Download Reflected wave

Field of the radiation from a foil of oscillating charges
We now will have to integrate over the whole plane of charges (the foil)
the fields that they irradiate. The electric field will be again along the “x”
axis for symmetry reasons. We will integrate
the field generated by all charges contained
in the circular ring drawn in the figure. The
field will be calculated in a “generic”
point P at a distance
“z” from the foil.
The integral will be
carried over the foil
surface.
The plane has a surface charge density N·e·Δ,
all charges are bound electrons. Their motion
under the radiation’s electric field is coherent,
i.e. according to the law
 x  A  eit
with
A
eE0
1
m (0 2   2 )  i
We therefore have a whole plane with a uniform distribution of charges
moving under the driving of the incident radiation a motion with an
acceleration of the form
a x  a   A 2  eit
and which will of course emit em radiation. All charges in an infinitesimal
ring of thickness dρ will contribute the same field in P: we will ignore the
small angle θ that the line from a point in the ring to the point
with the “z” axis. The field generated by that ring is
P makes
Ne 2 Ae i ( kR  t )
dEF ( z ) 
2 d
2
c R
and the field generated by the foil is the integral of this formula over ρ.
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A simplification of the integral comes by noticing that since
R2   2  z 2
And we can replace
 2R  dR  2  d
  d
by R·
dR in the integral, thus dropping
the R term in the denominator. We are now left with the integral:
2NeA 2 it  ikR
EF ( z ) 
e  e dR 
c2
z
2NeA 2

c2
( ik1 ) [e
 ik
e
 ikz
]
This result is not a result since it has a serious problem: the term

e
z
 ikR
ik
oscillates continuously for R ∞ and therefore
is
not defined. One argument that has been put forward is that
since it oscillates its contribution averages to zero and therefore:
e
2NeA 2 1 ikz
EF ( z ) 
 e
c2
ik
There is a better argument to obtain the same result , which is
found in Feynman 1, 30-11, and is given in the next two pages. We
assume this term to be zero, add the time dependence we obtain:
2 Ne A i i (kz ω t )
e
c
which is valid for Z>0. We now will replace A with is value from the eq.
EF ( z, t )  
of motion of the charges,
A  eE
ignoring for now the dissipative term γ.
{m(0   2 )}
2
We obtain for the field emitted by the foil electrons the value (which
we call EF , the field emitted by the foil) the formula
 i ( kz ω t )
2Ne 2 
2Ne2 
EF ( z , t )  
 i  E0 e

 i  ES
2
2
2
2
mc(0   )
mc(0   )
Where Es is as shown in lesson 21, p. 13.
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 ikR
z e dR mean? The symbol stands for
What does the integral
the sum of small segments (in the complex plane) of length dR and
phase θ= kR (see drawing). When we move away from the center of
the integration ring (ρ increases) the value of the integral turns
indefinitely around on the circumference shown in the next figure.
Real axis
When the infinitesimal
segments summed in the
integral add up, each
segment has an angle kΔR
with the preceding segment.
The curve traced by the
integral as the upper limit
tends to ∞ is a circle
with center 
Imaginary axis
i ikz
e
k
As it often happens in these cases, pushing the integral to infinity
without any convergence mechanism in the equations is a procedure
correct from the mathematical point of view but not physically. There
are physical reasons why the integral can not oscillate with the same
amplitude all the way to infinity; for example, a cos(θ) factor due to
the “transverse acceleration” causing the radiation field; and, the other
fact that there is not such a thing as a plane wave all the way to
infinity. All these factor cause the second term in the calculation of
the integral, i.e..
e
ik
to be zero.
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The integral, as R increases, goes around the circle and turns
around indefinitely. The value of the whole integral as we
increase the radius runs a spiral curve which eventually tends
to the circle center. But – as R increases the radius of the
integrated ring gets larger and larger and the contributed
field gets smaller and smaller and the integral runs along a
spiral, which eventually ends at the spiral’s center,
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We can, now that we have obtained it, compare this formula that we
have obtained for a foil of bound electrons (transparent material),
with the one we expected from the experimental study of the
behaviour of light incident on transparent matter (see lesson 21,
p.12): we compare
Light through transparent foil
radiation from sheet electrons
Eout  ES  EF  ES  i (n  1)kES
2Ne 2
EF  
 i  ES
2
mc(0   2 )
And obtain the result that the two equations are equal if
2 e 2 N
n  1
2
m(0   2 )
So, the model yields a formula for the effect of the foil that not only
gives the right effect ( a retardation of the phase wrt to the free
incident wave), but also explains its origin and allows us to calculate the
index of refraction. In fact, we have:
•Understood why the radiation moves with a phase velocity lower than c in
transparent materials.
•Obtained a formula which allows us to connect the index of refraction to
other measurable parameters of matter, such as electron densities and
emission frequencies.
•We understand also the dispersion of the index of refraction, i.e. its
dependence on the light’s frequency.
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Since bound electrons in matter have different resonance
frequencies, the equation for the index of refraction is modified
to take into account all those frequencies:
f
2 e 2 N
n 1 
 2 i 2
m
i (i   )
Where the fi are the fraction of the atom electrons that have
that resonance frequency. For most frequencies, n increases as
the frequency increases. Only near the resonance frequencies
does n decrease, an effect called anomalous dispersion. (as a
rule, a prism deflects more the higher frequencies. In presence
of anomalous dispersion it is the other way around).
Also, near the resonance frequencies, where the term (i  
can become very small, the term iγω becomes important. It has
the effect of making the index of refraction a complex number,
2
2Ne 2
n  1
2
m(0   2  i )
2
)
n = n’ -i·n”
As a consequence in the formula for the outgoing wave appears a
multiplying term of type exp( n" ) i.e. an absorption of the
incident wave inside the foil (of thickness Δ).
We have seen how the motion of the electrons radiates a field in
the forward direction. There is no reason why this radiation should
not be emitted in the backward direction as well: it is called the
“reflected wave”. The reflected wave is equal to the wave emitted
by the foil in the forward direction: only difference is the
propagation direction; in the formulas, the exponent sign of the
space dependence is: “+ikr”.
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In summary, the radiation fields are emitted by the foil, one wave
forward and the other backwards. They travel with velocity “c” and
have the same polarization as the incident wave.
Their amplitudes and phases are:
2 e 2 N

iE0 exp[ i (kz  t )]
2
2
mc(0   )
{
EF 
2 e 2 N

iE0 exp[ i (kz  t )]
2
2
mc(0   )
For z>0
For z<0
The two waves emitted by the foil are equal in amplitude and
frequency (and in phase at the origin). The one emitted forward
adds to the impinging external radiation with a phase difference
of 90 degrees. The wave emitted bacwards travels back but is far
less intense than the incoming wave: the motion of the electrons
under the effect of the radiation’s electric field is limited by the
electrons being bound to the nucleus.
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Rayleigh scattering and the diffusion of light
In the computing of the interaction of light with matter we made a
simplification which is not always valid: it was considered that the
surface density of electrons was so high that light was emitted only
forward and backward, but not at different angles because there the
fields emitted by the charges left or right, up or down were
compensated by equal fields out of the forward region emitted by the
other electrons. This is certainly the case in solids and liquids. It is less
so in gases, especially rarefied gases P. ex. high atmosphere.
The gas is a distribution of radiant systems (molecules, dust, ….) each
of small size compared to the wavelength of the incident radiation (ex.:
Sun’s light) which excites on that dust and on these molecules electric
and magnetic multipoles. Which multipoles are excited depends, of
course, on the characteristics of the molecules etc: in most cases in
fact it is dipoles. They oscillate in phase with the incoming radiation and
as consequence radiate em energy in all directions, with an angular
distribution that we know. If we have a regular and dense distribution
of the diffusion centers, which also oscillate and emit in phase, the
amplitudes of the emitted waves add up in phase only in the forward
direction. In the case instead of a completely random distribution of
the diffusion centers what adds up are not the amplitudes but the
intensities.
We have already calculated the intensity of the diffused radiation as a
function of the frequency for bound electrons
 scatt.
W
8 2
 rad 
r0
S in
3
(
)[
4
2
(0   2 ) 2   2 2
]
It turns out that in the atmosphere ω0²>>ω² and the cross-section gets:
 scatt
8 2  4 8 r0 (2c) 4

r0

4
3
3 0 4 4
0
2
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Owing to the dependence of the cross-section 
on the light
scatt
wavelength λ as 1/ λ4 the violet light (~410 nm) is diffused a factor 6
more than the red light(~600nm). This is the reason why the sky is blue
(when the sun is at the zenith): we see coming from all the sky the blue
light which has been diffused towards our eyes by the atmosphere all
around. And the reason why the sun and the sky around it are red at
sunset: the light from the sun crosses the atmosphere at grazing angle,
and goes through a much longer amount of atmosphere than when the
sun is above us. The blue light is totally scattered away, and only the red
light coming straight to the observer is left undisturbed. In NTP
conditions the absorption lengths of the atmosphere for respectively
violet, green and red light are 30, 70 and 188km.
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Interaction of light with matter:
the case of a foil of conducting material.
Consider again the foil of material perpendicular to the direction of an
incident plane wave, already treated for an insulator, but this time with
a conductor. The key point in the treatment is to model the movement of
the electrons in presence of an external electric field.
In the case of a conductor, what the electrons do is well modeled by an
old, well known law: Ohm’s law! The conductor has volume conductivity σ
and the current density is j   E .
For the case of our foil hit by this radiation the current density is (the
light being linearly polarized, with the electric field along the “x” axis).
jx     ei t
jy  0
jz  0
The vector potential
AF radiated by such current (reduced to surface
current jxΔ) is obviously aligned along the “x” axis A={AX,0, 0} and it
can be calculated with the formula for the retarded potentials – in this
case simplified because we are studying a case of harmonic time
dependence:

AF x ( z , t )   
z
E e
i ( t  R / c )
cR
AF x (  z , t )  AF x ( z , t )
2RdR 
2E i ( kz t )
e
i
Reflected wave
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It is easy to get EF and BF once AF is known (the e.s. potential Φ
since ρ=0):
is zero,
AF being parallel to the current, and therefore
along the “x” axis, so is EF). AF being moreover
B F    A F
only dependent on “z”, as well as parallel to “x” , the
only non-null component of B will be BFy.
E
B
F
F
x
y


2
i ( kzt )

E0 e
c
2
i ( kzt )

E0 e
c
{
{
for z  0
for z  0
2
i ( kzt )

E0 e
c
2
i ( kzt )

E0 e
c
for z  0
for z  0
SO FAR, we have not done anything really new wrt what done for the
glass foil. We have proceeded in a new way, that we did. We started by
computing the vector potential instead of writing the electric field.
And, we have calculated also the magnetic field. But we are basically
still in the same area. And also all the formulas we found for the case
of the glass foil can be derived from these ones by simply
remembering that
J  v
v  (v,0,0)
vi
eE
2
m(0   2  i )
The transparent case can therefore be dealt with by introducing an
imaginary or complex conductivity σ.
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Since in the formulas for the fields that we just obtained the conductivity
is a multiplying factor, the fields emitted by the insulating foil are 90° out
of phase with the incoming radiation –which explains the phase retardation..
Moreover, since the electrons are bound, their motion will be limited, and so
will be limited their acceleration and in the end the emitted fields will not
be large, much smaller actually than the incident field E0. In the case of a
conductor however, if the conductivity is sufficiently large, the field
emitted by the foil (of opposite direction wrt the incoming field), which
subtracts from it in the forward-going wave can be of comparable amplitude
after a very short distance Δ, thus ultimately reducing to zero the forwardgoing radiation: it is the absorption of light . So far, all is rather clear and
understandable. But, we still are considering only slabs, foils extremely thin.
Now we want to do the thing exactly, study the passage of radiation through
thick slabs of material. What we have to do is now to take into account, in
the calculation of the motion of the electrons, also the fields emitted by
the other electrons. There are two contributions from these electrons:
those upstream of the test point will contribute with their wave emitted
forward, while those downstream will contribute with the reflected wave.
The physical system we will study is that of a plane wave travelling towards
the right in vacuum, until it hits a semi-infinite slab of conducting material.
What we want to calculate is the field at a given point inside the material, at
a definite value of “z”.
The way to do this calculation is the following: we divide the material in thin
foils, for which we know the emitted field; in one of these foils is the point
P, with coordinate “z”, where we calculate the fields.
To the fields in P three terms contribute:
1.The incident field Es;
2.The sum (integral) of the fields radiated forward by the foils upstream
of P: an integral of the forward wave from 0 to z.
3.The sum (integral) of the fields radiated backwards by the foils
downstream of P: an integral of the fields from z to∞ .
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E ( z, t )  E ( z )  eiω t 
[ E(z')  e
2 i t
i ( kz ωt )
 E0e

e
c
z
ik ( z  z ')
]
dz ']
dz '  E ( z ' )  e ik ( z  z ') dz '
0
[

z

2 ikz
ikz
E ( z )  E0e ikz 
 e  E ( z ' )  eikz 'dz 'e  E ( z ' )  e ikz '
c
0
z
z
And a similar equation for B.
These equations look fairly horrible. To make them (the last one) easier,
much easier we need to derive both sides twice wrt “z”in order to
obtain:
[

[

]
]}
d 2 E( z)
2 k 2 ikz
4 k
2
ikz
ikz '
ikz
ikz '


k
E
e

e
E
(
z
'
)

e
dz
'

e
E
(
z
'
)

e
dz
'

iE ( z ) 
0
2


dz
c
c
0
z
{
 k 2 E0e ikz 
2
c
z

z
e ikz  E ( z ' )  eikz 'dz 'eikz  E ( z ' )  e ikz 'dz '
0
z

4 k
iE ( z ) 
c
d 2 E( z)
4 k
2


(

k

i) E ( z )
2
dz
c
And a similar equation for B. Now, this equation is very easy to solve. It
could have been found starting from the Maxwell equations, just
replacing J with σE. But this derivation has a much more direct physical
meaning. The equation has two possible solutions:
( {4c k i  k } z)
4 k
exp ( {
i  k }  z)
c
exp 
{
E (z ) 
2
2
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(
exp (
{
E (z ) 
4 k
i  k2] z
c
)
4 k
[
i  k ] z)
c
exp  [
2
The physics is in these two equations. In the complex plane, the square
roots in the exponents lay in the first quadrant, because the expression
under the root sign lay in the 2nd one, very near the imaginary axis.
Therefore both the imaginary and the real part of the exponent lay in
the first quadrant (for the case of the “plus” sign in the exponent), and
have nearly the same amplitude (see below).
Of the two solutions, the physical one is :
([
E  A exp 
4 ik
 k2
c
] z)
which corresponds to a wave being absorbed. The amplitudes A are found
by inserting the solutions in the differential equation and equating the
terms. It turns out that
E ( z , t )  E0eiω t 
(
4 k
i  k 2  ik
c
2 / c
)(exp  (
4 k
i  k2 z
c
))
At this point we may insert some number!
•The conductivity of aluminum is ~3.5x107 mho/m which, in Gaussian
units is 9x109 s-1. Putting all numbers together, for visible light
k
2


2
 1.2 105 cm 1
5
5 10 cm
To be compared with, for
aluminum,
4
 1.3  108 cm 1
c
Then:
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4 c
 103
k
14
For metals and visible lights we can neglect k wrt
~1000 between the two) and also
kc
2
4πσ/c, (a factor
kc
2
wrt
With these approximations we obtain:
E

E0 e  z

B  2 E0 e  z
2πσk c
2πσk c
e i ( z
2πσk c ω t  π/4)
e i ( z
2πσk c ω t )
•Both E and B decrease exponentially in amplitude entering in the
conductor with a decay constant sqrt(2πkσ), which is called the skin
depth and is smaller than a wavelength of visible light.
•In the conductor the magnetic field is larger than the electric field
by a factor sqrt(4πσ/ω).
•The reflected wave can be computed easily because now we know the
total electric field which acts on the electrons.
E
refl
( z, t )   E0
4ki c  k 2  ik
4ki c  k 2  ik
e i(kzt )
Nearly all the incident wave is reflected! The reason is that in the
conductors the conductivity σ is high, and the electrons move fast,
much faster than that small adjustment of the orbits in the case of
the bound electrons. They have sufficient time in a period to reach
high speeds, i.e. large average accelerations and therefore large
emitted fields –which have opposite sign wrt the incoming radiation.
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15
Interaction of radiation with matter
( i.e. with charges): last remarks
In the last lesson we have studied the passage of radiation through
matter. Three cases have been considered:
1.
We used the equations of motion of electrons bound in atoms to
calculate their velocity and acceleration when they are subject
to the electric field of an incoming radiation. The first case
studied was that of TRANSPARENT, homogeneous material,
which is also non-conducting. The radiation had been taken as a
sinusoidal function of time. It was found that the TOTAL field,
- i.e. the sum of the impinging radiation pus the wave generated
by the oscillating electrons - for a simple, linearly polarized
plane wave incident on a thin foil of transparent material
accounted for a slower phase velocity of the light in the foil
(which is seen experimentally) and also yielded a formula for
the material’s index of refraction.
2.
We then studied the case of a CONDUCTIVE thin foil. In such
case we have not used the equations of motion of the bound
electrons to calculate the emitted radiation but more simply the
Ohm’s law.
3.
We have then extended that treatment to account for the
passage of the radiation through a thick, semi-infinite layer of
conductor.
We have then found that inside the transparent, non-conducting
foil the field oscillates with the same angular frequency as the
incoming radiation, BUT that the wave generated by the small
oscillations of the bound electrons is 90 degrees out of phase wrt
the incident wave, therefore causing an apparent phase delay in the
transmitted one.
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We have also found that in the conductor THE FIELD from the
electrons can be much larger that what it is in an insulator
(transparent), In such case the wave emitted by the electrons has
opposite phase (or the same phase but negative amplitude) wrt the
incident one, thus subtracting from it and quickly bringing it to
nearly zero; and efficiently moving (nearly) all its energy to the
reflected wave.
The transmitted wave in a conductor has the form
(
exp 
4 k
i  k2  z
c
{
}
)
which corresponds to an absorbed wave (the exponent is a
complex number). In the formula,
• σ is the metal’s conductivity, and
• k is the radiation’s wave number.
For aluminum and visible light, at λ~500 nm we have:
k
2


2
 1.2 105 cm 1
5
5 10 cm
4
 1.3  108 cm 1
c
The coefficient of the exponent is then:
4πσ
 10 3  k
c
and therefore
4 k
 10 3  k 2
c
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We have seen in the previous page the dependence of the radiation
electric field on the depth “z” inside the conductor - which then, of
course, has to be multiplied by the time dependence.
That space dependence is determined by the differential equation:
d 2 E( z)
4 k
2

(

k

i) E ( z )
2
dz
c
4 k
(
i  k 2 ) , whose square
Now, that coefficient
c
root multiplied by “z” is the exponent in the field’s formula, is a
complex number. Its square root is therefore also a complex number.
The coefficient has a positive imaginary number and a negative (but
much smaller than the imaginary) real number. Its representation is
therefore in the second quadrant, very near the imaginary axis. And
its square root has an amplitude square root of its amplitude and a
phase of about 45 degrees: real and imaginary parts about equal, the
number is in the first quadrant. Its amplitude is approximately
4
c
with about equal real and imaginary parts
2
c
The solution of E(z) is then the product of two exponentials,
exp( 
2
2
z ) exp( i
z)
c
c
The first exponential is a decay with characteristic length
z0 
c
2k

1
500k 2

1
22k
Advanced EM - Master
in Physics 2011-2012
18
z0 
But…
4πσ
 10 3  k
c
Since

2
k
then
c
2 k
Then
z0 
z0 

44
c
2 k

1
1

500k 2 22k
and the field decays
exponentially inside the conductor, with a decay length much
shorter than one wavelength.
Then, if and how rapidly the incoming wave is absorbed inside the
conductor depends on the relative values of σ and k, the wavelength
of the incident radiation and on the conductivity, i.e. how many free
electrons are per unit volume and how fast they move.
Of course, the fact that the conductivity σ is so large in metals has
the consequence that the incoming wave is absorbed in a distance
much smaller than a wavelength.
Advanced EM - Master
in Physics 2011-2012
19
The index of refraction revisited
Having done the work of calculating the field in a thick slab of
conductor, we can recycle the obtained result for the case of a thick
slab of transparent material. In solving for the conductor case we
used the fact that the velocity of the electrons is proportional to the
electric field with coefficient σ . We can apply the results obtained
to the bound electron case by using the “conductivity” obtained from
the equation of motion of the bound electrons:
Jv
v  (v,0,0)
vi
eE
2
m( 2  0  i )
In this case the electrons velocity is also proportional to the
electric field, but with a complex conductivity. For a transparent
thick slab we can therefore use the formulas obtained for the
thick conductor using for the conductivity the value:
2

Ne 
i
2
2
m(0    i)
The two solutions found are:
4 e 2 k 2 N
exp [ z  k 
]
2
2
m(0    i)
2
{
E (z ) 
4 e 2 k 2 N
exp [ z  k 
]
2
2
m(0    i)
2
Advanced EM - Master
in Physics 2011-2012
20
Since the root will have a positive real part only the solution with the
minus sign in front of the square root is physical.
If we insert it in the differential equation we find the results for
E(z,T):
E ( z, t ) 
2 E0 i (nkzt )
e
n 1
Valid for z>0
i (nkzt ) Valid for z<0
E ( z , t )  1n  E0  e
1 n
In these equations the value of “n”, the index of refraction,
corresponds to the formula:
4 e 2 N
n  1
2
m( 2  0  i  )
This is a very useful formula. But it is not what we found a few pages
ago….
2
2 e N
n  1
2
m(0   2 )
It turns out that this second formula, which we got out from the
treatment of a thin foil, is an approximation – for thin foils - of
the exact formula.
Advanced EM - Master
in Physics 2011-2012
21