Transcript ppt
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
8V
3
U 2
d
d
3
3 0
0
c exp 1
hc exp 1
8V 1
4
hc 0
x
3
x3
8V
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
hc3 exp hc 1 2 5 exp hc 1
k T
k T
B
B
du
d
1
5
x 2 exp1 / x
const 5
const 6
5
0
2
df
dx x exp1 / x 1
x exp1 / x 1 x exp1 / x 1
5xexp1 / x 1 exp1/ 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
5.67 108 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 1034 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 4R2 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 1.51011 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
uT
cV
VT 3
c
T V
c T
16V
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 1034 3 108
max
1.1103 m 1.1mm
23
5 k BT 5 1.3810 2.7
(b)
k BT 2.8 1.381023 2.7
max 2.8
1.581011 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