A black body radiation explanation

1. Lecture 26. Blackbody Radiation
(Ch. 7)
Two types of bosons:
(a) Composite particles which contain an even
number of fermions. These number of these
particles is conserved if the energy does not
exceed the dissociation energy (~ MeV in the
case of the nucleus).
(b) particles associated with a field, of which the
most important example is the photon. These
particles are not conserved: if the total
energy of the field changes, particles appear
and disappear. We’ll see that the chemical
potential of such particles is zero in
equilibrium, regardless of density.

2. Radiation in Equilibrium with Matter
Typically, radiation emitted by a hot body, or from a laser is not in equilibrium: energy
is flowing outwards and must be replenished from some source. The first step towards
understanding of radiation being in equilibrium with matter was made by Kirchhoff,
who considered a cavity filled with radiation, the walls can be regarded as a heat
bath for radiation.
The walls emit and absorb e.-m. waves. In equilibrium, the walls and radiation must
have the same temperature T. The energy of radiation is spread over a range of
frequencies, and we define uS (,T) d as the energy density (per unit volume) of the
radiation with frequencies between  and +d. uS(,T) is the spectral energy
density. The internal energy of the photon gas:
    
 d
T
u
T
u S



0
,
In equilibrium, uS (,T) is the same everywhere in the cavity, and is a function of
frequency and temperature only. If the cavity volume increases at T=const, the
internal energy U = u (T) V also increases. The essential difference between the
photon gas and the ideal gas of molecules: for an ideal gas, an isothermal expansion
would conserve the gas energy, whereas for the photon gas, it is the energy density
which is unchanged, the number of photons is not conserved, but proportional to
volume in an isothermal change.
A real surface absorbs only a fraction of the radiation falling on it. The absorptivity 
is a function of  and T; a surface for which ( ) =1 for all frequencies is called a
black body.

3. Photons
The electromagnetic field has an infinite number of modes (standing
waves) in the cavity. The black-body radiation field is a superposition of
plane waves of different frequencies. The characteristic feature of the
radiation is that a mode may be excited only in units of the quantum
of energy h (similar to a harmonic oscillators) :
  
 h
ni
i
2
/
1


This fact leads to the concept of photons as quanta of the electromagnetic field. The
state of the el.-mag. field is specified by the number n for each of the modes, or, in other
words, by enumerating the number of photons with each frequency.
According to the quantum theory of radiation, photons are massless
bosons of spin 1 (in units ħ). They move with the speed of light :
c
h
c
E
p
cp
E
h
E
ph
ph
ph
ph
ph






The linearity of Maxwell equations implies that the photons do not
interact with each other. (Non-linear optical phenomena are
observed when a large-intensity radiation interacts with matter).
T
The mechanism of establishing equilibrium in a photon gas is absorption and emission
of photons by matter. Presence of a small amount of matter is essential for establishing
equilibrium in the photon gas. We’ll treat a system of photons as an ideal photon gas,
and, in particular, we’ll apply the BE statistics to this system.

4. Chemical Potential of Photons = 0
The mechanism of establishing equilibrium in a photon gas is absorption
and emission of photons by matter. The textbook suggests that N can be
found from the equilibrium condition:
0

ph

Thus, in equilibrium, the chemical
potential for a photon gas is zero:
0
,











V
T
N
F
On the other hand, ph
V
T
N
F












,
However, we cannot use the usual expression for the chemical potential, because one
cannot increase N (i.e., add photons to the system) at constant volume and at the same
time keep the temperature constant:
V
T
N
F
,










- does not exist for the photon gas
Instead, we can use 
N
G  PV
F
G 

 
V
V
T
F
V
F
P
T
,












- by increasing the volume at T=const, we proportionally scale F
0


 V
V
F
F
G
Thus,
- the Gibbs free energy of an
equilibrium photon gas is 0 !
0


N
G
ph

For  = 0, the BE distribution reduces to the Planck’s distribution:
 
1
exp
1
1
exp
1
,





















T
k
h
T
k
T
f
n
B
B
ph
ph


 Planck’s distribution provides the average
number of photons in a single mode of
frequency  = /h.

5. Density of States for Photons
ky
kx
kz
        2
3
2
3
3
6
6
3
/
4
8
1





 k
k
G
volume
k
L
L
L
k
k
N
z
y
x





extra factor of 2 due
to two polarizations:
     
 
 
 3
2
2
3
3
2
3
2
6 


c
g
c
G
k
c
cp
d
dG
g D
ph









 




     
  3
2
3
2
2
3
3 8
c
c
h
h
d
d
g
g D
ph
D
ph






 



  3
2
3 8
c
g D
ph

 
1
exp 










T
k
h
h
h
n
B




In the classical (h << kBT) limit: T
kB


The average energy in the mode:
In order to calculate the average number of photons per small energy interval d, the
average energy of photons per small energy interval d, etc., as well as the total
average number of photons in a photon gas and its total energy, we need to know the
density of states for photons as a function of photon energy.

6. Spectrum of Blackbody Radiation
 
  1
exp
8
,
3
3










T
k
hc
T
u
B




The average energy of photons with frequency
between  and +d (per unit volume):
u(,T) - the energy density per unit photon
energy for a photon gas in equilibrium with
a blackbody at temperature T.
      




 d
T
u
d
n
g
h S ,

- the spectral density of the
black-body radiation
(the Plank’s radiation law)
     
  1
exp
8
,
3
3











h
c
h
f
g
h
T
us
          h
T
h
u
d
d
T
u
T
u
d
T
u
d
T
u 


 ,
,
,
,
, 








u as a function of the energy:
3


h

g(

)

 =
n( )
average number
of photons
photon
energy
  3
2
3 8
c
g D
ph

 
 
h

g(

)
n(

)

7. Classical Limit (small f, large ), Rayleigh-Jeans Law
This equation predicts the so-called
ultraviolet catastrophe – an infinite
amount of energy being radiated at
high frequencies or short
wavelengths.
Rayleigh-Jeans Law
At low frequencies or high temperatures:   




 h
h
h 

 1
exp
1
 
 
T
k
c
h
c
h
T
u B
s 3
2
3
3
8
1
exp
8
,






 


- purely classical result (no h), can be
obtained directly from equipartition

8. Rayleigh-Jeans Law (cont.)
4
1

In the classical limit of large :
  4
large
8
,


 
T
k
T
u B

     
  1
exp
1
8
1
exp
8
,
,
, 5
2
3
3
2










































T
k
hc
hc
hc
T
k
hc
c
h
hc
T
u
hc
d
d
d
T
u
d
T
u
B
B 














u as a function of the wavelength:

9. High  limit, Wien’s Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature. The
position of maximum:
 
0
1
1
3
1
exp
2
3
2
3























































x
x
x
B
B
B
e
e
x
e
x
const
T
k
h
T
k
h
T
k
h
d
d
const
d
du




  8
.
2
3
3 


 x
e
x x
Wien’s displacement law
- discovered experimentally
by Wilhelm Wien
Numerous applications
(e.g., non-contact radiation thermometry)
- the “most likely” frequency of a photon in a
blackbody radiation with temperature T
u(

,T)

8
.
2
max

T
k
h
B

h
T
kB
8
.
2
max 

Nobel 1911
At high frequencies/low temperatures:    





 h
h
h exp
1
exp
1 


   




 h
c
h
T
us 
 exp
8
, 3
3

10. max  max
     
  1
exp
1
8
1
exp
8
,
,
, 5
2
3
3
2










































T
k
hc
hc
hc
T
k
hc
c
h
hc
T
u
hc
d
d
d
T
u
d
T
u
B
B 














8
.
2
max

T
k
h
B

- does this mean that
8
.
2
max


T
k
hc
B
?
max
 max

No!
 
   
 
   
 
 
0
1
/
1
exp
/
1
exp
1
/
1
exp
5
1
/
1
exp
1
2
5
2
6
5
























x
x
x
x
x
x
const
x
x
dx
d
const
df
du
 
   
x
x
x /
1
exp
1
/
1
exp
5 
  T
k
hc
B
5
max 

“night vision” devices
T = 300 K  max  10 m
 
T
u ,
  
T
u ,


11. Solar Radiation
The surface temperature of the Sun - 5,800K.
As a function of energy, the spectrum of
sunlight peaks at a photon energy of
m
5
.
0
5
max 
 

T
k
hc
B
eV
T
k
h
u B 4
.
1
8
.
2
max
max 

 
Spectral sensitivity of human eye:
- close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells

12. Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume:
The total energy of photons per unit volume :
(the energy density of a photon gas)    
 
 
 3
4
5
0 15
8
1
exp hc
T
k
d
g
V
U
T
u B









 

the Stefan-
Boltzmann Law
2
3
4
5
15
2
c
h
kB

 
the Stefan-Boltzmann
constant   4
4
T
c
T
u


    4
.
2
8
1
8
1
exp
8 3
3
0
2
3
3
0
2
3
0



























 





T
hc
k
e
dx
x
h
T
k
c
d
T
k
h
c
d
g
n
V
N
n B
x
B
B









- increases as T 3
The average energy per photon:
     
   
T
k
T
k
T
k
hc
hc
T
k
N
T
u
B
B
B
B
7
.
2
4
.
2
15
4
.
2
8
15
8 4
3
3
3
4
5










(just slightly less than the “most” probable energy)
The value of the Stefan-Boltzmann constant:  
2
4
8
/
10
76
.
5 m
K
W




Consider a black body at 310K. The power emitted by the body:
2
4
/
500 m
W
T 

While the emissivity of skin is considerably less than 1, it still emits a considerable
power in the infrared range. For example, this radiation is easily detectable by modern
techniques (night vision).
Some numbers:

13. Power Emitted by a Black Body
For the “uni-directional” motion, the flux of energy per unit area
c  1s
energy density u
1m2
T
4
2
4 T
R 


u
c

Integration over all angles
provides a factor of ¼:
u
c

4
1
area
unit
by
emitted
power
Thus, the power emitted by a unit-area
surface at temperature T in all directions:
  4
4
4
4
c
4
c
power T
T
c
T
u 





The total power emitted by a black-body sphere of radius R:
(the hole size must be >> the wavelength)

14. Sun’s Mass Loss
The spectrum of the Sun radiation is close to the black body spectrum with the
maximum at a wavelength  = 0.5 m. Find the mass loss for the Sun in one second.
How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the
Sun: 7·108 m, mass - 2 ·1030 kg.
max = 0.5 m  


















 


K
K
k
hc
T
T
k
hc
B
B
740
,
5
10
5
.
0
10
38
.
1
5
10
3
10
6
.
6
5
5 6
23
8
34
max
max


This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth
distance).
4
2
8
2
3
4
5
K
m
W
10
7
.
5
15
2 



c
h
kB


  4
2
4
sphere
a
by
emitted
power T
R
P 


      W
10
8
.
3
,740K
5
K
m
W
10
7
.
5
m
10
7
4
8
.
2
4 26
4
4
2
8
2
8
4
max
2















 




B
Sun
k
hc
R
P
 
kg/s
10
2
.
4
m
10
3
W
10
8
.
3 9
2
8
26
2






c
P
dt
dm
the mass loss per one second
1% of Sun’s mass will be lost in yr
10
1.5
s
10
7
.
4
kg/s
10
2
.
4
kg
10
2
/
01
.
0 11
18
9
28









dt
dm
M
t

15. Radiative Energy Transfer
Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask, a
container surrounded by a thin evacuated jacket. While the thermal
conductivity of gas at very low pressure is small, energy can still be
transferred by radiation. Both surfaces, cold and warm, radiate at a rate:
  2
4
/
1 m
W
T
r
J i
rad 


The net ingoing flux:
    rJ
T
r
J
J
r
T
r
J b
a 






 4
4
1
1 

Let the total ingoing flux be J,
and the total outgoing flux be J’:
i=a for the outer (hot) wall, i=b for the inner (cold) wall,
r – the coefficient of reflection, (1-r) – the coefficient of emission
 
4
4
1
1
b
a T
T
r
r
J
J 




 
If r=0.98 (walls are covered with silver mirror), the net flux is reduced to 1% of the
value it would have if the surfaces were black bodies (r=0).

16. Superinsulation
Two parallel black planes are at the temperatures T1 and T2 respectively. The
energy flux between these planes in vacuum is due to the blackbody
radiation. A third black plane is inserted between the other two and is allowed
to come to an equilibrium temperature T3. Find T3 , and show that the energy
flux between planes 1 and 2 is cut in half because of the presence of the third
plane. T1 T2
T3
 
4
2
4
1
0
T
T
J 
 
Without the third plane, the energy flux per unit area is:
The equilibrium temperature of the third plane can be found from the energy balance:
   
4
/
1
4
2
4
1
3
4
3
4
2
4
1
4
2
4
3
4
3
4
1
2
2 






 






T
T
T
T
T
T
T
T
T
T 

The energy flux between the 1st and 2nd planes in the presence of the third plane:
    0
4
2
4
1
4
2
4
1
4
1
4
3
4
1
2
1
2
1
2
J
T
T
T
T
T
T
T
J 









 



 

 - cut in half
Superinsulation: many layers of aluminized Mylar foil loosely
wrapped around the helium bath (in a vacuum space between the
walls of a LHe cryostat). The energy flux reduction for N heat
shields:
1
0


N
J
JN

17. The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption:
Emission:
4
2
4
out
Power E
E T
R 


the flux of the solar radiation energy
received by the Earth ~ 1370 W/m2
   
2
4
2
in
ower 








orbit
Sun
Sun
E
R
R
T
R
P 


Sun
orbit
Sun
E T
R
R
T
4
/
1
2
4 

















 = 1 – TEarth = 280K
Rorbit = 1.5·1011 m RSun = 7·108 m
In reality  = 0.7 – TEarth = 256K
To maintain a comfortable temperature on the
Earth, we need the Greenhouse Effect !
However, too much of the greenhouse effect
leads to global warming:

18. Thermodynamic Functions of Blackbody Radiation
4
4
4
3
4
3
16
4
VT
c
VT
c
VT
c
TS
U
F









 
N
PV
F
PV
TS
U
G 






 0
Now we can derive all thermodynamic functions of blackbody radiation:
the Helmholtz free energy:
the Gibbs free energy:
The heat capacity of a photon gas at constant volume:
  3
16
VT
c
T
T
u
c
V
V











This equation holds for all T (it agrees with
the Nernst theorem), and we can integrate it
to get the entropy of a photon gas:
    3
0
2
0
3
16
16
VT
c
T
d
T
c
V
T
d
T
T
c
T
S
T
T
V 








 

the pressure of a photon gas
(radiation pressure)
u
T
c
V
F
P
N
T 3
1
3
4 4
,













U
PV
3
1

For comparison, for a non-relativistic monatomic gas – PV = (2/3)U. The difference
– because the energy-momentum relationship for photons is ultra-relativistic, and
the number of photon depends on T.
In terms of the average
density of phonons:
T
k
N
T
k
n
V
n
V
U
PV B
B 9
.
0
7
.
2
3
1
3
1
3
1







 

19. Radiation in the Universe
The dependence of the radiated energy
versus wavelength illustrates the main
sources of the THz radiation: the
interstellar dust, emission from light and
heavy molecules, and the 2.7-K cosmic
background radiation.
In the spectrum of the Milky Way galaxy,
at least one-half of the luminous power is
emitted at sub-mm wavelengths
Approximately 98% of all the photons
emitted since the Big Bang are observed
now in the submillimeter/THz range.

20. Cosmic Microwave Background
In the standard Big Bang model, the radiation
is decoupled from the matter in the Universe
about 300,000 years after the Big Bang,
when the temperature dropped to the point
where neutral atoms form (T~3000K). At this
moment, the Universe became transparent
for the “primordial” photons. The further
expansion of the Universe can be
considered as quasistatic adiabatic
(isentropic) for the radiation:
Nobel 1978
A. Penzias
R. Wilson
  const
VT
c
T
S 
 3
3
16
Since V  R3, the isentropic expansion
leads to :
1

 R
T

21. CMBR (cont.)
Alternatively, the later evolution of the radiation temperature may be considered as a
result of the red (Doppler) shift (z). Since the CMBR photons were radiated at T~3000K,
the red shift z~1000.
At present, the temperature of the Planck’s
distribution for the CMBR photons is 2.735
K. The radiation is coming from all
directions and is quite distinct from the
radiation from stars and galaxies.
Mather, Smoot, Nobel 2006
“… for their
discovery of the
blackbody form and
anisotropy of the
CMBR”.

22. Problem 2006 (blackbody radiation)
The cosmic microwave background radiation (CMBR) has a temperature of
approximately 2.7 K.
(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λ,T) of the cosmic background radiation?
(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(,T) of the cosmic background radiation?
(c) (5) Do the maxima u(λ,T) and u(,T) correspond to the same photon energy? If
not, why?
mm
m
T
k
hc
B
1
.
1
10
1
.
1
7
.
2
10
38
.
1
5
10
3
10
6
.
6
5
3
23
8
34
max 









 



(a)
(b)
(c)
Hz
h
T
kB 11
34
23
max 10
58
.
1
10
6
.
6
7
.
2
10
38
.
1
8
.
2
8
.
2 






 

 meV
h 65
.
0
max 

meV
hc
1
.
1
max


the maxima u(λ,T) and u(,T) do not correspond to the same photon energy. The
reason of that is
     
1
exp
1
8
,
,
, 5
2




















T
k
hc
hc
T
u
c
d
d
d
T
u
d
T
u
B












23. Problem 2006 (blackbody radiation)
  2
16
23
6
2
2
10
3
7
.
2
10
38
.
1
7
.
2
10
3
m
s
photons
J
m
W
J
m
s
photons
N





















 


     
   
T
k
T
k
T
k
hc
hc
T
k
N
T
u
B
B
B
B
7
.
2
4
.
2
15
4
.
2
8
15
8 4
3
3
3
4
5










(d)     2
6
4
4
2
4
8
4
/
10
3
7
.
2
/
10
7
.
5 m
W
K
m
K
W
T
J CMBR









(d) (15) What is approximately the number of CMBR photons hitting the earth per
second per square meter [i.e. photons/(s·m2)]?