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Simple Radiating Systems

May 11, 20101

1J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Chapter 9Simple Radiating Systems

Fields and Radiation of a Localized Oscillating Source

For a system of charges and currents varying in time we can make aFourier analysis of the time dependence and handle each Fouriercomponent separately. The potentials, and fields from a localized systemof 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 suchexpressions while the EM potentials and fields assumed to have the sametime dependence. In the Lorentz gauge the solution (provided no

boundary surfaces are present) for the vector potential ~A(~x , t) is

~A(~x , t) =1

c

∫d3x ′

∫ ~J(~x ′, t ′)

|~x − ~x ′|δ

(t ′ +

|~x − ~x ′|c

− t

)dt ′ (2)

Using the sinusoidal dependence (1) we get (k = ω/c)

~A(~x) =1

c

∫d3x ′

~J(~x ′)

|~x − ~x ′|e ik|~x−~x′| (3)


Given a current distribution ~J(~x ′) the fields can be determined bycalculating the integral (3). The magnetic induction will be given by

~B = ~∇× ~A (4)

while outside the source the electric field is

~E =i

k~∇× ~B (5)

If the sources dimensions are of order d and the wavelength isλ = 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 ∼ λI The far (radiation) zone : d λ r

The fields have very different properties in the different zones.• In the near zone they behave as static and depend on the propertiesof the source.

• In the far zone the fields are transverse to the radius vector and fall

as 1/r (typical for radiation fields).


The Near Zone

The general solution is:

~A(~x) =1

c

∫d3x ′

~J(~x ′)

|~x − ~x ′|e ik|~x−~x′| (6)

For the near zone where r λ (or kr 1) the exponential in (6) can bereplaced by unity.Then by using:

1

|~x − ~x ′|= 4π

∞∑`=0

l∑m=−`

1

2`+ 1

r `<r `+1>

Y ∗`m(θ′, φ′)Y`m(θ, φ) (7)

The vector potential gets the form

limkr→0

~A(~x) =1

c

∑`,m

4π

2`+ 1

Y`m(θ, φ)

r `+1

∫~J(~x ′)r ′`Y ∗`m(θ′, φ′)d3x ′ (8)

This shows that the near fields are quasi-stationary, oscillating

harmonically as e−iωt , but otherwise static in character.


The Far Zone

In the far zone (kr 1) the exponential in (3) oscillates rapidly anddetermines the behavior of the vector potential.In this region is sufficient to approximate

|~x − ~x ′| ≈ r − ~n · ~x ′ (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 bereplaced by r and the vector potential is

limkr→∞

~A(~x) =e ikr

cr

∫~J(~x ′)e−ik~n·~x′

d3x ′ (10)

I.e. in the far zone the vector potential behaves as an outgoing sphericalwave.It is easy to show that the fields calculated from (4) and (5) aretransverse to the radius vector (how?) and fall like 1/r . Thus theycorrespond to radiation fields.


The Far Zone

If the source size is small compared to a wavelength one can expand inpowers of k:

limkr→∞

~A(~x) =e ikr

cr

∑n

(−ik)n

n!

∫~J(~x ′)(~n · ~x ′)nd3x ′ (11)

The magnitude of the n-th term is given by

1

n!

∫~J(~x ′)(k~n · ~x ′)nd3x ′ (12)

Since |~x ′| ≈ d and kd 1 by assumption, the successive terms in the

expansion of ~A fall off rapidly with n.

Consequently the radiation emitted from the source will come mainly

from the first nonvanishing term in the expansion (11).


The Intermediate Zone

In the intermediate zone neither kr 1 or kr 1 can be used; all powerof kr must be retained. For points outside the source, eqn (3) becomes

~A(~x) =4πik

c

∑`,m

h(1)` (kr)Ym`(θ, φ)

∫~J(~x ′)j`(kr

′)Y ∗`m(θ′, φ′)d3x ′ (13)

If the source dimensions are small compared to a wavelength, j`(kr′) can

be approximate as follows: 2

j`(x)→ x l

(2`+ 1)!!

(1− x3

2(2`+ 3)+ . . .

)(14)

Then the vector potential is of the form (8) but with the replacement

1

r `+1→ e ikr

r `+1

[1 + a1(ikr) + a2(ikr)2 + · · · al(ikr)l

](15)

The coefficients ai come from explicit expressions for the spherical

Hankel functions.2[(2` + 1)!! = (2` + 1)(2` − 1)(2` − 3) · · · 5 · 3 · 1]


Electric Monopole Fields

For a source varying with the time the analogue of (2) for a scalarpotential is

Φ(~x , t) =

∫d3x ′

∫dt ′

ρ(~x ′, t ′)

|~x − ~x ′|δ

(t ′ +

|~x − ~x ′|c

− t

)(16)

The electric monopole contribution is obtained by replacing in theintegral |~x − ~x ′| → |~x | ≡ r . The result is

Φmonopole(~x , t) =q(t ′ = t − r/c)

r

where q(t) is the total charge of the source. Since charge is conservedand a localized source has not charge flowing into or away from it, thetotal charge q is independent of time.? Thus the electric monopole part of the potential (and fields) oflocalized source is of necessity static.? The fields with harmonic time dependence e−iωt with ω 6= 0 haveno monopole terms.


Electric Dipole Fields and Radiation

If only the first term in (11) is kept, the potential is

~A(~x) =e ikr

c r

∫~J(~x ′)d3x ′ (17)

Examination of (13) and (15) shows that (17) is the ` = 0 part of theseries and that it is valid everywhere outside the source, not just in thefar zone. Integrating by part we get∫

~Jd3x ′ = −∫~x ′(~∇′ · ~J)d3x ′ = −iω

∫~x ′ρ(~x ′)d3x ′ (18)

since from the continuity equation

~∇ · ~J = iωρ (19)

Thus the vector potenial is:

~A(~x) = −ike ikr

r~p (20)

~p(~x) =

∫~x ′ρ(~x ′)d3x ′ (21)

is the electric dipole moment, as defined in electrostaticsSimple Radiating Systems

The electric dipole fields from (4) and (5) are (how?)

~B = k2(~n × ~p)

(1− 1

ikr

)e ikr

r(22)

~E = k2(~n × ~p)× ~n e ikr

r+ [3~n(~n · ~p)− ~p]

(1

r3− ik

r2

)e ikr(23)

Note that the magnetic induction is transverse to the radius vectorat all distances, but the electric field has components parallel andperpendicular to ~n.In the radiation zone (kr 1) the fields take the limiting form

~B = k2(~n × ~p)e ikr

r, ~E = ~B × ~n (24)

In the near zone (kr 1) the fields approach

~B = ik(~n × ~p)1

r2, ~E = [3~n(~n · ~p)− ~p]

1

r3(25)


? The electric field apart from its oscillation in time is just the staticelectric 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 innature.? 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 (26)

(~n × ~p)× ~n = (~n · ~n)~p − (~n · ~p)~n = ~p − (~p · ~n)~n (27)

? If we let ~p point in the z-direction, then ~p = p~ez also

~n = sin θ cosφ~ex + sin θ sinφ~ey + cos θ~ez

~p · ~n = p cos θ

~p − (~p · ~n)~n = −p sin θ (cos θ cosφ~ex + cos θ sinφ~ey − sin θ~z) = −p sin θ~eθ

~n × ~p = −p sin θ (− sinφ~ez + cosφ~ey ) = −p sin θ~eφ

and we get (prove it)

~B = −k2p sin θe ikr

r~eφ and ~E = −k2p sin θ

e ikr

r~eθ (28)


Electric Dipole : Power Radiated

The Poynting vector is

~S =c

8π~E×~B =

c

8π

(~B × ~n

)×~B =

c

8π

[(~B · ~B

)~n −

(~B · ~n

)~B]

=c

8π|~B|2~n

(29)

the blue term is vanishing because ~B · ~n = 0.The time-averaged power radiated per unit solid angle by the oscillatingdipole moment ~p is:

dP

dΩ=

c

8π<[r2~n ·

(~E × ~B∗

)](30)

where ~E and ~B are given by (24). Thus we find

dP

dΩ=

c

8πk4 |(~n × ~p)× ~n|2 (31)

If all the components of ~p have the same phase, the angular distributionis a typical dipole pattern (where θ is measured from the direction of ~p )

dP

dΩ=

c

8πk4 |~p|2 sin2 θ (32)


Electric Dipole : Power Radiated

The total power radiated, independent of the relative phases of thecomponents of ~p, is

P =1

3ck4 |~p|2 (33)

Figure: Power distribution.


Example I

For ~p = p~ez we can substitute (28) into (29) and we get (how?)

~S =c

8π2k4p2 sin2 θ

r2e2ikr~n (34)

Since the energy flux oscillates in time after averaging over a completewave cycle we get (〈cos2(kr)〉 = 1/2)

〈~S〉 =ck4p2

16π2

sin2 θ

r2~n (35)

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 closedsurface S . The averaged power is then

〈P〉 =

∮S

〈S〉 · d~a (36)

we can choose S to be a sphere of constant r then d~a = r2 sin θdθdφ~n

〈P〉 =ck4p2

16π2π

∫ π

0

sin3 θdθ =c

6πk4p2 (37)


Figure: Electric field lines of an oscillating dipole ~p(t) = p0 cos(ωt)~ez atselected moments of time for ω = 1. Witness the transition betweennear-zone behavior and wave-zone behavior which occurs nearr = λ = 2πc/ω ≡ 1.


Figure: Electric field lines of an rotating dipole~p(t) = p0 [cos(ωt)~ez + sin(ωt)~ey ] at selected moments of time, forω = 1. Witness the transition between near-zone behavior and wave-zonebehavior which occurs near r = λ = 2πc/ω ≡ 1.


Example II

A simple example of an electric dipole radiator is a centerfed linearantenna whose length d is small compared to a wavelength. The antennais assumed to be oriented along the z-axis, with the narrow gab at thecenter for purposes of excitation. The current is in the same direction ineach half of the antenna, with a value I0 at the gap and falling linearly tozero at the ends:

I (z)e−iωt = I0

(1− 2|z |

d

)e−iωt (38)

From the continuity eqn (19) the linearcharge density ρ′ (charge per unit length)along each arm is constant with value

ρ′(z) = ±2iI0ωd

(39)

The dipole moment (21) is parallel to the z-axis with magnitude

p =

∫ d/2

−d/2

zρ′(z)dz =iI0d

2ω(40)


The angular distribution of the power radiated is

dP

dΩ=

I 20

32πc(kd)2 sin2 θ (41)

while the total power radiated is

P =I 20

12πc(kd)2 (42)

We see that for a fixed input current the power radiated increases as thesquare of the frequency, at least in the long-wavelength domain (kd 1).

The coefficient of I 20 /2 has dimensions of a resistance and is called the

radiation resistance Rrad of the antenna.


Magnetic Dipole & Electric Quadrupole Fields

Electric-dipole radiation corresponds to the leading-order approximationof the EM field in an expansion in powers of k or v/c , where v is thetypical internal velocity of the field. In some case, however, the dipolemoment ~p either vanishes or does not depend on time, and the leadingterm 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 fieldsdepend on the dipole moment vector ~p, the magnetic momentvector term ~M and the electric quadrupole moment tensor Qab.


Magnetic Dipole & Electric Quadrupole Fields

The next term in the expansion (11) and (15) leads to a vector potential

~A(~x) =e ikr

cr

(1

r− ik

)∫~J(~x ′) (~n · ~x ′) d3x ′ (43)

This vector potential can be written as the sum of two terms, one ofwhich gives a transverse magnetic induction and the other who gives atransverse electric field.This can be achieved by writing the integrand as the sum of a partsymmetric in ~J and ~x ′ and a part that is antisymmetric.

1

c(~n · ~x ′)~J =

1

2c

[(~n · ~x ′)~J + (~n · ~J)~x ′

]+

1

2c

(~x ′ × ~J

)× ~n (44)

The 1st symmetric term will be shown to be related to the electricquadrupole moment density.The 2nd antisymmetric part is the so called magnetization due to thecurrent ~J:

~M =1

2c

(~x × ~J

)(45)


Magnetic Dipole

Considering only the magnetization term, we have the vector potential

~A(~x) = ik(~n × ~m)e ikr

r

(1− 1

ikr

)(46)

where ~m is the magnetic dipole moment

~m =

∫~Md3x =

1

2c

∫ (~x × ~J

)d3x (47)

The fields can be determined by noting that the vector potential (46) isproportional to the magnetic induction (23) for an electric dipole. Thismeans that the magnetic induction will be equal to the electric field forthe electric dipole, with substitution ~p → ~m

~B = k2(~n × ~m)× ~ne ikr

r+ [3~n(~n · ~m]− ~m]

(1

r3− ik

r2

)e ikr (48)

Similarly, the electric field for a magnetic dipole source is the negativeof the magnetic field for an electric dipole:

~E = −k2(~n × ~m)e ikr

r

(1− 1

ikr

)(49)


• All the arguments concerning the behavior of the fields in the nearand far zones are the same as for the electric dipole source, with theinterchange ~E → ~B, ~B → −~E , ~p → ~m.

• Similarly the radiation pattern and total power radiated are the samefor the two kinds of dipole. The only difference in the radiation fields isthe polarization.

• For an electric dipole the electric vector lies in the plane defined by ~nand ~p, while for a magnetic dipole it is perpendicular to the plane definedby ~n and ~m.

• The total power radiated, can be estimated (how?) by substituting~p → ~m in (33)

PM =c

3k4 |~m|2 (50)

ThusPM

PE=

(m0

p0c

)2

≈(vc

c

)2

(51)

Thus the magnetic dipole radiation power is by a factor(

vc

c

)2 1smaller compared with electric dipole radiation power.

Note that: p ∼ (ρr)(r3) ∼ ρr4 while m ∼ (Jr)(r3) ∼ Jr4 ∼ ρvc r4.


Electric Quadrupole Fields

The integral of the symmetric term in (44) can be transformed by anintegration by parts and some rearrangement (how?):

1

2c

∫ [(~n · ~x ′)~J +

(~n · ~J

)~x ′]d3x ′ = − ik

2

∫~x ′ (~n · ~x ′) ρ(~x ′)d3x ′ (52)

where the continuity eqn (19) has been used to replace ~∇ · ~J by iωρ.Since the integral involves second moments of the charge density,this symmetry part corresponds to an electric quadrupole source. Thevector potential is

~A(~x) = −k2

2

e ikr

r

(1− 1

ikr

)∫~x ′(~n · ~x ′)ρ(~x ′)d3x ′ (53)

Since the complete fields are complicated to write down, we will studyfields in the radiation zone. Then it is easy to see that

~B = ik(~n × ~A) and ~E = ik(~n × ~A)× ~n (54)

Consequently the magnetic induction is

~B = − ik3

2

e ikr

r

∫(~n × ~x ′)(~n · ~x ′)ρ(~x ′)d3x ′ (55)


Using the definition (??) for the quadrupole moment tensor

Qab =

∫(3xaxb − r2δab)ρ(~x)d3x (56)

the integral (55) can be written (how?)

~n ×∫~x ′(~n · ~x ′)ρ(~x ′)d3x ′ =

1

3~n × ~Q(~n) (57)

The vector ~Q(~n) is defined as having components

Qa =∑

b

Qabnb (58)

The magnetic induction will be written as:

~B = − ik3

6

e ikr

r~n × ~Q(~n) (59)

The time-averaged power radiated per unit solid angle

dP

dΩ=

c

288πk6∣∣∣[~n × ~Q(~n)

]× ~n∣∣∣2 (60)

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 powerradiated can be calculated in a straightforward way. We can write theangular dependence as∣∣∣[~n × ~Q(~n)

]× ~n∣∣∣2 = ~Q∗ · ~Q − |~n · ~Q|2

=∑a,b,c

A∗abQacnbnc −∑

a,b,c,d

Q∗abQcdnanbncnd(61)

The necessary angular integrals over products of the rectangularcomponents of ~n are:∫

nbncdΩ =4π

3δbc ,

∫nanbncdΩ = 0 (62)∫

nanbncnddΩ =4π

15(δabδcd + δacδbd + δadδbc)

Then∫ ∣∣∣[~n × ~Q(~n)]× ~n∣∣∣2 dΩ =

4π

3

∑a,b

|Qab|2−4π

13

∑a

Q∗aa∑

c

Qcc + 2∑a,b

|Qab|2

(63)


As we have mentioned earlier the trace of Qab is zero and thus the totalpower radiated by a quadrupole source is

P =c

360k6∑a,b

|Qab|2 (64)

Notice that the radiated power varies as the 6th power of the frequencyfor fixed quadrupole moments, compared to the 4th power for dipoleradiation.

In orders of magnitude Qab ∼ k6(ρr2)(r3) ∼ k6ρr5 then

PE−Q

PE−D∼(v

c

)2

(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.


Example : Electric Quadrupole Fields

An oscillating spheroidal distribution of charge is a simple example of aradiating quadrupole source.The off-diagonal elements of Qab vanish (why?).

The diagonal terms may written

Q11 = Q22 = −1

2Q33 = −1

2Q0 (66)

Then the angular distribution ofradiated power is

dP

dΩ=

ck6

128πQ0 sin2 θ cos2 θ (67)

The total power radiated by thisquadrupole is:

P =c

128πk6Q0 (68)

Figure: Quadrupole radiationpattern


Example : Pulsar spin-down

As an application of the magnetic-dipoleradiation we consider the oblique-rotatormodel of a pulsar, a rotating neutron star thatemits pulses of EM radiation at regular intervals.The pulsars (e.g. Crab) is observed to spindown and if we associate the spin frequencywith the pulse frequency then we can interpret itin terms of pulsar losing rotational energy

Erot =1

2IΩ2 (69)

where I is the star’s moment of inertia and Ω its angular velocity. If Pdenotes the pulses period, then Ω = 2π/P. If we model the neutron staras a solid sphere then I = 2

5MR2. The loss of rotational energy translatesinto a decrease of Ω

Erot = IΩΩ = −(2π)2IP

P3(70)

For the Crab pulsar the observed values are P ≈ 4× 10−13s/s andP ≈ 0.03s.


If we assume M = 1.4M and R = 12km thenthe rate of loss of rotational energy is

Erot ≈ −7× 1031J/s (71)

This is comparable to the energy required topower the Crab nebula. The energetics of theCrab can therefore be explained by the pulsarlosing rotational energy.The energy carried away by the radiation willthen come to the expense of the star’srotational energy. For this we need the field’sorientation to differ from star’s rotational axis(oblique rotator model).If the NS maintains a magnetic dipole moment

~m(t) = m0 (sinα cos Ωt~ex + sinα sin Ωt~ey + cosα~ez)(72)

Because the NS is located within the near zone

~B(t) =1

R3[3(~m · ~n)~n − ~m] (73)


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-dipoleradiation that takes energy away from the star at a rate given by eqn (50)

Erad =1

3c3| ~m|2 (75)

and by substitution of (72) we get

Erad =1

3c3(m0Ω2 sinα)2 = ... =

1

12c3(BmaxΩ2R3 sinα)2 (76)

and if we setErad = Erot

we getBmax sinα ≈ 5× 1012Gauss

which is large but not unreasonable. A main sequence star typicallysupports a magnetic field of 103 G if the field is frozen in the star duringthe collapse the magnetic flux 4πR2B is conserved. Thus because Rdecrease by a factor 105 then B increases by a factor 1010.


Simple Radiating Systems - uni-tuebingen.dekokkotas/Teaching/Field...1J.D.Jackson, "Classical...

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Transcript of Simple Radiating Systems - uni-tuebingen.dekokkotas/Teaching/Field...1J.D.Jackson, "Classical...