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
______________________________________________________________________