me c2 (1
− cos Θ)
(3)
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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
# 0.0243 Å
Compton wavelength
(5)
me c
Thus, the wavelength change is of order λC . For long wavelengths (λ $ λC ) or equivalently,
ε % me c2 , the scattering is approximately elastic (i.e. ε1 # ε). This is the Thomson regime.
λC ≡
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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:
1 ε2
dσKN
= r02 12
dΩ
2 ε
#
$
ε1
ε
+
− sin2 Θ
ε1
ε
(6)
= e2 /(4π'0 me c) = 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 = 12 r02 (1 + cos2 Θ). The total cross section is obtained by integrating over
% +1
solid angle, σKN = 2π −1 (dσKN /dΩ) d cos Θ:
&
'
)
(
1
1 + 3x
3 1 + x 2x(1 + x)
σKN = σT
− ln(1 + 2x) +
ln(1 + 2x) −
4
x3
1 + 2x
2x
(1 + 2x)3
where r0
(7)
where x
2
≡ hν/me c .
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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.
For x
!
"
$ 1, the asymptotic solution is σKN ∼ 38 σT x−1 ln 2x + 12
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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
= (1 − β 2 )−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
relative velocity βc defines the Lorentz factor γ
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 = ε"1 γ(1 + β cos Θ"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 γ 2 in the lab frame, implying that Compton scattering by relativistic
electrons can be quite efficient.
γ 2 is only possible for scatterings that are in the Thomson regime in the
rest frame (i.e. ε"1 # ε" ) and which have Θ, Θ"1 >
∼ π/2. The condition for Thomson scattering
A maximum gain of ∼
in the rest frame is
ε" % me c2 =⇒ γhν % me c2
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10.2 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
dE1"
= cσT
dt"
*
ε"1 n" (ε" ) dε"
(11)
Now we know that the emitted power is an invariant. Another invariant is the quantity
n(ε)dε/ε. So
*
*
dE1
ndε
" " "
= cσT ε n dε = cσT ε"2
dt
ε
"
Now we substitute ε = εγ(1 − β cos Θ) from eqn. (8) to get
*
dE1
2
(1 − β cos Θ)2 εn dε
= cσT γ
dt
(12)
(13)
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PHYS 4011 – HEA
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which now only contains quantities in frame K . For an isotropic distribution of photons, we
have
1
)(1 − β cos Θ)2 * = 1 + β 2
3
giving
#
$
1 2
dE1
2
= cσT γ 1 + β Uγ
dt
3
where Uγ
=
%
(14)
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
dErad
dt
= σT cUγ . So
' #
$
(
1 2
dE1
dε
2
=
−
= cσT Uph γ 1 + β − 1
dt
dt
3
(15)
which gives the following for the inverse Compton power for a single electron:
Pic =
This has used γ 2
4
σT cγ 2 β 2 Uγ
3
inverse Compton power
− 1 = γ2β2.
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PHYS 4011 – HEA
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Emitted power for a distribution of electrons
For a nonthermal power law distribution of electrons N (γ)
= Ke γ −p , we can obtain the total
power per unit volume from
Pic,tot =
This gives, for β
Pic,tot =
*
γ2
Pic N (γ)dγ
γ1
# 1,
4
σT cUγ Ke (3 − p)−1 (γ23−p − γ13−p )
3
For a thermal distribution of electrons, γ
nonthermal power-law electrons
(17)
= 1 and )β 2 * = 3kTe /me c2 . 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 = σT cUγ Ne
4kTe
me c2
(18)
thermal electrons
where Ne is the total electron number density.
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10.3 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
=
σT cε1 Ke
16π
*
where the function f is defined by f (x)
n(ε)
dε
ε
*
γ2
dγ γ
−(p+2)
f
γ1
= 2x ln x + x + 1 − 2x2 .
110
#
ε1
4γ 2 ε
$
(19)
PHYS 4011 – HEA
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For sufficiently large limits on the γ integral, we have
jεic1
3
p2 + 4p + 11
−(p−1)/2
= 2p−2
Ke
σT cε1
π
(p + 1)(p + 3)2 (p + 5)
*
ε(p−1)/2 n(ε) dε
(20)
Thus, inverse Compton scattering also predicts a power law spectrum with a spectral index
α=
1
(p − 1)
2
(21)
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
n(ε) =
.−1
, ε 8π 2 +
−
1
exp
ε
(hc)3
kT
(22)
Inserting this into the expression jε1 above, and solving the integrals gives
jεic,bb
=
1
σT
−(p−1)/2
fbb (p)(kT )(p+5)/2 Ke ε1
3
2
h c
where
2p+1 (p2 + 4p + 11)
fbb (p) = 3
γ
(p + 1)(p + 3)2 (p + 5)
where ζ is the Riemann zeta function.
112
#
p+5
2
$ #
$
p+5
ζ
2
(23)
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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
Uγ (ε0 )
n(ε) =
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
jνssc
1
= f (p)σT cKe Uγν0 ln
where the the relation jν1
#
εmax
εmin
$#
ν1
ν0
$−(p−1)/2
(25)
= hjε1 has been used and where
f (p) =
p2 + 4p + 11
3 p−2
2
π
(p + 1)(p + 3)2 (p + 5)
The term ln(εmax /εmin ) is known as the Compton logarithm.
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10.4 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
The mean number of scatterings is determined by the optical depth, τ
the size of the scattering region. A value of τ
(26)
= σNe r , where r is
∼ 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
y<
∼1
y$1
negligible spectral changes
power law spectrum, with exponential cut-off
saturated Comptonisation
114
(28)
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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
y=
me c2
#
$
4kTe
2
1+
max(τT , τT )
me c2
(29)
y parameter for thermal Comptonisation
(30)
where τT
= σT Ne r 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 =
0.7me c2 . The incident seed photons
are injected at the centre with a blackbody spectrum at kT ! kTe . The
Compton y parameter thus ranges
from y " 0.5 for the τT = 0.05
spectrum, to y "
0.05 spectrum.
116
103 for the τT =