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Simple Radiating Systems
May 11, 20101
1
J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Chapter 9
Simple Radiating Systems
Fields and Radiation of a Localized Oscillating Source
For a system of charges and currents varying in time we can make a
Fourier analysis of the time dependence and handle each Fourier
component separately. The potentials, and fields from a localized system
of charges and currents which vary sinusoidally in time become:
ρ(~x , t) = ρ(~x )e iωt ,
~J(~x , t) = ~J(~x )e iωt
(1)
The physical quantities will be taken from the real part of such
expressions while the EM potentials and fields assumed to have the same
time dependence. In the Lorentz gauge the solution (provided no
~ x , t) is
boundary surfaces are present) for the vector potential A(~
Z
Z ~ 0 0 J(~x , t )
|~x − ~x 0 |
1
3 0
0
~
d x
δ t +
− t dt 0
(2)
A(~x , t) =
c
|~x − ~x 0 |
c
Using the sinusoidal dependence (1) we get (k = ω/c)
~ x) = 1
A(~
c
Z
d 3x 0
~J(~x 0 )
0
e ik|~x −~x |
|~x − ~x 0 |
Simple Radiating Systems
(3)
Given a current distribution ~J(~x 0 ) the fields can be determined by
calculating the integral (3). The magnetic induction will be given by
~ =∇
~
~ ×A
B
(4)
while outside the source the electric field is
~ = i∇
~
~ ×B
E
k
(5)
If the sources dimensions are of order d and the wavelength is
λ = 2πc/ω and if d λ then there are 3 spatial regions of inderest
I
The near (static) zone :
d r λ
I
The intermediate (induction) zone :
d r ∼λ
I
The far (radiation) zone :
d λr
The fields have very different properties in the different zones.
• In the near zone they behave as static and depend on the properties
of the source.
• In the far zone the fields are transverse to the radius vector and fall
as 1/r (typical for radiation fields).
Simple Radiating Systems
The Near Zone
The general solution is:
~ x) = 1
A(~
c
Z
d 3x 0
~J(~x 0 )
0
e ik|~x −~x |
|~x − ~x 0 |
(6)
For the near zone where r λ (or kr 1) the exponential in (6) can be
replaced by unity.
Then by using:
∞ X
l
`
X
r<
1
1
=
4π
Y ∗ (θ0 , φ0 )Y`m (θ, φ)
`+1 `m
|~x − ~x 0 |
2` + 1 r>
(7)
`=0 m=−`
The vector potential gets the form
X 4π Y`m (θ, φ) Z
~ x) = 1
~J(~x 0 )r 0` Y ∗ (θ0 , φ0 )d 3 x 0
lim A(~
`m
kr →0
c
2` + 1 r `+1
`,m
This shows that the near fields are quasi-stationary, oscillating
harmonically as e −iωt , but otherwise static in character.
Simple Radiating Systems
(8)
The Far Zone
In the far zone (kr 1) the exponential in (3) oscillates rapidly and
determines the behavior of the vector potential.
In this region is sufficient to approximate
|~x − ~x 0 | ≈ r − ~n · ~x 0
(9)
where ~n is the unit vector in the direction of ~x .
If only the leading term in kr is desired, the inverse distance can be
replaced by r and the vector potential is
ikr Z
~J(~x 0 )e −ik~n·~x 0 d 3 x 0
~ x) = e
(10)
lim A(~
kr →∞
cr
I.e. in the far zone the vector potential behaves as an outgoing spherical
wave.
It is easy to show that the fields calculated from (4) and (5) are
transverse to the radius vector (how?) and fall like 1/r . Thus they
correspond to radiation fields.
Simple Radiating Systems
The Far Zone
If the source size is small compared to a wavelength one can expand in
powers of k:
Z
ikr X
(−ik)n ~ 0
~ x) = e
lim A(~
J(~x )(~n · ~x 0 )n d 3 x 0
(11)
kr →∞
cr n
n!
The magnitude of the n-th term is given by
Z
1
~J(~x 0 )(k~n · ~x 0 )n d 3 x 0
n!
(12)
Since |~x 0 | ≈ d and kd 1 by assumption, the successive terms in the
~ fall off rapidly with n.
expansion of A
Consequently the radiation emitted from the source will come mainly
from the first nonvanishing term in the expansion (11).
Simple Radiating Systems
The Intermediate Zone
In the intermediate zone neither kr 1 or kr 1 can be used; all power
of kr must be retained. For points outside the source, eqn (3) becomes
Z
X (1)
∗
~ x ) = 4πik
(θ0 , φ0 )d 3 x 0 (13)
h` (kr )Ym` (θ, φ) ~J(~x 0 )j` (kr 0 )Y`m
A(~
c
`,m
If the source dimensions are small compared to a wavelength, j` (kr 0 ) can
be approximate as follows: 2
xl
x3
j` (x) →
1−
+ ...
(14)
(2` + 1)!!
2(2` + 3)
Then the vector potential is of the form (8) but with the replacement
1
r `+1
→
e ikr 1 + a1 (ikr ) + a2 (ikr )2 + · · · al (ikr )l
`+1
r
The coefficients ai come from explicit expressions for the spherical
Hankel
functions.
2
[(2` + 1)!! = (2` + 1)(2` − 1)(2` − 3) · · · 5 · 3 · 1]
Simple Radiating Systems
(15)
Electric Monopole Fields
For a source varying with the time the analogue of (2) for a scalar
potential is
Z
Z
x 0, t 0)
|~x − ~x 0 |
3 0
0
0 ρ(~
Φ(~x , t) = d x
dt
δ t +
−t
|~x − ~x 0 |
c
(16)
The electric monopole contribution is obtained by replacing in the
integral |~x − ~x 0 | → |~x | ≡ r . The result is
Φmonopole (~x , t) =
q(t 0 = t − r /c)
r
where q(t) is the total charge of the source. Since charge is conserved
and a localized source has not charge flowing into or away from it, the
total charge q is independent of time.
? Thus the electric monopole part of the potential (and fields) of
localized source is of necessity static.
? The fields with harmonic time dependence e −iωt with ω 6= 0 have
no monopole terms.
Simple Radiating Systems
Electric Dipole Fields and Radiation
If only the first term in (11) is kept, the potential is
ikr Z
~ x) = e
~J(~x 0 )d 3 x 0
A(~
cr
(17)
Examination of (13) and (15) shows that (17) is the ` = 0 part of the
series and that it is valid everywhere outside the source, not just in the
far zone. Integrating by part we get
Z
Z
Z
~Jd 3 x 0 = − ~x 0 (∇
~ 0 · ~J)d 3 x 0 = −iω ~x 0 ρ(~x 0 )d 3 x 0
(18)
since from the continuity equation
~ · ~J = iωρ
∇
(19)
Thus the vector potenial is:
ikr
~ x ) = −ik e ~p
A(~
r
Z
~p (~x ) = ~x 0 ρ(~x 0 )d 3 x 0
is the electric dipole moment, as defined in electrostatics
Simple Radiating Systems
(20)
(21)
The electric dipole fields from (4) and (5) are (how?)
ikr
1
e
2
~ = k (~n × ~p ) 1 −
B
(22)
ikr
r
e ikr
1
ik
2
~
E = k (~n × ~p ) × ~n
+ [3~n(~n · ~p ) − ~p ] 3 − 2 e ikr(23)
r
r
r
Note that the magnetic induction is transverse to the radius vector
at all distances, but the electric field has components parallel and
perpendicular to ~n.
In the radiation zone (kr 1) the fields take the limiting form
~ = k 2 (~n × ~p ) e
B
ikr
r
,
~ =B
~ × ~n
E
(24)
In the near zone (kr 1) the fields approach
~ = ik(~n × ~p ) 1 ,
B
r2
~ = [3~n(~n · ~p ) − ~p ] 1
E
r3
Simple Radiating Systems
(25)
? The electric field apart from its oscillation in time is just the static
electric dipole field seen earlier
? The magnetic induction is smaller than the electric field by a factor
(kr ) in the region where kr 1.
? Thus the fields in the near zone are dominantly electric in
nature.
? The magnetic induction vanishes, in the static limit k → 0.
? The following identities might be useful for future derivations
[~p − (~p · ~n)~n] · (~n × ~p ) = ~p · (~n × ~p ) − (~p · ~(n))~n · (~n × ~p ) = 0
(~n × ~p ) × ~n = (~n · ~n) ~p − (~n · ~p ) ~n = ~p − (~p · ~n) ~n
?
(26)
(27)
If we let ~p point in the z-direction, then ~p = p~ez also
~n
~p · ~n
~p − (~p · ~n) ~n
~n × ~p
=
sin θ cos φ~ex + sin θ sin φ~ey + cos θ~ez
= p cos θ
= −p sin θ (cos θ cos φ~ex + cos θ sin φ~ey − sin θ~z ) = −p sin θ~eθ
= −p sin θ (− sin φ~ez + cos φ~ey ) = −p sin θ~eφ
and we get (prove it)
~ = −k 2 p sin θ e
B
ikr
r
~eφ
~ = −k 2 p sin θ e
and E
Simple Radiating Systems
ikr
r
~eθ
(28)
Electric Dipole : Power Radiated
The Poynting vector is
h
i
~S = c E
~ ×B
~ = c B
~ × ~n ×B
~ = c
~ ·B
~ ~n − B
~ · ~n B
~ = c |B|
~ 2~n
B
8π
8π
8π
8π
(29)
~ · ~n = 0.
the blue term is vanishing because B
The time-averaged power radiated per unit solid angle by the oscillating
dipole moment ~p is:
c h 2 ~ ~ ∗ i
dP
=
< r ~n · E × B
(30)
dΩ
8π
~ and B
~ are given by (24). Thus we find
where E
dP
c 4
2
=
k |(~n × ~p ) × ~n|
(31)
dΩ
8π
If all the components of ~p have the same phase, the angular distribution
is a typical dipole pattern (where θ is measured from the direction of ~p )
c 4 2 2
dP
=
k |~p | sin θ
dΩ
8π
Simple Radiating Systems
(32)
Electric Dipole : Power Radiated
The total power radiated, independent of the relative phases of the
components of ~p , is
1
2
(33)
P = ck 4 |~p |
3
Figure: Power distribution.
Simple Radiating Systems
Example I
For ~p = p~ez we can substitute (28) into (29) and we get (how?)
2
~S = c k 4 p 2 sin θ e 2ikr ~n
(34)
8π 2
r2
Since the energy flux oscillates in time after averaging over a complete
wave cycle we get (hcos2 (kr )i = 1/2)
ck 4 p 2 sin2 θ
~n
h~Si =
(35)
16π 2 r 2
From the angular dependence of this expression we see that:
• most of the energy is emitted near the equatorial plane (θ = π/2), and
• none of the energy propagates along the z-axis (θ = 0 or θ = π).
The total power radiated is obtained by integrating ~S over any closed
surface S. The averaged power is then
I
hPi =
hSi · d~a
(36)
S
we can choose S to be a sphere of constant r then d~a = r 2 sin θdθdφ~n
Z π
c 4 2
ck 4 p 2
2π
sin3 θdθ =
k p
(37)
hPi =
16π
6π
0
Simple Radiating Systems
Figure: Electric field lines of an oscillating dipole ~p (t) = p0 cos(ωt)~ez at
selected moments of time for ω = 1. Witness the transition between
near-zone behavior and wave-zone behavior which occurs near
r = λ = 2πc/ω ≡ 1.
Simple Radiating Systems
Figure: Electric field lines of an rotating dipole
~p (t) = p0 [cos(ωt)~ez + sin(ωt)~ey ] at selected moments of time, for
ω = 1. Witness the transition between near-zone behavior and wave-zone
behavior which occurs near r = λ = 2πc/ω ≡ 1.
Simple Radiating Systems
Example II
A simple example of an electric dipole radiator is a centerfed linear
antenna whose length d is small compared to a wavelength. The antenna
is assumed to be oriented along the z-axis, with the narrow gab at the
center for purposes of excitation. The current is in the same direction in
each half of the antenna, with a value I0 at the gap and falling linearly to
zero at the ends:
2|z|
I (z)e −iωt = I0 1 −
e −iωt
(38)
d
From the continuity eqn (19) the linear
charge density ρ0 (charge per unit length)
along each arm is constant with value
ρ0 (z) = ±
2iI0
ωd
(39)
The dipole moment (21) is parallel to the z-axis with magnitude
Z d/2
iI0 d
p=
zρ0 (z)dz =
2ω
−d/2
Simple Radiating Systems
(40)
The angular distribution of the power radiated is
dP
I2
= 0 (kd)2 sin2 θ
dΩ
32πc
(41)
while the total power radiated is
P=
I02
(kd)2
12πc
(42)
We see that for a fixed input current the power radiated increases as the
square of the frequency, at least in the long-wavelength domain (kd 1).
The coefficient of I02 /2 has dimensions of a resistance and is called the
radiation resistance Rrad of the antenna.
Simple Radiating Systems
Magnetic Dipole & Electric Quadrupole Fields
Electric-dipole radiation corresponds to the leading-order approximation
of the EM field in an expansion in powers of k or v /c, where v is the
typical internal velocity of the field. In some case, however, the dipole
moment ~p either vanishes or does not depend on time, and the leading
term is actually zero. In such cases or when higher accuracy is required,
we need to compute the next term in the expansion.
We will see that at the next-to-leading-order, the wave-zone fields
depend on the dipole moment vector ~p , the magnetic moment
~ and the electric quadrupole moment tensor Qab .
vector term M
Simple Radiating Systems
Magnetic Dipole & Electric Quadrupole Fields
The next term in the expansion (11) and (15) leads to a vector potential
Z
ikr
1
~J(~x 0 ) (~n · ~x 0 ) d 3 x 0
~ x) = e
(43)
A(~
− ik
cr r
This vector potential can be written as the sum of two terms, one of
which gives a transverse magnetic induction and the other who gives a
transverse electric field.
This can be achieved by writing the integrand as the sum of a part
symmetric in ~J and ~x 0 and a part that is antisymmetric.
i
1
1 h
1 0 ~
~x × J × ~n
(~n · ~x 0 )~J =
(~n · ~x 0 )~J + (~n · ~J)~x 0 +
(44)
c
2c
2c
The 1st symmetric term will be shown to be related to the electric
quadrupole moment density.
The 2nd antisymmetric part is the so called magnetization due to the
current ~J:
~ = 1 ~x × ~J
M
(45)
2c
Simple Radiating Systems
Magnetic Dipole
Considering only the magnetization term, we have the vector potential
1
e ikr
~
~)
1−
(46)
A(~x ) = ik(~n × m
r
ikr
~ is the magnetic dipole moment
where m
Z
Z 1
3
~
~ = Md x =
~x × ~J d 3 x
m
2c
(47)
The fields can be determined by noting that the vector potential (46) is
proportional to the magnetic induction (23) for an electric dipole. This
means that the magnetic induction will be equal to the electric field for
~
the electric dipole, with substitution ~p → m
e ikr
1
ik
~ = k 2 (~n × m
~]−m
~ ] 3 − 2 e ikr
~ ) × ~n
B
+ [3~n(~n · m
(48)
r
r
r
Similarly, the electric field for a magnetic dipole source is the negative
of the magnetic field for an electric dipole:
e ikr
1
~ = −k 2 (~n × m
~)
E
1−
(49)
r
ikr
Simple Radiating Systems
• All the arguments concerning the behavior of the fields in the near
and far zones are the same as for the electric dipole source, with the
~ → B,
~ B
~ → −E
~ , ~p → m
~.
interchange E
• Similarly the radiation pattern and total power radiated are the same
for the two kinds of dipole. The only difference in the radiation fields is
the polarization.
• For an electric dipole the electric vector lies in the plane defined by ~n
and ~p , while for a magnetic dipole it is perpendicular to the plane defined
~.
by ~n and m
• The total power radiated, can be estimated (how?) by substituting
~p → m
~ in (33)
c
2
m|
(50)
PM = k 4 |~
3
Thus
2 m0
vc 2
PM
=
≈
(51)
PE
p0 c
c
2
Thus the magnetic dipole radiation power is by a factor vcc 1
smaller compared with electric dipole radiation power.
Note that: p ∼ (ρr )(r 3 ) ∼ ρr 4 while m ∼ (Jr )(r 3 ) ∼ Jr 4 ∼ ρvc r 4 .
Simple Radiating Systems
Electric Quadrupole Fields
The integral of the symmetric term in (44) can be transformed by an
integration by parts and some rearrangement (how?):
Z h
Z
i
ik
1
0 ~
0
3 0
~
~x 0 (~n · ~x 0 ) ρ(~x 0 )d 3 x 0 (52)
(~n · ~x ) J + ~n · J ~x d x = −
2c
2
~ · ~J by iωρ.
where the continuity eqn (19) has been used to replace ∇
Since the integral involves second moments of the charge density,
this symmetry part corresponds to an electric quadrupole source. The
vector potential is
Z
2 ikr
1
~ x) = − k e
~x 0 (~n · ~x 0 )ρ(~x 0 )d 3 x 0
A(~
1−
(53)
2 r
ikr
Since the complete fields are complicated to write down, we will study
fields in the radiation zone. Then it is easy to see that
~ = ik(~n × A)
~
~ = ik(~n × A)
~ × ~n
B
and E
(54)
Consequently the magnetic induction is
3 ikr Z
~ = − ik e
(~n × ~x 0 )(~n · ~x 0 )ρ(~x 0 )d 3 x 0
B
2 r
Simple Radiating Systems
(55)
Using the definition (??) for the quadrupole moment tensor
Z
Qab = (3xa xb − r 2 δab )ρ(~x )d 3 x
the integral (55) can be written (how?)
Z
1
~ n)
~n × ~x 0 (~n · ~x 0 )ρ(~x 0 )d 3 x 0 = ~n × Q(~
3
~ n) is defined as having components
The vector Q(~
X
Qa =
Qab nb
(56)
(57)
(58)
b
The magnetic induction will be written as:
3 ikr
~ = − ik e ~n × Q(~
~ n)
B
(59)
6 r
The time-averaged power radiated per unit solid angle
h
2
i
c
dP
~ n) × ~n
=
k 6 ~n × Q(~
(60)
dΩ
288π
and the direction of the radiated electric field is given by the vector inside
the absolute value.
Simple Radiating Systems
The general angular distribution is complicated. But the total power
radiated can be calculated in a straightforward way. We can write the
angular dependence as
h
2
i
~ n) × ~n = Q
~∗·Q
~ − |~n · Q|
~ 2
~n × Q(~
X
X
∗
=
A∗ab Qac nb nc −
Qab
Qcd na nb nc nd(61)
a,b,c
The necessary angular integrals
components of ~n are:
Z
nb nc dΩ =
Z
na nb nc nd dΩ =
a,b,c,d
over products of the rectangular
Z
4π
δbc ,
na nb nc dΩ = 0
3
4π
(δab δcd + δac δbd + δad δbc )
15
(62)
Then


Z h
2
i
X
X
X
X
4π 
∗
~ n) × ~n dΩ = 4π
|Qab |2 −
Qaa
Qcc + 2
|Qab |2 
~n × Q(~
3
13
a
c
a,b
a,b
(63)
Simple Radiating Systems
As we have mentioned earlier the trace of Qab is zero and thus the total
power radiated by a quadrupole source is
c 6X
P=
k
|Qab |2
(64)
360
a,b
Notice that the radiated power varies as the 6th power of the frequency
for fixed quadrupole moments, compared to the 4th power for dipole
radiation.
In orders of magnitude Qab ∼ k 6 (ρr 2 )(r 3 ) ∼ k 6 ρr 5 then
v 2
PE −Q
∼
PE −D
c
(65)
For slowly-moving distributions, the power emitted in electric-quadrupole
radiation is smaller than the power emitted in electric-dipole radiation by
a factor of order (v /c)2 1.
Simple Radiating Systems
Example : Electric Quadrupole Fields
An oscillating spheroidal distribution of charge is a simple example of a
radiating quadrupole source.
The off-diagonal elements of Qab vanish (why?).
The diagonal terms may written
1
1
Q11 = Q22 = − Q33 = − Q0
2
2
(66)
Then the angular distribution of
radiated power is
dP
ck 6
=
Q0 sin2 θ cos2 θ
dΩ
128π
(67)
The total power radiated by this
quadrupole is:
P=
c
k 6 Q0
128π
Figure: Quadrupole radiation
pattern
(68)
Simple Radiating Systems
Example : Pulsar spin-down
As an application of the magnetic-dipole
radiation we consider the oblique-rotator
model of a pulsar, a rotating neutron star that
emits pulses of EM radiation at regular intervals.
The pulsars (e.g. Crab) is observed to spin
down and if we associate the spin frequency
with the pulse frequency then we can interpret it
in terms of pulsar losing rotational energy
1
(69)
Erot = I Ω2
2
where I is the star’s moment of inertia and Ω its angular velocity. If P
denotes the pulses period, then Ω = 2π/P. If we model the neutron star
as a solid sphere then I = 52 MR 2 . The loss of rotational energy translates
into a decrease of Ω
Ėrot = I ΩΩ̇ = −(2π)2 I
Ṗ
P3
For the Crab pulsar the observed values are Ṗ ≈ 4 × 10−13 s/s and
P ≈ 0.03s.
Simple Radiating Systems
(70)
If we assume M = 1.4M and R = 12km then
the rate of loss of rotational energy is
Ėrot ≈ −7 × 1031 J/s
(71)
This is comparable to the energy required to
power the Crab nebula. The energetics of the
Crab can therefore be explained by the pulsar
losing rotational energy.
The energy carried away by the radiation will
then come to the expense of the star’s
rotational energy. For this we need the field’s
orientation to differ from star’s rotational axis
(oblique rotator model).
If the NS maintains a magnetic dipole moment
~ (t) = m0 (sin α cos Ωt~ex + sin α sin Ωt~ey + cos α~ez )
m
(72)
Because the NS is located within the near zone
1
~
~]
m · ~n)~n − m
B(t)
= 3 [3(~
R
(73)
Simple Radiating Systems
~
The field is maximum at the magnetic pole, where ~n is aligned with m
Bmax =
2m0
R3
(74)
The time-changing magnetic moment produces a magnetic-dipole
radiation that takes energy away from the star at a rate given by eqn (50)
Ėrad =
1 ¨ 2
~|
|m
3c 3
(75)
and by substitution of (72) we get
Ėrad =
1
1
(m0 Ω2 sin α)2 = ... =
(Bmax Ω2 R 3 sin α)2
3
3c
12c 3
(76)
and if we set
Ėrad = Ėrot
we get
Bmax sin α ≈ 5 × 1012 Gauss
which is large but not unreasonable. A main sequence star typically
supports a magnetic field of 103 G if the field is frozen in the star during
the collapse the magnetic flux 4πR 2 B is conserved. Thus because R
decrease by a factor 105 then B increases by a factor 1010 .
Simple Radiating Systems