magnetic dipole.

Multipole expansion of radiation fields: the second term
Let us now investigate the next term in the series expansion.
 i ( kr  t )
 j(r' )(n  r' )dV '
With a long calculation (see Schwartz, and shown in next page) it is
found that A2 can be written as the sum of two terms:
A2 r, t   A2m  A2q
called, respectively
Potential of
( A2m )
( A2q )
The “magnetic dipole” term of the potential turns out to be:
A 2m
 i ( kr  t )
 ik e i ( kr  t )
n   r ' j'dV '  ik
n  μ0
where the magnetic moment of the current distribution is defined as:
μ0 
r' j'dV '
2c V
we recognize here the
term in the expansion
r’ of the second
Once A2m is known (i.e. once the magnetic moment is known) the
fields are computed by taking its curl and, of its curl, considering
only the terms in 1/r, which are the radiation field. Being it the
field of a dipole, one can expect an angular dependence of the fields
similar to that of the electric dipole |S| ~ (sinθ)2
Advanced EM - Master
in Physics 2011-2012
To see how the first term (the magnetic quadrupole) comes about, we
use the vector formula:
a  (b  c)  b(a  c)  c(a  b)
which we turn around:
b(a  c)  a  (b  c)  c(a  b)
And apply to the previous formula:
 i ( kr  t )
 j(r' )(n  r' )dV '
We write the function to be integrated as
j(n  r ' )  [ j(n  r ' )  r ' ( j  n)]  [ j(n  r ' )  r ' ( j  n)] 
 [n  ( j  r ' )]  [ j(n  r ' )  r ' ( j  n)]
Now the formula for A2 can be written as the sum of two
terms, of which the first is, as was written in the previous page,
the magnetic dipole and the second is the electric quadrupole.
A 2m
 i ( kr  t )
 ik
A 2e  ik
 i ( kr  t )
r ' j(r ' )
V 2 dV '
[ j(n  r ' )  r ' ( j  n)]
dV '
Note that we have taken the vector n out of the integral since it
does not vary with r’ (in first approximation), and inverted the
order of r’ and j’ in the integral which caused the change of sign in
the formulas.
Advanced EM - Master
in Physics 2011-2012
What is found – in much a similar way to how we found the fields for
the electric dipole – is:
B 2m
E 2m
 μ   e i (kr  t )
 n B 2m
Which is very similar , but not equal to what we found for the electric
dipole. In this case it is the magnetic field which is equal to the dipole
component orthogonal to the observation direction (i.e. in the plane
containing the line of sight and the dipole).
A loop of radius “d” with a current
i t
I  I0  e
has a magnetic moment
I 0d 2
0 
The average irradiated power is computed as for the electric dipole.
One finds:
 W2 m
I 0 (kd ) 4
To be
compared with
( I 0 kd ) 2
 W1 
At equal size d of the systems, the power irradiated by a magnetic
dipole is a factor (kd )2 smaller that that of the electric dipole. It was
to be expected, since the magnetic dipole comes from the second term
in the series expansion.
Advanced EM - Master
in Physics 2011-2012
The electric quadrupole
The second term of A2, called A2q, is the electric quadrupole. The fields
are products of a tensor (the quadrupole moment of the charge
distribution) times the unitary vector of the line of observation.
B 2q
3  i ( kr  t )
 e
n   [3r ' (n  r ' )  nr '2 ] (r ' )dV ' 
3  i ( kr  t )
 e
n  (Q  n)
where the second equation contains the definition of the quadrupole
tensor of the charge density ρ (electric quadrupole). The derivation of
this expression from the formula given in page 2 is long but not difficult,
and is given in the book by Schwartz, pp. 283-284
The quadrupole tensor of the charge density is a 9-component quantity.
In the case – that we examined when we discussed the multipoles in
electrostatics – of a system symmetric for rotations around an axis, the
quadrupole tensor takes the diagonal form
 Q0
Q 0
 0
 Q0
0 
0 
2Q0 
if we choose the symmetry axis of the system as the third (“z”)
Q0 is the quadrupole moment of the charge distribution,
that we had called B2. Note that the system is symmetric for
exchange “z” to “-z”. And as a consequence the P2(cosθ) is – like all
even polynomials – an even function of cosθ.
Advanced EM - Master
in Physics 2011-2012
Let us now examine the term to be integrated, i.e.
[3r' (n  r' )  nr '2 ]   (r' )
in order to show that it is the product of a tensor times n. The
product term is (has to be) a vector. Its components are linear
functions of the three components of vector n, and quadratic
functions of the components of r’ (x’,y’,z’). The vector properties
of this functions do not depend on ρ, so we study only the first part
(between square brackets) of this formula.
[3r (n  r )  nr 2 ] 
i[3x( xnx  yny  znz )  nx r 2 ] 
 j[3 y ( xnx  yny  znz )  ny r 2 ] 
 k[3z ( xnx  yny  znz )  nz r 2 ] 
 nx [i (3x 2  r 2 )  j(3xy)  k( 3xz)] 
 ny [i (3xy)  j(3 y 2  r 2 )  k( 3 yz)] 
 nz [i (3xz)  j(3 yz)  k( 3z 2  r 2 )]
This last formula shows that the term under the integral is linear in the
components of n. The quantities which multiply them are the 9 integrals
of products of the components of r’. A quantity that, multiplied by a
vector gives another vector is a tensor. Therefore the 9 integrals are
the components of such tensor.
Advanced EM - Master
in Physics 2011-2012
The irradiated power has an angular dependence of the type
~ (cos  sin  ) 2
while the total emitted power –for equal current, system size and
oscillation frequency- is of the same order of magnitude as the
magnetic dipole.
The vector potential in the case of a charge and current distributions
oscillating harmonically in time, with a wavelength of the generated
radiation much longer than the physical dimensions of the system, can
be expanded in a series of multipolar radiation fields, of emitted
power decreasing as powers of (d/λ)2.
Each term of the series is characterized by the angular distribution
of the emitted power, and the amplitude of each multipole is
determined by the multipole expansion of the ρ and j.
Advanced EM - Master
in Physics 2011-2012
Interaction of radiation with matter
( i.e. with charges)
Let the radiation be a plane wave, with its frequency, its energy
density and its Poynting vector. In its path there is a free
electron at rest. What happens?
The wave has its transverse electric field E; the electron will be
accelerated by the Lorentz force
F eE
m m
And, being accelerated, but with speed near zero, will emit
radiation in the form of spherical wave. The total power is given
by the Larmor formula.
2 e2a 2 2 e4 E 2
3 c
3 m 2c 3
On the other hand the incident radiation has its Poynting’s vector
which is its energy flow.
The ratio
2 e4 E 2
W 3 m 2c 3 8 e 4
8 2
r0   Thomson
2 4
c 2
has the dimensions of a surface, and is called the
Advanced EM - Master
in Physics 2011-2012
The absorption and re-emission of radiation are condensed in one parameter,
called the Thomson cross section: It is as though in the way of the radiation
was an absorbing disk, of area σ, that absorbs the incident radiation and
reemits it with its own angular distribution. The radius r0 of that virtual disk
is called the ELECTRON CLASSICAL RADIUS. Its value is
r0 
 2.821013 cm  2.82 fermi
A remark on the expression classical electron radius.
With these words we do not mean to say that the electron is a uniformly
charged sphere, of radius 2.82 fermi. This expression is derived from
classical theories (before the advent of quantum mechanics) in which the
mass of the electron was interpreted as due to its electrostatic energy.
All things that required the electron to have a finite volume. While now
the electron size is known to be no larger than 10-16cm.
The next step – in the study of the interaction of radiation with matter – is
the study of absorption and re-emission of radiation by the electrons
orbiting around a nucleus.
To calculate the interaction of radiation with matter we will employ a very
simplified model of matter; more precisely, a simple model of how the
charged particles –in practice, only the electrons- which constitute matter
act under the force exerted on them by the radiation. In this model only
the electrons are considered because their mass is ~2000 times smaller
than that of the protons; and in the Thomson cross section the mass is in
the denominator of the formula for the acceleration. In most materials –
the exception being the conductors – the electrons are not free, but bound
to the nuclei. In the model it is this binding of electrons that is described
in an approximate way: they are considered to be in a sort of equilibrium
position, with an elastic force to bring them back to it.
F  kr  m0 r
where the resonant frequency of the electrons has been inserted explicitly.
Advanced EM - Master
in Physics 2011-2012
This approximation can seem incongruous if applied in a situation dealing with
atoms and bound electrons, realm of quantum mechanics. But, if we apply it to
an em radiation of wavelength of the order of visible light or longer, we are
talking of wavelengths three orders of magnitude larger than the atom size.
The atom sees basically a uniform electric field, rising and decreasing so
slowly that it can be considered constant. The electron displacement we are
discussing here can be interpreted -now that we know about quantum
mechanics - as the displacement of the whole orbit under the effect of a very
slowly varying external electric field. This model and its results are usually
referred to as the classical electron theory of matter.
We shall now study in more detail the scattering of radiation by these
electrons. The incident radiation will be a simple linearly polarized plane wave,
of angular frequency ω, and electric field E [E0cosωt, 0,0] . The electron’s
equation of motion is: 2
eE0 it
d x  dx
(e  e it )
m dt
We have added to the forces in equilibrium also a dissipative force
to keep the equation still finite at and near a resonant frequency.
The solution of that differential equation is a driven oscillation with
angular frequency ω and amplitude:
 dx
m dt
cos( t   )
m (0   )   
Where the phase φ is: t an( ) 
0 2   2
2 2
NOW, we have the electron’s equation of motion ;
•We derive it twice wrt time and have the acceleration
•We replace the acceleration in Larmor’s formula, average it over
an oscillation period and obtain the total irradiated power
•We divide that power by the power per unit area of the incident
plane wave and obtain the scattering cross section.
Advanced EM - Master
in Physics 2011-2012
8 2
 scatt.  rad 
(0   2 ) 2   2 2
Remark how this formula becomes equal to that for the free electron
(Thomson cross section) in case the electron is free (γ=ω0=0).
Remark also that a consequence of the electrons being bound is that the
cross section now depends on the driving frequency (it was independent
for a free electron).
Now, in the classical electron theory of matter the radiation wave which
propagates through matter is the SUM (for the principle of
superposition) of the incident plane wave PLUS the wave radiated by the
electrons in their driven oscillatory motion.
Advanced EM - Master
in Physics 2011-2012
Radiation propagation in matter:
the case of a transparent foil
One of the main concepts (principles) in electromagnetism is the
principle of superposition. It states that fields are the vector sum
of all the fields generated by all charges, near or far away. And,
another principle (of relativity, this one) is that the velocity of light
in vacuum is “c”.
There are no problems with these principles in vacuum; there are
however in matter:
•In transparent materials, the light travels at a different speed:
c/n, where n is the index of refraction, which also turns out to be
a function of the light frequency.
•In most materials, which are not transparent, the light does simply
not travel, i.e. gets absorbed. Not only; in some materials
(conductors), it even is reflected.
Now, how can all this be compatible with the previous principles?
After all, matter consists of many charges in vacuum.
To these questions gives an answer the classical electron theory.
We have already seen the basic interaction of radiation – which we
have taken to be a simple harmonic function, i.e. a sin(ωt) - with a
free electron and with an elastically bound electron. The next step
is to consider matter as a dense –to the extent that it may be
treated as a continuous charge distribution – a dense collection of
charges, be they light electrons (bound or, in same cases, free) or
heavy nuclei, whose motion under the effect of a field is however
very slow owing to their large mass and can be neglected.
Advanced EM - Master
in Physics 2011-2012
To study the interaction of light with dense matter we shall study a
well defined physical system, but greatly simplified.
•We consider radiation to impinge at right angle on a thin sheet (foil)
of material, causing the electrons in it to oscillate and as a
consequence to generate a radiation.
•The incident radiation we consider is a plane wave. Since it hits the
sheet at right angle, all the electrons in the sheet see the same value
for the fields at any given time.
•The charges, although discrete, are so many and so close and evenly
distributed that they can be treated as a continuous charge density
(NB: this condition may not be applicable to gases. Example: the high
•The material considered is in the form of a thin sheet. How thin? And…
why is it thin? The answers: the first is is that the electrons in the
matter are subject to an oscillating field with a finite wavelength.
Which field? The field incident on the material plus the field generated
by all other electrons; and at this point things get complicated: better
to start with a simple system, a very thin sheet, so that we do not have
to consider the effect of the other electrons. The second answer is
that all the electrons in the sheet must be subject (for simplicity) to
the same field. Then Δ<<λ.
Advanced EM - Master
in Physics 2011-2012
The incident radiation is a linearly polarized plane wave, with its electric
field aligned along the “x” axis, while the wave propagates along the “z”
axis. The sheet contains a charge surface density of N·e·Δ, where N is
the number of electrons per unit of volume. All electrons are subjected
to the incident field Es=iE0exp(iωt) (where i is the unitary vector
along the “x” axis, and Es is the “source incoming radiation).
REMARK that at this point we spend the request that the sheet of
material be very thin: we approximate the real field acting on the
charges with the incident field only.
Under the influence of the electric field the charges will all move
-with equations given in the previous pages for free and bound electrons
-all electrons will move coherently (i.e. in phase).
As a consequence, all electrons will have the same acceleration at the
same time and will irradiate the same electric field directed, for
symmetry reasons of course, along the “x” axis.
Advanced EM - Master
in Physics 2011-2012
The total electric field which emerges forward will be directed along
the “x” axis and be the sum of Es and Ef (generated by the foil of
Now… WHAT DO WE EXPECT –from 1st grade physics-TO HAPPEN
The light which has gone through a glass foil of refraction index “n”
and thickness Δ is retarded –wrt the incident wave- of the phase δφ
  k(n  1)
(it still oscillates with angular frequency ω, but through the foil with
speed c/n). Therefore, if we take
the incident wave at the foil entrance to be
the incident wave (unaltered by the material) at the foil exit is
~ exp(iωt)
~ exp(-i[kΔ-ωt])
the wave measured at the exit of the foil is exp(-i[nkΔ-ωt])=
 exp(i[n  1]k  ik  it ) 
 exp(i[n  1]k)  exp(i[k  t ])
i.e., we expect –from experimental classical optics- the wave at the
foil’s exit to be multiplied by a phase factor
On the other hand, we know that Eout=ES+EL: the sum of two fields,
not the product of the incoming wave with a phase factor. But,the
refracted wave can also be written as a sum (Taylor series expansion of
the phase factor):
exp(i[n  1]k  1  i(n  1)k
And the exiting wave becomes:
Eout  Es  i(n 1)kEs
Advanced EM - Master
in Physics 2011-2012
Eout  Es  i(n 1)kEs
Which is what optics tells us.
We therefore expect – from the
experimental knowledge of the
effect of the passage of light
through a thin layer of transparent
foil of matter- that the wave at its
exit will be the incident wave
slightly delayed in phase or –which
amounts to the same- a wave that has passed through the foil unaffected
plus a small wave retarded in phase by 90 degrees.
Let us now see what our approximated model calculates! We already know
the motion of the charges, i.e. the motion of the whole sheet of charges:
x  A  eit
m (0 2   2 )  i
Advanced EM - Master
in Physics 2011-2012
Related flashcards


39 cards


44 cards


36 cards

Geometrical optics

16 cards

Create Flashcards