RADIATION OF ELECTROMAGNETIC WAVESphysics.usask.ca/~hirose/EP464/ch11-2012-c.pdf · 2012-10-17 ·...
Transcript of RADIATION OF ELECTROMAGNETIC WAVESphysics.usask.ca/~hirose/EP464/ch11-2012-c.pdf · 2012-10-17 ·...
Chapter 11
RADIATION OF
ELECTROMAGNETIC WAVES
11.1 Introduction
We know that a charge q creates the Coulomb field given by
Ec =1
4πε0
q
r2er,
but a stationary charge cannot radiate electromagnetic waves which are necessarily accompanied by
energy flow in the form of the Poynting vector S = E×H whose magnitude is E2/Z or ZH2. For a
stationary charge, the magnetic field is absent. Even a charge drifting with a constant velocity (not
speed) cannot radiate, since the electric field due to a drifting charge is still of Coulombic nature
being proportional to 1/r2. Because of energy conservation, the radiation electric field due to a
localized source (such as point charge) must be proportional to 1/r, so that the radiation power
through a spherical surface with radius r is independent of the radius r,
P =1
Z
∫E2r2dΩ = constant
where dΩ is the differential solid angle. Therefore, the radiation electric field should be inversely
proportional to the distance r and it is fundamentally different from the familiar Coulomb field
(∝ 1/r2).
1
Electromagnetic radiation in free space occurs when charges are under acceleration or deceler-
ation. In antennas, electrons are forced to oscillate back and forth by a generator, and they are
under periodic acceleration and deceleration. In this Chapter, radiation of electromagnetic waves
from an accelerated charge will be discussed first. This is followed by analysis on radiation from a
macroscopic object (such as antennas) in which many charges are collectively involved. In material
media, a charged particle can have a velocity larger than the velocity of electromagnetic waves. In
this case, Cherenkov radiation, which does not require acceleration on charges, occurs.
11.2 Qualitative Picture of Radiation from an Accelerated Charge
Let a charge q be accelerated from rest with an acceleration a (m/sec2) for a short duration ∆τ .
The charge acquires a velocity v = a∆τ after the acceleration, and starts drifting with the velocity.
After t seconds, Before and after the acceleration, the electric field due to the charge is of Coulombic
nature and radially outward from the charge. However, the electric field lines before acceleration are
radially outward from the original stationary position of the charge, while those after acceleration
originate from the position at vt = (a∆t)t from the origin. From the continuity of the electric
flux (Gauss’law), the electric field lines before and after acceleration must be somehow connected.
The only way to make such a connection is to bend the field line at the radial position r ' ct, the
distance travelled by the disturbance in the electric field lines at the speed of light, c. At the kink,
there is indeed an electric field component perpendicular to the distance r, as well as the radial
Coulomb field, Ec. The tangential component, Et, is the desired radiation electric field ER. The
ratio between the two fields isEREC
=vt
c∆τsin θ =
at
csin θ (11.1)
Since the Coulomb field at the kink is given by
EC =q
4πε0
1
(ct)2=
1
4πε0
q
r2(11.2)
we find the following for the radiation electric field
ER =qa
4πε0c2rsin θ (11.3)
The radiation field is maximum in the direction perpendicular to the acceleration a. Indeed, at
θ = 0 and π, there are no kinks in the electric field lines. Vectorially, the radiation electric field ER
2
due to an acceleration a can be written as
ER =q
4πε0c21
rn× (n× a) (11.4)
where
n =r
r
is the unit vector in the radial direction. Note that
n× (n× a) = (n · a) n− (n · n) a = a‖ − a = −a⊥
where a⊥ is the component of acceleration perpendicular to the radius r. The magnitude of a⊥
is a sin θ where θ is the angle between the acceleration and radial vector r. It should be cautioned
that the radiation field given in Eq. (11.3) is valid only if the charge is nonrelativistic, v c. Also,
the acceleration a appearing in Eqs. (11.3) and (11.4) is the acceleration r/c seconds earlier than
the observing time t, because it takes the electromagnetic disturbance r/c seconds to travel over
the distance r. The acceleration at t − (r/c) is called the acceleration at the retarded time and
denoted by
aret
Similarly, other variables, r and n, are, to be precise, those at the retarded time.
If a charge undergoes harmonic oscillation (continuous acceleration and deceleration), the radi-
ation field is also harmonic with the same frequency. (Again, this is valid only in non-relativistic
cases.) The radiation magnetic field associated with the radiation electric field is perpendicular to
both ER and r which is the direction of the Poynting vector or energy flow,
BR =1
cn×ER (11.5)
The magnitude of the Poynting vector is
Sr =1
ZE2R
=√ε0/µ0
q2a2
(4πε0c2)
sin2 θ
r2
=1
4πε0
q2a2
4πc
sin2 θ
r2(W/m2) (11.6)
3
Then, the total radiation power can be readily found,
P = r2∫SrdΩ (11.7)
=1
4πε0
q2a2
4πc3
∫sin2 θdΩ (11.8)
where dΩ = sin θdθdφ is the differential solid angle. Performing integration, we find
P =1
4πε0
q2a2
4πc3
∫ π
0sin3 θdθ
∫ 2π
0dφ
=1
4πε0
2q2a2
3c3(W) (11.9)
This is known as the Larmor’s formula for radiation power emitted by nonrelativistic charge v c
under acceleration.
Example 1 Short Dipole Antenna
In antennas used for broadcasting and communication, a large number of conduction electrons
are collectively accelerated by a harmonic generator. In fact, any unshielded transmission lines
can effectively become an antenna and they radiate electromagnetic waves. For signal and power
transmission purposes, this is an undesirable feature since energy loss inevitably occurs. If, how-
ever, a transmission line is carefully shielded except at its end, the open end becomes an effective
antenna. At an open end, current standing waves are formed. For an antenna much shorter than
the wavelength, the radiation electric field can be found from that due to an accelerated charge
E =q
4πε0c2rn× (n× a) (11.10)
The magnitude of the radiation electric field is
E =qa sin θ
4πε0c2r(11.11)
where θ is the angle between the radial vector n = r/r. Consider a cylindrical conductor of length
l ( λ) and cross-section A. The total charge in the conductor is denoted by q. If the charge move
collectively at a velocity v along l, the current is
I =qv
l
4
or
qv = Il
If the velocity and current are oscillating at frequency ω, the time derivative is
qdv
dt= l
dI
dt
qa = jωIl
Magnitude-wise, we have
qa = Iωl
that is, the quantity qa can be replaced by Iωl for a short antenna. The radiation power can then
be found readily from the Larmor’s formula,
P =1
4πε0
2
3
q2a2
c3
=1
4πε0
2
3
(Iωl)2
c3
=1
6πZ0 (kl)2 I2 (W) (11.12)
The radiation resistance may be defined by
P = RradI2, (11.13)
and in the case of short antenna kl 1 under consideration, it is given by
Rrad =1
6πZ0 (kl)2 , (Ω) (11.14)
11.3 Wave Equations for the Scalar and Vector Potentials Φ and
A
The four Maxwell’s equations
∇ ·E =ρ
ε0or ∇ ·D = ρfree (11.15)
∇×E = −∂B
∂t(11.16)
∇ ·B = 0 (11.17)
5
∇×B =µ0
(J + ε0
∂E
∂t
)(11.18)
are suffi cient to describe electromagnetic fields under any circumstances. In electrostatics, a scalar
potential Φ was introduced
E = −∇Φ (static) (11.19)
and in magnetostatics, the vector potential was introduced
B = ∇×A (11.20)
The static potentials satisfy Poisson’s equations
∇2Φ = − ρ
ε0(11.21)
∇2A = −µ0J (11.22)
whose solutions were
Φ (r) =1
4πε0
∫ρ (r′)
|r− r′|dV′ (static)
A (r) =µ04π
∫J (r′)
|r− r′|dV′ (static)
For time varying sources, the electric field has additional term due to Faraday’s law
E = −∇Φ− ∂A
∂t(11.23)
The curl of the above equation yields
∇×E = −∂B
∂t
since ∇×∇Φ ≡ 0 identically. Substituting E = −∇Φ− ∂A
∂tinto ∇ ·E = ρ/ε0,
∇2Φ +∂
∂t∇ ·A = − ρ
ε0(11.24)
Substituting both B = ∇×A and E = −∇Φ− ∂A
∂tinto
∇×B =µ0
(J + ε0
∂E
∂t
)we obtain
∇ (∇ ·A)−∇2A =µ0J−1
c2∇∂Φ
∂t− 1
c2∂2A
∂t2
6
or
∇2A− 1
c2∂2A
∂t2= −µ0J +∇ (∇ ·A) +
1
c2∇∂Φ
∂t(11.25)
At this stage, let us recall the Helmholtz’s theorem: For a vector to be uniquely defined, both
its divergence and curl must be specified. The curl of A is B, B = ∇×A. which satisfies ∇·B = 0
since ∇ · ∇ ×A = 0 identically. What about the divergence of A? There are no known physical
laws to define ∇·A and we have a freedom to assign any scalar function for ∇·A without affecting
the electromagnetic fields E and B. The choice
∇ ·A+1
c2∂Φ
∂t= 0, (Lorenz gauge) (11.26)
is called Lorenz gauge and the choice
∇ ·A =0, (Coulomb gauge) (11.27)
is called Coulomb gauge. The merit of the Lorenz gauge is that both potentials Φ and A satisfy
similar wave equations
∇2Φ− 1
c2∂2Φ
∂t2= − ρ
ε0(11.28)
∇2A− 1
c2∂2A
∂t2= −µ0J (11.29)
In Coulomb gauge, such separation cannot be achieved. The scalar potential continues to satisfy
the Poisson’s equation even in time varying fields,
∇2Φ (r, t) = −ρ (r, t)
ε0(11.30)
while the vector potential is coupled to the scalar potential
∇2A− 1
c2∂2A
∂t2= −µ0J +
1
c2∇∂Φ
∂t(11.31)
The scalar potential in Coulomb gauge is not subject to retardation and propagation is instanta-
neous. Of course, all electromagnetic fields must be retarded and this puzzling aspect of Coulomb
gauge has been a subject of controversy.
The solutions for the wave equations can be constructed as follows. We know the solution for
static vector Poisson’s equation for the vector potential
∇2A = −µ0J (11.32)
7
is
A (r) =µ04π
∫J (r′)
|r− r′|dV′ (11.33)
For time varying current J (r, t) , the solution for the wave equation
∇2A− 1
c2∂2A
∂t2= −µ0J (11.34)
introduces retardation, that is, the field to be observed at distance |r− r′| from the current is due
to the current at the instant
t− |r− r′|c
(11.35)
namely, |r− r′| /c seconds earlier than the observing time t. Then
A (r,t) =µ04π
∫ J(r′, t− |r−r
′|c
)|r− r′| dV ′
This intuitive solution is in fact correct and agrees with that based on the Green’s function for
wave equation. Likewise, the solution for the scalar potential is
Φ (r, t) =1
4πε0
∫ ρ(r′, t− |r−r
′|c
)|r− r′| dV ′ (11.36)
Example 2 Find the vector potential and radiation magnetic field due to a short antenna carrying
a current I0ejωt in z direction.
Since the current is in z direction, so is the vector potential. For a filamentary short current,
JdV can be replaced by Idz′. Then
Az =µ04π
I0l
rejω
(t−r
c
)=µ04π
I0l
rej(ωt−kr) (11.37)
provided the antenna length l is much shorter than the wavelength λ, kl 1. Note that for r l,
t− |r−r′|
c = t− r
c+
n · z′c
(11.38)
where −l/2 < z′ < l/2. The factor
ωn · z′c
= kz′ cos θ (11.39)
8
is ignorable for short antenna. The magnetic field is
B = ∇×Az
=µ0I0le
jωt
4π∇(e−jkr
r
)× ez
=µ0I0le
jωt
4π
(jk
r+
1
r2
)e−jkr sin θeφ (11.40)
In source free region (far away from the antenna), the electric field can be found from
1
c2∂E
∂t= ∇×B
as
Er =c2
jω
µ0I0lej(ωt−kr)
4π2
(jk
r2+
1
r3
)cos θ
= Z0I0e
j(ωt−kr)
2π
(1
r2− j 1
kr3
)cos θ (11.41)
Eθ = Z0I0le
j(ωt−kr)
4πj
(k
r− j 1
r2− 1
kr3
)sin θ
= Z0I0le
j(ωt−kr)
4π
(jk
r+
1
r2− j
kr3
)sin θ (11.42)
Derivation is left for exercise.
In both the magnetic field and electric field, the dominant radiation terms are those proportional
to 1/r. Higher order terms proportional to 1/r2 and 1/r3.do not contribute to radiation of energy.
However, in radiation of angular momentum, the radial electric and magnetic field proportional to
1/r2 play a crucial role as we will see later.
11.4 Lienard-Wiechert Potentials for Single Charge and Conse-
quent Fields
If the particle velocity is large and approaches the speed of light, the radiation electric field in
Eq. (11.4) will be modified significantly. To see how relativistic effects modifies the radiation field,
we need to find how the scalar and vector potentials are affected by relativistic velocity. As a
preparation, let us first convince ourselves that a rod moving toward (away from) us appears longer
9
(shorter) than its actual length l. (This has nothing to do with the celebrated relativistic length
contraction, which was formulated by Lorentz well after the work by Lienard-Wiechert.) This is
because light emitted from the rear end of the rod takes a longer time than that emitted from the
front end to reach an observer, and for an observer to be able to measure the length of the rod,
he/she needs the two signals arriving at the same instant. Let the rod move toward an observer
with a velocity v. Light emitted by the front end at the instant when the front end is at a distance
x from the observer reaches the observer after x/c sec. Light emitted by the rear end at the same
instant reaches the observer at (x+ l)/c sec, that is l/c sec later. If we denote the apparent length
seen by the observer by l′, the extra time needed for the light leaving the rear end is
l′
c
During this interval, the rod has moved a distance l′ − l with a velocity v. Therefore,
l′
c=l′ − lv
.
Solving for l′, we find
l′ =l
1− v
c
(11.43)
In general, if an object is moving with a velocity v, its dimension toward an observer appears to
change by a factor1
1− n · β (11.44)
where β = v/c, and n is the unit vector toward the observer. If the object has a volume dV , the
apparent volume is
dV ′ =dV
1− n · β (11.45)
For a charge density ρ, the apparent differential charge is therefore
dq′ =ρdV
1− n · β =dq
1− n · β (11.46)
The current density J = ρv is also modified as
JdV ′ =JdV
1− n · β (11.47)
10
vt = a∆τt
θr = ct
EE C
q
Rc∆τ
z
Figure 11.1: The Coulomb field EC which is in radial direction and the radiation field ER due to
an accelerated charge which is transverse to the radius r. The asymmetric shell radially expands
at speed c.
v
l'l
A B
pulse A
pulse B
Figure 11.2: A rod of length l moving toward an observer appears to be longer l′ =l
1− β This
should not be confused with relativity effect.
11
Therefore, the scalar and vector potentials due to a charge moving at velocity v (t) are to be
evaluated according to
Φ =1
4πε0
q
κ |r− rp(t′)|(11.48)
A =µ04π
qv (t′)
κ |r− rp(t′)|(11.49)
where rp(t′) is the instantaneous location of the charge at the retarded time and v (t′) = drp(t
′)/dt′
is the particle velocity at the retarded time t′. The dimensionless quantity κ
κ(t′)≡ 1− n · β (11.50)
also depends on time t′. The potentials given in Eqs. (11.48, 11.49) are called the Lienard-Wiechert
potentials which were formulated in 1898-1900. The electromagnetic fields to be derived from these
potentials properly contain all relativistic effects. It should be noted that all variables in those
potentials are to be evaluated at the retarded time which is related to the observing time t through
t′ = t− |r− rp(t′)|
c(11.51)
where r′(t′) is the instantaneous position of the charge. For example, the gradient operation ∇
with respect to the observing coordinates r is to be performed as follows:
∇ = ∇R +∇t′ ∂∂t′
= ∇R +∇t′ ∂t∂t′
∂
∂t(11.52)
where R = r− rp(t′), R = |R|,n = R/R. Similarly, the time derivative follows the chain rule,
∂
∂t=∂t′
∂t
∂
∂t′(11.53)
where
∂t′
∂t=
∂
∂t
(t− |r− rp(t
′)|c
)= 1 +
n · vc
∂t′
∂t
In this derivation, note that
v(t′) =∂r′
∂t′
12
Then, we find an important relationship between t and t′,
∂t′
∂t=
1
1− n · β (11.54)
Figure 11.3: The change in the unit vector n is caused by the perpendicular velocity v⊥.
The gradient of the retarded time ∇t′ can be calculated similarly,
∇t′ = ∇(t− |r− rp (t′) |
c
)= −1
c∇|r− rp
(t′)|
= −n
c+ (n · β)∇t′
from which it follows
∇t′ = − 1
1− n · βn
c= − n
cκ(11.55)
The transformations in Eqs. (11.54) and (11.55) allow us to evaluate the electric and magnetic
fields due to a moving charge. For example, the electric field is to be found from
E = −∇Φ− ∂A
∂t(11.56)
13
The gradient of the scalar potential becomes
∇Φ =q
4πε0
(∇R +∇t′
∂
∂t′
)1
1− n · β1
R
=q
4πε0
[∇R
(1
1− n · β1
R
)+∇t′ ∂
∂t′
(1
1− n · β1
R
)]
=q
4πε0
β − n(n · β)
(1− n · β)2R2− n
1− n · β1
R2− n
c (1− n · β)
∂n
∂t′· β + n·∂β
∂t′
(1− n · β)2R+
1
1− n · βn · vR2
=q
4πε0
[β − n− n(n · β)
(1− n · β)2R2− 1
(1− n · β)3n
cR
∂n
∂t′· β + n·∂β
∂t′
](11.57)
However, the time variation of the unit vector n = R/R can be caused only by the velocity
component perpendicular to n, as seen in Fig. 11.3. From the two similar triangles in the figure,
we find
dn = −v⊥dt′
R
or∂n
∂t′= −v⊥dt
′
R(11.58)
Therefore,
∇Φ = − q
4πε0
[(n− β) (1− n · β) + n (1− n · β) (n · β)− nβ2⊥
κ3R2+
n
cκ3R
(n · β
)](11.59)
where
β =∂β
∂t′(11.60)
The time derivative of the vector potential is
∂A
∂t=
∂t′
∂t
∂A
∂t′
=µ0q
4π
1
κ
∂
∂t′
( v
κR
)=
q
4πε0
1
cκ
(β
κR− β
(κR)2∂
∂t′(κR)
)
=q
4πε0
1
cκ
[β
κR+
1
κ2Rβ(n · β)− cβ
(κR)2(β2 − n · β)
](11.61)
where µ0 has been eliminated in favour of ε0 through c2 = 1/ε0µ0. Substituting Eqs. (11.59) and
(11.61) into (11.56), we thus find
E (r, t) =q
4πε0
[1
κ3R2(n− β)
(1− β2
)]ret
+q
4πε0
[1
cκ3Rn×
[(n− β)× β
]]ret
(11.62)
14
and the vector identity
A× (B×C) = B (A ·C)−C (A ·B) (11.63)
has been used. Eq. (11.28) is the desired general expression for the electric field due to a moving
charge, and valid for arbitrarily large particle velocity. The electric field above was first formulated
by Heaviside. [· · ·]ret means that all quantities in the brackets are retarded, that is, for evaluation
of the electric field at time t, [· · ·] evaluated at
t′ = t− 1
c
∣∣r− rp(t′)∣∣
should be used.
The reason the formula, Eq. (11.28), remains valid at relativistic velocities even though it has
been derived from the Lienard-Wiechert potentials formulated before the discovery of the relativity
theory (Einstein 1905) is due to the obvious fact that electromagnetic waves propagate at the speed
c irrespective of the source speed once they are emitted. Sound waves in air also propagate with
the sound velocity after being emitted irrespective of source speed. A major difference between
electromagnetic waves in vacuum and sound waves in air is that the speed of electromagnetic waves
remains c even when the observer is moving, while the sound speed appears to change if the observer
is moving relative to the wave medium, that is, the air.
The magnetic field is to be calculated from
B = ∇×A =1
cn×E (11.64)
where E is the electric field given in Eq. (11.28). Derivation of Eq. (11.32) is left for an exercise.
The electric field in Eq. (11.28) has two terms. The first term is inversely proportional to
R2, and does not contain the acceleration β. This is essentially the Coulomb field corrected for
relativistic effects (β). The second term is inversely proportional to R and proportional to the
acceleration β. This term is the desired radiation electric field. Note that at large R, the radiation
field (∝ 1/R) becomes predominant over the Coulomb field (∝ 1/R2).
In nonrelativistic limit, β 1, we recover the radiation electric field worked out in Eq. (11.4)
from qualitative arguments where all quantities, n, R, v are to be evaluated at the retarded time.
We will return to radiation problems associated with relativistic particles in Sec. 11.5.
15
11.5 Radiation from a Charge under Linear Acceleration
If the acceleration is parallel (or anti-parallel) to the velocity, β ‖ β, the particle trajectory remains
linear, as in linear accelerators. (The reason high energy electron accelerators are linear rather
than circular as for proton accelerators is because radiation loss in circular electron accelerators
becomes intolerably large.) The radiation electric field in this case is given by
E (r, t) =1
4πε0
q
cr
n×(n× β‖
)κ3
t′
(11.65)
where all quantities at the retarded time should be used. If the angle between β and n is θ, the
Poynting flux is
S (r, t) =1
Z0|E (r, t)|2
=1
4πε0r2q2β
2‖ sin2 θ
4πc (1− β cos θ)6
∣∣∣∣∣∣t′
(11.66)
If the Poynting flux is integrated over the spherical surface of radius r, one gets a radiation power
at the observing time t. However, what is more meaningful is the radiation power at the retarded
time. Since
dt′ =dt
1− β cos θ
the Poynting flux at the retarded time t′ is
S(r, t′)
= (1− β cos θ)S (r, t)
=1
4πε0r2q2β
2‖ sin2 θ
4πc (1− β cos θ)5. (11.67)
Integration over the solid angle yields
P(t′)
=1
4πε0
q2β2‖
4πc2π
∫ π
0
sin2 θ
(1− β cos θ)5sin θdθ
=1
4πε0
2
3
q2β2‖
cγ6 (11.68)
where
γ =1√
1− β2(11.69)
16
r
θv, dv/dtq
E
Figure 11.4: Linear acceleration β ‖ β with nonrelativistic velcoity β 1. In highly relativistic
case, γ 1, the radiation is emitted predominantly along the beam velocity. (Fig. 17)
is the relativity factor. Relevant integral is∫ 1
−1
1− x2
(1− βx)5dx =
4
3
1(1− β2
)3 (11.70)
The radiation power is independent of the particle energy γmc2 since the parallel acceleration is
inversely proportional to γ3 as can be seen from the equation of motion
mcd
dt
β‖√1− β2‖
= F
mc
β‖√1− β2‖
+β2‖β‖(
1− β2‖)3/2
= F
mcγ3β‖ = F
Therefore, the radiation power in Eq. (11.68) is independent of the particle energy. This is the
main advantage of linear electron accelerators.
The angular distribution of the radiation power is proportional to the function
f (θ) =sin2 θ
(1− β cos θ)5(11.71)
In nonrelativistic limit β 1, the radiation intensity peaks in the direction θ = π/2 (perpendicular
to the velocity β and acceleration β. In relativistic case β → 1, γ 1, the radiation intensity profile
becomes narrow with an angular spread about the velocity of order ∆θ ' 1/γ 1. The angle at
which the radiation intensity peaks can be found from
df (θ)
dθ= 0
17
which yields
3β cos2 θ + 2 cos θ − 5β = 0
cos θ =1
3β
(√15β2 + 1− 1
)In the limit β ' 1 (γ 1) ,
cos θ =1
3β
(√15β2 + 1− 1
)=
1
3
(√15
(1− 1
γ2
)+ 1− 1
)' 1− 5
8γ2
Since θ 1, cos θ ' 1− θ2/2, we find
θ '√
5
4
1
γ 1
The radiation is essentially confined in the angle ∆θ ' 1/γ about the velocity vector. This align-
ment with the velocity is in fact independent of the acceleration direction and ∆θ ' 1/γ 1 about
the velocity holds even for acceleration perpendicular to the velocity.
sin2 θ
(1− 0.1 cos θ)5
0.5
1.0
0.5
0.5
1.0
x
y
sin2 θ
(1− 0.9 cos θ)5
18
500 1000 1500
400
200
0
200
400
x
y
β = 0.9.
sin2 θ
(1− 0.99 cos θ)5
5e+6 1e+7 1.5e+71e+6
1e+6
x
y
Figure 11.5: Radiation pattern when, from top, β = 0.1, β = 0.90 and β = 0.99. Note the large
radiation intensity as β appoaches unity. .
11.6 Radiation due to Acceleration Perpendicular to the Velocity
β ⊥ β
In this case, the radiation electric field is
E (r, t) =1
4πε0
e
cκ3rn×
[(n− β)× β⊥
](11.72)
19
We consider a charge undergoing circular motion with radius ρ and normalized velocity β in the
x− z plane. At t = 0, the charge passes the origin. At this instant,
β = βez
and
β⊥ = β⊥ex
Substituting
ex = ∇x = ∇ (r sin θ cosφ)
= sin θ cosφn + cos θ cosφeθ − sinφeφ (11.73)
and
ez = ∇ (r cos θ)
= cos θn− sin θeθ (11.74)
into
n×[(n−βez)× β⊥ex
]we obtain
n×[(n−βez)× β⊥ex
]=∣∣∣β⊥∣∣∣ [(β − cos θ) cosφeθ + (1− β cos θ) sinφeφ] (11.75)
and
E (r, t) =1
4πε0
e∣∣∣β⊥∣∣∣cκ3r
[(β − cos θ) cosφeθ + (1− β cos θ) sinφeφ] (11.76)
The retarded differential power is
dP (t′)
dΩ= r2
1
Z0|E|2
=1
4πε0
e2∣∣∣β⊥∣∣∣24πc
1
(1− β cos θ)5
[(1− β cos θ)2 − 1
γ2sin2 θ cos2 φ
](11.77)
and the radiation power is
P(t′)
=
∫dP (t′)
dΩdΩ
=1
4πε0
e2∣∣∣β⊥∣∣∣24πc
γ4 (11.78)
20
Relevant integrals are ∫ 1
−1
dx
(1− βx)3=
2
(1− β2)2= 2γ4 (11.79)
∫ 1
−1
1− x2(1− βx)5
dx =4
3
1
(1− β2)3=
4
3γ6 (11.80)
Example: Radiation power emitted by 3 GeV electron undergoing circular motion with radius
R = 10 m.
The relativity factor is
γ =E
mc2=
3 GeV0.512 MeV
= 5900
The acceleration is
a⊥ =v2
R' c2
R= 9× 1015 m/s2
Then the radiation power is
P =1
4πε0
e2∣∣∣β⊥∣∣∣24πc
γ4
=1
4πε0
e2 |a⊥|2
4πc3γ4
= 9× 109(1.6× 10−19
)2 (9× 1015
)24π (3× 108)3
(5900)4
= 6.66× 10−8 W = 4.17× 1011 eV/s/electron
Electron rapidly loses its energy to radiation (synchrotron radiation). To reduce radiation power,
the orbit radius R must be increased and γ must be decreased. Circular particle accelerators are
therefore practical only for protons. (For proton energy of 100 GeV, the relativity factor is rather
mild, γ = 106.)
11.7 Synchrotron Radiation
Charged particles emit radiation whenever they are subjected to acceleration. Synchrotron ra-
diation is emitted by relativistic electrons. The classical radiation mechanism is simply bending
electron trajectory by a magnetic field. The acceleration due to trajectory bending is
a⊥ 'c2
R
21
where R is the curvature radius of the trajectory. The radiation power due to perpendicular
acceleration is
P =1
4πε0
2
3
(ea⊥)2
c3γ4
with peak frequency components around
ω ' γ3 eBγme
= γ2eB
me
where
ωce =eB
γme
is the relativistic cyclotron frequency. If B = 0.5 T and γ = 5000, the dominant radiation frequency
is 3. 5× 1017 Hz.
Modern synchrotron light sources are equipped with wigglers and undulators to cover wider
radiation spectrum and provide higher radiation intensities. In wigglers, pairs of NS magnets are
placed periodically and electron beam going through such structure experiences periodic kicks in
the direction perpendicular to the beam. The wavelength of emitted radiation is approximately
given by
λ =λw2γ2
where λw is the spatial period of the wiggler. In contrast to the radiation by bending magnet, wiggler
radiation is a result of maser or laser action, namely, amplification of electromagnetic waves in an
electron beam. Wiggler radiation is thus more coherent than bending magnet radiation.
11.8 Radiation by Macroscopic Sources (Antennas, Apertures)
Radiation by antennas can be analyzed by solving the wave equation for the vector potential A(∇2 − 1
c2∂2
∂t2
)A = −µ0J (11.81)
provided the Lorenz gauge is adopted,
∇ ·A +1
c2∂Φ
∂t= 0 (11.82)
The scalar potential Φ obeys a similar wave equation(∇2 − 1
c2∂2
∂t2
)Φ = − 1
ε0ρ (11.83)
22
Figure 11.6: In a wiggler, an electron beam is modulated by a periodic magnetic field. Electrons
acquire spatially oscillating perpendicular displacement x(z) and velocity vx(z) which together with
the radiation magnetic field BRy produces a ponderomotive force vx(z)×BRy(z) directed in the z
direction. The force acts to cause electron bunching required for coherent radiation (as in lasers).
Since the charge density ρ and the current density J are related through the charge conservation
law,∂ρ
∂t+∇ · J = 0 (11.84)
for a given current J, the charge density can be found in principle and the scalar potential found
from the wave equation should be consistent with that calculated through the Lorenz condition in
terms of the vector potential A. In practice, finding the vector potential alone should be suffi cient,
since all electric field and magnetic fields can be derived from the vector potential as follows. The
magnetic field is
B = ∇×A (11.85)
In source free region (ρ = 0, J = 0) ,
∇×B =1
c2∂E
∂t(11.86)
yields the electric field
E =c2
jω∇×B =
c2
jω∇×∇×A
= −j c2
ω[∇(∇ ·A)−∇2A] (11.87)
Since in the Lorenz gauge,
∇ ·A +1
c2∂Φ
∂t= 0
23
and in source free region, the vector potential satisfies the wave equation(∇2 − 1
c2∂2
∂t2
)A = 0 (11.88)
the electric field above reduces to
E = −∇Φ− ∂A
∂t
11.8.1 Short Dipole Antenna (revisit)
In Section 11.2, we applied the Larmor’s formula to calculate the radiation power emitted by a short
dipole antenna kl =2πl
λ 1. In this case, electrons oscillate up and down collectively without
any phase difference along the antenna. Here we directly solve the wave equation for the vector
potential to see if the same result can be recovered.
We assume an antenna of length l carries a current I0ejωt in z direction. The vector potential
has only z component since the current density is unidirectional in z direction. The wave equation
for Az (∇2 − 1
c2∂2
∂t2
)Az = −µ0Jzejωt (11.89)
can be integrated as
Az (r, t) =µ04πr
∫Jz(r′)ejω(t−
|r−r′|c )dV ′ (11.90)
where retardation due to finite propagation velocity c is taken into account. The observation
position r is much larger than the antenna length and we have
Az (r, t) =µ04πr
ej(wt−kr)I0l (11.91)
where for a filamentary current, JdV has been replaced with Idl. The radiation magnetic field is
B = ∇×A ' −jk×A
= −j µ0I0l4πr
ej(wt−kr)ker × (− sin θeθ + cos θer)
= jµ0I0l
4πrej(wt−kr)k sin θeφ (11.92)
The Poynting vector is
Sr = Z0 |H|2 = Z0(I0l)
2
16π2r2k2 sin2 θ (11.93)
24
I cos(kz')0
I(z')dz'r
r z'cosθ
z'
feeder
dAz
θ
Figure 11.7: Center-fed half wavelength dipole antenna.
and the radiation power is
P = r2∫SrdΩ
= Z0(I0l)
2
16π2k2∫ π
0sin3 θdθ
∫ 2π
0dφ
=
√µ0/ε06π
(kl)2 I20 (W) (11.94)
in agreement with the earlier result based on equivalent acceleration. Remember this is subject to
the condition of short antenna, kl 1.
In practice, a stand alone current segment cannot exist physically. What is happening is that the
charge oscillates along the antenna and at the top and bottom, opposite charges (dipole) appear
to satisfy charge conservation law. Charges create scalar potential Φ. However, as long as the
radiation power is concerned, the scalar potential does not contribute and can be ignored. In
analyzing fields near the antenna, scalar potential does play roles as we will see in the case of half
wavelength dipole antenna.
11.8.2 Half Wavelength (λ/2) Dipole Antenna
Figure 11.7 shows the case of center-fed half wavelength long (l = λ/2) antenna. Since the antenna
length is comparable with the wavelength, the radiation field (vector potential) should be calculated
taking into account the phase difference of the antenna current. To find the radiation field, we
25
assume the following standing wave,
I (z, t) = I0 cos (kz) ejωt, − λ
4< z <
λ
4. (11.95)
The retarded radiation vector potential at kr 1 is
Az (r, θ) =µ04π
I0ej(ωt−kr)
r
∫ λ/4
−λ/4ejkz
′ cos θ cos kz′dz′
=µ04π
I0ej(ωt−kr)
r2
∫ λ/4
0cos(kz′ cos θ
)cos(kz′)dz′
=µ0I02πr
cos(π
2cos θ
)k sin2 θ
ej(ωt−kr) (11.96)
and the radiation magnetic field is
H ' − 1
µ0jk×Az
which yields
Hφ = jI0
2πkr
cos(π
2cos θ
)sin θ
ej(ωt−kr) (11.97)
The radiation Poynting flux is
Sr = Z0 |H|2
= Z0I20
(2πr)2
cos2(π
2cos θ
)sin2 θ
, W/m2 (11.98)
and the radiation power is
P = Z0I20
(2π)2
∫ π
0
cos2(π
2cos θ
)sin θ
dθ
∫ 2π
0dφ
= Z0I202π
∫ π
0
cos2(π
2cos θ
)sin θ
dθ (11.99)
The integral is approximately 1.22 and we find the radiation resistance of the half wavelength dipole
antenna,
Rrad =Z02π× 1.22 = 73.2 Ω (11.100)
If the Poynting flux on the antenna surface is directly integrated, the reactive power can be
estimated as well. When the antenna length is λ/2, the reactance is about j40 Ω (inductive).
However, it sensitively depends on the length. For l = 0.49 λ, the reactance vanishes.
26
Figure 11.8: The Poynting flux on the surface of λ/2 antenna of finite radius a is approximated by
that on the cylindrical surface of radius a surrounding a thin antenna.
The calculation presented above entirely ignores the reactive power which may exist due to stor-
age of electric and/or magnetic energy in the vicinity of the antenna. To account for such reactive
power, we must deviate from the far-field analysis and integrate the Poynting flux directly on a
surface close to the antenna surface. Instead of calculating the Poynting flux on the antenna surface
of finite radius a, we calculate the Poynting flux on a cylindrical surface of radius a surrounding
an ideally thin antenna of length λ/2 as shown in Fig. 11.8. In this approximation, the magnetic
field on the antenna surface may be replaced by the static form without retardation,
Hφ(z) =I(z)
2πa=
I02πa
cos(kz), −λ4< z <
λ
4. (11.101)
The electric field on the antenna surface is zero within our assumption of ideally conducting antenna
except at the gap at the midpoint. However, the electric field due to the current filament assumed
at the axis is finite at the surface a distance a away. It can be calculated from
1
c2∂
∂tE = ∇×B = ∇(∇ ·A)−∇2A, (11.102)
where A is the vector potential on the surface. It can be found from the integral,
Az(z) =µ04π
∫ λ/4
−λ/4
I0 cos kz′
R(z, z′)e−jkR(z,z
′)dz′, (11.103)
27
where
R(z, z′) =√
(z − z′)2 + a2, (11.104)
is the distance between a point on the surface (ρ = a, z) and a current segment I(z′)dz′ at z′.With
this approximation, Eq. (11.102) reduces to
jω
c2Ez(z) =
∂2Az∂z2
+ k2Az, (11.105)
since in current-free region the vector potential satisfies the Helmholtz equation
(∇2 + k2
)Az = 0. (11.106)
The integral in
Ez(z) = −j c2µ0I04πω
∫ λ/4
−λ/4cos(kz′)
(∂2
∂z2+ k2
)e−jkR(z,z
′)
R(z, z′)dz′, (11.107)
can be performed by noting∂
∂z
e−jkR(z,z′)
R(z, z′)= − ∂
∂z′e−jkR(z,z
′)
R(z, z′), (11.108)
and by integrating by parts twice with the result
Ez(z) = −j Z0I04π
(e−jkR1
R1+e−jkR2
R2
), (11.109)
where
R1 =
√(z − λ
4
)2+ a2, R2 =
√(z +
λ
4
)2+ a2. (11.110)
The radial outward Poynting flux on the antenna surface is therefore given by
Sρ = −EzH∗φ
= jZ0I
20
8π2a
(e−jkR1
R1+e−jkR2
R2
)cos(kz), (11.111)
and the power leaving through the antenna surface is
P = 2πa
∫ λ/4
−λ/4Sρdz
= jZ0I
20
4π
∫ λ/4
−λ/4
(e−jkR1
R1+e−jkR2
R2
)cos(kz)dz. (11.112)
The integral has to be performed numerically. Introducing
x =4z
λ, A =
4a
λ,
28
we rewrite the integral in the form
f(A) = Re
∫ 1
−1
exp(−j π
2
√(1− x)2 +A2
)√
(1− x)2 +A2+
exp(−j π
2
√(1 + x)2 +A2
)√
(1 + x)2 +A2
cos(π
2x)dx
(11.113)
which is shown in Fig. 11.9 as a function of the normalized antenna radius A = 4a/λ. For A < 0.01,
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
3
2
1
0
1
2
x
y
Figure 11.9: Real part (red) and imaginary part (black) of the integral in Eq. 11.113 as functions
of A = 4a/λ.
the radiation impedance is constant and approximately equal to
Zrad ' 73.1 + j42.0 (Ω)
which is inductive. The real part agrees with the radiation resistance calculated earlier. The
reactive part of the impedance is inductive due to dominant magnetic energy compared with the
electric capacitive energy. However, the reactance is a very sensitive function of the antenna length.
It vanishes if the antenna length is chosen at l ' 0.49λ and further decrease in the length makes the
reactance capacitive. Radiation from a center-fed antenna can be analyzed by assuming a standing
wave form,
I(z) = I0 sin[k(l − |z|)],
and is left for exercise.
29
The axial electric field Ez(z) in Eq. (11.109) can be alternatively (perhaps more conveniently)
found from
Ez = − ∂
∂zΦ− ∂
∂tAz, (11.114)
where Φ is the retarded scalar potential given by
Φ (z) =1
4πε0
∫e−j|r−r
′|
|r− r′| ρ(r′)dV ′
=1
4πε0
∫e−jR
Rρl(z′)dz′, (11.115)
with
R =
√(z − z′)2 + a2,
and ρl the linear charge density that can be found as
∂ρl∂t
+∂I (z)
∂z= 0,
ρl (z) = −jcI0 sin kz, (C m−1). (11.116)
Then,
Ez = − ∂
∂zΦ− ∂
∂tAz
= − jI04πε0c
∫ λ/4
−λ/4
∂
∂z′
(e−jkR
R
)sin(kz′)dz′ − jωµ0I0
4π
∫ λ/4
−λ/4
cos kz′
Re−jkRdz′
= − jI04πε0c
(e−jkR1
R1+e−jkR2
R2
)= −j Z0I0
4π
(e−jkR1
R1+e−jkR2
R2
), (11.117)
which agrees with Eq. (11.109). In radiation zone, the scalar potential is irrelevant but in near
field region, it should be considered together with the vector potential in a self consistent manner.
The radiation impedance Zrad can be defined by
Zrad = jZ04π
∫ λ/4
−λ/4
(e−jkR1
R1+e−jkR2
R2
)cos(kz)dz (11.118)
Remember that this is for a center-fed λ/2 antenna. For an antenna of arbitrary length 2l with a
current standing wave
I (z) = Im sin [k (l − |z|)] (11.119)
30
the impedance is modified as
Zrad = jZ04π
(ImI (0)
)2 ∫ l
−l
(e−jkR1
R1+e−jkR2
R2− 2 cos (kl)
e−jkR
R
)sin [k (l − |z|)] dz
= jZ04π
1
sin2 (kl)
∫ l
−l
(e−jkR1
R1+e−jkR2
R2− 2 cos (kl)
e−jkR
R
)sin [k (l − |z|)] dz (11.120)
where
R1,2 =
√(z ∓ l)2 + a2, R =
√z2 + a2 (11.121)
Note that the current seen by the generator is I (0) = Im sin (kl) .
11.9 Radiation by Small Sources (Multipole Radiation)
The retarded vector potential due to a nonrelativistic (β 1, κ ' 1) charged particle
A (r, t) =µ04π
ev(t′)
|r− rp(t′)|
∣∣∣∣ret
(11.122)
can be generalized for a collection of moving charges as
A (r, t) =µ04π
∫ J
(r′, t− |r− r′|
c
)|r− r′| dV ′ (11.123)
where J (r, t) is the current density. If the current is oscillating at ω, we have
A (r, t) =µ04πr
ej(ωt−kr)∫
J(r′)ejk·r
′dV ′ (11.124)
where the following approximation is used,
ejω
(t−|r−r
′|c
)' ej(ωt−kr)+jk·r′
Note that
k|r− r′| ' kr − k · r′, r r′
If the radiation source is small compared with the wavelength λ, that is, if ωr′/c 1, or kr′ 1
Eq. (11.124) may be approximated by
A (r, t) =µ04πr
ej(ωt−kr)∫
J(r′)(
1 + jk · r′ − 1
2
(k · r′
)2)dV ′ (11.125)
31
Note that in the limit of dc (or very low frequency) current (ω → 0, k → 0), Eq. (11.125) does
reduce to the vector potential in magnetostatics. By assumption, |k · r′| 1. As we will see, each
term in this series expansion can be identified as electric and magnetic multipole radiation fields.
The lowest order radiation vector potential is
A (r, t) =µ04πr
ej(ωt−kr)∫
J(r′)dV ′ (11.126)
The integral of the current density can be calculated as follows. The x component of the integral is∫JxdV =
∫∇x · JdV
=
∫∇ · (xJ) dV −
∫x∇ · JdV (11.127)
The first integral vanishes, ∫∇ · (xJ) dV =
∮S
xJ · dS =0
because the closed surface S is at infinity where all sources vanish. Then∫JxdV = −
∫x∇ · JdV =
∫x∂ρ
∂tdV =
d
dtpx
where
px =
∫xρdV
is the x component of the electric dipole moment. In general,∫JdV =
d
dtp (11.128)
The term of next order ∫J (r) (k · r)dV (11.129)
can be calculated in a similar manner.∫J (r) (k · r)dV =
∫r (k · J) dV − k×
∫r× JdV
and ∫r (k · J) dV = k ·
∫rr∂ρ
∂tdV −
∫(k · r) JdV
= k · ddt
←→Q −
∫(k · r) JdV
32
we find ∫J (r) (k · r)dV = −k×m +
1
2k · d
dt
←→Q (11.130)
where←→Q =
∫rrρdV (11.131)
is the quadrupole moment and
m =1
2
∫r× JdV (11.132)
is the magnetic dipole moment. Therefore, to order kr′ 1, the vector potential is given by
A (r,t) =µ04πr
ej(ωt−kr)(
p− jk×m + j1
2k · d
dt
←→Q
)(11.133)
and the radiation magnetic field H is
H =1
µ0∇×A ' −j 1
µ0k×A
=1
4πcrej(ωt−kr)
(p× n+
1
cn× (n× m)− 1
2cn×
(n ·←→...Q
))(11.134)
where
n =r
r
is the radial unit vector.
Example 3 Radiation by a Nonrelativistic Charge undergoing Circular Motion: Electric dipole
radiation
We assume a charge q is undergoing circular motion with a radius a and angular frequency ω,
as shown in Fig. . The dipole moment is
p (t) = qa (cos (ωt) ex + sin (ωt) ey)
Then the radiation magnetic field is
H (r, t) =1
4πcrej(ωt−kr)p× n
=ω2
4πcrej(ωt−kr)n× p
33
qa
ωt
x
y
z
Figure 11.10: Charge in circular motion.
and the radiation power is given by
P = Z0r2
∫|H|2 dΩ
=1
4πε0
2
3c3ω4 |p|2
=1
4πε0
2
3c3|p|2 (11.135)
The radiation magnetic field
H (r, t) =ω2
4πcrej(ωt−kr)n× p
may be written as
H (r, t) = jaqω2
4πcrej(ωt−kr)
1√2
(eθ − j cos θeφ) (11.136)
The field is plane polarized at θ = π/2 and circularly polarized at θ = 0. The radiation magnetic
filed due to an electric dipole is proportional to p, that is, acceleration of the charge.
Example 4 Radiation by a Small Current Loop: Magnetic dipole radiatio
If the loop radius is a (ka 1) , the magnetic dipole moment is
mz = πa2I0ejωt
The radiation magnetic field is
H =1
4πc2rej(ωt−kr)n× (n× m)
= − ω2
4πc2rej(ωt−kr)n× (n×m)
34
The radiation power is
P = r2Z0
∫|H|2 dΩ
=Z0k
4
(4π)2(πa2I0
)2 ∫sin2 θdΩ
= Z0I20
(ka)4
6π
The radiation resistance is
Rrad = Z0(ka)4
6π
This is smaller than the radiation resistance of electric dipole antenna of length l ( λ)
Rrad = Z0(kl)2
6π
if l and a are comparable.
11.10 Radiation of Angular Momentum
The Poynting vector
S = E×H (11.137)
is the energy flux density. Since the electromagnetic energy is carried at the velocity c, the mo-
mentum flux density may be defined by1
cE×H (11.138)
and the momentum density by1
c2E×H (11.139)
Similarly, the angular momentum flux density is given by
1
cr× (E×H) (11.140)
and the angular momentum density by
1
c2r× (E×H) = ε0r× (E×B)
= ε0r× (E× (∇×A)) (11.141)
35
The vector E× (∇×A) can be expanded as
E×∇×A =Ei∇Ai − (E · ∇)A (11.142)
and we have
r× (E×B) = r× (Ei∇Ai)− r× [(E · ∇)A] (11.143)
However,
r× [(E · ∇)A] = ∇i (Eir×A)−E×A− (∇·E) (r×A)
= ∇i (Eir×A)−E×A
since ∇·E =0 in source free region. Then the total angular momentum is
ε0
∫r× (E×B)dV = ε0
∫r× (Ei∇Ai) dV + ε0
∫(E×A)dV (11.144)
where use is made of ∫∇i (Eir×A) dV =
∮r×A(E·dS) = 0
It is evident that in Eq. (11.144) the first term containing the factor r× can be identified as the
orbital angular momentum. Then the last term can be interpreted as the spin angular momentum.
Consider a circularly polarized plane wave propagating in the z direction. If the field has positive
helicity, the electric fields components are
Ex (z, t) = E0 cos (ωt− kz)
Ey (z, t) = E0 sin (ωt− kz)
Corresponding vector potentials are
Ax (z, t) = −E0ω
sin (ωt− kz)
Ay (z, t) = +E0ω
cos (ωt− kz)
The spin momentum density is
ε0E×A =1
ωε0E
20ez (11.145)
For negative helicity wave,
Ex (z, t) = E0 cos (ωt− kz)
Ey (z, t) = −E0 sin (ωt− kz)
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Ax (z, t) = −E0ω
sin (ωt− kz)
Ay (z, t) = −E0ω
cos (ωt− kz)
the spin direction is reversed,
ε0E×A = − 1
ωε0E
20ez (11.146)
as expected.
Example 5 Radiation of angular momentum by an electric dipole
Electric multipoles radiate Transverse Magnetic (TM) modes having no radial component of
magnetic field, Hr = 0. Then the angular momentum flux density becomes
1
cr× (E×H) = −1
c(r ·E) H
The radiation vector potential of an electric dipole is
A =µ04πr
ej(ωt−kr)p (11.147)
Corresponding magnetic field is
H =1
µ0∇×A ' −j e
j(ωt−kr)
4πrk× p
= −ej(ωt−kr)
4πcrn× p (11.148)
where
jk = njω
c=
n
c
∂
∂t
The electric field can be found from the Maxwell’s equation
ε0µ0∂E
∂t= ∇×B = ∇× (∇×A)
= ∇ (∇ ·A)−∇2A
= ∇ (∇ ·A) + k2A
∇ ·A is
∇ ·A =µ04π
d
dr
(ej(ωt−kr)
r
)n · p
=µ04π
(−j k
r− 1
r2
)ej(ωt−kr)n · p
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Then the radial component of the electric field is
Er =ej(ωt−kr)
2πε0cr2n · p (11.149)
The rate of angular momentum radiation is
−r2
c
∫(r ·E) H∗dΩ
=1
8π2ε0c3
∫(n · p) (n× p) dΩ (11.150)
When applied to a charge undergoing circular motion,
px = eρ cos (ωt) , py = eρ sin (ωt)
we find
n · p =eρω sin θ sin (φ− ωt)
and
ez · (n× p) =eρω2 sin θ sin (φ− ωt)
Then
−dLdt
=1
8π2ε0c3
∫(n · p) (n× p) dΩ
=e2ρ2ω3
8π2ε0c3
∫sin2 θ sin2 (φ− ωt) dΩ
=e2ρ2ω3
8π2ε0c34
3× π
=e2ρ2ω3
6πε0c3ez =
P
ωez (11.151)
where
P =e2ρ2ω4
6πε0c3=e2(ρω2)2
6πε0c3
is the radiation power with a = ρω2 the acceleration.
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