Traduccion 1 2016-I-p2 (1).pdf

8/16/2019 Traduccion 1 2016-I-p2 (1).pdf

1/19

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.


2/19

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


3/19

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.


5/19

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


8/19


9/19

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)


10/19

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


11/19

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.


13/19

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


14/19

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,


15/19

28

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)


16/19

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)


17/19

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


18/19

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


19/19

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

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