1

Radiation field Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton (Date: May 06, 2014)

We discuss phenomena on the interaction between atoms and electromagnetic field, in terms

of the quantum mechanics. The electromagnetic field is classically treated, while the state of atoms is quantum mechanically treated. Such a method is called semi-classical treatment. 1. Lagrangian (Goldstein, Classical Mechanics)

We start with the Lorentz force given by

)1

( BvEF c

q .

We use the vector potential A and scalar potential as

AB , tc

AE

1 .

The Newton’s second law:

)()1

( AvA

Fv

c

q

tcq

dt

dm

Here we note that

)()(

)()(

)()()]([

z

Av

y

Av

x

Av

x

Av

x

Av

x

Av

z

Av

y

Av

x

Av

x

Av

x

A

z

Av

x

A

y

Av

xz

xy

xx

zz

yy

xx

xz

xy

zz

yy

zxz

yxyx

Av

We also note that

)(z

Av

y

Av

x

Av

t

A

dt

dA xz

xy

xx

xx

,

or

t

A

dt

dA

z

Av

y

Av

x

Av xxx

zx

yx

x

.

2

Then we get

t

A

dt

dA

x

Av

x

Av

x

Av xxz

zy

yx

xx

)()]([ Av .

In considering the Lagrange equation, it is supposed that there are two kinds of independent variables, r and v. In this case

)()( Av

xx

Av

x

Av

x

Av z

zy

yx

x .

Then we have

t

A

dt

dA

xxx

x

)()]([ AvAv ,

or

tdt

d

AA

AvAv )()]([ ,

Using this formula, we get

dt

d

c

q

cq

tdt

d

c

q

tcq

dt

dm

AAv

AAAv

A

Fv

)1

(

])([)1

(

Then we have

)1

()( AvAv c

qc

qm

dt

d .

This equation takes the form of Newton’s second law. The rate of change of a quantity that looks like momentum is equal to the gradient of a quantity that looks like potential energy. It therefore motivates the definition of the canonical momentum

Avpc

qm ,

3

and an effective potential energy experienced by the charged particle,

)1

( Av c

q ,

which is velocity-dependent. The force on the charge can be derived from the velocity-dependent potential energy

)1

( Av c

qU .

So the Lagrangian L is defined by

)1

(2

1 2 Avv c

qmUTL

The canonical momentum is defined as

Avv

pc

qm

L

.

Then the mechanical momentum (the measurable quantity) is given by

Apvπc

qm .

The Hamiltonian H is given by

qc

q

m

qm

Lc

qm

LH

2

2

)(2

12

1

)(

Ap

v

vAv

vp

.

The Hamiltonian formalism uses A and , and not E and B, directly. The result is that the description of the particle depends on the gauge chosen. In conclusion we have two kinds of momentum.

Mechanical momentum vπ m

Canonical momentum: Avpc

qm .

4

or

Apvπc

qm

2. Hamiltonian

The Hamiltonian of the classical radiation field ( p : momentum operator of the system, Quantum mechanical operator) is given by

]ˆˆˆ[2

1

ˆˆ2

1

ˆ2

1ˆ

22

22

2

AppAAp

ApAp

Ap

c

e

c

e

m

c

e

c

e

m

c

e

mH

where q = -e is the charge of electron (e>0) and = 0.

ArrA

ArArrA

rArArAppA

)()(2

)()()(

)()()(ˆˆ

ii

ii

ii

Thus

ApAAp

ic

e

c

e

c

e

mH

ˆ

2ˆ

2

1ˆ 22

22 .

We use the Coulomb gauge A 0. Then we have the perturbations such that

22

2

2"ˆ

ˆ'ˆ

A

pA

mc

eH

mc

eH

.

where we use the vector potential A for the classical case. 3. Classical radiation field Maxwell's equation:

5

tcc

tc

EjB

B

BE

E

140

14

The equation of the continuity

0

t

j ,

tc

AE

AB1 ,

where A is the vector potential and is the scalar potential.

ctctc

tcj

AA

A

A

411

41

2

2

22

2

Gauge transformation:

tc

1

'

' AA

4. Coulomb gauge We start any pair of A and . Using the Gauge transformation we have a pair of A’and ’, where

0' A or

0)( A or

A 2

6

This is a Poisson equation with known value of A . The solution of is uniquiely determined. Therefore we can always choose the Coulomb gauge with 0' A . 5. Vector potential A in the Coulomb gauge

Here we assume that

0 A (Coulomb gauge) In the vacuum, we have

0 , 0j

011

0

2

2

22

2

tctc

AA

From the first equation, we have = 0 or

( 0 A , 0 ) Then we have

01

2

22

tc

AA , with 0 A

The plane wave monochromatic solution for the wave equation is

t rkAA cos2 0 ,

where

2 k2c2 or ck (Dispersion relation) Note that

0sin2 0 trkAkA

since

)sin(

)cos()]cos([))cos((

0

000

t

ttt

rkAk

ArkArkrkA.

7

Then we have

00 Ak

In other words, 0A is perpendicular to the wavevector k.

e

n

k

Fig. n; propagation vector of the light. ε is the polarization vector. The vector potential A is

parallel to the polarization vector. A must lie in a plane perpendicular to the direction of the propagation vector ( n ).

]2Re[cos2 000tititi eeet rkrkrk AεεArkεAA ,

]2Re[

sin2

sin2

sin)1

(2

1

0

0

0

0

tieik

tk

tc

tcc

tc

rkεA

rkεA

rkεA

rkεA

AE

8

and

])(2Re[

sin)(2

sin)(2

0

0

0

tieki

tc

t

rkεnA

rkεnA

rkεkA

AB

,

where n is the unit vector defined by kkn / and is the polarization vector (unit vector). The

direction of the magnetic field and electric field is perpendicular to the propagation direction, forming the transverse wave. 6. Poynting vector

The electromagnetic energy is given by

tc

tcc

22cos12

sin1

41

48

1

8

1

2

2

02

222

2

02

2

2

022

rkA

rkAABE

The time average is

)(erg/cm 28

11 32

2

02

0

22 uc

dtT

T

A

BE ,

where u is the energy density. The Poynting vector is given by

BES 4

c.

Since

tc

tcc

22cos12

14

sin)ˆ(2ˆ2

2

22

0

200

rknA

rkεnAεABE

the time average of the Poynting vector is obtained as

9

)cm(erg/s 2

1 222

0

0

cuc

dtT

T

nnA

S

.

area 1

photons

ct

The intensity s; the energy flow per unit area per unit time.

ccus

2

2

02 A

(erg /s cm2).

The flux of photons (the number of photons per unit area per unit time)

c

sf

2

2

0A .

7. Application interaction with the classical radiation field

We consider the absorption and emission of light which is caused through the interaction between atoms and electromagnetic fields. The light is the electromagnetic field which periodically varies with time. Here we discuss the absorption and stimulated emission, where the electromagnetic field is semi-classically treated and the atoms are quantum-mechanically treated. There is another emission, so-called the spontaneous emission, where the electromagnetic field should be quantum-mechanically treated. ((Classical radiation field))

The electric or magnetic field is derivable from a classical radiation field as opposed to quantized field,

pArp ˆ)ˆ(ˆ2

1ˆ 2 mc

ee

mH

which is justified if

0 A . (Coulomb gauge) We work with a monochromatic field of the plane wave

t rkεAA cos2 0

10

nkc

, 0kε

(ε and n are the (linear) polarization and propagation directions.) or

titi ee rkrkεAA 0

The Hamiltonian is given by

10ˆˆˆ HHH ,

where 1H is the time dependent perturbation

titi

titi

eHeH

eemc

eH

11

01

ˆˆ

)ˆ]([ˆ

pεA rkrk

where

)ˆ(ˆ01 pεA rk ie

mc

eH , )ˆ(ˆ

01 pεA rk iemc

eH

(i) Absorption

The first term: responsible for absorption

ii

ffi emc

eH pε

A rk ˆˆ 01

and

ifii

ffi EEecm

eW

22

022

2

ˆ2

pεA rk

((Fermi’s golden rule))

where the energy is conserved during the process; Ef Ei

11

Ei

E f

Absorption

Ñw

Fig. Absorption process. if EE .

(ii) Stimulated emission

The second term: responsible for the stimulated emission

ii

ffi emc

eH pε

A rk ˆˆ 01

and

ifii

ffi EEecm

eW

22

022

2

ˆ2

pεA rk

12

E f

Ei

Stimulated Emission

Ñw

Fig. Stimulated emission process. if EE .

8. The semi-classical form of 2

0A

As is derived above, the transition probability rate is given by

ifii

ffi EEecm

eW

22

022

2

ˆ2

pεA rk

The semiclassical form of 2

0A can be obtained using the expressions of the energy density u,

and the magnitude of the Poynting vector, s,

s

1

cm

serg )()

cm

erg(

2 33

2

02

2

dWc

u A

)(cm

serg

3 dIcus

Note that the units of 0A is G.cm, since erg =G2 cm3, dI )( is the intensity and dW )( is

the energy density between and + d,

)()( WcI , )(1

)( Ic

W

13

We assume that these quantities are dependent on the angular frequency . We note that the

units of )(W and )(I are given by

]cm

serg[ )(

3

W , ]

cm

erg[ )(

3 dW ,

23 cm

erg

s

cm

cm

serg)(I ,

sdI

2cm

erg)( .

Using the relation

dWc

dIc

)(2

)(2

2

2

2

2

0 A ,

we get the transition probability as

ifi

iffi EEedW

c

cm

eW

2

2

2

22

2

ˆ)(22

pεrk ,

where 0

1

if EE and 0 if EE .

(i) Absorption:

2

020

22

22)( ˆ)(

4i

if

afi eW

m

eW

pεrk

.

(ii) Stimulated emission:

2

020

22

22)( ˆ)(

4i

if

efi eW

m

eW

pεrk

9. Electric dipole approximation

The vector potential periodically change over the order of the distance (wavelength, 600 nm = 6000 Å). The radius of electron in atoms is much smaller than the wavelength. In this case, we can use the approximation,

11 rkrk iei . Here

14

2

kc

,

atom2 r

c

rnrnrnrk

« 1.

This approximation is valid for » ratom (atomic dimension). ((Note))

atom

2

0

2

/ r

Ze

Za

Ze , a0/Z: atomic level spacing

atom2atom 137

2r

ZZe

rcc

where

(

e2

c

1

137)

In other words

137

1atom

Zr

« 1

for the light atoms (small Z). In this approximation, we get

2

020

22

22)( ˆ)(

4if

afi W

m

eW

pε

.

(ii) Stimulated emission

2

020

22

22)( ˆ)(

4if

efi W

m

eW

pε

.

Next we need to calculate the matrix element if pε ˆ . For simplicity we take

xeε ( zen )

Then we get the matrix element as

15

ixf p ˆ .

Suppose that the Hamiltonian is given by

)ˆ,ˆ,ˆ()ˆˆˆ(2

1ˆ 2220 zyxVppp

mH zyx .

Then we have

xx pm

ipx

mHx ˆ]ˆ,ˆ[

2

1]ˆ,ˆ[ 2

0

,

where i and i are the eigenkets of 0H ,

iii EH 0ˆ , fff EH 0

ˆ .

We have

iffiifififif xxEExHHxHx ˆˆ)(]ˆˆˆˆ]ˆ,ˆ[ 000

or

iffiixf xpm

i ˆˆ

or

iffiixf ximp ˆˆ

In the direction of the electric polarization vector, we have

ififif rmm ˆˆ 00 rεpε

Using this expression, we get the final form

16

)(

ˆ)(4

ˆ)(4

ˆ)(4

012

2

02

22

220

202

022

22

2

020

22

22

WB

rWe

rmWm

e

Wm

eW

if

if

iffi

pε

or

2

02

22

ˆ)(4

iffi rIc

eW

where we use the electric dipole approximation

)(1

)( 00 Ic

W

B12 and B21 are called the Einstein B-coefficient. We have

2

2

22

2

2

22

2112

ˆ3

4

ˆ4

if

if

re

re

BB

(Average)

The factor 1/3 arises from the random distribution of since the radiation is isotropic.. Note that

cosrr and is the angle between and r. We need to take an average over the random

orientations of the electric dipole moments.

3

1sincos

2

1)sin2(cos

4

1cos

0

2

0

22

dd .

where

17

q

rre

e

Fig. cosrr . The angle is variable, since the orientation of the dipole moment is not

fixed. 12. Relation between Einstein’s A & B co-efficients

The energy density in thermal equilibrium between and d is given by dW T )( . We know that the Planck’s law for the radiative energy density is given by

1)(

32

2

ecW T .

from the Black-body problem (see the Black body problem in the APPENDIX)

Suppose that a gas of N identical atoms is placed in the interior of the cavity:

E2 E1 . Two atomic levels are not degenerate. N1, N2 are the level population.

18

We assume that

W() WT ( )WE ( ) where

W(): cycle-average energy density of radiation at

WT( ): thermal part

WE ( ): contribution from some external source of electromagnetic radiation

E2

E1

A21 B12W B21W

Spontaneous emission

Absorption

Stimulated emission

N1

N2

We set up the rate equations for N1 and N2

19

)()(

)()(

2121212212

2121212211

WBNWBNNAdt

dN

WBNWBNNAdt

dN

Note that the spontaneous emission is independent of )(W . In the case of thermal equilibrium, we have

dN1

dt

dN2

dt 0 ,

or

N2 A21 N1B12W() N2B21W( ) 0 . For thermal equilibrium with no external radiation introduced into the cavity

W() WT ( ) with

21122

1

21)(

BBN

N

AWT

The level populations N1 and N2 are related in thermal equilibrium by Boltzman’s law

N1

N2

eE1

eE2 exp() , ( = 1/kBT)

Then

12

21

21

21

2112

21)(

BB

e

BA

BeB

AW T

,

which is compared with the Planck’s law,

1)(

32

3

e

cW T

20

with

32

3

12

21

2112

cB

ABB

nB

AW T

12

21)( ,

where

1

1

e

n

or

1)(21

21

e

WB

A

T

((Example)) TkBT ( T , TT 2 )

For T = 300 K, T = 6.25 1012 Hz = 6.25 THz

For TkB , )(2121 TWBA ( << T)

For TkB , )(2121 TWBA ( >> T) For optical experiments that use electromagnetic radiation in the near-infrared, we have visible, ultraviolet region of the spectrum ( >> 5 THz). We have

(i) )(2121 TWBA

A21: spontaneous emission rate

B21 )(TW : rate of thermally stimulated emission

(ii) )()()()( EET WWWW Therefore the radioactive process of interest involve the absorption and stimulated emission associated with the external source.

21

E2

E1

A21 B12WE B21WE

Spontaneous emission

Absorption

Stimulated emission

N1

N2

)()(

))()(

2121212212

2121212211

EE

EE

WBNWBNNAdt

dN

WBNWBNNAdt

dN

13. Spontaneous emission (quantum mechanics) From the above discussion, we get

32

3

12

21

2112

cB

ABB

We note that

2

2

222

2

22

2112 ˆ3

4ˆ

4ifif r

er

eBB

Thus we have

2

3

23

2

2

22

32

3

2132

3

21

ˆ3

4

ˆ3

4

if

if

rc

e

re

c

Bc

A

22

The radiative lifetime is given by

21

1

A

or

23

23

322

3

232

3

23

21

2

3

4

3

4

3

4ˆ

3

41

xcx

cc

ecx

c

er

c

eA if

(1) or

23

18

]][[)]([

102354.71Ax

A

s-1.

with

2

c

, c

e

2

.

Setting x = 1 Å and 400 nm = 4000 Å (typical value), we get

23

18

14000

102354.71

= 1.13 x 108 s-1.

((Note))

281061

s-1

for the transition 2P to 1S in atomic hydrogen. It is interesting to compare Eq.(1) with the result obtained from the classical radiation theory (see later). The power radiated by accelerated particle of the charge (-e) is given by the Larmor formula

3

22

3

)(2

c

veP

,

where v is the acceleration. If we assume that the particle undergoes a circular motion of radius r, with uniform angular velocity. The centripetal acceleration is

23

rr

vv 2

2

.

We can argue that the time requires for the classical system to radiate energy 2/ is equivalent to the lifetime . Thus

3

232

3

242

3

22

3

4

3

4

3

)(421

c

re

c

re

c

veP

(classical)

Note that the qualitative agreement between the classical and quantum mechanical results is a manifestation of the correspondence principle. However, the mechanism for the emission of the radiation is completely different in the two cases, and the classical argument can never produce the discrete spectrum of the radiation. 14. Larmor’s formula (classical)

The classical electrodynamics tells us that an accelerating charge radiates an electromagnetic field with far-field electric and magnetic field values. The instantaneous electromagnetic energy flow is given by the Poynting vector

nE

EnEnE

EnE

BES

2

2

4

])([4

)(4

4

c

c

c

c

The electric field is given by

retRc

e]

)([

2

vnnE

v is evaluated at the retarded time cRttret / . This radiation has the characteristic dipole

pattern and causes the electric dipole to lose energy, that is, to be damped. The pointing vector is then obtained as

nvnvnnvnnS )(sin1

4)(

1

4)]([

1

422

23

22

23

22

23

2

Rc

e

Rc

e

Rc

e ,

where is the angle between n and v .

24

2222 sin)()]([ vvnvnn .

The total power radiated is given by the integration of S over a sphere surrounding the charge,

2

3

2

0

32

3

2

232

2

3

2

22

2

3

2

3

2

sin2

)2(sin1

4

sin1

4

v

v

v

anvaS

c

e

dc

e

dRRc

e

dRc

ed

where

3

4sin

0

3

d .

15. Absorption cross section

The absorption cross section is defined as

abs = absorption cross section

][cm

)(ergs/cm2

1

(erg/s)

s)(erg/cm field radiation theofflux Energy

)( atom by the absorpted it time)(energy/un

2

22

0

2

2

Ac

W

fi

fi

ˆ4

2

1

ˆ2

22

2

2

2

0

2

22

022

2

abs

ifii

f

ifii

f

EEec

e

m

c

EEecm

e

pε

A

pεA

rk

rk

In the electric dipole approximation,

25

fiif

fiif

fiii

fabs

m

c

e

m

ec

e

m

2

2

2

22

2

2

22

2

2

ˆ4

ˆ4

ˆ4

pε

pε

pεrk

When xeε , we have

iffiixf ximp ˆˆ

Then

fiiffi

fiiffiabs

x

xmm

22

222

2

2

ˆ4

ˆ)(4

where

c

e

2

. (fine structure constant).

In atomic physics, we define oscillator strength ffi

2ˆ

2if

fifi x

mf

Thomas-Reiche-Kuhn sum rule indicates that

ffif 1.

Using this rule

26

2

22

2

2

22

22

2

24

24

ˆ2

24

ˆ4)(

mc

ec

m

fm

xm

m

xd

ffi

fif

fi

fiffiabs

16. Thomas-Reiche-Kuhn sum rule

We consider a particle in one dimension whose Hamiltonian is given by

)ˆ(ˆ2

1ˆ 2 xVpm

H x

We have the Thomas-Reiche-Kuhn rule that

mEEnx

nn 2

)(ˆ02

0

2

where

nEnH nˆ

27

m

mi

pxm

i

xm

pi

xm

p

pi

xxm

p

xxxVxxm

pxxH

2

2

2

2

2

1

ˆ,ˆ1

]ˆ,ˆ

[

]ˆ,2

ˆ

ˆ[

]ˆ],ˆ,2

ˆ[[

]ˆ],ˆ),ˆ([[]ˆ],ˆ,2

ˆ[[]ˆ],ˆ,ˆ[[

Then

mnxxH

2

]ˆ],ˆ,ˆ[[0

(1)

On the other hand

nn

nnn

n

nxEE

xnnxEExnnxEE

xHnnxxnnxH

xHxxxHxxH

2

0

00

02

0ˆˆ00ˆˆ0

0]ˆ,ˆ[ˆ00ˆ]ˆ,ˆ[0

0]ˆ,ˆ[ˆ00ˆ]ˆ,ˆ[00]ˆ],ˆ,ˆ[[0

(2) Combining Eqs. (1) and (2), we obtain

n

n mnxEE

2ˆ0

22

0

.

or

n

n mnx

2ˆ0

2

0

28

where

00 nn EE

17. Electric dipole transition selection rule: hydrogen atom

2ˆ ifI r

zifyifxifif zyx eeer ˆˆˆˆ

is a vector. Then we have

2222ˆˆˆˆ ifififif zyxI r

Here we note that

iffiififif yixyixyixyixyix ˆˆˆˆˆˆˆˆˆˆ*2

or

222ˆˆˆˆ iffiif yxyix

Similarly we have

222ˆˆˆˆ iffiif yxyix

Then we have

2222ˆˆˆ

2

1ˆˆ

2

1ˆ ifififif zyixyixI r

Spherical tensor of rank 1

2

ˆˆ)1(1

yixT

zT ˆ)1(

0

29

2

ˆˆ)1(1

yixT

From the Wigner-Eckart theorem,

mlnTmln q ,,ˆ',',' )1( ≠ 0 for qmm ' and for 1,,1' llll

)1(

qT is the odd parity operator,

)1()1( ˆˆˆˆ qq TT

and

mlnmln l ,,)1(,,ˆ

Then the matrix element mlnTmln q ,,ˆ',',' )1( is equal to zero for ll ' .

Then we have

(i) 0,,ˆˆ',','2

1,,ˆ',',' )1(

1 mlnyixmlnmlnTmln

for 1' mm and for 1' ll

(ii) 0,,ˆˆ',','2

1,,ˆ',',' )1(

1 mlnyixmlnmlnTmln

for 1' mm and for 1' ll .

(iii) 0,,ˆ',',',,ˆ',',' )1(0 mlnzmlnmlnTmln

for mm ' and for 1' ll .

______________________________________________________________________________ REFERENCES J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley,

New York, 2011). E. Merzbacher, Quantum Mechanics, third edition (John Wiley & Sons, New York, 1998). L. Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955). G. Baym, Lectures on Quantum Mechanics (Westview Press, 1990).

30

D. Bohm, Quantum Theory (Prentice Hall, New York, 1951). H. Lüth, Quantum Physics in the Nanoworld (Springer, 2013). D. Park, Introduction to the Quantum Theory, 2nd-edition (McGraw-Hill, New York, 1974). E. Hecht, Optics, 4-th edition (Addison Wesley, 2002). J.D.Jackson, Classical Electrodynamics (John Wiley & Sons, 1962) A. Das and A.C. Melissinos, Quantum Mechanics A Modern Introduction (Gordon and Breach

Science Publishers, New York, 1986). A. Das, Lectures on Quantum Mechanics, 2nd edition (World Scientific Publishing Co.. Ltd,

2012). H.A.Bethe and R. Jackiw, Intermediate Quantum Mechanics (Westview, 1997). ___________________________________________________________________________ APPENDIX Black body problem A.1 Maxwell’s equation

We start with the Maxwell’s equation (in cgs units)

tc

tc

EB

BE

B

E

1

1

0

0

We assume that

]~

Re[

]~

Re[)

0

0

ti

ti

e

e

BB

EE

Then we have

0~

0~

0

0

B

E

00

~~BE

ci

00

~~EB

ci

EBEEEE2

2

222 1

)()()(tct

or

31

EE2

2

22 1

tc

or

0~~

02

02 EE k

with ck . Similarly, we have

tc

B

B2

2 1

0~~

0

2

202 BB

c

or

0~~

02

02 BB k

We now consider an electromagnetic wave in the closed cube with side L.

Fig. Boundary condition for the electric field (red) (tangential component continuous) and the

magnetic field (green) (normal component continuous). From the boundary conditions we have

32

)sin()sin(

)cos(

)sin( 32

1

11

zkyk

xk

xkEEx

)sin(

)cos(

)sin()sin( 3

2

212

zk

yk

ykxkEEy

)cos(

)sin()sin()sin(

3

3213 yk

zkykxkEEz

where

xnL

k

1 , ynL

k

2 , znL

k

3

(nx, ny, nz = 1, 2, 3, …)

)cos()sin()cos(

)sin(11

1

1 xkBxkAxk

xk

.

Note that

Ex = 0 for y = 0 and y = L planes and z = 0 and z = L planes. Ey = 0 for z = 0 and z = L planes and x = 0 and x = L planes. Ez = 0 for x = 0 and x = L planes and y = 0 and y = L planes.

From the condition

0~ E

we have

)sin()sin()cos( 3211 zkykxkEEx ,

)sin()cos()sin( 3212 zkykxkEEy ,

)cos()sin()sin( 3213 ykykxkEEz

From the condition

00

~~BE

ci

,

33

we have

)cos()cos()sin( 3211 zkykxkBBx ,

)cos()sin()cos( 3212 zkykxkBBy ,

)sin()cos()cos( 3213 zkykxkBBz

where

Bx = 0 for x = 0 and x = L planes By = 0 for y = 0 and y = L planes. Bz = 0 for z = 0 and z = L planes.

We note that

0)sin()sin()sin()( 321332211 zkykxkkEkEkEE

This means that the vector (E1, E2, E3) is perpendicular to the wave vector k = (k1, k2, k3). For each k, there are two independent directions for (E1, E2, E3); polarization.

A.2. Density of states for the modes

Since 0332211 kEkEkE , only one of k1,k2, k3 can be zero at a time. Since if two or three

are zero, E1 = E2 = E3 = 0. There is no electromagnetic field in the cavity. Each set of integers (nx, ny, nz) defines a mode of the radiation field and corresponds to two degrees of freedom of the field when two polarization directions are taken into account.

34

There are 2 states per 3

L

.

222

zyx kkkcck

or

222

2

2

zyx kkkc

The density of states (k to k +dk)

2

2

3

2

24

8

1

dkVk

L

dkkdkk

.

where V = L3.

35

Since ck ,

32

2

2

2

c

dVcd

cV

d

of modes having their frequencies between and +d,

)(32

2

VDc

V (density of modes)

where c is the velocity of light and

32

2

)(c

D .

We have the following formula;

dkkk

,

or

36

dDVdk

)( .

For single mode k , the energy is given by

kk )2

1(. kn nE .

We use the Planck distribution. The total energy is given by

duVndc

VnEtot )(

32

2

kk

k ,

or the energy density by

00

)()( duduV

Etot ,

where

1)exp(1)exp(

1

1)exp(

1)()(

3

322

33

32

3

32

3

x

x

c

Tk

xcTk

cWu B

B

T

.

(Planck’s law for the radiation energy density). It is clear that

1)exp()(

)( 3

322

33

x

xxf

c

Tk

u

B

,

is dependent on a variable x given by

Tkx

B

.

(the scaling relation). The experimentally observed spectral distribution of the black body radiation is very well fitted by the formula discovered by Planck.

(1) Region of Wien ( 1Tk

xB

),

37

xBW ex

c

Tku 3

322

33

)(

.

(2) Region of Rayleigh-Jeans ( 1Tk

xB

),

2322

333

322

33

1)exp()( x

c

Tk

x

x

c

Tku BB

RJ

.

0 2 4 6 8 10x=

Ñw

kB T0.0

0.5

1.0

1.5

u wkB3 T3

c3 p2 Ñ2

Rayleigh-Jean wave

Wien particle

Planck

Fig. Scaling plot of f(x) vs x for the Planck's law for the energy density of electromagnetic

radiation at angular frequency and temperature T. Planck (red). Wien (blue, particle-like). Rayleigh-Jean (green, wave-like).

38

0.2 0.5 1.0 2.0 5.0 10.0 20.0x=

Ñw

kB T0.0

0.5

1.0

1.5

u wkB3 T3

c3 p2 Ñ2

RJ

W

P

Fig. Scaling plot of Planck's law. Wien's law, and Rayleigh-Jean's law. A.3 Deivation of u(, T)

032

3

0 1)exp(

1)(

d

Tkc

du

B

Since c2

, 2

2 d

cd

0232

3

032

3

0

21)

2exp(

12

1)exp(

1)(

dc

Tk

cc

c

d

Tkc

du

B

B

or

05

2

00 1)2

exp(

1116)()(

d

Tk

ccdudu

B

Then we have

39

1)2

exp(

116)(

5

2

Tkc

cu

B

where

= 1.054571596 x 10-27 erg s, kB = 1.380650324 x 10-16 erg/K c = 2.99792458 x 1010 cm/s. J = 107 erg

A.4 Wien’s displacement law

u() has a maximum at

96511.42

Tk

c

B

, (dimensionless)

or

)(

28977.0

KT ( in the units of cm)

or

610)(

897768551.2

KT . ( in the units of nm)

T is the temperature in the units of K. is the wave-length in the unit of nm

T(K) (nm)

40

1000 2897.771500 1931.852000 1448.892500 1159.113000 965.9243500 827.9354000 724.4434500 643.9495000 579.5545500 526.8676000 482.9626500 445.8117000 413.9677500 386.3698000 362.2218500 340.9149000 321.9759500 305.02910 000 289.777

1000 2000 3000 4000 5000 6000 7000TK0

500

1000

1500

2000lmaxnm

Wien's displacement law

Fig. Wien's displacement law. The peak wavelength vs temperature T(K). A.5 Rate of the energy flux density

It is assumed that the thermal equilibrium of the electromagnetic waves is not disturbed even when a small hole is bored through the wall of the box. The area of the hole is dS. The energy which passes in unit time through a solid angle d, making an angle with the normal to dS is

41

dSd

dTcudSddTJ

4

cos),(),,(

,

where c is the velocity of light. The right hand side is divided by 4, because the energy density u comprises all waves propagating along different directions. The emitted energy unit time, per unit area is

4

),(4

4

),(

4cos),(),,(

c

dTuc

dTcud

dTcuddTJ

where

dTu ),( ,

4

1|)]2cos([

2

1

4

1

)2sin(2

12

4

1

sincos4

1

4cos

2/0

2/

0

2/

0

2

0

d

ddd

(only for the half upper plane).

42

x

y

z

q

dq

f df

r

drr cosq

r sinqr sinq df

rdq

er

ef

eq

dS

Fig. Radiation intensity is used to describe the variation of radiation energy with direction.

43

Fig. Geometrical factor. The photons pass from the lower half plane to the upper half plane in

a straight way through a pin hole with the effective area (dS cos). Since the area dS is small enough, the form of the wave changes from plane wave to spherical wave. The spherical wave propagates in all directions (the total solid angle 4) after passing through the pin hole. The fraction of the photons propagating over the solid angle (d) is d/4

In other words, the geometrical factor is equal to 1/4. Then we have a measure for the intensity of radiation (the rate of energy flux density);

1)2

exp(

14

4

),(),(

5

22

Tk

ccTcu

TS

B

where

S (λ ,T)dλ = power radiated per unit area in ( , + d) Unit

][101

)10(

101.]

1[

331

32

7

32

2

552

m

W

m

W

sm

J

scm

erg

s

cm

cm

sergc

The energy flux density ),( TS is defined as the rate of energy emission per unit area. ((Note)) The unit of the poynting vector <S> is [W/m2]. S is the energy flux (energy per

unit area per unit time). (1) Rayleigh-Jeans law (in the long-wavelength limit)

452 2

211

4)(4

1)(

Tk

Tk

cccuS B

B

RJRJ

for

12

c

TkB

(2) Wien's law (in short-wavelength limit)

44

)2

exp()(4

1)(

5

2

Tk

cccuS

BWW

12

c

TkB

We make a plot of ),( TS as a function of the wavelength, where ),( TS is in the units of

W/m3 and the wavelength is in the units of nm.

0 5000 10000 15000 20000l nm10-5

0.001

0.1

10

1

4c ul 1014 Wm3

T =2000 K

Fig. cu()/4 (W/m3) vs (nm). T = 2 x 103 K. Red [Planck]. Green [Wien]. Blue [Rayleigh-

Jean]. Wien's displacement law: The peak appears at = 1448.89 nm for T = 2 x 103 K. This figure shows the misfit of Wien's law at long wavelength and the failure of the Rayleigh-Jean's law at short wavelangth.

100 200 500 1000 2000 5000 1μ104l nm10-4

0.001

0.01

0.1

1

10

1

4c ul 1014 Wm3

45

3000 K3500 K

4000 K

4500 K

T=5000 K

0 500 1000 1500 2000 2500l nm0

1

2

3

4

1

4c ul 1014 Wm3

Fig. (a) and (b) cu()/4 (W/m3) vs (nm) for the Plank's law. T = 1000 K (red), 1500 K, 2000

K, 2500 K, 3000 K (blue), 3500 K, 4000 K (purple), 4500 K, and 5000 K. The peak shifts to the higher wavelength side as T decreases according to the Wien's displacement law.

Fig. Power spectrum of sun. cu()/4 (W/m3) vs (nm). T = 5778 K. The peak wavelength is

501.52 nm according to the Wien's displacement law.

46

1 2 3 4 5 6lmm

0.5

1.0

1.5

2.0

1

4cul 10-2 Wm3

T=2.726 K

1.063 mm Cosmic Radiation

Fig. Power spectrum of cosmic blackbody radiation at T = 2.726 K. The peak wavelength is

1.063 mm (Wien's displacement law. ________________________________________________________________________ A.6. Stefan-Boltzmann radiation law for a black body (1879). Joseph Stefan (24 March 1835 – 7 January 1893) was a physicist, mathematician and poet of Slovene mother tongue and Austrian citizenship.

http://en.wikipedia.org/wiki/Joseph_Stefan _______________________________________________________________________ Ludwig Eduard Boltzmann (February 20, 1844 – September 5, 1906) was an Austrian physicist famous for his founding contributions in the fields of statistical mechanics and statistical thermodynamics. He was one of the most important advocates for atomic theory at a time when that scientific model was still highly controversial.

47

http://en.wikipedia.org/wiki/Ludwig_Boltzmann ____________________________________________________________________ The total energy per unit volume is given by

33

42

0

3

332

4

15

)(

1)exp(

)()()(

c

Tk

x

x

c

Tkdudu

V

E BBtot

((Mathematica))

0

x3

x 1x

4

15 A spherical enclosure is in equilibrium at the temperature T with a radiation field that it contains. The power emitted through a hole of unit area in the wall of enclosure is

4423

42

604

1TT

c

kcP B

where is the Stefan-Boltzmann constant

423

42

105670400.060

c

kB

erg/s-cm2-K4 = 5.670400 x 10-8 W m-2 K-4

and the geometrical factor is equal to 1/4. The application of the Stefan-Boltzmann law is discussed in lecture notes of Phys.131 (Chapter 18) (see URL at http://bingweb.binghamton.edu/~suzuki/GeneralPhysLN.html

48

A.7 Duality of wave and particle Region of Rayleigh-Jeans: wave-like nature Region of Wien: particle-like nature The mean energy contained in a volume V in the frequency range between and +, is given by

1

1)()()()()(

eVDnVDWVEE T

where

1

1

e

n ,

and

32

2

)(c

D

The mean-square of the fluctuation in energy is obtained as

)()()]([)]([ 2222 ET

TkEEE B

from the general theory of thermodynamics, or

]1

1

1

1[)(

1

1)()]([ 2

2222

ee

DVeT

TkDVE B

or

2222222 )()(][)()]([ nDVnnDVE

where

22 )( nnn

49

(See the Appendix for the detail). Note that

2222 )()( nnnnn (from the definition). (i) Rayleigh-Jean (wave-like)

For 1

TkB

, nn 2

22)( nn , or nn )( (wave-like, Rayleigh-Jeans)

2222 )()]([ nDVE

Then we have

2

3

2

222

2

21

)(

1

))((

)(

)(

)]([

c

VDVnDV

nDV

E

E

(ii) Wien (particle-like)

For 1

TkB

, nn 2

nn 2)( (particle-like, corpuscle, Wien)

)())(()()]([ 222 EnDVnDVE

or

)(

)]([ 2

E

E

(iii) Planck

2

2

32 )(

1)()]([

E

c

VEE

______________________________________________________________________