Download Radiating systems in free space

Chapter 3
Radiating systems in free
space
Electromagnetic waves are always generated by temporal changes of charge
and current distributions. This chapter deals with the primary fields of such
sources, i.e. there are no boundaries between different materials. Chapter
4 introduces scattering of primary waves from material bodies. Chapter
5 deals basically with scattering too, but at a quasi-static limit. A useful reference for Chapters 3-4 is Jackson. As mathematical tools, we need
some experience with spherical harmonics and spherical Bessel and Neumann functions. Again, mathematical methods are common to several other
branches of physics, for example quantum mechanics.
3.1
Multipole expansion of the vector potential
Assume that in an otherwise empty space there are charge and current
distributions ρ(r)e−iωt and J(r)e−iωt . A general time dependence follows
from the inverse Fourier transform. The vector potential in the Lorenz
gauge is
Z
J(r0 , ω) ik|r−r0 | 3 0
µ0
e
d r
(3.1)
A(r, ω) =
4π
|r − r0 |
where k = ω/c. The magnetic field is B = ∇ × A and outside of the
source region the electric field is E = ic 2 ∇ × B/ω. So it is not necessary to
determine the scalar potential. Note that it would be obtained readily from
the Lorenz gauge condition ∇ · A − iωµ 0 0 ϕ = 0.
Next we study sources whose length scales (d) are much smaller than
the wavelength λ = 2π/k = 2πc/ω = c/f . Then the space can be divided
17
18
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
into three different regions:
near (static) zone:
intermediate (induction) zone:
far (radiation) zone:
drλ
dr∼λ
dλr
For the near zone the exponential in the integral of the vector potential
can be replaced by unity. Consequently, apart from the harmonic time
dependence, the spatial behaviour of the vector potential is identical to the
static case, and can be expanded into a series of spherical harmonics:
A(r) =
X
Clm
l,m
Ylm (θ, φ)
r l+1
(3.2)
In the far zone kr 1, so the exponential term in the integral oscillates
rapidly. The distance |r − r0 | can be replaced by |r − r0 | ≈ r − n · r0 , where
n = r/r. The vector potential is now
A(r) =
µ0 eikr
4πr
Z
0
J(r0 , ω)e−ikn·r d3 r0
(3.3)
If the source dimensions are small compared to the wavelength then it is
reasonable to expand this expression into a power series with respect to k:
A(r) =
Z
∞
(−ik)m
µ0 eikr X
J(r0 )(n · r0 )m d3 r0
4πr m=0 m!
(3.4)
So the leading behaviour is A ∼ eikr /r. This is a spherical wave with an
angular dependent coefficient. It is an exercise to show that the fields are
transverse to r and fall off as 1/r, corresponding to radiation fields.
The intermediate region is difficult, since the approximations made above
are not possible. Leaving mathematical details as an exercise, we give the
result
∞
l
X
X
eik|r−r |
(1)
∗
=
ik
j
(kr
)h
(kr
)
Ylm
(θ 0 , φ0 )Ylm (θ, φ)
<
>
l
l
4π|r − r0 |
l=0
m=−l
0
(3.5)
(1)
where jl and hl are spherical Bessel and Hankel functions, respectively,
and r< = min(r, r 0 ), r> = max(r, r 0 ). Then the expansion of the vector
potential, valid for all r outside the source, is
A(r) = iµ0 k
X
l,m
(1)
hl (kr)Ylm (θ, φ)
Z
∗
J(r0 )jl (kr 0 )Ylm
(θ 0 , φ0 )d3 r0
(3.6)
The mathematical usefulness of this expansion is the explicit separation of
the observation and the source points in spherical coordinates.
3.1. MULTIPOLE EXPANSION OF THE VECTOR POTENTIAL
3.1.1
19
Electric dipole field
Consider the 0th term of Eq. 3.4:
µ0 eikr
4πr
A(r) =
Z
J(r0 )d3 r0
(3.7)
Remembering the continuity equation ∇ · J = iωρ and taking into account
that J is non-zero only in a finite volume, we obtain by integration by parts
A(r) = −
iωµ0 eikr
4πr
Z
rρ(r)d3 r
(3.8)
Using
the definition of the electric dipole moment from electrostatics (p =
R
rρ(r)d3 r), this can be written as
A(r) = −
iωµ0 eikr
p
4πr
(3.9)
An inspection of Eq. 3.6 shows that this is the exact form of the first term
everywhere outside the source, not only in the far region.
An exercise is to calculate the fields
B(r) = k 2 (n × p)
µ0 ceikr
1
(1 −
)
4πr
ikr
(3.10)
E(r) = k 2 (n × p) × n
eikr
4π0 r
+ (3n(n · p) − p)(
eikr
ik
1
− 2)
3
r
r 4π0
where n = r/r. At the static limit, the magnetic field vanishes and the electric field takes its familiar static dipole form. In the far zone, the radiation
fields are
B(r) = k 2 (n × p)
µ0 ceikr
4πr
E(r) = cB × n
(3.11)
The fields are transverse to the radius vector from the source to the observation point, and they are also transverse to each other, and |E| = c|B|.
The radiated power is calculated using the Poynting vector. The meaningful quantity is the time-average, which for the harmonic time-dependence
leads to the power per solid angle
µ0 c3 k 4
dPrad
1
Re(r 2 n · E × B∗ ) =
|n × p|2
=
dΩ
2µ0
32π 2
The total power is
Prad =
µ0 c3 k 4 2
|p|
12π
(3.12)
(3.13)
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3.1.2
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
Magnetic dipole field
The next term of the multipole expansion is
A(r) =
µ0 eikr 1
( − ik)
4πr r
Z
J(r0 ) n · r0 d3 r0
(3.14)
which is valid everywhere outside the source region. It is useful to separate
the integrand into symmetric and antisymmetric parts with respect to r 0
and J:
1
1
(n · r0 )J = [(n · r0 )J + (n · J)r0 ] + (r0 × J) × n
(3.15)
2
2
An exercise is to show that
1
iµ0 keikr
(1 −
) n×m−
4πr
ikr
Z
1
µ0 ck 2 eikr
(1 −
) r0 (n · r0 )ρ(r0 )d3 r0
8πr
ikr
A(r) =
(3.16)
where m is the magnetic dipole moment of the current system:
m=
1
2
Z
r × J(r) d3 r
(3.17)
The first term of the vector potential has the same form as the magnetic
field of the electric dipole field in the previous subsection. So the fields are
obtained from the fields of the previous subsection leading to
1
µ0 ceikr
(1 −
)
4πr
ikr
eikr
+
B(r) = k 2 (n × m) × n
4π0 cr
ik eikr
1
(3n(n · m) − m)( 3 − 2 )
r
r 4π0 cr
E(r) = −k 2 (n × m)
(3.18)
The second term of the vector potential is more complicated. Since the
integral involves second moments of the charge density, this term corresponds to an electric quadrupole. We will not study it further, and we
also neglect all higher multipoles for which the present technique is tedious. A generally more powerful method deals with vector multipole fields
(Sect. 3.3).
3.2
Examples of radiating systems
The previous section was somewhat abstract in considering the multipole
expansion of the vector potential and related fields. Now we present some
simple concrete radiating systems.
21
3.2. EXAMPLES OF RADIATING SYSTEMS
z
q
L/2
y
x
–q
–L/2
Figure 3.1: Simple dipole antenna.
3.2.1
Radiating dipole antenna
Consider an electric dipole consisting of two small spheres in the z axis at
points z = ±L/2 (Fig. 3.1). They are connected by a wire whose capacitance
is negligible. If the charge of the upper sphere is q(t) then the charge of the
lower one is −q(t). Conservation of charge yields the current density
J(r, t) = I(t)δ(x)δ(y)θ(L/2 − z)θ(z + L/2) e z
(3.19)
where I = +dq/dt. The vector potential has only the z component
µ0
Az (r, t) =
4π
Z
L/2
−L/2
I(t − |r − z 0 ez |/c) 0
dz
|r − z 0 ez |
(3.20)
Before continuing, the reader should think qualitatively, which components
the field has. Although this is dynamic system, impressions from electroand magnetostatics are quite useful in this case.
At a large distances (r L) we can approximate
|r − z 0 ez | = (r 2 − 2z 0 ez · r + z 02 )1/2 ≈ r − z 0 cos θ
(3.21)
where θ is the angle between the radius vector r of the observation point and
the z axis. In the denominator of the vector potential, z 0 cos θ can be ignored
at large distances. In the retardation term it is negligible if z 0 cos θ/c is small
compared to the time scale of the current, for example, to the period T of
a harmonically varying current. Since z 0 cos θ ≤ L/2, we can omit z 0 cos θ/c
only if
L/2 cT = λ
(3.22)
If this is the case then at large distances
Az (r, t) =
µ0 L
I(t − r/c)
4π r
(3.23)
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CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
The simplest way to determine the scalar potential is to apply the Lorenz
gauge
1 ∂ϕ
∇·A+ 2
=0
(3.24)
c ∂t
yielding
∂ϕ
∂t
L ∂ 1
= −
I(t − r/c)
4π0 ∂z r
L
z
z 0
=
I(t − r/c) + 2 I (t − r/c)
4π0 r 3
r c
(3.25)
where I 0 refers to differentiation of I with respect to t − r/c. On the other
hand I = +q 0 , so
L z q(t − r/c) I(t − r/c)
+
ϕ(r, t) =
4π0 r 2
r
c
(3.26)
Consider a harmonically oscillating dipole
q(t − r/c) = q0 cos ω(t − r/c)
I(t − r/c) = I0 sin ω(t − r/c) = −ωq0 sin ω(t − r/c)
(3.27)
In spherical coordinates
µ0 I0 L
cos θ sin ω(t − r/c)
4π r
µ0 I0 L
sin θ sin ω(t − r/c)
= −
4π r
= 0
Ar =
Aθ
Aφ
(3.28)
The magnetic field B = ∇ × A has now only a φ component
ω
1
µ0 I0 L
sin θ
cos ω(t − r/c) + sin ω(t − r/c)
Bφ =
4π r
c
r
(3.29)
The components of the electric field E = −∂A/∂t − ∇ϕ are
Er
Eθ
Eφ
2I0 L cos θ sin ω(t − r/c) cos ω(t − r/c)
=
−
4π0
r2 c
ωr 3
I0 L sin θ
1
ω
1
= −
−
cos ω(t − r/c) − 2 sin ω(t − r/c)
4π0
ωr 3 rc2
r c
= 0
(3.30)
It is easy to show that at large distances E = cB × e r .
The radial component of the Poynting vector expresses the power radiated per a unit solid angle:
1 I0 Lω 2 2
dP
= R2 Eθ Bφ /µ0 =
(
) sin θ cos2 ω(t − R/c)
dΩ
20 c 4πc
(3.31)
23
3.2. EXAMPLES OF RADIATING SYSTEMS
Integration over a spherical surface yields the total radiated power:
P =
I
1 2
R
µ0
S · n da =
Z
π
0
Eθ Bφ 2π sin θ dθ
(3.32)
When R → ∞, only the radiation fields proportional to 1/r contribute, so
Prad =
I
S · n da =
(I0 L)2 ω 2
cos2 ω(t − R/c)
6π0 c3
(3.33)
This is the instantaneous radiated power. Integration over the period T =
2π/ω provides the average power, which is generally a more relevant quantity:
r
L2 ω 2 I02
2π µ0 L 2 I02
hPave i =
=
(3.34)
6π0 c3 2
3
0 λ
2
This is analogous to the average power RI 02 /2 dissipated in an AC circuit
whose resistance is R and current I0 cos ωt. So it is meaningful to define the
radiation resistance
2π
Rr =
3
r
µ0
0
2
L
λ
L
≈ 789
λ
2
Ω
(3.35)
A magnetic dipole is analysed in the same way. It can be modelled as a
circular loop carrying a time-harmonic current I 0 cos ωt. The dipole vector
is perpendicular to the plane of the loop. Because the current has only the
φ component, the only non-zero component of the vector potential is
µ0 I0 a
Aφ (r, t) =
4π
Z
2π
0
cos ω(t − |r − r0 |/c)
cos φ dφ
|r − r0 |
(3.36)
where a is the radius of the loop. The dipole approximation presumes that
r a and ωa c. The rest of the calculation is left as an exercise. A
difference to an electric dipole is that the electric field is now tangential to
any spherical surface centered at the current loop.
3.2.2
Half wavelength antenna
The result obtained in the previous section does not yield the correct radiation power of a true radio antenna, because an antenna is usually not short
compared to the wavelength, and the current is typically fed into the centre,
not to the ends.
Consider an antenna whose length is exactly equal to a half wavelength.
This is a realistic example, since for example, the wavelength of a 100 MHz
wave is 3 m. The antenna can be formally constructed of infinitesimal dipoles
24
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
considered in the previous section. Assume that the antenna is in the z-axis
in (−λ/4, +λ/4) and that its current is
2πz 0
I(z , t) = I0 sin ωt cos
λ
0
(3.37)
This is zero at both ends. The element at z 0 produces a radiation electric
field whose θ-component is
sin θ
2πz 0
dEθ = I0
ω
cos
ω(t
−
R/c)
cos
4π0 Rc2
λ
dz 0
(3.38)
Here R is the distance from dz 0 to the observation point, and terms of the
order of 1/R2 are ignored. The φ component of the magnetic field is
dBφ =
µ0 I0 ω
2πz 0
sin θ cos ω(t − R/c) cos
4π Rc
λ
dz 0
(3.39)
To get the total fields Eθ and Bφ , we must calculate the integral
Z
K=
π/2
−π/2
1
cos ω(t − R/c) cos u du
R
(3.40)
where u = 2πz 0 /λ. Again, R = r − z 0 cos θ and at large r we can replace
1/R by 1/r. The cosine term requires more care:
K≈
1
r
Z
π/2
−π/2
cos[ω(t − r/c) + u cos θ] cos u du
(3.41)
This is equal to
1
K = Re (eiω(t−r/c)
r
Z
π/2
eiu cos θ cos u du)
−π/2
yielding
cos[(π/2) cos θ]
2
cos ω(t − r/c)
r
sin2 θ
Substituting this gives the fields
K=
I0
cos[(π/2) cos θ]
cos ω(t − r/c)
2π0 rc
sin θ
cos[(π/2) cos θ]
µ0 I0
cos ω(t − r/c)
Bφ =
2πr
sin θ
An exercise is to calculate the time-averaged radiated power:
Eθ =
1
hP i =
4π
r
µ0 2
I
0 0
Z
π
0
cos2 [(π/2) cos θ]
sin θ dθ
sin2 θ
(3.42)
(3.43)
(3.44)
(3.45)
The integral is approximately 1,219, so the radiating power of a half-wavelength antenna is
I2
(3.46)
hP i ≈ 73 Ω 0
2
25
3.2. EXAMPLES OF RADIATING SYSTEMS
3.2.3
Centre-fed linear antenna
The third example is a thin linear antenna of length d excited by a coaxial
cable across a small gap at its midpoint. The current density is
J(r) = I sin(kd/2 − k|z|)θ(|z| − d/2)δ(x)δ(y)e z
(3.47)
and the time-dependence is again harmonic. The vector potential in the far
zone (kr 1) is according to Eq. 3.3
A(r) =
µ0 Ieikr
4πr
Z
d/2
−d/2
sin(kd/2 − k|z|)e−ikz cos θ dz ez
(3.48)
resulting in
A(r) =
kd
µ0 Ieikr cos( kd
2 cos θ) − cos( 2 )
ez
2πkr
sin2 θ
(3.49)
An exercise is to calculate the time-averaged radiation power per unit solid
angle:
kd
cos( kd
I2
dP
2 cos θ) − cos( 2 ) 2
= 2
|
|
(3.50)
dΩ
8π 0 c
sin θ
3.2.4
Radiation due to a moving charged particle
The electromagnetic field due to a moving charged particle is familiar from
the course of electrodynamics. We give a short overview of the results here.
The charge and current densities of a charged particle are
ρ(r, t) = qδ(r − rq (t))
(3.51)
J(r, t) = q ṙq (t)δ(r − rq (t))
(3.52)
A straightforward way is to solve the inhomogeneous wave equations of the
vector and scalar potentials in the Lorenz gauge using the method of Green’s
functions. The potentials are
q
1
ϕ(r, t) =
4π0 (1 − n · β)R
β
q
A(r, t) =
4π0 c (1 − n · β)R
ret
ret
q
1
=
4π0 R − R · β
β
q
=
4π0 c R − R · β
(3.53)
ret
(3.54)
ret
where R = r − rq , n = R/R and β = v/c. The subscript ret refers to the
evaluation of the expressions at the retarded time t 0 defined by
t0 + |r − rq (t0 )|/c = t
(3.55)
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CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
The observer measures the fields at r at time t. A tedious differentiation
yields the electric field
"
(1 − β 2 )(R − Rβ) + R × ((R − Rβ) × β̇)/c
q
E(r, t) =
4π0
(R − β · R)3
#
(3.56)
ret
and the magnetic field
B(r, t) =
1 R
c R
ret
× E(r, t)
(3.57)
The electric field is basically a sum of a Coulomb field and a field due
to acceleration of the particle. The former is of no further interest here,
since its Poynting flux vanishes at infinity. The latter term is the radiation
field vanishing as 1/R. So the corresponding Poynting flux remains finite at
infinity. In other words, radiation carries the field energy (and momentum
and angular momentum) away from the particle.
The radiation fields of a slowly moving particle (β 1) are
q
n × (n × v̇)/R
4π0 c2
(3.58)
q
1R
×E=
v̇ × n/R
cR
4π0 c3
(3.59)
Erad (r, t) =
Brad (r, t) =
Poynting’s vector is
S=
1
|R × v̇|2
q2
Erad × Brad =
R
µ0
16π 2 0 c3
R5
(3.60)
This behaves as 1/R2 , so the Poynting flux does not vanish. The radiation
power per unit solid angle dΩ is
dP
q 2 v̇2
sin2 θ
=
dΩ
16π 2 0 c3
(3.61)
where θ is the angle between v̇ and n. Integration yields the total power
(Larmor’s formula)
q 2 v̇2
P =
(3.62)
6π0 c3
For particles moving at high velocities, the difference between t and t 0 is
significant. The radiated energy during the interval t 1 = t01 + R(t01 )/c...t2 =
t02 + R(t02 )/c is
W =
Z
t2
t1
[S · n]ret dt =
Z
t02
t01
S·n
dt 0
dt
dt0
(3.63)
3.2. EXAMPLES OF RADIATING SYSTEMS
27
It is meaningful to define the power radiated per unit area in terms of the
charge’s own time: S·n dt/dt0 = S·n (1−n·β). The power radiated per unit
solid angle is then obtained in a straightforward manner using Poynting’s
vector resulting in
dP (t0 )
dΩ
=
q2
16π 2 0 c
2
n × ((n − β) × β̇)
(1 − n · β)5
(3.64)
To interpret this, we may imagine that the particle is accelerated only for
a short interval during which the velocity and acceleration vectors remain
nearly constant. If we are far away so that n and R do not practically
change during the acceleration interval then this formula gives the angular
distribution of the radiated power.
If β → 1 the effect of the denominator of dP/dΩ increases and the
radiated energy flux concentrates more parallel to the velocity. The total
radiated power is
q2
2
γ 6 (β̇ − (β × β̇)2 )
(3.65)
P =
6π0 c
This result can be obtained in two ways, by a direct integration, or by using
the relativistic formulation. However, we ignore the detailed calculation
here.
3.2.5
Radiation due to a system of moving charged particles
Consider a set of slowly moving charges v c which are assumed to be
far away from the observation point. More explicitly, all charges are within
a volume V1 for the time when the wave reaches the observer. Further,
the scales of V1 are assumed small compared to the wavelength and to the
distance to the observer.
Let the origin be inside V1 , denote coordinates of the charges by r 0 , of
the observation point by r, and define R = r − r 0 . Now
R = |r − r0 | ≈ r −
r · r0
r
(3.66)
and the retarded scalar potential is
ϕ(r, t) =
≈
ρ(r0 , t − R/c)
1
dV 0
4π0 V1
R
Z
1
ρ(r0 , t − r/c + r · r0 /cr)
dV 0
4π0 V1
r − (r · r0 )/r
Z
(3.67)
Use of the binomial series
(r − r · r0 /r)−1 = r −1 + r −2 (r · r0 /r) + . . .
(3.68)
28
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
and the Taylor expansion
r r · r0
ρ r ,t − +
c
cr
0
r
= ρ r ,t −
c
0
gives
1
4π0 r
r · r0 ∂ρ + ...
+
cr ∂t r0 ,t−r/c
r
r
1
ϕ(r, t) =
ρ r ,t −
r·
r0 ρ r0 , t −
dV 0 +
3
c
4π0 r
c
V1
V1
Z
r
d
1
dV 0 + . . .
r0 ρ r0 , t −
r·
+
4π0 r 2 c dt V1
c
Z
0
Z
(3.69)
dV 0
This is the familiar multipole expansion:
ϕ(r, t) =
1
Q r · p(t − r/c) r · ṗ(t − r/c)
+
+
4π0 r
r3
cr 2
(3.70)
The retarded vector potential is
A(r, t) =
=
µ0
J(r0 , t − r/c + r · r0 /cr)
dV 0
4π V1
r − (r · r0 )/r
Z
r
µ0
0
dV 0 + . . .
J r ,t −
4πr V1
c
Z
It is an exercise to show that for a finite volume V
A(r, t) =
µ0
ṗ(t − r/c)
4πr
R
V
(3.71)
J dV = dp/dt, so
(3.72)
Next we only consider radiation fields with the spatial dependence r −1 .
Because ṗ is a function of (t − r/c), then ∂ ṗ/∂r = −(1/c)p̈, and
E(r, t) = −
µ0
1 r · p̈(t − r/c)
p̈(t − r/c) +
r
4πr
4π0
c2 r 3
(3.73)
An exercise is to show that the magnetic radiation field is
B(r, t) = ∇ × A(r, t) = −
µ0
r × p̈
4πcr 2
(3.74)
and that
c
E(r, t) = − r × B(r, t)
(3.75)
r
All derivatives are evaluated at the retarded time t − r/c, which can be
assumed identical to all particles in V 1 .
The radiation field is a transverse electromagnetic wave, whose Poynting
vector
cB 2
1
S=
r=
(r × p̈)2 r
(3.76)
2
µ0 r
16π 0 c3 r 5
3.2. EXAMPLES OF RADIATING SYSTEMS
29
Setting z-axis parallel to p̈ yields
S=
p̈2 sin2 θ r
16π 2 0 c3 r 2 r
(3.77)
The maximum intensity is obtained in the perpendicular direction to p̈. To
determine the radiated power Prad , consider a spherical surface at a large
distance:
I
1 p̈2
(3.78)
S · n da =
Prad =
6π0 c3
∂V
This result shows that a set of charged particles radiates if the related dipole
moment has ”acceleration”. It may also happen that the dipole moment is
independent of time, although particles have acceleration. In such a case, it
would be necessary to consider higher terms in the multipole expansion.
3.2.6
Auroral kilometric radiation
Auroral kilometric radiation (AKR) is emitted by electrons in the auroral
acceleration region at heights 2000-20000 km 1 . The frequency is equal to
the Larmor frequency of electrons (30-600 kHz) and its wavelength is a few
kilometres. The maximum radiation power during magnetic storms is about
1 GW, which is about 1 % of the power of particle precipitation into the
ionosphere. The solar radiation power incident at the Earth is in turn about
108 GW. This is of the same order of magnitude as the infrared radiation
emitted by the Earth. So AKR is not significant from the energy viewpoint.
However, the number of photons per unit time can be larger for AKR
than for infrared radiation. This is remarkable, since the photon flux determines how distant objects it is possible to observe, assuming that the
detector technology is sophisticated. It might be possible to observe AKR
emitted up to a distance of about 100 light years (numerical argumentation
is an exercise). In other words, if there are any earth-like exoplanets within
the distance of 100 ly they could be observed due to their AKR emission.
There are a few problems in such an observation. First, the detector
should be in space, because AKR is dissipated in the ionosphere due to its
low frequency. It is evidently possible to construct such a space instrument
with the present technology. Another problem is that the Sun is a more intense radiation source at AKR frequencies than the Earth, at least most of
the time. Concerning exoplanets, their host stars could disturb the observation of planetary AKR. A solution is interferometry: signal must be received
at two sites simultaneously. An exercise is to show that this measurement
could be possible to perform in our solar system. The third difficulty may
be scintillation due to interplanetary plasmas. Finally, the reader should
1
AKR section and its exercises are based on ideas presented by Pekka Janhunen (FMI).
30
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
use her/his imagination to think of what we could infer from an observation
of AKR from an Earth-like exoplanet.
3.3
Vector multipole fields
The spherical harmonic expansion is useful in electro- and magnetostatic
problems having some symmetry with respect to the origin of the coordinate system. For time-varying fields the scalar spherical harmonic expansion
can be generalised to a vector form. This is also more powerful than the approach based on the expansion of the vector potential (Sect. 3.1). The vector
multipole approach is useful in boundary value problems with a spherical
symmetry and in studies of radiation from a localised source.
3.3.1
Basic spherical wave solutions
We start with the scalar wave equation
∇2 ψ(r, t) −
1 ∂ 2 ψ(r, t)
=0
c2 ∂t2
(3.79)
Fourier transform with respect to time leads to the Helmholtz equation
∇2 ψ(r, ω) + k 2 ψ(r, ω) = 0
(3.80)
Separation in spherical coordinates leads to the familiar expansion
ψ(r, ω) =
X
fl (r)Ylm (θ, φ)
(3.81)
l,m
where Ylm are spherical harmonics. The reader may (should) show that the
general solution can be written as
ψ(r, ω) =
X
(1) (1)
(2) (2)
(Alm hl (kr) + Alm hl (kr))Ylm (θ, φ)
(3.82)
l,m
(1,2)
Here hl
(1,2)
hl
is the spherical Hankel function defined by
(x) = (
π 1/2
) (Jl+1/2 (x) ± iNl+1/2 (x)) = jl (x) ± inl (x)
2x
(3.83)
where J and N are the Bessel and Neumann functions, and j and n are
the spherical Bessel and Neumann functions, respectively. Their explicit
expressions are easily computed from
1 d l sin x
)
x dx
x
cos
x
1
d
)l
nl (x) = −(−x)l (
x dx
x
jl (x) = (−x)l (
(3.84)
31
3.3. VECTOR MULTIPOLE FIELDS
For other useful formulas, see Jackson or Arfken and Weber, for example.
Next we consider more closely the spherical harmonics, which satisfy the
equation
−(
∂
1 ∂2
1 ∂
(sin θ ) +
)Ylm = l(l + 1)Ylm
sin θ ∂θ
∂θ
sin2 θ ∂φ2
(3.85)
It is convenient to define an operator
L=
1
r×∇
i
(3.86)
Readers familiar with quantum mechanics may recognize that this is basically the orbital angular momentum operator. Its components can be
written as
∂
∂
+ i cot θ )
∂θ
∂φ
∂
∂
+ i cot θ )
= Lx − iLy = e−iφ (−
∂θ
∂φ
∂
= −i
∂φ
L+ = Lx + iLy = eiφ (
L−
Lz
(3.87)
The operator L affects only angular variables and is independent of r. A
straightforward calculation shows that L · L = L 2 = L2x + L2y + L2z is the
operator on the left hand side of Eq. 3.85:
L2 Ylm = l(l + 1)Ylm
(3.88)
The proof of the following results is an exercise:
r · L = 0q
(l − m)(l + m + 1)Yl,m+1
L+ Ylm =
L− Ylm =
q
(l + m)(l − m + 1)Yl,m−1
Lz Ylm = mYlm
L2 L = LL2
L × L = iL
Lj ∇2 = ∇ 2 Lj
L2
1 ∂2
r
−
∇2 =
r ∂r 2
r2
i∇ × L = r∇2 − ∇(1 + r
L · F = i∇ · (r × F)
L · (∇ × F) = i∇2 (r × F) −
∂
)
∂r
i ∂ 2
(r ∇ · F)
r ∂r
(3.89)
32
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
3.3.2
Multipole expansion of the electromagnetic fields
Next we consider the electromagnetic field in a source-free region. Assuming
time-harmonic fields (e−iωt ) and denoting k = ω/c, the magnetic field satifies
the Helmholtz equation and the divergence-free condition, and the electric
field is obtained from the magnetic field:
(∇2 + k 2 )B = 0
∇·B = 0
ic
E =
∇×B
k
(3.90)
Alternatively,
(∇2 + k 2 )E = 0
∇·E = 0
B = −
i
∇×E
kc
(3.91)
The goal is to represent the fields as multipole expansions with explicitly
separated radial and angular dependences. One possibility is to first note
that for any well-behaved vector function F,
∇2 (r · F) = r · (∇2 F) + 2∇ · F
(3.92)
Consequently, outside of the source region the scalar functions r · B and r · E
satisfy the Helmholtz equation
(∇2 + k 2 )(r · B) = 0, (∇2 + k 2 )(r · E) = 0
(3.93)
The general solution for these scalar functions is given by Eq. 3.82.
Now we define a magnetic multipole field of order (l, m) by
(M )
r · Blm
(M )
r · Elm
where
l(l + 1)
gl (kr)Ylm (θ, φ)
kc
= 0
=
(1) (1)
(2) (2)
gl (kr) = Al hl (kr) + Al hl (kr)
(3.94)
(3.95)
The magnetic and electric fields are related by
kcr · B =
1
1
r · (∇ × E) = (r × ∇) · E = L · E
i
i
(3.96)
The electric field of the magnetic multipole must then satisfy the equation
(M )
L · Elm (r, θ, φ) = l(l + 1)gl (kr)Ylm (θ, φ)
(3.97)
33
3.3. VECTOR MULTIPOLE FIELDS
(M )
and r · Elm = 0. Because the operator L acts only on the angular variables,
(M )
the radial dependence of Elm is given by gl (kr). As shown in the previous
subsection, the effect of L on Ylm is to raise or lower m, but not to modify l
at all. Remembering that L2 Ylm = l(l + 1)Ylm , we see rather easily that the
electric field must be
(M )
Elm (r, θ, φ) = gl (kr)LYlm (θ, φ)
(3.98)
Since
i
(M )
∇ × Elm
(3.99)
kc
the fields of a magnetic multipole of order (l, m) are now specified apart
(1,2)
from the coefficients Al .
(M )
Blm = −
In the same way, we define an electric multipole of order (l, m) by
(E)
= −
(E)
= 0
r · Elm
r · Blm
l(l + 1)c
fl (kr)Ylm (θ, φ)
k
(3.100)
The electric multipole fields are
(E)
Blm (r, θ, φ) = fl (kr)LYlm (θ, φ)
ic
(E)
(E)
Elm =
∇ × Blm
k
(3.101)
The radial function fl (kr) has a similar expression to gl (kr).
We are now ready to write the general solution of the Maxwell equations in the source-free region. For more convenient notations, we define a
normalized form of the vector spherical harmonic:
Xlm (θ, φ) = p
1
LYlm (θ, φ)
l(l + 1)
(3.102)
It has the following orthogonality properties:
Z
Z
X∗l0 m0 · Xlm dΩ = δll0 δmm0
X∗l0 m0 · (r × Xlm ) dΩ = 0
(3.103)
Consequently, the fields are
B =
X
lm
E =
aE (l, m)fl (kr)Xlm −
X ic
lm
k
i
aM (l, m)∇ × (gl (kr)Xlm )
kc
aE (l, m)∇ × (fl (kr)Xlm ) + aM (l, m)gl (kr)Xlm (3.104)
34
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
The unknown coefficients are obtained from
k
∗
Ylm
r · B dΩ
l(l + 1)
Z
k
∗
aE (l, m)fl (kr) = − p
Ylm
r · E dΩ
l(l + 1)
aM (l, m)gl (kr) =
p
Z
(3.105)
Thus, we only have to know r · B and r · E at two different radii r 1 and
r2 in the source-free region for a complete solution, including the relative
(1,2)
magnitudes of Al .
3.3.3
Near and far zone multipole fields
To study the near zone fields (kr 1), we note that the radial dependence
(1,2)
is given by hl (kr) = jl (kr) ± inl (kr). With a small kr, the spherical
Hankel function behaves like (kr)−(l+1) , and the magnetic field for an electric
multipole (l, m) is
k Ylm
(E)
(3.106)
Blm ∼ − L l+1
l r
where the specific proportionality coefficient is chosen for convenience. The
electric field is
(E)
Elm =
ic
Ylm
i
(E)
∇ × Blm ∼ − ∇ × L( l+1 )
k
l
r
(3.107)
Using an operator identity, the electric field takes the form
(E)
Elm ∼ −(r∇2 − ∇(1 + r
∂
Ylm
))
∂r lr l+1
(3.108)
The first term vanishes, since Ylm /r l+1 is a solution of the Laplace equation.
It follows that
Ylm
(E)
Elm ∼ −∇( l+1 )
(3.109)
r
From the basic course of electrodynamics we remember that this is exactly
the electrostatic multipole field. The magnetic multipole fields are treated
in the same way just by interchanging E (E) → −B(M ) , B(E) → E(M ) .
For the far zone (kr 1), we investigate outgoing waves from a local(1)
ized source, corresponding to the radial dependence given by h l (kr). Its
asymptotic behaviour implies that the magnetic field of an electric multipole
is
eikr
(E)
LYlm
(3.110)
Blm ∼ (−i)l+1
kr
and the electric field is
(E)
Elm ∼
eikr
(−i)l c
eikr
(∇(
)
×
LY
+
∇ × LYlm )
lm
k2
r
r
(3.111)
35
3.3. VECTOR MULTIPOLE FIELDS
Keeping only the leading terms and using the same operator identity as for
the near zone calculation, this becomes
(E)
Elm ∼ −(−i)l+1
1
ceikr
(n × LYlm − (r∇2 − ∇)Ylm )
kr
k
(3.112)
where n = r/r. The second term is clearly some angular function times 1/r,
so it is ignorable. Not surprisingly, the radiation zone fields have an exact
relationship
(E)
(E)
Elm = cBlm × n
(3.113)
Again, the magnetic multipole fields are obtained by a similar interchange
like with the near zone case.
3.3.4
Energy of multipole radiation
We start the energy consideration by studying only a linear superposition of
electric multipoles (l, m) with a varying m but a fixed l. Then the outgoing
fields are
Bl =
X
m
El =
(1)
aE (l, m)Xlm hl (kr)
ic
∇ × Bl
k
(3.114)
The time-averaged energy density for time-harmonic fields is
u=
0
(E · E∗ + c2 B · B∗ )
4
(3.115)
In the radiation zone E = cB × n, so the energy dU in a spherical shell
[r, r + dr] is
Z
2π0 c2 dr X ∗
0
dU =
X∗lm0 · Xlm dΩ
aE (l, m )aE (l, m)
k2
(4π)
0
m,m
(3.116)
where the asymptotic expression of the spherical Hankel function is applied.
Using the orthogonality integral of vector spherical harmonics, this yields
dU
2π0 c2 X
|aE (l, m)|2
=
dr
k2
m
(3.117)
For a general superposition of electric and magnetic multipoles the result is
dU
2π0 c2 X
(|aE (l, m)|2 + |aM (l, m)|2 )
=
dr
k 2 l,m
(3.118)
36
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
Next we discuss the power of multipole radiation. At the limit kr 1
the fields are
B =
eikr X
(−i)l+1 (aE (l, m)Xlm + aM (l, m)n × Xlm )
kr lm
E = cB × n
(3.119)
The average power radiated per unit solid angle is calculated from Poynting’s
vector yielding
X
dP
c
=
|
(−i)l+1 (aE (l, m)Xlm × n + aM (l, m)Xlm )|2
dΩ
2µ0 k 2 lm
(3.120)
The electric and magnetic multipoles of a given (l, m) have the same angular
dependence, but perpendicular polarizations. So the multipole order can
be determined by measuring the angular distribution of radiated power. To
distinguish between the electric and magnetic nature of the radiating source,
polarization must be detected. The angular power distribution of a single
multipole is
c
dP (l, m)
=
|a(l, m)Xlm |2
(3.121)
dΩ
2µ0 k 2
3.3.5
Multipole moments
The next task is to relate the multipole expressions directly to the source
terms. We assume that the charge density ρ(r)e −iωt and current density
J(r)e−iωt are known. We could also consider the intrinsic magnetization as
a source term, but we neglect it here for brevity (see Jackson for a complete
discussion).
Again, we start with the Maxwell equations
∇ · E = ρ/0
∇·B = 0
∇ × E − ikcB = 0
∇ × B + (ik/c)E = µ0 J
(3.122)
Defining E0 = E+iJ/(ω0 ) and using the equation of the current continuity,
we obtain
∇ · E0 = 0
0
∇·B = 0
∇ × E − ikcB = i∇ × J/(ω0 )
∇ × B + (ik/c)E0 = 0
(3.123)
37
3.3. VECTOR MULTIPOLE FIELDS
If we had included magnetization, its curl would appear in the right side of
the last equation. Note that E0 = E outside of sources.
The inhomogeneous Helmholtz wave equations follow from the curl equations:
(∇2 + k 2 )B = −µ0 ∇ × J
(∇2 + k 2 )E0 = −i∇ × (∇ × J)/(0 ck)
(3.124)
We use the same trick as earlier and take the scalar product with vector r,
and use the vector identity r · (∇ × F) = (r × ∇) · F = −iL · F. This yields
scalar equations
(∇2 + k 2 )r · B = −iµ0 L · J
(∇2 + k 2 )r · E0 = L · (∇ × J)/(0 ck)
(3.125)
These equations can be solved using the method of Green’s functions (see
the course of electrodynamics). Since we are interested in outgoing waves,
the solution is
r · B(r) =
iµ0
4π
Z
0
eik|r−r | 0
L · J(r0 ) d3 r0
|r − r0 |
1
r · E (r) = −
4π0 ck
0
Z
0
eik|r−r | 0
L · ∇0 × J(r0 ) d3 r0
|r − r0 |
(3.126)
The multipole coefficients aE , aM are obtained from
k
∗
Ylm
r · B dΩ
l(l + 1)
Z
k
∗
aE (l, m)fl (kr) = − p
Ylm
r · E0 dΩ
l(l + 1)
aM (l, m)gl (kr) =
p
Z
(3.127)
as derived in Sect. 3.3.2. The radial dependence must be f l (kr) = gl (kr) =
(1)
hl (kr), since this corresponds to outgoing radiation. Next, we need Eq. 3.5:
∞
l
X
X
eik|r−r |
(1)
∗
=
ik
j
(kr
)h
(kr
)
Ylm
(θ 0 , φ0 )Ylm (θ, φ)
< l
>
l
4π|r − r0 |
l=0
m=−l
0
(3.128)
where now r< = r 0 , r> = r, because we study the region outside of the
source. It follows that
Z
0
∗
dΩ Ylm
(θ, φ)
eik|r−r |
(1)
∗
= 4πikhl (kr)jl (kr 0 )Ylm
(θ 0 , φ0 )
4π|r − r0 |
(3.129)
38
CHAPTER 3. RADIATING SYSTEMS IN FREE SPACE
Consequently, the multipole coefficients are
aM (l, m) = − p
aE (l, m) =
µ0 k 2
l(l + 1)
Z
ik
0 c l(l + 1)
p
Z
∗
jl (kr)Ylm
(θ, φ) L · J(r) d3 r
(3.130)
∗
jl (kr)Ylm
(θ, φ) L · ∇ × J(r) d3 r
A centre-fed linear antenna (Sect. 3.2.3) provides an illustrating concrete
exercise.