Transcript Slide 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.
i t
J(r,t )  J(r )  e
i t
(r,t )  (r )  e
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:
i t  (r ' ) exp(ik r  r ' )
(r, t )  e 
dV '
r

r
'
V
i t J (r ' ) exp(ik r  r ' )
A(r, t )  e
dV
V
c r  r'
With the usual notation for the wave number k=ω/c
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The magnetic field is given by:
B   A
and, in the region where there are neither charges nor currents, the
4th Maxwell law reads:
  (  A)    B 
1 E i

E
c t
c
E
From this equation we obtain:
1
  (  A)
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
1 r  r' , while in this case (quantities
varying in time) it is :
exp(ik r  r' )
r  r'
2.
In electrostatics we had to expand a scalar. Now, it is a vector.
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exp(ik r  r' )
We will now also restrict somewhat the system under study: the system
size must be much smaller than the oscillation wavelength.

2
 d
k
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:
1.
2.
3.
d  r  
d  r  
d    r
is the proximity zone.
is the intermediate zone
is the remote (or radiation) region.
Before diving into the calculation, a question: what does it mean the
request that
  d ?
• 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.
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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:
  J(r)  i(r)  0
Oscillations of charge and current are out of phase by 90o.
An example:
Linear antenna fed at the center
The current is:
I  I ( z)  e
i t
(
 I0 1
2z
d
)e 
i t
And the linear charge
density:
2iI 0 i t
i t
   ( z)  e

d
e
Where the + (-) sign is applies to
positive (negative) z.
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We return now to looking for the solutions of our “general” case.
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(ik ( r  r' )
With the constant value of “1”. This is due to the fact that the
distance r  r' 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 1 r  r' with 1/r and take it out from the
integral over dV’. And, it is also possible to replace
exp(ik ( r  r' )
If we define
with
n
r
r
r
exp[ ik (r   r ' )]
r
We then have the ideal function to
approximate with a Taylor’s series
1
exp( ik ( r  r ' )  e ikr  eik (r 'n)  e ikr (1  ikn  r ' k 2 (r 'n) 2 )  .......
2
And A(r) becomes a series expansion:
A  A1  A2  A3  .....
in which each term Ai corresponds to
the power ( n·r’)(i-1). Since kr’<<1
(remember that d/λ<<1!) the series converges rapidly.
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Let us now take A1 (corresponding to the constant value 1 in the
expansion)
 i ( kr   t ) V
We need to compute this integral. It can
A1 (r, t ) 
e
 J(r' )dV '
cr
be written as a parts integral.
V
This passage is a parts integral, see next page
 J (r' )dV '    r' ('J( x' ))dV '   r'
V
V'
A1 (r, t ) 
V'
ike
 i ( kr   t )
r
d (r ' )
dV '  i  r ' (r' )dV '  iP0
dt
V'
 r' (r' )dV ' 
ike
 i ( kr   t )
P0
r
V'
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.
P0   r ' (r ' )dV '
A1 (r, t ) 
P(t )  P0  ei t
V'
ik e
 i ( kr   t )
cr
P0
The fields are obtained via the curls of the potential: the unitary vector
defined in the previous page, appears, it is minus the direction of
observation from point P.
 i ( kr   t )
e
i
2
B  k (n  P0 )
(1  ) 
r
kr
 i ( kr   t )
e
 k 2 (n  P )
0
Erad  k
2
r
0
This term has a 1/r2
dependence on r: it is
the velocity field, not
interesting to us here.
This is the field of radiation
 i ( kr   t )
e
(n  P )  n
n,
This is only the field of
radiation part of the
r
electric field
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Calculation of  J (r ' ) dV '
V
That requires the integration of a vector, i.e. three scalar
integrals. Let us do the calculation for Jx only.
 j (r' )dV '  J(r' )  idV '  J(r' )  ' x' dV ' 
x
V
V
V
  ['( x' J )  x' 'J ]   x'J  ndA   x''JdV 
V
S
V
  x' 'JdV '
V
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.
We can extend to the three components of the vector potential
what we have shown only for Jx , and then obtain:
 J(r' )dV '    r' ('J)dV '
V
V'
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•From the formulas of the previous page it is evident that
the fields:
E  Bn
• 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:
Erad  k
2
e i(kr  t ) [P
0
r
 (n  P0 )n] 
 i ( kr  t )
e
 k2
P
r

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:
qa ' 
E(r, t )   2
c r'
-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.
+q
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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:
a'  
2
q
P'
where the primed quantities indicate that they are retarded. And,
by looking at the figure of the previous page, we easily deduce that:
P  qz  Aq sin t
d 2z
a  2   A 2 sin t
dt
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.
Dipole radiation
Let us take a look at the formulas which give the electric dipole
radiation fields. They determine the directions and sizes of the fields.
B Rad  k
2
 i ( kr   t )
e
(n  P)
E Rad  B Rad  n  P
r
e i (kr   t )
r
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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 signalsis 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.
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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.
2 q 2 a 2 2  4 P 2 (t )
W1 


3 c3
3 c3
2
2
4
4

P
ck
P
2 4
2
0
0
W1 
  P0 cos2 (t ) 

3
3
3 c
3c
3
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:
P0  k (
d d
)   q d  I0d k
2 2
2 2i
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 !!!):
( I 0 kd ) 2
 W 
12c
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
A2
 (ikr  t )
e
(r, t ) 
ik
r
 j(r' )  (n r' )dV '  A
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2m
 A 2q
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