Advanced EM - Master in Physics 2011-2012 1 Multipole expansion of the radiation fields Consider an electrical system –charges and currents- varying in

Advanced EM - Master in Physics 2011-2012

1

Multipole expansion of the radiation fields

Consider an electrical system –charges and currents- varying in time. And, for ease of treatment, with both the charge and the current distributions with a sinusoidal (harmonic) time dependence: oscillating charges.

Such limitations in our scope do not entail any loss of generality: any arbitrary time dependence – which satisfies the condition of charge conservation, of course - can be expanded into a Fourier series. Then the problem is solved (in principle) for any frequency; and the solutions are convoluted again with the original Fourier spectrum.

ti(,t(

ti(,t(

ee

))

))

rr

rJrJ

Where, as usual, the physical values of the measurable quantities are the real parts of the complex variables.

In such situation the scalar and vector potentials, as well as the E and B fields, will also have a harmonic dependence on time.

We have already seen the general solution of the retarded

potentials as a function of their sources, i.e. of ρ and J . In this case of a harmonic dependence on time they take the form:

dVc

iktit

dViktit

V

V

e

e

'

)'exp()'(),(

''

)'exp()'(),(

rr

rrrJrA

rr

rrrr

With the usual notation for the wave number k=ω/c


2

The magnetic field is given by:

and, in the region where there are neither charges nor currents, the 4th Maxwell law reads:

AB

EE

BAc

i

tc

1

)( From this equation we obtain:

)(1

AE ik

• It is not necessary to calculate the scalar potential. And

• It is not necessary to know the charge density ρ.

The problem of solving the previous 3 equations (for A, B and

E) given a current distribution J(r’)·e(iωt) is mathematically analogous to the problem of electrostatics, for which a general solution has already been found in the multipole expansion.

In these lectures we have not insisted on the most general solution (it can be found in the book by Jackson), but limited ourselves to the case of systems with symmetry for rotations around an axis.

The are two differences wrt the electrostatic case:

1. In electrostatics the function which had to be expanded in multipoles was , while in this case (quantities varying in time) it is :

'

)'exp(

rr

rr

ik

'1 rr

2. In electrostatics we had to expand a scalar. Now, it is a vector.

)'exp( rr ik


3

We will now also restrict somewhat the system under study: the system size must be much smaller than the oscillation wavelength.

dk

2

where λ is the wavelength of the emitted radiation and d is the “size” of the charge system.

We can then divide the space free of charges (“outside of the charge region”) in three zones:

rd

rd

rd

1. is the proximity zone.

2. is the intermediate zone

3. is the remote (or radiation) region.

Before diving into the calculation, a question: what does it mean the

request that ?

• Well, even if this condition is not satisfied, it is always possible (in principle) to subdivide the charge region into smaller volumes where the condition is satisfied.

• A very common application of this treatment is in the field of antennas, especially of medium and long waves.

•A natural application is the emission of light from atoms and molecules (as long as it can be treated with classical theory). Visible light has wavelengths around 500 nm = 5·10-5 cm, the typical size of atoms is around 5·10-8 cm.

d


4

A last remark: the conditions that both currents and charges oscillate at the same frequency does not imply that they do so with the same phase, which could raise problems with charge conservation. There can be of course a phase factor between the two. And, as a matter of fact, the charge conservation condition in this case is:

0)()( rrJ iOscillations of charge and current are out of phase

by .o90

An example:

Linear antenna fed at the center

And the linear charge density:

The current is:titi ee

d

zIzII )( 2

1)( 0

titi eed

iIz

02

)(

Where the + (-) sign is applies to positive (negative) z.


5

Solutions of the equation for the vector potential A(r), once the current density are known:

•In the nearest region, the fields are quasi-static, in the sense that we can approximate the exponential

)'(exp( rr ik

With the constant value of “1”. This is due to the fact that the distance is in that region much smaller than the wavelength.

The potentials are the static potentials ( no measurable retardation effect), slowly varying in time.

• In the “radiation region” the exponential above oscillates rapidly, and determines the shape of the vector potential.

Moreover, one can approximate the factor with 1/r and

take it out from the integral over dV’. And, it is also possible to replace

'rr

'1 rr

)'(exp( rr ik with )]'(exp[ rrr

rik

If we define r

rn We then have the ideal function

to approximate with a Taylor’s series

.......)22 )'(

2

1'1()'(exp( )'(

nrrnrr nr kikik ikrikikr eee

And A(r) becomes a series expansion:

.....321 AAAA in which each term Ai corresponds

to the power (( nn·r’·r’))(i-1)(i-1).. Since Since kr’<<kr’<<1 1 (remember that d/λ<<1!) the series converges rapidly.

We return now to looking for the solutions of our “general” case.


6

Let us now take A1 (corresponding to the constant value 1 in

the expansion)

')'(),()(

1 dVcr

tAV

V

tkrie

rJr

0

0

PrrrA

Prrr

rJrr

r

ikdV

r

ikt

idVidVdt

ddVxdVJ

tkritkri eeV

VVVV

)()(

')'('),(

')'('')'(

'')'(''')'(

'

1

'''

)(

We need to compute this integral. It can be written as a parts integral.

This passage is a parts integral, see next page

This integral is an old friend: it is the momentum of ELECTRIC DIPOLE of the charge density distribution. Note how it has become a constant, since it does not depend any longer on r or on t.

The fields are obtained via the curls of the potential: the unitary vector

n, defined in the previous page, appears, it is minus the direction of observation from point P.

')'(''

dVrrP0 V

tiet 0PP )( 0Prcr

ekitA

tkri )(

),(1

This term has a 1/r2

dependence on r: it is the velocity field, not interesting to us here.

This is the field of radiationr

k

kr

i

rk

tkri

tkri

e

e

)(

)(

)(

1)(

2

2 )(

0

0

Pn

PnB

rk

tkrie )(

)(2

nPnE 0rad

This is only the field of radiation part of the electric field


7

Calculation of

')'( dVV

rJ

That requires the integration of a vector, i.e. three scalar

integrals. Let us do the calculation for Jx only.

V

VSV

VVV

x

dVx

dVxdAxxx

dVxdVdVj

'''

''']'')'('[

''')'(')'(')'(

J

JnJJJ

rJirJr

We can extend to the three components of the vector

potential what we have shown only for Jx , and then obtain:

'''')'('

)( dVdVVV JrrJ

In the last passage we have assigned to the surface integral the value “zero”. This is because the surface encompasses all the volume of charge and current and, of course, outside that volume the current distribution is null.


8

•From the formulas of the previous page it is evident that the fields:

• Are orthogonal to each other

• Are orthogonal to the direction of propagation

• Have a 1/r dependence on distance

• |E|= |B| We write again the formula for the electric field:

nBE

P

nPnPE 00rad

rk

rk

tkri

tkri

e

e

)(

)(

2

2 ])([

We do recognize here something very similar to the radiation electric field generated by a charge at rest that we had treated time ago, in the derivation of the Larmor formula:

-q

+q

An OSCILLATING DIPOLE can be seen in its simplest mode as a single charge oscillating along an axis (z-axis here) . Another charge –q would reset the total charge of the system, and turn it into a better looking dipole. It has, of course, no influence on the radiation: the dipole field is generated by the oscillating charge, not by the fixed one.

'

'),(

2rc

qt

arE


9

From the comparison of the two formulas for the dipole radiation of an electric field shown in the previous page we recognize that the radiation emitted by an oscillating dipole is the same as that generated by a charge oscillating with an acceleration equal to:

''2

Paq

where the primed quantities indicate that they are retarded. And, by looking at the figure of the previous page, we easily deduce that:

tAdt

zda

tAqqzP

sin

sin

22

2

We will remark explicitly that in the exact formulas for the fields also terms in r-2, r-3 …... are present. They belong of course to the fields of velocity. We shall not discuss them here any further.

r

rk

tkri

tkri

e

e

)(

)(

)(2

PnBE

PnB

RadRad

Rad

Dipole radiationLet us take a look at the formulas which give the electric dipole radiation fields. They determine the directions and sizes of the fields.


10

What about the direction of the fields? Let us resort to the image of the dipole oscillating up and down in the Earth center: up and down means on the line connecting the two poles. And let us examine how the fields are directed along any spherical surface centered on the dipole. Of course they lay on the sphere surface, since they are perpendicular to the line of sight, which is also the sphere radius.

The system in which these directions are along the coordinate axes is that of the spherical coordinates [r,θ,φ] which, however, is not easy to visualize – until we realize that it is the first system we have learnt – in primary school!

Well then, the radiation of ELECTRIC PIPOLE has the electric field aligned with the meridians, and the magnetic field is aligned with the parallels. The radiation of MAGNETIC DIPOLE

(we shall talk about it later) instead has the B field along the meridians, the electric field along the parallels.

This picture shows the electric field as a function of

θ. Note that its amplitude –

and the amplitude of B as well, of course – is proportional to sinθ. The irradiated power – of the utmost importance in a world full of electromagnetic signals- is proportional to E2. Given that the electric dipole term is the first in a series expansion, it is obviously the term which will allow the largest power emitted

per power spent in the emitting antenna.


11

The emitted power can be easily computed with the Poynting’s vector or, simpler, with the Larmor formula, using the equivalence formula between charge acceleration and oscillating dipole moment given two pages ago.

33)(cos

3

2

)(

3

2

3

2

20

4

3

20

422

03

4

1

3

24

3

22

1

Pck

c

PtP

cW

c

tP

c

aqW

Where W is the power irradiated averaged over a full period

τ=2π/ω.

If we want to calculate directly the dipole momentum of the antenna shown in page 4:

kkP0

i

dIdq

dd

22220)(

Where the “i” factor in the denominator indicates that the oscillation of the dipole is out of phase by 90º wrt the current in the antenna. The emitted power is –as a function of the current peak and of the physical dimensions of the radiating system (and for k·d<<1 !!!):

c

kdIW

12

)( 20

We have so far only dealt with the first term in the series

expansion of the electric field. The next term, A2, will be the sum

of two contributions, the magnetic dipole and the electric quadrupole

2q2m2 AArnrjrA

')'()'(),()(

dVikr

ttikre

Documents

Advanced EM - Master in Physics 2011-2012 1 Multipole expansion of the radiation fields Consider an electrical system –charges and currents- varying in