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8/16/2019 Traduccion 1 2016-I-p2 (1).pdf
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14
2
2
1oocE S (16)
is the average energy per square meter per unittime passing through a surface normal to the
direction of propagation.
2
2
1oocE S
t
o
je E E
is also called the intensity ”I”
E o is the amplitude of the electric field of the incident wave
Fig. 3.10 Incident plane-wave radiation of frequency .
When an atom characterized by a resonance frequency o, is placedin a region where there is a bath of electromagnetic radiation, the
radiation’s electric fieldt j
e E E o will drive the atom’s charge eq up
and down; that is, it will accelerate the charge thus causing the atomto re-emit electromagnetic radiation. This process, which occurs at
any frequency , is called scatter ing . That is,
scattering is the process by which energy isabsorbed by an atom from the incident radiation (17) field and re-emitted in all directions.
t jωe E E o
t jωee x x jo ][ ,
Fig. 3.11 Incident light is absorbed and (re-emitted) scattered by an atom in
all directions.
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15
Let’s calculate the energy absorbed (hence re-emitted) by the charge
Actually the energy emitted by the accelerated charge has already been calculated
in expression (14) above, except that we need to find out the amplitude of oscillation
o x ; the latter will depend on the electric field amplitude E o, as well as the frequency
of the incident radiation. In other words, let’s calculate the relationship between E o, ,
and o x .
q e , m e ,o
Scattered (re-emitted)
light
Atom model as anoscillator of natural
resonance frequency o.
Incidentradiation
()
The spring constant ischosen according to k
2 ≡
me o2
4
3
22
12
c
xq P
o
oe
Fig. 3.12 Atom modeled as an oscillator of natural frequency o. The ability of the
oscillator to absorb energy from the incident radiation depends on .
To find o x , let’s model the atom as a damped harmonic oscillator. Accordingly the
equation of motion for the charge eq
is given by,
t je E qkx
dt
dxm
dt
xd m oeee
2
2
(18)
Here the termdt
dxme accounts for the presence of a dissipation energy
source, which, in our case, can be identified in the loss of energy due to the
electromagnetic radiation by the accelerated charge.
A stationary solution of (18) is given by,
t jωee x x jo ][ (19)
where
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16
2/1
22222 )(
)/()(
o
oeeoo
E mqω x x (20)
Amplitude of oscillation
as a function of fre-
quency
and
tan
22
1
o
(20)’
Expression (20) indicates that the amplitude of oscillation xo (and hence the
acceleration) of the charge depends on the incident radiation’s frequency .
Let’s proceed now to calculate the total power radiate by the accelerated charge under
the influence of an electric filed of amplitude E0 and frequency . Replacing the value of
xo given in (20) for xo into the expression for the radiation power4
3
22
12
c
xq P
o
oe
given in (14), one obtains,)()(
)/(
12 2222
224
3
2
ωωω
E mqω
c
q P
o
oee
o
e
. Rearranging
terms,
)()(43
8
2
12222
42
2
22
)(
oeo
eoo
cm
qcE P =
Expression (21) gives the total average energy emitted by the charge qe when subjectedto a harmonic electric field (given in expression (15) ) of amplitude E 0 and frequency .
Notice the expression2
2
1oocE (incident energy per unit area per second, i.e. the
incident intensity Io) has been factored out in expression (21). This is convenient, for it
allows to interpret (21) the following way: Out of the incident intensity Io present in the
cavity, a ‘fraction’ of it equal to)()( 222
2
42
2
2
)(43
8
oeo
e
cm
q is present in the
form of scattered power. We say ‘fraction’ because the units of that last expression is
area (not a simple fraction number). Hence, it is better to interpret (21) in terms of
“scattering cross section).
Note (dated 09-2012).
Expression (21) quantifies the amount of energy that the the atom is able to
re-radiate (due to the fact that is is a charge) upon the incidence of an
harmonic electric field of amplitude Eo and frequency .
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17
It has nothing to do with the ability of the atom to capture energy from
radiation in the cavity (as the concept of scattering cross section may
erroneously suggest). Hence, be careful with the proper interpretation of
the “scattering cross section concept.) Andres
The concept of scattering cross section
If we considered a hypothetical cross section of area intersecting the incident
radiation, the amount of energy per second hitting that area would be
I cE P oo ]2
1[
2
(22)
2
2
1oocE S
cross section
of area
is called the
light intensity I
Fig. 3.13 Pictorial representation of scattering cross section. Notice,
this has nothing to do with the size of the atoms nor the spatial
distribution of atoms inside the cavity. It is just a measurement of the
ability of the atom (once radiation impinges on it) to radiate energy
in all directions.
One can use the analogy of an affective area being intercepted by the incident radiation
to define how effectively the radiation is absorbed and scattered ( i. e. re-emitted) by an
atom. In effect, comparing expressions (21) and (22), the total power scattered by an
atom is numerically equal to the energy per second incident on a “surface” of cross-
section area scattering ,
scatterin g oo scattering cE P ]2
1[
2 (23)
where
2222
42
2
2
)()()
4(
3
8
oeo
e scattering
cm
q (24)
scattering has units of area.
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18
scattering
Fig.3.14 Sketch of the atom’s scattering cross section.
Note (Dated 09-2012) scattering is an indicator of the ability of the atom scatterlight once it is excited by an harmonic electric field of amplitude Eo and
frequency w. We cannot ask, we cannot expect, the atom to scatter more
(or less) than Io (where Io.is the intensity associated to the filed incident
on the atom).
Expression (24) indicates that the closer is to o, the higher the re-
emitted energy.
3.1.B.c Electromagnetic Radiation Damping: What is the value of ?)
We address here the fact that
the term
dt
dxme in Eq. (18) (the term in the differential equation that takes into
account the energy dissipation,)
should be compatible with
expression (21) (that gives the electromagnetic energy dissipated by the oscillator.)
We should require then that these two expressions be consistent with each other.
Indeed,
On one hand, the power dissipate by a oscillator is given by
[ force]x(velocity ) = [dt dxme ](
dt dx ) = [ )( x jωme ] ( x jω ) = =
22 xωme . Here we
use the expression for )(t x given in (19)t jω
ee x x j
o ][ .
The average value of the dissipated power will be,
22 )2/1( oe xωm .
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19
On the other hand, according to (14), the emitted electromagnetic power is,
4
3
22
12
c
xq P
o
o
The last two expressions should be equal.
22 )2/1( o xm 4
3
22
12
c
xq
o
o
This allows to identify2
3
2
6ω
mc
q
o . Rearranging terms,
2
2
2
43
2 cm
q
ceo
e
electromagnetic
radiation damping
constant
(25)
For practical purposes, however (given the very narrow bandwidth of the cross section
( ) shown in Fig 3.10 above,) will be typically end up being evaluated at = 0 ,(i.e.
the narrow bandwidth of ( ) tell us that most of the physics happens around = 0 .)
Rate at which the oscillator looses energy
(A more detailed description of this section is given in the supplementary Appendix-3 of
this chapter.)
Let
)(t W W be the average energy of an oscillating (26) charge at a given time t .
If the oscillating charge is left alone to oscillate, its amplitude of vibration will die out
progressively as the oscillator looses its mechanical energy by emitting electromagnetic
radiation.
If the motion of the oscillator is alternatively modeled by a mechanics equation of
motiont j
e E qkxdt
dxm
dt
xd m oeee
2
2
, it can be calculated that the rate at which
the oscillating charge looses energy is given by,
(27)
with its corresponding solution
t eW t W o
)( (28)
W dt
dW
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20
As an example, an atom that has a resonance frequency corresponding to = 600 nm,
would have a damping constant of ~ 108 s-1. That is, the radiation will effectively dye
out after ~10-8
s (or after ~ 107 oscillations.)
3.1.C Radiation and thermal equilibrium
Let’s consider an atom enclosed in a box made of mirror walls which contains
electromagnetic radiation. Radiation re-emitted by the atom remains inside the box
undergoing multiple bounces on the mirror walls. Let’s further assume that the
temperature of the whole system is T .
q
Scattered (re-emitted)
light
Bo x at temperature T
Atom
(modeled as an oscillator )
Incident
radiation
Fig. 3.15 Schematic representation of an atom as an harmonic oscillator that radiates
energy. The atoms absorb energy from the electromagnetic radiation existent inside the
box (the latter assumed to be made of perfectly reflecting walls.)
How to make the temperature T intervene in an expression like (14) that gives
the power scattered by an atom in the form scattering P 4
3
22
12
c
xq
o
oe
?
It is plausible to assume that the equilibrium temperature should correspond
to proper value of the amplitude of the electric field, o E , since the higher the
value of o E , the higher the charge’s amplitude of vibration xo, the greatertemperature to be associated with the atom (i.e. the amplitude of the oscillator
should increase with temperature.)
If our assumption were correct, how to find then the proper value of o E corresponding to a given temperature T ?
Aiming to find a proper answer let’s outline some considerations:
- If an atomic oscillator had no charge, it would oscillate forever. It would havean average energy W compatible with the temperature in the box; that is
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22
It is an integral because the atom is exposed to all the frequencies existent in the cavity. However, the maximum emission of power
occurs when is close to to (because )( has a sharp peak at
.)
On the other hand, this same amount of emitted power (i. e.
0)()( d w I ), can be seen from the perspective of a system loosing
energy due to a damping process ][[
)()()](
W W
dt
d
characterized by a damping constant . On the other hand, therequirement of compatibility between i ) power re-radiated by the atom,and ii ) a simple damping harmonic oscillator model
t je E qkx
dt
dxm
dt
xd m oeee
2
2
, lead to expression (25)
2
2
2
43
2
cm
q
c eo
e)( . But since all the dynamics occurs at , that
is W ()~0 for ≠ , we can use W W
odt
d )( , with the
interpretation that W is the total energy of the atom.
Formalization of the thermal equilibrium condition:
How much light intensity spectral density I( ) there must be inside the box attemperature T for,
the electromagnetic energy re-emitted by the oscillator (which should
come from the radiation bath in the cavity) per unit time
00
)()()( d I d P dt
dW (29)
to be equal to
the energy lost by the oscillator per unit time
W dt
dW
Average energy of the
oscillator at temperature T
(30)
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23
Bo x at temperature T
0 qe
Fig. 3.16 Atom of natural frequency 0 in
a bath of electromagnetic radiation of
spectral density I( ).
Let’s evaluate the integral that appears in (29). Since the expression for )(ω peaks at o then extending the integral
down to does not cause any significant change (this is done just tofacilitate the calculation )
d ω I d ω I )()()()(
0
where 2222
42
2
2
)()()
4(
3
8
oeo
e scattering
cm
q
For the same reason that only the values of ω very close to oω willsignificantly contribute to the integral we can picture in our mind that
0
0
)()()()( d ω I d ω I
Fig.3.17 Sketch of the atom’s scattering cross section and the spectral
density light intensity present in side the cavity
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24
Accordingly the following approximations can be considered appropriate,
22
oωω ))(( oo ωωωω ))(2( oo ωωω
2222
4
)()( ωωω
ω
o 2222
4
)(][ oo
o
ωωω
ω
22
4
)()])(2[( ooo
o
ωωωω
ω
22
2
)(4
o
o
ωω
ω
)()( o I I
All these approximations lead to
d ω I d ω I )()()()(
0
d
cm
q I
oeo
e
2222
42
2
2
)()()
4(
3
8)(
d
ωω
ω
cm
q I
o
o
eo
e
22
2
2
2
2
)(4)
4(
3
8)(
d
ωωω
cmqω I
o
o
eo
eo
22
22
2
2
)2/()(1)
4(
32)(
(usinga
x
aa x
dxarctan
1
22
)
)( )2
(2
2/
1)
4(
3
2)(
22
2
2
o
eo
eo ω
cm
qω I
2
)4
(3
2)(
22
2
2
o
eo
eo ω
cm
qω I
222
22
0)(
43
2)()()( o
eo
eo ω
cm
qω I d ω I
(31)
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25
At equilibrium we should have,
0
)()( d w I dt
dW = W
dt
dW , which leads to
W ω
cm
qω I o
eo
eo
222
22
)(43
2)(
W
ωq
cmω I oe
eoo 2
22
2
2
2)(4
2
3)(
Using expression (25) for the value of 22
2
43
2ω
cm
q
c o evaluated at 0ωω
one obtains,
W ωcm
q
cq
cmω I o
oe
eoo
22
2
22
2
2
2)
43
2(
4
2
3)(
)(
, or
W ω
cω I oo
22
2)
3
2(
2
3)(
,
W ωc
ω I oo 3
1)(
2
22 (32)
Average energy
of the oscillator
I( ) is the light spectral density at = .
Here is the natural frequency of the oscillator we were focusing in. Had we used an
oscillator of a different natural frequency, let’s say ’, we would have obtained a
similar expression (32) but with ’ instead of . Hence, in general,
Average energy of the
oscillator at temperature T
W ωc
ω I 3
1)( 2
22
light intensity
spectral density
I( )
(32)’
Required light intensity spectral density I( ) inside thebox at temperature T , in order to maintain equilibriuminside the cavity.
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26
I( )
I( ) =
Notice
Units of [I( ) d ] = ][ Intensity = sm
J 2
Units of [I( ) ] =][
][
Intensity
=2
2
/1
)(/
m
J
s
sm J
It is worth to highlight that,
Expression (32)’ has remained undisputed. That is, it is still considered correct even
when the new quantum mechanics concepts are introduced.
It is in the calculation of the average energy W where the classical and quantum
approaches fundamentally diverge.
3.1.C.b Classical calculation of the atom’s average energy W .
In classical statistical mechanics there exists a very general result so called
“equipartition theorem,” which states that the mean value of a quadratic term in the
energy is equal to ½ k BT. Here k B is the Boltzmann’s constant and T is the absolute
temperature.
The Boltzmann distribution
The equipartition theorem can be obtained from the Boltzmann’s probability
distribution for a small system A in equilibrium with a (huge) reservoir at temperature
T. The Boltzmann distribution states that the probability that the system S be found in a
state of energy E is proportional toT k E
Be/
; that is ,
T k E Be E P /
)(
T k E BCe / probability to find the system (33)A in a state of energy E
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27
EReservoir at
temperature
T
Energyexchange
Energy E
P(E)
Boltzmann
dis tr ibut ion
Smallsystem A
Fig 3.18 Left: A system interacting with a thermal reservoir. Right: Boltzmann’s
distribution to find the system in a state of energy E.
The values of E could go from 0 to infinity (the reservoir being in charge of keeping thetemperature constant); but, as the expression above indicates, the states of lower
energy have a higher probability.
Since for a given energy there may be several states characterized by the same energy,
it is usual to define,dE E g )( number of states with energy (34)
E , within an interval dE ,
thus giving
dE E g C T k E Be )(/
probability to find the system A in a state
of energy between E and E + dE ,
which suggests to rather identify a probability-density )( E P defined as follows ,
dE E g C dE E T k E Be )()( /P probability to find the (35)
system in a state of energy betweenE and E + dE ,
with C being a constant to be determined.
Since the probabilities added over all the possible states should be equal to 1, we must
require, 1')'(0
/'
dE E g C T k E Be , which gives,
1C = 0
/' T k E Be ')'( dE E g (36)
A self-consistent expression for )( E P is therefore given by,
0'
)()(
)'(/'
/
dE E g T k E
T k E
B
B
e
e dE E g dE E P (37)
(Notice in the denominator we are using a ‘dummy’ variable ' E .)
From expression (37) we can formally calculate the average energy of the system,
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0
0
')'(
)(
/'
/
dE E g
dE E g E E
T k E
T k E
B
B
e
e (38)
The Equipartition Theorem
It turns out, very often the energy of the system may contain a quadratic term.
Consider for example
...2
1
2
2
2
kxm
p E x ,
and we would like to calculate, for example, the average value of the kinetic energy
termm
p x
2
2
alone. As we know, being the system in contact with a heat reservoir, the
value ofm
p x
2
2
is sometimes high, sometimes it is low because it gains or looses energy
from the heat reservoir; we would like to know what would be its average value
m
p x
2
2
.
0
2
0
2
2
2
/
/
2
2
2
2 x
Bm x
x
Bm
x x
x
dp
dpm
p
m
p
T k
T k
p
p
e
e (39)
Let’s call
T k B
1 (40)
In terms of expression (39) becomes,
0
2
0
2
0
2
0
2
2
2
2
2
2
2
2
2
xm
x
xm
x
xm
x
xm
x x
x
dp
dp β
dp
dpm
p
m
p
p
p
p
p
β
β
β
β
e
e
e
e
0
2
0
2
2
2
xm
x
xm
x
dp
dp β
p
p
β
β
e
e
0
2
22
ln2
xm
x x dp
β m
p p
β e (41)
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29
Defining the variable x pm
β u
2 ,
00
222
21
ln2
ln2
dum β
dum
β m
p uu ee x
0
22
2ln1
ln2
dum β m
p ue x
Term independent of
2
1ln
2
1
2
2
β m
p x
T k
m
p B
x
2
1
2
2
(41)’
Had we chosen any other quadratic term of the energy we would have obtainedthe same result. This is the equipartition theorem. It states,
If the energy W of the system has the form
... 22
x kx2
1
2m
p E (42)
the average value of each independent
quadratic term is equal to T k B2
1 .
... kT 2
1 kT
2
1 E W
The ultraviolet catastrophe
Let’s assume there are f different quadratic terms in the expression for the total
energy W (translation motion, rotational motion, …, etc.). The equipartition theorem
leads to T k 2
1 f W B . Using this result in expression (32)’ W I ω
c3π 1ω 222)( , one
obtains T k 2
1 f ω
c3π
1ω B
2
22 I )( , or,
6
)( 2
22 ω
c
T k f ω B I
classical prediction (43)
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30
(that assumes ~ kT )
I ( )
Frequency
Classical
prediction Experimental
results
Fig. 3.19 The serious discrepancy between the experimental results
and the theoretical prediction is called the ultraviolet catastrophe.
3.1.D The birthday of Quantum Physics: Planck’s Hypothesis to calculate
the atom’s average energy W
To bring the theoretical prediction closer to the experimental results Planckconsidered the possibility of a violation of the law of equipartition of energy described
above, expression (42).
The starting expression would still be expression (32) W ωc3π
1 I 2
22)( , BUT with the
average energy of the oscillator W not being constant (as the equipartition theorem
predicts) but rather being a function of the frequency, )(ωW , with the following
requirements,
00
)(2
ω
W
and (44)
0
)(2
ω
W
For the statistical calculation of W ,
Planck did not question the classical Boltzmann’s
Statistics described in the section above; that (45)
Theory would still be considered valid.
Planck realized that he could obtain the desired behavior expressed in (44) if,
rather than treating the energy of the oscillator as a continuousvariable,
the energy states of the oscillator should take only discrete step
values:
0 , … (46)
the energy steps would be different for each frequency
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31
= () (47)
where the specific dependence of in terms of to be determined
q
Incident
radiation
Planck postulated that
the energy of the
oscillator is quantized
Fig. 3.20 An atom receiving radiation of frequency , can be
excited only by discrete values of energy 0 , , 2, 3, …
According to Planck, in the classical integral expression (37)
0
0
')'(
)(
/'
/
dE E g
dE E g
T k E
T k E
B
B
classical
e
e E E W , one would have to replace:
g(E)dE [ g(E)dE gives the number of states with energy E
within an interval dE ]
dE
00
n
dE
thus obtaining,
0
0
/
/
)(
n
Bn
Bn
n
n
Planckl T k
T k
E
E
e
e E
E ωW (48)
where n E = n () ; n= 1 2, 3, …
A graphic illustration can help understand why this hypothesis could indeed work:
First we show how classical physics evaluate the average energy.
0
)( dE E P E E classical
][ = area under the curve of P(E) E vs E
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32
Energy E
P(E)
Boltzmann
dis tr ibut ion
Energy E
E P(E)
=
=Area
Classic calculation:
continuum addition
(integral)
Fig. 3.21 Schematic representations to calculate the average energy of
the oscillator under a classical physics approach.
Using Plank’s hypothesis Planck
E )(0
n
n
n E P E
Case: Low frequency values of
For this case, Planck assumed should have a small value (for the reasonsexplained in Fig. 3.17).
≈ small value (for small values of (49)
Energy E
P(E)
Boltzmann
dis t r ibut ion
Energy E
E P(E)
Quantum calculatio
discrete addition
Fig. 3.22 Schematic representations to calculate the average energy of the
oscillator assuming the oscillator can admit only discrete values of energy,
for the case where the separation between contiguous energy levels beingof relatively low value.
Indeed, comparing Fig. 3.21 and Fig 3.22 one notices that if is small then the
value of )(0
n
n
n E P E
will be very close to the classical value.
It is indeed desirable that Planck’s results agree with the classic results at low
frequencies, since the classical predictions and the experimental results agree
well at low frequencies (see Fig. 3.19 above.)
Case: High frequency values of