Lecture 24. 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.
Ultraviolet Catastrophe (classical E&M)
i  i
Normal modes (standing waves) of EM wave in a box:

L
,
i  1, 2, 3,
1
2
Each mode has degeneracy 2 degrees of freedom
(polarization). From equipartition theorem:
 i  2  k BT  k BT

Total energy of the EM standing waves:
U   i    k BT ! Catastrophe!
i 1
Solution: Light consist of quantized energy units,
hc
h
2
   pc, p   k k 
called photons with energy  and momentum p:



i
And equipartition theorem is only applicable when
k BT

The partition function of photon gas in a box: Z   exp   j   
i
i
j 0
1
1  exp    i 
Average photon energy on ith level:
 i exp   i 
 ln Zi
i

Ei  

ln 1  exp   i  



1  exp   i  exp   i   1
Since Ei  ni  i  mean occupancy of photon: ni 
1
exp   i   1
Chemical Potential of Photons = 0
Recall Bose-Einstein distribution:
1
nBE   
exp        1
For photon:
n
1
exp     1
Photons are bosons with =0.
Why ph=0?
Photons are not conserved because they can be emitted or
absorbed by matters. Therefore, free energy cannot
depends on total number (Nph) of photons, i.e.
 G 
ph  
0
 N 
 ph T , P
or matter +ph =matter  ph =0
The Planck spectrum (photon gas)
The total energy of photon gas at temperature T:
U   ni  i  

0
i

g  
d
exp     1
3D
ph
 using continuum approx.
dk
V 2
 L 
3D
3D
2 dk
g ph    g ph  k 
 2
 2
  4 k
3
d
2

d




 c
3
  ck
Spin (polarization)
 V
2

8V 
3
U   2

d 
d
3
3 0
0 
 c  exp     1
 hc  exp     1

8V 1

4
 hc   0
 x   
3
x3
8V
x3
4 
dx 
k T
dx
3  B  0
x
x
e 1
e 1
 hc 
The energy density per unit photon energy :
8 V
3
u   
(Planck spectrum)
3
 hc  exp     1
The Planck spectrum (photon gas)
Planck spectrum
Wien’s law
x
1.0
3
 hc 
3
3
exp     1
x /(e -1)
u  , T  
8 V
1.5
 peak  2.82kBT
Numerous applications
(e.g., non-contact radiation thermometry)
0.5
0.0
0
Total energy of photo gas (T, V)
U
8 5V
15  hc 
3
 k BT 
4
32 5V
3
 U 
CV  

k
T
kB


B

3
 T V 15  hc 
4
6
x = 
8
10
energy density (unit volume)
 U
8 5
4

k T
 Pr. B19 
3  B 
 V 15  hc 
  x3
4
 0 e x  1 dx  15

Heat capacity
2
Entropy
T
CV T ' 
0
T'
S
dT ' 
32 5V
45  hc 
k T
3  B 
3
kB
max  max
u  , T 
u  , T 
 max
max
h max
 2.8 - does this mean that
k BT
hc
No!
 2.8 ?
k BT max
3
u  , T d  u  , T d
 c
h 
hc 
8
1
 d
 hc  8 hc
 




u

,
T




 d
2 
hc3 exp hc   1  2  5 exp hc   1
k T 
k T 
 B 
 B 




du
d 
1
5
 x 2 exp1 / x  
 const  5
 const  6
 5
0

2
df
dx  x exp1 / x   1
 x exp1 / x   1 x exp1 / x   1 
5xexp1 / x 1  exp1/ x
T = 300 K
“night vision” devices


max 
hc
5 k BT
max  10 m
Solar Radiation
The surface temperature of the Sun - 5,800K.
max 
As a function of energy, the spectrum of
sunlight peaks at a photon energy of
hc
 0.5 m
5 k BT
umax  h max  2.8kBT  1.4 eV
- close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells
Spectral sensitivity of human eye:
Blackbody radiation (photons escape from a hole)
Blackbody: absorb all EM radiation falls on it. Blackbody also
emits photons with Planck spectrum. (Blackbody could be very
bright, e.g. Sun)
A physical mode of a blackbody with
surface area A  a hole of area A of a
very large box (so that incoming
photo get absorbed before escape).
A
The photons that come out of the hole (i.e. emitted by a blackbody
with area A) come from the photo gas inside the box that is in
thermal equilibrium of the wall of box at temperature T.
Blackbody radiation = photon flux (EM energy) of a photon gas at
temperature T through surface area A (the hole).
Blackbody radiation (photons escape from a hole)
energy density u
d
c  1s
power
energy flux 
unit area

A
dE  d , dt , A 

U
A cos 
  cdt  R 2 d   
V
4 R 2
1 dU
U
sin  d d cU
   c  cos  

A dt
V
4
4V
Blackbody radiation (Stefan’s law )
Real materials
5
power  e AT 4
Stefan-Boltzmann constant
Emissivity: e
Reflectivity: 1-e
1 dE
2
4
4

k
T


T


B
A dt 15h 3c 2
2 5 k B4

15h 3c 2
1m2
  5.67  108 W/  m 2  K 4 
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 
hc
max 
5 k BT
 T
hc
5 k B max


6.6 1034  3 108
 

K

5
,
740
K
 23
6
5

1
.
38

10

0
.
5

10


2 5 k B
W
8


5
.
7

10
15h3c 2
m2K 4
4
P power emittedby a sphere  4R2 T 4
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
 hc 
W
2
4
8
8
26
P  4  RSun   

4

7

10
m

5.7

10

5,740K

3.8

10
W





2 4
5
k

m
K
 B max 
the mass loss per one second
1% of Sun’s mass will be lost in
dm P 3.8 1026 W
9
 2 

4
.
2

10
kg/s
2
8
dt c
3 10 m


0.01M
2 1028 kg
t 

 4.7 1018 s  1.51011 yr
9
dm / dt 4.2 10 kg/s
Dewar
Radiative Energy Transfer
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:
J rad  1  r  Ti
4
W / m2
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
Let the total ingoing flux be J,
and the total outgoing flux be J’:
The net ingoing flux:
J  1  r  Ta4  rJ  J   1  r  Tb4  rJ
J  J 
1 r
 Ta4  Tb4
1 r


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).
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.

J 0   T1  T2
Without the third plane, the energy flux per unit area is:
4
T1
4
T3 T2

The equilibrium temperature of the third plane can be found from the energy balance:

 
 T14  T3 4   T3 4  T2 4

1/ 4
T1  T2  2T3
4
4
 T14  T2 4 

T3  

2


4
The energy flux between the 1st and 2nd planes in the presence of the third plane:

J   T1  T3
4
4



 4 T14  T2 4  1
1
   T14  T2 4  J 0
   T1 
 2
2
2


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:
J0
JN 
N 1
- cut in half
The Greenhouse Effect
Absorption:

Power in  e  RE
2


4  R
 TSun   Sun 
 Rorbit 
2
the flux of the solar radiation energy
received by the Earth ~ 1370 W/m2
Emission:
Power out  4 RE  TE
2
4
1/ 4
 e  R 2 
TE    Sun  
 4  Rorbit  
Rorbit = 1.5·1011 m
Transmittance of the Earth atmosphere
e = 1 – TEarth = 280K
In reality
e = 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:
TSun
RSun = 7·108 m
Thermodynamic Functions of Blackbody Radiation
The heat capacity of a photon gas at constant volume:
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:
16
 uT 
cV  

VT 3

c
 T V
c T 
16V
16
2
3


S T    V
dT  
T
d
T

VT
T
c 0
3c
0
T
T
Now we can derive all thermodynamic functions of blackbody radiation:
the Helmholtz free energy:
the pressure of a photon gas
(radiation pressure)
F  U  TS 
4
16
4
VT 4 
VT 4  
VT 4
c
3c
3c
4 4 1
 F 
P  

T  u

3
 V T , N 3c
1
PV  U
3
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.
the Gibbs free energy:
G  U  TS  PV  F  PV  0
 N 
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?
(a)
hc
6.6 1034  3 108
max 

 1.1103 m  1.1mm
 23
5 k BT 5 1.3810  2.7
(b)
k BT 2.8 1.381023  2.7
 max  2.8

 1.581011 Hz
34
h
6.6 10
(c)
hc
max
 1.1 m eV
h max  0.65 meV
the maxima u(λ,T) and u(,T) do not correspond to the same photon energy. The
reason of that is
u , T d  u , T d
c
8 hc
 d




u

,
T

 d
2 
5
1
 hc 
  1
exp

k
T
 B 