Thermal Radiation Radiation in thermal equilibrium
with its surroundings
E0
B0
k
rk·r
i(
r
r
E = E0 e r−ωt)
ω = c|rk|
r = Br0 ei(rk·rr−ωt)
B
Br0 = r
1k × Er0 /c
8.044 L15B1
Time average energy flux
- 0 |2
u = 12 E0|E
-jE = (cu) 1k
Time average pressure (⊥ to -k)
P =u
Time average energy density
Thermal radiation has a continuous distribution of
frequencies.
u( ν,T)
Peaks near hν = 3kB T
(h/kB ∼ 5 × 10−11 K-sec)
ν
8.044 L15B2
Spectral Region
ν (Hz)
T (K)
Radio
106
1.7 × 10−5
Microwave
1010
0.17
cosmic background
Infrared
1013
1.7 × 102
room temp.
× 1015
8.5 × 103
sun’s surface
Ultraviolet
1016
1.7 × 105
X ray
1018
1.7 × 107
γ ray
1021
1.7 × 1010
Visible
1
2
Thermal Rad.
black holes
8.044 L15B3
ABSORPTIVITY
α( ν,T)
ENERGY ABSORBED
ENERGY INCIDENT
ISOTROPIC
EMISSIVE POWER
e( ν,T)
ENERGY EMITTED
AREA
ISOTROPIC
8.044 L15B4
THERMAL RADIATION: PROPERTIES
2 ENERGY FLUXES,
IN AND OUT OF CAVITY B
TB
CAVITY A
TA
CAVITY B
FILTER: FREQUENCY
OR POLARIZATION
ASSUME TA = T B AND THERMAL EQUILIBRIUM
8.044 L15B5
CONCLUSIONS:
• u(ν, T ) is independent of shape and wall material
• u(ν, T ) is isotropic
• u(ν, T ) is unpolarized
8.044 L15B6
CONSIDER AN OBJECT IN THE CAVITY,
IN THERMAL EQUILIBRIUM
COMPUTE THE ENERGY FLUX
T
dA
t
c∆
θ
n
∆A
8.044 L15B7
∆E =
=
(E in cylinder) p(θ, φ) dθ dφ
sin θ 1
(u ∆A cos θ c∆t)
2 2π
= c u ∆A ∆t
π/2 cos θ sin θ
0
2
1/4
dθ
2π
0
dθ dφ
1
dφ
2π
1
⇒ energy flux onto dA = 14 c u(ν, T )
8.044 L15B8
Momentum Flux
Plane wave momentum density
|∆p| = 2|p⊥|
since
p = uc 1k
p⊥ in = −p⊥ out
8.044 L15B9
|∆p|ν =
e
2 cos θ
(E in cylinder) p(θ, φ) dθ dφ
c
= u(ν, T ) ∆A ∆t
e π/2
0
cos2 θ sin θ dθ
1/3
=
e 2π
0
1
dφ
2π
1
1 u(ν, T ) ∆A ∆t
3
⇒ P (T ) =
1
3
e ∞
0
u(ν, T ) dν
8.044 L15B10
Apply detailed balance to the object in the cavity.
Eout = Ein
e dA = α ( 14 c u(ν, T )) dA
e(ν, T )
⇒
=
α(ν, T )
1 c u(ν, T )
4
This ratio has a universal form for all materials.
The result is known as KIRCHOFF’S LAW.
8.044 L15B11
Black Body Radiation
If α ≡ 1 ≡ “Black”
Then e(ν, T ) = 14 c u(ν, T )
OVEN
CAVITY
AT
T
Measure e(ν, T )
and obtain u(ν, T )
8.044 L15B12
Thermodynamic Approach
u(T ) ≡
∞
0
u(ν, T ) dν
Then
E(T, V ) = u(T )V
P (T, V ) =
1 u(T )
3
8.044 L15B13
This is enough to allow us to find u(T ).
dE = T dS − P dV
∂E
∂S
∂P
= T
−P =T
−P
∂V T
∂V T
∂T V
=
1 T u' (T )
3
− 13 u(T )
also = u(T )
⇒
u'(T )
4
=
u(T )
T
u(T ) = AT 4
8.044 L15B14
Emissive Power of a Black (α = 1) Body
e(ν, T ) = 14 c u(ν, T ) ⇒ e(T ) = 14 c u(T ) = 14
AcT 4
e(T ) ≡ σT 4
This is known as the STEFAN-BOLTZMANN LAW.
σ = 56.7 × 10−9 watts/m2K 4
8.044 L15B15
Statistical Mechanical Approach
H? Single normal mode (plane standing wave) in a
rectangular conducting cavity.
Ex
0
z
L
�
�
E
(�
r
,
t)
=
E(t)
sin(nπz/L)
1x
�
0,0,n,1x
2/L)−1Ė(t) cos(nπz/L)�
�
B
(�
r
,
t)
=
(nπc
1y
0,0,n,�
1
y
8.044 L15B16
r ·E
r +11B
r ·B
r
Energy density = 12 E0E
2µ
0
V
H =
2

 1 E E 2(t)
2 0
[no r or t average]

1
(nπc2/L)−2Ė 2(t)
+
µ0
1
2
V E0
=
E 2(t) + (nπc/L)−2 Ė 2(t)
2 2
⇒ Each mode corresponds to a harmonic oscillator.
8.044 L15B17
Count the modes.
E nx,ny ,nz = |E|Ej sin(nxπx/L) sin(ny πy/L) sin(nz πz/L)eiωt
E
The unit polarization vector Ej has 2 possible orthog­
onal directions and ni = 1, 2, 3 · · ·.
E
∂ 2E
∂t2
E 2E
− c2 ∇
=0
⇒
ω2
πc
=
L
2
2 + n2
)
(n2
+
n
x
y
z
8.044 L15B18
If the radian frequency < ω nz
�
R
R=
ny
nx
GRID SPACING
1 UNIT
n2
x
+ n2
y
+ n2
z
L
=
ω
πc
# modes (freq. < ω)
4 πR3
=2×1
×
8
3
=
π
3
L 3 ω3
πc
8.044 L15B19
d#
L 3 2
V
D(ω) =
=π
ω = 2 3 ω 2
dω
πc
π c
D( ω)
ω
8.044 L15B20
Classical Statistical Mechanics
< �(ω) >= kB T
u(T ) =
∞
0
D(ω)
kB T 2
⇒ u(ω, T ) =< �(ω) >
= 2 3ω
V
π c
u(ω, T ) dω = ∞
u( ω, T)
CLASSICAL
MEASURED
ω
8.044 L15B21
Quantum Statistical Mechanics
h̄ω
< E(ω) >= h̄ω/kT
+ h̄ω/2
e
−1
D(ω)
h̄
ω3
u(ω, T ) =< E(ω) >
= 2 3 h̄ω/kT
+
π c e
V
−1
z. p. term
du(ω, T )
To find the location of the maximum, set
= 0.
dω
The maximum occurs at h̄ω/kT ≈ 2.82.
8.044 L15B22
Z=
Zi = e−h̄ω/2kT (1 − e−h̄ω/kT )−1
Zi
states i
The first factor in the expression for Zi comes from
the zero-point energy.
F (V, T ) = −kT ln Z = −kT
�
ln Zi
states i
= −kT
� ∞
0
D(ω) [ln Zi] dω
8.044 L15B23
F (V, T ) = −kT
∞
0
�
�
D(ω) − ln(1 − e−h̄ω/kT ) dω + · · · kT V ∞ 2
= 2 3
ω ln(1 − e−h̄ω/kT ) dω
π c 0
∞ 2
V
4
x ln(1 − e−x) dx
= 2 3 3 (kT )
0
π c h̄
v_
4
π
− 45
1 π2
4V
= −
(kT
)
45 c3 h̄3
8.044 L15B24
∂F
1 π2
4
P = −
=
(kT
)
∂V T
45 (ch̄)3
∂F
4 π2
4T 3 V
S = −
=
k
∂T V
45 (ch̄)3
1
4
1 π2
4 V
E = F + TS = −
+
(· · ·) =
(kT
)
45
45
15 (ch̄)3
1 u(T ) independent of V .
Note: P = 1
E/V
=
3
3
8.044 L15B25
NOTE: THE ADIABATIC PATH IS T3 V=CONSTANT T
ADIABATIC
T / T 0 = (V/V 0 ) - 1 / 3 V
8.044 L15B26
MIT OpenCourseWare
http://ocw.mit.edu
8.044 Statistical Physics I
Spring 2013
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