# Chapter 9. Radiation - Department of Physics | Oregon State University ```Chapter 9. Radiation
We discuss electromagnetic wave generation in this chapter. We consider the emission of
radiation by localized systems of oscillating charge and current densities. Approximations are
made for fields produced by slowly moving (nonrelativistic) charges. The results are widely
applicable from the emission of radio waves from antennas and to the emission of light from
atoms.
9.1 Fields and Radiation of a Localized Oscillating Source
We can define 𝐁 in terms of a vector potential:
𝐁=𝛁&times;𝐀
And define a scalar potential Φ satisfying
𝜕𝐀
𝐄 = −∇Φ −
𝜕𝑡
Imposing the Lorenz condition
1 𝜕Φ
𝛁⋅𝐀+ 2
=0
𝑐 𝜕𝑡
(9.1)
(9.2)
(9.3)
we have the wave equations for the scalar and vector potentials:
1 𝜕 2Φ
𝜌
𝛁 Φ− 2 2 =−
𝑐 𝜕𝑡
𝜀0
(9.4)
1 𝜕 2𝐀
= −𝜇0 𝐉
𝑐 2 𝜕𝑡 2
(9.5)
2
and
𝛁2𝐀 −
Using the Green functions for the wave equations, we can express the retarded scalar potential
as
[𝜌(𝐱 ′ , 𝑡 ′ )]𝑟𝑒𝑡 3 ′
1
Φ(𝐱, 𝑡) =
∫
𝑑 𝑥
(9.6)
|𝐱 − 𝐱 ′ |
4𝜋𝜀0
and the retarded vector potential as
𝜇0 [𝐉(𝐱 ′ , 𝑡 ′ )]𝑟𝑒𝑡 3 ′
𝐀(𝐱, 𝑡) =
∫
𝑑 𝑥
|𝐱 − 𝐱 ′ |
4𝜋
(9.7)
Where [ ]𝑟𝑒𝑡 means that 𝑡 ′ = 𝑡 − |𝐱 − 𝐱 ′ |/𝑐. Eqs. 9.6 and 9.7 indicate that at a given point 𝐱
and a given time 𝑡 the potentials are determined by the charge and current that existed at
other points in space 𝐱 ′ at earlier times 𝑡 ′ . The time appropriate to each source point is earlier
than 𝑡 by |𝐱 − 𝐱 ′ |/𝑐, the time required to travel from source to field point 𝐱 with velocity 𝑐. In
the above procedure it is essential to impose the Lorentz condition on the potentials; otherwise
it would not be the simple wave equations.
1
We consider charge and current distributions which are harmonically periodic in their time
dependence:
𝜌(𝐱 ′ , 𝑡′) = 𝜌(𝐱 ′ )𝑒 −𝑖𝜔𝑡′
(9.8)
𝐉(𝐱 ′ , 𝑡′) = 𝐉(𝐱 ′ )𝑒 −𝑖𝜔𝑡′
The source is localized in the area of which dimension is of order 𝑑 as shown in Fig 9.1.
Fig 9.1 Arbitrarily moving charges localized
in a volume of size ~𝒅. The fields are to be
calculated at 𝐱.
Substituting Eq. 9.8 into the solution for the vector potential 𝐀(𝐱, 𝑡) in the Lorenz gauge
𝐀(𝐱, 𝑡) =
𝜇0
∫
4𝜋
|𝐱 − 𝐱 ′ |
𝑐 ) 3 ′
𝑑 𝑥 = 𝐀(𝐱)𝑒 −𝑖𝜔𝑡
|𝐱 − 𝐱′|
𝐉 (𝐱 ′ , 𝑡 −
(9.9)
we have
′
𝜇0 𝐉(𝐱 ′ )𝑒 𝑖𝑘|𝐱−𝐱 | 3 ′
𝐀(𝐱) =
∫
𝑑 𝑥
|𝐱 − 𝐱′|
4𝜋
(9.10)
where 𝑘 = 𝜔/𝑐 = 2𝜋/𝜆 is the wave number. The magnetic induction is given by
𝐁=𝛁&times;𝐀
(9.11)
and making use of the Maxwell’s equation outside of the source
𝛁&times;𝐁=
1 𝜕𝐄
𝑐 2 𝜕𝑡
(9.12)
we obtain
𝐄=
𝑖𝑐
𝑖𝑐
𝛁 &times; 𝐁 = 𝛁 &times; [𝛁 &times; 𝐀]
𝑘
𝑘
(9.13)
We shall take the point of view that 𝑑, i.e., 𝑟 ′ = |𝐱 ′ |, is infinitesimal compared with 𝑟 = |𝐱|.
Then, there are three spatial regions of interest:
Near (static) zone:
𝑑≪𝑟≪𝜆
Intermediate (induction) zone:
𝑑 ≪ 𝑟~𝜆
𝑑≪𝜆≪𝑟
Near zone
In the near zone for 𝑘𝑟 ≪ 1, Eq. 9.10 reduces to
𝐀(𝐱) ≅
𝜇0
𝐉(𝐱 ′ ) 3 ′
∫
𝑑 𝑥
4𝜋 |𝐱 − 𝐱′|
2
(9.14)
This is the form of the magnetostatic vector potential. Therefore, the near fields are quasistationary, oscillating harmonically as 𝑒 −𝑖𝜔𝑡 , but otherwise static in character. Using the
addition theorem, we can expand Eq. 9.14 as
∞
𝑙
𝜇0
4𝜋 𝑌𝑙𝑚 (𝜃, 𝜙)
𝑙 ∗
(𝜃 ′ , 𝜙′) 𝑑 3 𝑥 ′
𝐀(𝐱) ≅
∑ ∑
∫ 𝐉(𝐱 ′ )𝑟 ′ 𝑌𝑙𝑚
4𝜋
2𝑙 + 1 𝑟 𝑙+1
(9.15)
𝑙=0 𝑚=−𝑙
for the points outside the source (𝑟 &gt; 𝑟′).
Far zone
In the far zone for 𝑘𝑟 ≫ 1, we apply the condition 𝑑 ≪ 𝑟
|𝐱 − 𝐱′| ≅ 𝑟 − 𝐧 ⋅ 𝐱 ′
where 𝐧 is a unit vector in the direction of 𝐱. Then, Eq.9.10 reduces to
𝜇0 𝑒 𝑖𝑘𝑟
′
𝐀(𝐱) ≅
∫ 𝐉(𝐱 ′ )𝑒 −𝑖𝑘𝐧⋅𝐱 𝑑3 𝑥 ′
4𝜋 𝑟
(9.16)
(9.17)
This demonstrates that in the far zone the vector potential behaves as an outgoing spherical
wave with an angular dependent coefficient. The fields are transverse to the radius vector and
fall off as 1/𝑟. Since 𝑑 ≪ 𝜆 (or 𝑘𝑑 ≪ 1),
∞
(−𝑖𝑘)𝑛
𝜇0 𝑒 𝑖𝑘𝑟
𝐀(𝐱) ≅
∑
∫ 𝐉(𝐱 ′ )(𝐧 ⋅ 𝐱 ′ )𝑛 𝑑 3 𝑥 ′
4𝜋 𝑟
𝑛!
(9.18)
𝑛=0
and the magnitude of the 𝑛th term,
(𝑘𝑑)𝑛
1
′ )(𝑘𝐧
′ )𝑛 3 ′
∫ 𝐉(𝐱
⋅𝐱 𝑑 𝑥 ~
𝑛!
𝑛!
(9.19)
falls off rapidly with 𝑛.
Electric monopole
In the far zone Eq. 9.6 reduces to
|𝐱 − 𝐱 ′ | 3 ′
1 1
1 𝑄(𝑡 − 𝑟/𝑐)
′
Φ(𝐱, 𝑡) ≅
∫ 𝜌 (𝐱 , 𝑡 −
)𝑑 𝑥 =
4𝜋𝜀0 𝑟
𝑐
4𝜋𝜀0
𝑟
(9.20)
where 𝑄 is the total charge of the source. Since charge is conserved, 𝑄 must be independent of
time. Thus electric monopole fields are intrinsically static. In other words, harmonic fields
cannot include electric monopole terms.
9.2 Electric Dipole Fields and Radiation
As a first approximation we keep the first term in Eq. 9.18. We then find
𝜇0 𝑒 𝑖𝑘𝑟
𝐀(𝐱) =
∫ 𝐉(𝐱 ′ ) 𝑑 3 𝑥 ′
4𝜋 𝑟
3
(9.21)
Note that here we apply the limit 𝑑 ≪ 𝑟 independent of 𝜆, so that this approximation is valid
both in the far and near zone. We evaluate the 𝑥 component of ∫ 𝐉(𝐱 ′ ) 𝑑 3 𝑥 ′ :
∫ 𝐽𝑥 (𝐱 ′ ) 𝑑 3 𝑥 ′ = ∫ 𝐉(𝐱 ′ ) ⋅ 𝐞𝑥 ′ 𝑑3 𝑥 ′ = ∫ 𝐉(𝐱 ′ ) ⋅ 𝛁 ′ 𝑥 ′ 𝑑 3 𝑥 ′
= ∫ 𝛁 ′ ⋅ [𝑥 ′ 𝐉(𝐱 ′ )] 𝑑 3 𝑥 ′ − ∫ 𝑥 ′ 𝛁 ′ ⋅ 𝐉(𝐱 ′ ) 𝑑 3 𝑥 ′
= ∫[𝑥 ′ 𝐉(𝐱 ′ )] ⋅ 𝑑𝐚 − ∫ 𝑥 ′ 𝛁 ′ ⋅ 𝐉(𝐱 ′ ) 𝑑 3 𝑥 ′ = − ∫ 𝑥 ′ 𝛁 ′ ⋅ 𝐉(𝐱 ′ ) 𝑑3 𝑥 ′
(9.22)
Generalizing three dimensions, we obtain
∫ 𝐉(𝐱 ′ ) 𝑑3 𝑥 ′ = − ∫ 𝐱 ′ 𝛁 ′ ⋅ 𝐉(𝐱 ′ ) 𝑑 3 𝑥 ′
(9.23)
Using the continuity equation, 𝛁 ⋅ 𝐉 = 𝑖𝜔𝜌,
we have
𝐀(𝐱) = −
𝑖𝜔𝜇0 𝑒 𝑖𝑘𝑟
𝐩
4𝜋 𝑟
(9.24)
where
𝐩 = ∫ 𝐱 ′ 𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
(9.25)
is the electric dipole moment, as defined in electrostatics. In order to determine 𝐁, we take the
curl of 𝐀,
𝑖𝜔𝜇0
𝑒 𝑖𝑘𝑟
𝜇0 𝑐𝑘 2 𝑒 𝑖𝑘𝑟
1
(9.26)
𝐁=−
𝛁(
)&times;𝐩=
(1 −
)𝐧 &times; 𝐩
4𝜋
𝑟
4𝜋 𝑟
𝑖𝑘𝑟
where 𝐧 = 𝐱/𝑟 is a radial unit vector. 𝐁 is transverse to 𝐧. From Eq. 9.13, we obtain
𝐄=
𝑖𝑐
𝑖𝑐
𝜇0 𝑐𝑘 2 𝑒 𝑖𝑘𝑟
1
𝛁&times;𝐁= 𝛁&times;[
(1 −
) 𝐧 &times; 𝐩]
𝑘
𝑘
4𝜋 𝑟
𝑖𝑘𝑟
1 𝑒 𝑖𝑘𝑟 2
1 𝑖𝑘
=
{𝑘 (𝐧 &times; 𝐩) &times; 𝐧 + [𝟑𝐧(𝐧 ⋅ 𝐩) − 𝐩] ( 2 − ) }
4𝜋𝜀0 𝑟
𝑟
𝑟
(9.27)
The electric field has components parallel and perpendicular to 𝐧.
Far zone
The fields take on the limiting forms,
1 𝑒 𝑖𝑘𝑟 𝑘 2
𝐁 = 4𝜋𝜖 𝑟 𝑐 (𝐧 &times; 𝐩)
0
1 𝑒 𝑖𝑘𝑟 2
𝐄 = 4𝜋𝜀 𝑟 𝑘 (𝐧 &times; 𝐩) &times; 𝐧 = 𝑐𝐁 &times; 𝐧
0
showing the typical behavior or radiation fields. We can rewrite the electric field as
4
(9.28)
1 𝑒 𝑖𝑘𝑟 2
1 𝑒 𝑖𝑘𝑟 2
𝐄=
𝑘 [𝐩 − (𝐩 ⋅ 𝐧)𝐧] =
𝑘 𝐩𝑝
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟
(9.29)
where 𝐩𝑝 is the component of the dipole moment perpendicular to 𝐧. Therefore,
1 𝑘2
𝑟
𝐄(𝐱, 𝑡) = 4𝜋𝜀 𝑟 𝐩𝑝 (𝑡 − 𝑐 )
0
1
𝐁(𝐱, 𝑡) = 𝑐 𝐧 &times; 𝐄(𝐱, 𝑡)
(9.30)
where 𝐩𝑝 (𝑡 − 𝑟/𝑐) is 𝐩𝑝 evaluated at 𝑡 − 𝑟/𝑐.
Near zone
The fields are expressed as
1 𝑖𝑘
𝐁 = 4𝜋𝜀
2 (𝐧 &times; 𝐩)
0 𝑐𝑟
1 1
𝐄 = 4𝜋𝜀 3 [𝟑𝐧(𝐧 ⋅ 𝐩) − 𝐩]
0𝑟
(9.31)
The electric field, apart from its oscillations in time, is just the static electric dipole field. The
magnetic induction 𝑐𝐵 ≈ 𝑘𝑟𝐸 ≪ 𝐸, since 𝑘𝑟 ≪ 1 in the near zone. Thus the fields in the near
zone are dominantly electric in nature.
In the far zone the time-averaged Poynting vector is
𝐒=
1
1
𝑐
𝑘4
2
𝐄 &times; 𝐇 ∗ = 𝑐𝜀0 |𝐄|2 𝐧 =
|𝐩𝑝 | 𝐧
2
2
2
2
32𝜋 𝜀0 𝑟
(9.32)
Then the time-averaged power radiated per unit solid angle by the oscillating dipole 𝐩 is
𝑑𝑃
𝑐 2 𝑍0 4 2 2
= 𝑟 2𝐒 ⋅ 𝐧 =
𝑘 |𝐩| sin 𝜃
𝑑Ω
32𝜋 2
(9.33)
where 𝜃 is the angle between 𝐩 and 𝐧 , and 𝑍0 = √𝜇0 /𝜀0 ≅ 376.7 Ω is the vacuum impedance.
Integrating over a sphere, we find the average amount of energy radiated by the dipole per unit
time:
𝑐 2 𝑍0 4 2 2𝜋 𝜋 3
𝑐 2 𝑍0 𝑘 4 2
(9.34)
|𝐩|
|𝐩|
𝑃=
𝑘
∫
∫
sin
𝜃
𝑑𝜃𝑑𝜙
=
32𝜋 2
12𝜋
0
0
Dipole antenna
We calculate the radiation field due to the simple short dipole antenna shown in Fig. 9.2. We
assume that the overall length of antenna is 𝑎 and that 𝑎 ≪ 𝜆. The current in the antenna is
taken to be
2|𝑧 ′ | −𝑖𝜔𝑡
(9.35)
𝐼(𝑧 ′ , 𝑡) = 𝐼0 (1 −
)𝑒
𝑎
5
In this model, the current is in the same direction in each half of the antenna and falls off
linearly as we approach the ends.
Fig 9.2 Short, center-fed dipole antenna
of length 𝑎.
To obtain the dipole moment, we can make use of Eq. 9.21 and Eq. 9.24.
0
𝑖
𝑖𝐼0 𝑎/2
2𝑧 ′
2𝑧 ′
𝑖𝐼0 𝑎
′) 3 ′
′
𝐩 = ∫ 𝐉(𝐱 𝑑 𝑥 =
[∫ (1 −
) 𝑑𝑧 + ∫
(1 +
) 𝑑𝑧 ′ ] 𝐞𝑧 =
𝐞
𝜔
𝜔 0
𝑎
𝑎
2𝜔 𝑧
−𝑎/2
(9.36)
The angular distribution of radiated power is
𝑑𝑃
𝑍0
(𝑘𝑎)2 𝐼02 sin2 𝜃
=
2
𝑑Ω 128𝜋
(9.37)
and the total power radiated is
𝑃=
𝑍0
(𝑘𝑎)2 𝐼02
48𝜋
(9.38)
Thus for a given maximum current the radiation power increases as the square of the frequency
in the situation where 𝑎 ≪ 𝜆. A resistance 𝑅 carrying a current 𝐼0 𝑒 −𝑖𝜔𝑡 dissipates energy at an
1
average rate 𝑃 = 2 𝑅𝐼02 . Comparing this with Eq. 9.38, we see that it is sensible to define the
radiation resistance of a dipole by
𝑍0
(𝑘𝑎)2 ≅ 5(𝑘𝑎)2
24𝜋
(9.39)
In the next order of approximation we consider the second term in the expansion in Eq. 9.18:
𝐀(𝐱) =
𝜇0 𝑒 𝑖𝑘𝑟
(−𝑖𝑘) ∫ 𝐉(𝐱 ′ )(𝐧 ⋅ 𝐱 ′ ) 𝑑 3 𝑥 ′
4𝜋 𝑟
6
(9.40)
It is convenient to break up the integrand as follows:
1
1
𝐉(𝐧 ⋅ 𝐱 ′ ) = [𝐉(𝐧 ⋅ 𝐱 ′ ) − 𝐱 ′ (𝐉 ⋅ 𝐧)] + [𝐉 (𝐧 ⋅ 𝐱 ′ ) + 𝐱 ′ (𝐉 ⋅ 𝐧)]
2
2
(9.41)
1
1
= [𝐧 &times; (𝐉 &times; 𝐱 ′ )] + [𝐉(𝐧 ⋅ 𝐱 ′ ) + 𝐱 ′ (𝐉 ⋅ 𝐧)]
2
2
The part of the vector potential corresponding to the first term is called the magnetic dipole
potential, and the part corresponding to the second term is called the electric quadrupole
potential.
The magnetic dipole potential is expressed as
𝐀(𝐱) =
𝑖𝑘𝜇0 𝑒 𝑖𝑘𝑟
1
𝑖𝑘𝜇0 𝑒 𝑖𝑘𝑟
𝐧 &times; ∫ [𝐱 ′ &times; 𝐉(𝐱 ′ )] 𝑑 3 𝑥 ′ =
𝐧&times;𝐦
4𝜋 𝑟
2
4𝜋 𝑟
(9.42)
Where 𝐦 is the magnetic dipole moment,
1
𝐦 = ∫ [𝐱 ′ &times; 𝐉(𝐱 ′ )] 𝑑 3 𝑥 ′
2
In the far zone we find the magnetic induction
𝜇0 2 𝑒 𝑖𝑘𝑟
(𝐧 &times; 𝐦) &times; 𝐧
𝐁=𝛁&times;𝐀≅
𝑘
4𝜋
𝑟
𝜇0 2 𝑒 𝑖𝑘𝑟
𝜇0 𝑘 2
𝑟
[𝐦 − (𝐦 ⋅ 𝐧)𝐧] =
=
𝑘
𝐦𝑝 (𝑡 − )
4𝜋
𝑟
4𝜋 𝑟
𝑐
(9.43)
(9.44)
where 𝐦𝑝 is the component of the dipole moment perpendicular to 𝐧 evaluated at 𝑡 − 𝑟/𝑐.
Note the similarity to the electric dipole field in Eq. 9.29. The electric field takes the form
𝐄=
𝑖𝑐
𝛁 &times; 𝐁 = 𝑐𝐁 &times; 𝐧
𝑘
(9.45)
Circular current loop
The circular current loop shown in Fig. 9.3 is a magnetic dipole. The loop of circumference 𝑎 has
the current 𝐼(𝑡) = 𝐼0 𝑒 −𝑖𝜔𝑡 . We assume that 𝑎 ≪ 𝜆.
Fig 9.2 Circular current loop of
circumference 𝑎.
7
The magnetic moment of the loop is
𝑎2
𝐦 = 𝐼0
𝐞
4𝜋 𝑧
(9.46)
Hence the angular distribution of radiated power (see Eq. 9.33) is
𝑑𝑃
𝑍0 4
𝑍0
2
2
|𝐦|
(𝑘𝑎)4 𝐼02 sin2 𝜃
=
𝑘
sin
𝜃
=
𝑑Ω 32𝜋 2
512𝜋 3
(9.47)
It is instructive to compare this result with Eq. 9.37. For comparable currents and a comparable
size, the ratio of magnetic dipole radiation to electric dipole is
𝑑𝑃
)
𝑎 2
𝑑Ω MD (𝑘𝑎)2
=
=
(
)
2
𝑑𝑃
4𝜋
𝜆
( )
𝑑Ω ED
(
(9.48)
The intensity of magnetic dipole radiation is thus characteristically smaller by a factor of (𝑎/𝜆)2
than that of electric dipole radiation.
Referring to Eqs. 9.40 and 9.41, we can write the electric quadrupole potential as
𝐀(𝐱) =
𝜇0 𝑒 𝑖𝑘𝑟
(−𝑖𝑘) ∫[𝐉(𝐧 ⋅ 𝐱 ′ ) + 𝐱 ′ (𝐉 ⋅ 𝐧)] 𝑑3 𝑥 ′
4𝜋 2𝑟
(9.49)
We first examine the 𝑥 component of the first part of the integral.
∫ 𝐽𝑥 (𝐧 ⋅ 𝐱 ′ ) 𝑑3 𝑥 ′ = ∫(𝐧 ⋅ 𝐱 ′ ) 𝐉 ⋅ 𝐞𝑥 𝑑3 𝑥 ′ = ∫(𝐧 ⋅ 𝐱 ′ ) 𝐉 ⋅ 𝛁 ′ 𝑥 ′ 𝑑 3 𝑥 ′
= ∫ 𝛁 ′ ⋅ [𝑥 ′ (𝐧 ⋅ 𝐱 ′ )𝐉] 𝑑 3 𝑥 ′ − ∫ 𝑥 ′ 𝛁 ′ ⋅ [(𝐧 ⋅ 𝐱 ′ )𝐉] 𝑑3 𝑥 ′
= ∫[𝑥 ′ (𝐧 ⋅ 𝐱 ′ )𝐉] ⋅ 𝑑𝐚 − ∫ 𝑥 ′ 𝛁 ′ ⋅ [(𝐧 ⋅ 𝐱 ′ )𝐉] 𝑑 3 𝑥 ′
𝑆
We eliminate the first integral by noting that 𝐉 = 0 on the surface. Hence
∫ 𝐽𝑥 (𝐧 ⋅ 𝐱 ′ ) 𝑑3 𝑥 ′ = − ∫ 𝑥 ′ 𝛁 ′ ⋅ [(𝐧 ⋅ 𝐱 ′ )𝐉] 𝑑 3 𝑥 ′
= − ∫ 𝑥 ′ 𝛁 ′ (𝐧 ⋅ 𝐱 ′ ) ⋅ 𝐉 𝑑 3 𝑥 ′ − ∫ 𝑥 ′ (𝐧 ⋅ 𝐱 ′ )𝛁 ′ ⋅ 𝐉 𝑑 3 𝑥 ′
Since 𝛁 ′ (𝐧 ⋅ 𝐱 ′ ) = 𝐧 and 𝛁 ′ ⋅ 𝐉(𝐱 ′ ) = −𝑖𝜔𝜌(𝐱 ′ ), we have
∫ 𝐽𝑥 (𝐧 ⋅ 𝐱 ′ ) 𝑑 3 𝑥 ′ = − ∫ 𝑥 ′ (𝐉 ⋅ 𝐧) 𝑑 3 𝑥 ′ − 𝑖𝜔 ∫ 𝑥 ′ (𝐧 ⋅ 𝐱 ′ )𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
Generalizing to three dimensions, we obtain
∫[𝐉 (𝐧 ⋅ 𝐱 ′ ) + 𝐱 ′ (𝐉 ⋅ 𝐧)] 𝑑 3 𝑥 ′ = −𝑖𝜔 ∫ 𝐱 ′ (𝐧 ⋅ 𝐱 ′ )𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
8
(9.50)
Substituting back into Eq. 9.49, we find
𝐀(𝐱) = −
𝜇0 𝑐𝑘 2 𝑒 𝑖𝑘𝑟
∫ 𝐱 ′ (𝐧 ⋅ 𝐱 ′ )𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
8𝜋 𝑟
(9.51)
Taking the curl of 𝐀 and ignoring terms of order 1/𝑟 2 , we find the magnetic induction in the
𝐁 = 𝑖𝑘𝐧 &times; 𝐀 = −
𝑖𝜇0 𝑐𝑘 3 𝑒 𝑖𝑘𝑟
∫ 𝐧 &times; 𝐱 ′ (𝐧 ⋅ 𝐱 ′ )𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
8𝜋
𝑟
(9.52)
2
We add to the integrand a term, 𝐧 &times; 𝐧 𝑟 ′ 𝜌(𝐱 ′ ), which is zero, yet helps us put the field in a
more conventional form. Then we have
𝐁 = 𝑖𝑘𝐧 &times; 𝐀 = −
𝑖𝜇0 𝑐𝑘 3 𝑒 𝑖𝑘𝑟
2
𝐧 &times; ∫[3𝐱 ′ (𝐧 ⋅ 𝐱 ′ ) − 𝐧 𝑟 ′ ]𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
24𝜋 𝑟
(9.53)
The integral on the right side of Eq. 9.53 can be written as a vector 𝐐(𝐧) which is the product of
⃡ and the vector :
2
𝐐(𝐧) = ⃡𝐐 ⋅ 𝐧 = ∫[3𝐱 ′ (𝐧 ⋅ 𝐱 ′ ) − 𝐧 𝑟 ′ ]𝜌(𝐱 ′ ) 𝑑 3 𝑥 ′
(9.54)
where the tensor elements are
𝑄𝛼𝛽 = ∫(3𝑥𝛼 𝑥𝛽 − 𝑟 2 𝛿𝛼𝛽 )𝜌(𝐱) 𝑑 3 𝑥
(9.55)
And the vector 𝐐(𝐧) is defined as having components,
𝑄𝛼 = ∑ 𝑄𝛼𝛽 𝑛𝛽
(9.56)
𝛽
The magnetic induction is thus
𝐁=−
𝑖𝜇0 𝑐𝑘 3 𝑒 𝑖𝑘𝑟
𝐧 &times; 𝐐(𝐧)
24𝜋 𝑟
and, as usual the electric field is
𝐄 = 𝑐𝐁 &times; 𝐧
(9.57)
(9.58)
Charge distribution of cylindrical symmetry
When the charge distribution is cylindrically symmetrical, the off-diagonal elements of 𝐐 vanish
(𝑄𝛼𝛽 = 0 for 𝛼 ≠ 𝛽) and the diagonal elements can be written as
𝑄33 = 𝑄0 ,
1
𝑄11 = 𝑄22 = − 𝑄0
2
(9.59)
𝑄0 = ∫(3𝑧 2 − 𝑟 2 )𝜌(𝐱) 𝑑 3 𝑥
9
(9.60)
The angular distribution of radiated power is
𝑑𝑃
𝑐𝑟 2 |𝐵|2 𝑐 2 𝑍0 𝑘 6
|𝐧 &times; 𝐐(𝐧)|2
= 𝑟 2𝐒 ⋅ 𝐧 =
=
𝑑Ω
2 𝜇0
1152𝜋 2
1
(9.61)
1
Since 𝐐(𝐧) = (− 2 𝑄0 𝑛𝑥 , − 2 𝑄0 𝑛𝑦 , 𝑄0 𝑛𝑧 ),
1
9
9
2
|𝐧 &times; 𝐐(𝐧)|2 = 𝑄02 |(−3𝑛𝑦 𝑛𝑧 , 3𝑛𝑧 𝑛𝑥 , 0)| = 𝑄02 (𝑛𝑥2 + 𝑛𝑦2 )𝑛𝑧2 = 𝑄02 sin2 𝜃 cos 2 𝜃
4
4
4
Thus we have
𝑑𝑃 𝑐 2 𝑍0 𝑘 6 2 2
=
𝑄 sin 𝜃 cos2 𝜃
𝑑Ω 512𝜋 2 0
(9.62)
The radiation pattern is shown in Fig. 9.3.
𝑃=
𝑐 2 𝑍0 𝑘 6 2
𝑄
960𝜋 0
(9.63)
As a simple example of a quadrupole radiator we consider the assembly of point charges shown
in Fig. 9.4. The charge −2𝑞 is stationary at the origin, and two positie charges harmonically
oscillate, each with amplitude 𝑎, about the origin. The two positive charges are always on
opposite sides of the origin and exchange positions every half-cycle.
10
Assuming that the charges are at their maximum amplitude at 𝑡 = 0, we calculate the
(9.64)
2
𝑄0 = ∫(3𝑧 − 𝑟
2 )[−2𝑞𝛿(𝐱)
3
+ 𝑞𝛿(𝐱 − 𝑎𝐞𝐳 ) + 𝑞𝛿(𝐱 + 𝑎𝐞𝐳 )] 𝑑 𝑥 = 4𝑞𝑎
2
The angular distribution of radiated power is
𝑑𝑃 𝑐 2 𝑍0 𝑞 2 𝑎4 𝑘 6 2
=
sin 𝜃 cos2 𝜃
𝑑Ω
32𝜋 2
(9.65)
and the total radiation power is
𝑃=
𝑐 2 𝑍0 𝑞 2 𝑎4 𝑘 6
60𝜋
(9.66)
It is instructive to compare this result with the electric dipole radiation power (Eq. 9.34). For
electric dipole moment 𝑝 = 𝑞𝑎,
𝑐 2 𝑍0 𝑞 2 𝑎2 𝑘 4
(9.67)
𝑃ED =
12𝜋
𝑃QD (𝑘𝑎)2
𝑎 2
=
≅ 8( )
𝑃ED
5
𝜆
(9.68)
The intensity of electric quadrupole radiation is thus characteristically smaller by a factor of
(𝑎/𝜆)2 than that of electric dipole radiation. This result is similar to Eq. 9.48, indicating that
11
9.4 Center-Fed Linear Dipole Antenna
The restriction to source dimension small compared with 𝜆 can be removed for certain
radiating systems. As an example of such a system we consider a thin, linear antenna of length
𝑎 which is excited across a small gap at its midpoint (Fig 9.5).
Fig 9.5 Center-fed, linear antenna
The current along the antenna can be taken as sinusoidal in time and space with wave number
𝑘 = 𝜔/𝑐 and is symmetric on the two arms of the antenna. The current vanishes at the ends of
the antenna.
𝑘𝑎
𝐼(𝑧) = 𝐼 sin ( − 𝑘|𝑧|)
2
for
|𝑧| &lt;
𝑎
2
(9.69)
Then the vector potential in the far zone (Eq. 9.17) takes the form
𝑎
𝜇0 𝐼𝑒 𝑖𝑘𝑟 2
𝑘𝑎
′
𝐀(𝐱) = 𝐞𝑧
∫ sin ( − 𝑘|𝑧 ′ |) 𝑒 −𝑖𝑘𝑧 cos 𝜃 𝑑𝑧 ′
𝑎
4𝜋 𝑟 −
2
(9.70)
2
The integration is straightforward:
𝑘𝑎
𝑘𝑎
𝜇0 2𝐼𝑒 𝑖𝑘𝑟 cos ( 2 cos 𝜃) − cos ( 2 )
𝐀(𝐱) = 𝐞𝑧
[
]
4𝜋 𝑘𝑟
sin2 𝜃
(9.71)
Since the magnetic induction in the radiation zone is given by 𝐁 = 𝑖𝑘𝐧 &times; 𝐀, its magnitude is
|𝐁| = 𝑘 sin 𝜃 |𝐴𝑧 |. Thus the time averaged power radiated per unit solid angle is
2 |𝐵|2
𝑑𝑃
𝑐𝑟
= 𝑟 2𝐒 ⋅ 𝐧 =
𝑑Ω
2 𝜇0
𝑘𝑎
𝑘𝑎 2
cos
(
cos
𝜃)
−
cos
(
𝑍0 𝐼
2
2 )|
=
|
8𝜋 2
sin 𝜃
2
12
(9.72)
For the special values 𝑘𝑎 = 𝜋 (𝑎 = 𝜆/2, half-wave antenna) and 𝑘𝑎 = 2𝜋 (𝑎 = 𝜆, full-wave
antenna), the angular distributions are
2
𝑑𝑃 𝑍0 𝐼
=
𝑑Ω 8𝜋 2
𝜋
cos 2 ( 2 cos 𝜃)
{
,
sin2 𝜃
𝜋
4 cos 4 ( 2 cos 𝜃)
sin2 𝜃
𝑎=
𝜆
2
(9.73)
, 𝑎=𝜆
Fig 9.6 Normalized radiation pattern of a center-fed linear dipole antenna with sinusoidal current distribution on
the 𝒛-axis. Three of the sides of the cube correspond to intersections with the coordinate planes.
13
```