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)

36

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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