PHYS 4011 – HEA Lec. 5 Lecture 5: Radiation from Moving Charges 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. 40 PHYS 4011 – HEA Lec. 5 5.1 Overview of the Radiation Field of Single Moving Charges 5.1 Overview of the Radiation Field of Single Moving Charges Consider a radiating charge moving along a trajectory r 0 (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 r 0 (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) - r 0 (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) - r 0 (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 time t. At most, only one source point on the trajectory is in communication with P at time t. 41
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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.
40
PHYS 4011 – HEA Lec. 5
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.
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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
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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PHYS 4011 – HEA Lec. 5
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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PHYS 4011 – HEA Lec. 5
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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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
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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PHYS 4011 – HEA Lec. 5
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.
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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 = 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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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 =∑
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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PHYS 4011 – HEA Lec. 5
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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PHYS 4011 – HEA Lec. 5
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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PHYS 4011 – HEA Lec. 5
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.
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)
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PHYS 4011 – HEA Lec. 8
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).
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PHYS 4011 – HEA Lec. 8
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.
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PHYS 4011 – HEA Lec. 8
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.
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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:
〈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).