Chapter 18
Bose-Einstein Gases
18.1 Blackbody Radiation
1. The energy loss of a hot body is attributable to the
emission of electromagnetic waves from the body.
2. The distribution of the energy flux over the
wavelength spectrum does not depend on the
nature of the body but does depend on its
temperature.
3. Electromagnetic radiation can be regarded as a
photon gas.
The wavelength spectrum of blackbody
radiation energy for three temperatures:
T1>T2>T3
4. Photons emitted by one energy level can be
absorbed at another, so the total number of
photons is not constant, i.e. ΣNJ = N does not apply.
(For phonon gas, N is not independent!)
5. The Lagrange multiplier α that was determined by
ΣNJ = N becomes 0, so that
e- a= e-0 = 1
6. Photons are bosons of spin 1 and thus obey BoseEinstein Statistics
For a continuous spectrum of energy
The energy of a photon is hv
So:
Recall that f(v)dv is the number of quantum
states with frequency v in the range v to v + d
v
For phonon gas (chapter 16)
In a photon gas, there are two states of
polarization corresponding to the two
independent directions of polarization of an
electromagnetic wave. As a result
4V 2
f (v)  dv  2  3 v  dv
c
Where c is the speed of electromagnetic wave
(i.e. light)
The energy within the range v to v + dv equals
the number of photons within such a range
times the energy of each photon:
• The above is the Planck radiation formula,
which gives the energy per unit frequency.
• When expressed in terms of the wavelength
Since
Hereμ(λ)is the energy per unit wavelength
The Stephan-Boltamann Law states that the
total radiation energy is proportional to T4
The total energy can be calculated
Setting
one has
3
 x dx
U 8
 3 3 (kT ) 4  x
0 e 1
V hc
The integral has a value of
with
Particle flux equals
, where is the mean
speed and n is the number density.
In a similar way, the energy flux e can be
calculated as
Assuming
with
this is the Stephan-Boltzmann Law
the Stephan-Boltzmann constant.
The wavelength
at which
is a
maximum can be found by setting the
derivative of
equal to zero!
Or equivalently, set
(minimum of
One has
)
is known as Wien’s
Displacement Law
For long wavelengths:
(Rayleigh- Jeans formula)
For short wavelength:
Sun’s Surface T≈ 6000K, thus
Sketch of Planck’s law, Wien’s law and the
Rayleigh-Jeans law.
• Using
• The surface temperature of earth equals 300K,
which is in the infrared region
• The cosmic background
microwave radiation is a
black body with a
temperature
of 2.735± 0.06 K.
18.2 Properties of a Photon Gas
The number of photons having frequencies
between v and v + d v is
 2

8V  v dv 
N (v)dv  3  hv 
c  kT

e

1


The total number of photons in the cavity is
determined by integrating over the infinite
range of frequencies:
Leads to
Where T is in Kelvins and V is in m3
• The mean energy of a photon
• The ratio of
unity.
is therefore of the order of
• The heat capacity
• Substitute
• Entropy
• Therefore, both the heat capacity and the
entropy increase with the third power of the
temperature!
• For photon gas:
• It shows that F does not explicitly depends on
N. Therefore,
1 4 1U
 F 
P
  aT 
3V
 V T , N 3
• 18.1 a) calculate the total electromagnetic
energy inside an oven of volume heated to a
temperature of 400 F
• Solution
Use equation 18.8
4
U  aT  V
a  7.55 1016 Jm3 k 4
 400  32

400F  
 273k
 18

plug in
4
 368

16
U  7.55  10  
 273  1
 18

 3.9  105 J
• 18.1b) Show that the thermal energy of the air
in the oven is a factor of approximately 1010
larger than the electromagnetic energy.
• Solution:
Thermal energy  nRT
 400  32

 (1000 / 22.4)  8.314 103  
 273 
 18

 1.77 106 J
which is 1010 times l arg er than a 