Chapter 1 Thermal radiation and Planck’s postulate
1.2 thermal radiation

Thermal radiation: The radiation emitted by a body as a result of temperature.

Blackbody : A body that surface absorbs all the thermal radiation incident on
them.

Spectral radiancy RT ( ): The spectral distribution of blackbody radiation.
RT ( )d : represents the emitted energy from a unit area per unit time
between  and   d at absolute temperature T.
1899 by Lummer and
Pringsheim
Chapter 1 Thermal radiation and Planck’s postulate

The spectral radiancy of blackbody radiation shows that:
(1) little power radiation at very low frequency
(2) the power radiation increases rapidly as ν increases from very
small value.
(3) the power radiation is most intense at certain  max for particular
temperature.
(4)    max , RT ( ) drops slowly, but continuously as ν increases
, and RT (  )  0.
(5)  max increases linearly with increasing temperature.

(6) the total radiation for all ν ( radiancy RT   RT ( )d )
0
increases less rapidly than linearly with increasing temperature.
Chapter 1 Thermal radiation and Planck’s postulate

4
8
2 o
4
Stefan’s law (1879):RT  T ,   5.67 10 W / m  K
Stefan-Boltzmann constant

Wien’s displacement (1894):  max  T
1.3 Classical theory of cavity radiation

Rayleigh and Jeans (1900):
(1) standing wave with nodes at the metallic surface
(2) geometrical arguments count the number of standing waves
(3) average total energy depends only on the temperature

one-dimensional cavity:
one-dimensional electromagnetic standing wave
2x
E ( x , t )  E0 si n (
) si n (2 t )

Chapter 1 Thermal radiation and Planck’s postulate

for all time t, nodes at 2x /   n , n  0,1,2,3.......
x0
x  a  2a  n    2a / n    nc / 2a
standing wave
N ( )d : the number of allowed standing wave between ν and ν+dν
n  ( 2a / c )  dn  ( 2a / c )d
N ( )d  2  dn  (4a / c )d
two polarization states
d  (2a / c )(  d )
d  (2a / c )
0
n
Chapter 1 Thermal radiation and Planck’s postulate

for three-dimensional cavity
r  (2a / c )  dr  (2a / c )d
the volume of concentric shell r  r  dr
2a 2 2 2a
2a
) v ( )d  4 ( ) 3 2 d
c
c
c
1
8a 3 2
8V 2
2
N ( )d  2   4r dr 

d


 d
3
3
8
c
c
The number of allowed electromagnetic standing wave in 3D
4r 2 dr  4 (
Proof:
( x / 2) cos   / 2
( y / 2) cos    / 2
λ/2
propagation
direction
( z / 2) cos   / 2
E ( x , t )  E0 x si n (2x /  x ) si n (2 t )
E ( y , t )  E0 y si n (2y /  y ) si n (2 t )
E ( z , t )  E0 z si n (2z /  z ) si n (2 t )
λ/2
nodal
planes
Chapter 1 Thermal radiation and Planck’s postulate

for nodes:
x  0, a ,2 x /  x  n x , n x  1,2,3.....
y  0, a ,2 y /  y  n y , n y  1,2,3.....
z  0, a ,2 z /  z  nz , nz  1,2,3.....
( 2a /  ) cos  n x , ( 2a /  ) cos   n y , ( 2a /  ) cos  nz
 ( 2a /  ) 2 (cos2   cos2   cos2  )  n x2  n 2y  nz2
 2a /  
n x2  n 2y  nz2
  c /   (c / 2a ) n x2  n 2y  nz2  (c / 2a )r
r
n x2  n 2y  nz2  ( 2a / c )  dr  ( 2a / c )d
N ( r )dr  (1 / 8)4r 2dr  r 2dr / 2  N ( )d
 N ( )d  ( / 2)(2a / c )3 2d  4 (a / c )3 2d
considering two polarization state
N ( )d / V  2  4 (1 / c)3 2d
N ( )  8 2 / c 3 : Density of states per unit volume per unit frequency
Chapter 1 Thermal radiation and Planck’s postulate

the law of equipartition energy:
For a system of gas molecules in thermal equilibrium at temperature T,
the average kinetic energy of a molecules per degree of freedom is kT/2,
k  1.38 1023 joule/ oK is Boltzmann constant.

average total energy of each standing wave :   2  KT / 2  KT

the energy density between ν and ν+dν:
8  2
T ( )d  3 kTd Rayleigh-Jeans blackbody radiation
c
ultraviolet catastrophe
Chapter 1 Thermal radiation and Planck’s postulate
1.4 Planck’s theory of cavity radiation
Planck’s assumption:    (T , ) and   kT ,   0
 0
 
 the origin of equipartition of energy:

Boltzmann distribution P( )  e  / kT / kT
P ( )d : probability of finding a system with energy between ε and ε+dε

P ( )d

 
 P ( )d
0

0
e  / kT
1
  / kT 
P
(

)
d


d


(

kT
)
e
|0  1
0
0 kT
kT
  / kT

 e
0 P ( )d  0 kT d

1

[ (  kT )e  / kT |0   (  kT )e  / kT ]  kT
0
kT
   kT


Chapter 1 Thermal radiation and Planck’s postulate

Planck’s assumption:   0,  ,2 ,3 ,4 ..............
(1)   0    kT
small ν
(2)  large    0
large ν
      h
34
h  6.63 10
joul  s
  kT,   kT
  kT,   kT
Planck constant
Using Planck’s discrete energy to find 
  nh , n  0,1,2,3......
nh  nh  / kT
 n
e
n

e


n 0
n  0 kT
  
 
 kT n 0
1  nh  / kT
P
(

)
e
e  n



n 0
n  0 kT
n 0
  h / kT

 p( )


  kT,   kT
Chapter 1 Thermal radiation and Planck’s postulate

d

ln  e  n 
d n 0
  kT[

e
 n

d

d


e
n 0
e

 n

d  n

e
n  0 d

 n
e

 n
n 0



n 0
d
d
ln e  n ]   h
ln e  n
d n 0
d n 0
 1  e   e  2  e  3  .....
n 0
X  e 
 1  X  X 2  X 3  ....... (1  X )1  (1  e  )1
d
d
   h
l n (1  e  ) 1  (  h )
[ l n (1  e  )]
d
d
1
h
h

 h (
)
e


1  e 
e   1 e h / kT  1
h  kT  e h / kT  1  h / kT    kT
h  kT  e h / kT  h  1    0
 n
n

e

n 0

 n
e

n 0
Chapter 1 Thermal radiation and Planck’s postulate

energy density between ν and ν+dν:
8  2
h
T ( )  3   h / kT
c
e
1
T ( )d   T ( )d
d
c
8hc
1
 T ( )   T ( )
 T ( ) 2 
d

5 e hc / kT  1
dV
Ex: Show T ( )  (4 / c ) RT ( )

dA  rˆ dAcos

solid angle expanded by dA is  
4r 2
4r 2
spectral radiancy:
dAcos
RT ( )   T ( )dV (
) /(dA t )
4r 2
2
 /2
ct
cos 2
  d  d  T ( )
r si n2 dr
2
0
0
0
4r t
c
 T ( )
4

dA
r
Chapter 1 Thermal radiation and Planck’s postulate
Ex: Use the relation RT ( )d  (4 / c )T ( )d between spectral radiancy
and energy density, together with Planck’s radiation law, to derive
Stefan’s law RT  T 4 ,   2 5 k 4 / 15c 2 h3
c 
2  h 3
RT   RT ( )d   T ( )d  2  h / kT
d
0
0
0
4
c
e
1
2 ( kT )4  x 3
 2
dx
x  h / kT
c
h3 0 e x  1

3
x
4
2 ( kT )4  4
4
x
/(
e

1
)
dx


/ 15
 2

 T
0
3
c
h
15


2 5 k 4
 
15c 2 h3
Chapter 1 Thermal radiation and Planck’s postulate

Ex: Show that
I


0
(1  e

0
x 3 (e x  1)1 dx   4 / 15
1
x ( e  1) dx 
3
 x 1
)
x
 1 e
x
e


x 3 e  x (1  e  x ) 1 dx
0
2 x

 .....   e  nx
n 0
I


0
3
x e
x

e
 nx
n 0


dx    x e
n 0
3
 ( n1) x
0


1
3 y
y
e dy
4 0
n  0 ( n  1)
dx  
3
3
 ( n1) x
 e y
Set y  (n  1) x  dx  dy /(n  1)  x  y /(n  1) , e


0
y 3e  y dy  6
by consecutive partial integration


1
1
I  6

6

4
4
(
n

1
)
n 0
n 1 n
1 2
F(x )   2 
6
n 1 n
2

1
?

4
n
n 1
x  


1
1
1 4
F(x )   
 8  2  48 4   4 
5
90
n 1 n
n 1 n
n 1 n
4
x 
4
4
F : Fourier series expansion

2
Chapter 1 Thermal radiation and Planck’s postulate
Ex: Derive the Wien displacement law (  max  T ), maxT  0.2014hc / k .
T ( ) 
8
hc
5 e hc / kT  1
 max  T
d T ( )
5
hc
e hc / kT
 0  hc / kT 
0
hc / kT
2
d
e
kT (e
 1)

x
 ex  1
5
x  hc / kT
Solve by plotting: find the intersection point for two functions
x
y1  1  , y2  e  x
Y
5
y1  1  x / 5
intersection points:
x  0, x  4.965
 maxT  0.2014hc / k
y2  e  x
X
5
Chapter 1 Thermal radiation and Planck’s postulate
1.5 The use of Planck’s radiation law in
thermometry
optical pyrometer
(1) For monochromatic radiation of wave length λ the ratio of the spectral
intensities emitted by sources at T1 o K and T2 o K is given by
e hc / kT1  1
e hc / kT2  1
T1 : standard temperature ( Au Tmelting  1068oC )
T2 : unknown temperature
(2) 3o K blackbody radiation supports the big-bang theory.
Chapter 1 Thermal radiation and Planck’s postulate
1.6 Planck’s Postulate and its implication
Planck’s postulate: Any physical entity with one degree of freedom whose
“coordinate” is a sinusoidal function of time
(i.e., simple harmonic oscillation can posses
only total energy  nh
Ex: Find the discrete energy for a pendulum of mass 0.01 Kg suspended
by a string 0.01 m in length and extreme position at an angle 0.1 rad.
1

2
g
1

l 2
9.8
 1.6(1 / se c)
0.1
mgh  mg(1  cos )  0.01 9.8  0.1  (1  cos0.1)  5  10 5 ( J )
 33

E
10
E  h  6.63 10 34  1.6  10 33 ( J ) 

 2  10 29
5
E
5  10
The discreteness in the energy is not so valid.