space science : atmosphere part-5 planck radiation law local thermodynamic equilibrium: let...
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Space Science : Atmosphere Part-5
Planck Radiation LawLocal Thermodynamic Equilibrium: LETRadiative TransportApproximate Solution in Grey AtmosphereSkin TemperatureGreenhouse EffectRadiative BalanceRadiative Time Constant
Reading Ionosphere for Previous partRadiation TransportGreenhouse Effect
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Windows and Absorptions in the Solar Spectrum
0.2 0.6 1.0 1.4 1.8 2.2 2.6 3.0m
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Radiation: Solar and Earth Surface
€
Bλ (T) Planck Black Body Emission
Atmosphere is mostly transparent in visible but opaque in UV and IR;
IR window 8-13um
B(T)
Fraction absorbed
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Before Discussing Radiation Define Solid Angle
r sin d
r sin d r d
x
z
y
d
dr sin
s)(steradian ddr r
dAdr dV
2 0
d dcos r ddsin r d r
) d sinr ( )dr ( dA
2
222
Ω=
=
≤≤≤≤
→=Ω=
=
πϕπ
ϕϕϕ
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€
EMISSION of RADIATION
Ideal Emitter at Temperature T
Photon energy = hν ; ν = c/λ
Planck's Law (written in ν or λ ; Bν dν = Bλdλ ]
Energy flux per unit solid angle between ν and ν + dν from a surface at T
Bν (T) = 2 hν
(c /ν )2
1
exph ν
k T
⎛
⎝ ⎜
⎞
⎠ ⎟−1
;
Peak λmax ≈ 3000μm/T(K) ; ν max ≈ 0.57c/λmax
Brightness peak in ν is at longer wavelengths than peak in λ
Energy peak (hν max ) → λ ≈ 5100μm/T(K)
Bv (T) ν →∞ ⏐ → ⏐ ⏐ 2 c k T
λ4
Bλ (T) λ→∞ ⏐ → ⏐ ⏐ 2 ν 3 h
c2exp[−hν/kT] (like a Boltzmann distribution)
Bλ (T) dλ = Bν (T) dν = 0
∞
∫ 1
π0
∞
∫ σ T4 Total energy flux per unit solid angle
* Net flux across a flat surface :
Integrate over solid angle : dΩ = sinθdθdϕ → dcosθ dϕ
∫ [1
π σ T4 ] cosθ dΩ =
1
π0
1
∫ σ T4 2π cosθ dcosθ
= σ T4
Planck’s Law for Thermal Emission of Photons
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GREY ATMOSPHERE Chap 3 G+W , H p10-17
Gray vs. Black vs. Transparent Also, absorption independent of frequencyover the range of relevant frequencies
Processes Surface heated by visible Warm Surface emits IR ~ 3 – 100 m peak ~ 15 m
IR absorbed by CO2, O3, H2O, etc.
Remember why not O2 and N2 ?
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Vibrational Bands CO2 (IR active?)
Symmetric Stretch O C O 7.46 m (N)Asymmetric Stretch
O C O 4.26 m (Y) Bending
O C O 15.0 m (Y)
H2O Symmetric Stretch O 2.73 m (Y) H H
Asymmetric Stretch O 2.66 m (Y) H H
Bending O 6.27 m (Y) H H
You can have combination bands
or 2 vib. levels
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IR Emission and Absorption
Ground Emits
Primarily Triatomc Molecules Absorb and Re-emit: vibrational and rotational states
To determine T we assume excited molecules heat locally by collisions.
CO2(v=1) + M --> CO2(v=0) + M + K.E.
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Slab of Atmosphere Absorption
I + dI
dz
I
€
I = Intensity of the radiation
≡ energy flux per unit solid angle (through the atmosphere)
= power / area / solid angle;
I dΩ∫ = F ; Ω is solid angle
Only Absorbing (notation in C +H)
dI = - k ( I ρ dz )
= - σ abs nabs I dz
Absorption coef.
k = fabs σ abs / m ;
m = average mass of molecules in atmosphere
fabs = fraction of species absorbing
mean free path of a photon for absorption =
= (σ abs nabs )-1 = (k ρ)-1
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Solution: Absorption Only
(did earlier; new notation)
€
I(z) = I0 exp[− k ρ dz]∫
k ρ dz∫ = σ abs ∫ nabs dz ≈ σ abs Nabs [ ]
Nabs = column density of
absorbing species
Optical Depth for IR
kρ dzz
∞
∫ ≡ τ IR
Note : k ρ dzz
∞
∫ ≈ k
gdp
z
∞
∫ ≈ k
gp
Therefore, z τ p
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What about EMISSIONSlab of atmosphere has a T
emits IRAssume LTE LTE Local Thermodynamic Equilibrium molecular motion and the population
of the vibrational and rotational states are all described by Boltzmann distribution and photons by Planck’s law ---using the same T
Kirchhoff’s LawIn LTE the emissivity of a body (or surface) equals its absorptivity.
€
Probability of absorption in dz ≈ k ρ dz ∝ Emission
Therefore, using the Planck flux for emission from a thick material
when the material is in LTE, one writes :
emission from the slab of thickness dz = [ k ρ dz ] B(T)
with B(T) = 1
πσ T4 (power/area/solid angle)
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Radiative Transport with Emission + Absorption
∫∫
=Ω
Ω=
−=
+=+−
4T d cos B(T) Also
d cos I surface a acrossFlux
angle solidunit per are B and I :Note
)(simple! B(T)Id
dI
substitute and ) dz (k -=d Use
B(T) ) dz (k I dz) k ( - dI
emission absorption =ChangeIntensity
σθ
θ
τ
ρτ
ρρ
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Flux (cont)
€
I [r2 dΩ] = Energy per unit time
across surface area dA
Therefore : I(θ, ϕ ) ⇒ Watts
m2 ster
If isotropic I(θ, ϕ ) = I
Flux across a surface
I(contains speed, c)
Flux = I cosθ dΩ = = ∫ I cosθ dcosθ dϕ∫ = π I∫∫ = F
I
dA = r2dΩ
r
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Radiative Transport (cont.)include angles
dz = cos
↓
−
↑
=−=Ω
==Ω
Ω
+
−=
=
+=
∫ ∫∫ ∫
F- I d I
F I d I
d dcos = d and cos= :I isotropican For
fluxes downward upward into Divide
BId
dI
dz k - d
B) cos
dzk (I)
cos
dz k ( - dI
2
0
0
1
2
0
1
0
π
π
πμ
πμ
ϕθθμ
τμ
ρτθ
ρθ
ρ
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Radiation Transport (cont.) I and B are isotropic
€
Upward I Downward I
I dΩ = 0
1
∫0
2π
∫ I dΩ = −1
0
∫0
2π
∫ 2 π I
same for B
Now : Use these integrals to integrate
the radiative transport Equation.
μ dI
dτ = I - B
0
2π
∫ →0
1
∫ dF↑
dτ= 2 F↑ - 2 π B Upward flux
0
2π
∫ →-1
0
∫ − dF↓
dτ= 2 F↓ - 2 π B Downward flux
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Not Quite Isotropic * 5/3 Use * not in transport eq.
dz
↑+F ↓
+F
↓−
↑− F F
e.unit volumper energy in change dt
dT c
EquationHeat
T ,F ,F unknowns 3but Eqs. 2
T B : Remember
(2) B F d
dF
(1) B F d
dF Solve
p
4
*
*
=
=
−=−
−=∴
↓↑
↓↓
↑↑
ρ
σπ
π
π
Need
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Third Equation is the Heat Equation
(3) CF F
)F F (dz
d 0
T state)(steady mequilibriu Find
0 = )F F ( dz
d
dt
dT c
EquationHeat
1
p
=−
−−=
−−=
↓↑
↓↑
↓↑ρ
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Radiative Transport Solution (cont.)Use Eqs. (1) and (2) with (3).
€
(1) dF↑
dτ * = F↑ − π B ; (2) −dF↓
dτ * = F↓ − π B
ADD (1) and (2) M SUBTRACT (1) and (2)
d(F↑ − F↓ )
dτ * = F↑ + F↓ − 2πB M
d(F↑ + F↓ )
dτ * = F↑ − F↓
Use Eq. (3) (dz ∝ - dτ*) Use solution to (3)
0 = F↑ + F↓ − 2 πB M d(F↑ + F↓ )
dτ * = C1
(1a) F↑ + F↓ = 2 π B M (2a) F↑ + F↓ = C1τ* + C2
Combine (1a) and (2a)
2 π B(T) = C1 τ* + C2
σ T4 = 1
2[C1 τ
* + C2] (4)
Now need C1 and C2 in order to get T vs. τ (z)
Therefore,
Apply boundary conditions at surface
and the top of the atmosphere
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Radiative Transport Solution (cont.)Use Eq. (3) and (4)
€
Solutions
F↑ + F↓ = C1τ* + C2 (4) ; F↑ − F↓ = C1 (3)
Add F ↑ = 1
2C1(τ
* +1) +1
2C2 (5)
Sub. F ↓ = 1
2C1(τ
* −1) +1
2C2 (6)
Boundary conditions on IR radiation
Top : τ * = 0 assume F↓ (0) ≈ 0 WHY ??
Use Eq. (6) to get
C1 = C2
Therefore,
F ↑ = 1
2C1(τ
* + 2) (5) ; F ↓ = 1
2C1τ
* (6)
Bottom : τ ≡ τ g use the ground temperature, Tg, for emission :
i.e. surface flux in IR is F↑ (τ g*) = π B(Tg ) = π Bg
Therefore, evaluate (5) at τ g * and use C1 = C2
π Bg = 1
2C1[τ g
*+2]
or
C1 /2 = π Bg / [ τ g* + 2 ]
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Radiative Transport: Solution use C1 in (5) and (6)
€
F ↑ = π Bg [τ * + 2]
[τ g* + 2]
F ↓ = π Bg
τ *
[τ g* + 2]
Use earlier result
[F↑ + F↓ ]/2 = π B = σ T4 = π Bg [τ * + 1]
[τ g* + 2]
Also use : π Bg = σ Tg4
Therefore, we get solution for T
T4 = Tg4
[τ * + 1]
[τ g* + 2]
Note : T = temp. of air
τ g* = ground ,
T(z = 0) = To = air temp at surface
To4 = Tg
4 τ g
* +1
τ g* + 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
In radiative equilibrium no conductive contact
between the air and ground, only radiative; so there is a
discontinuity! Obviously not realistic; need conductive also.
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Solution to the radiative transfer equations for a grey atmosphere
ConductiveTransport(Adiabatic Lapse Rate)Radiative
TransportBecomes radiative dominated near tropopause
* Optical Thickness in IR
g*
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Finally: we do not know Tg
we know only Te for emission to space!
This is the Green House EffectGround T exceeds T for
emission to space
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
=
+=
=
↑
↑
2 1 T T
T B using Therefore,
]2[
2 B (0)F :solutionour From
T of Definition ;T (0) F
*g4
e4
g
4gg
*g
g
e4
e
τ
σπ
τπ
σ
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A Real Green House
How do you get IR out equal to Visible light absorbed inside: RAISE T
Note: For a real green house convection may be as important:
i.e. glass a thermal barrier
IR
Visible
OutsideInside
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Greenhouse Effect is Complex
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PLANETARY ENERGY BALANCE G+W fig 3-5
113 86Convective30
IR Radiation To Space67
GROUND
€
Absorbed
Clouds 21
Atmos. 22
Ground 24
67
Back to space
Reflect Ground 7
Clouds 26
Albedo 33
Incoming solar radiation
mesopause
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Radiation Transport (Review)
dz Atmospheric Slab
IR
4
abs
T B
BId
dI *
dzn dz k d
IR in thedepth Optical
Bcos
k I
cos
k
dz
dI
Emission Absorption Intensity of Change
Equations General
Flux downward F
Flux upward F
intersity I Review
σπ
σρ
ρ
ρ
=
−=
−=−=
+−=
+−=
=
=
=
↓
↑
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Integrate (*) for upward moving and downward moving IR photons
4g
*g
4e
p
*
*
*
T )(F
T 0)=( F :Top (3)
right). quite(not in coming IR no 0 0)= ( F :Top )2(
ConditionsBoundary
T , F , F
unknowns Three (1)
mEquilibriuin 0 ]F[Fdz
d
dt
dT c .3
EquationHeat
B Fd
dF 2.
B Fd
dF 1.
) 5/3 ( Equations 2 :Result
σ
σ
ρ
π
π
=
=
=
=−−=
−=−
−=
≈
↑
↑
↓
↓↑
↓↑
↓↓
↑↑
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(Review continued)
fluxsolar the
with balance thefrom comes T and
z offunction a is *remember
3. ]1[2
T T
T of instead T of in terms rature)(air tempe T Write
2. ][2
T F
1. ]2[2
T F
T of in terms T and Fluxes Can write
T B Remember
e
*4
e4
ge
*4
e
*4
e
e
4
σπ
+=
=
+=
=
↓
↑
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Ground T (review)
€
In terms of Te (determined from solar flux and albedo)
Tg4 = Te
4 [ 1 + τ g *
2]
τ g * = optical depth of atmosphere + IR
This is the Green house effect
Tg > Te
T in terms of Tg
(rearrange the equations )
T4 = Tg4
[1+ τ*]
[2 + τ g*]
This is air temperature relative to ground temperature
T(z = 0) ≠ Tg for radiative only solution
What have we ignored :
absorption is in bands which have a width
molecule motion causes doppler shifts affecting the absorption efficiencies
collisions also affect the absorption efficiences
population of the levels affects absorption efficiency
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G + W (simple version; 4 layers)
0
1
2
3
4
Ground
SpaceF1
F1
F2
F2
F3
F3
F4
F4
Fg
agreement closely Fortuitous
layers ofnumber =
] (5/6) 1 [ T= ] /2* 1 [ T T
assolution wour
F ] layers ofnumber 1 [ F can write :Note
F 5 F F 2 F F F F 2 4
F 4 F F 2 F F F F 2 3
F 3 F F 2 F F F F 2 2
F 2 F 2 F F F 2 1
T F F 0
1) = ( s thicknesoptical one layer Each
layers into atmosphere Divide
g
g4
eg4
eg
outg
out34gg34
out234423
out123312
out1221
4eout1
σ
++=
+=
=−=→+==−=→+==−=→+=
==→===
=
FVIS=Fout
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Earth g 2
€
Te = 250K
Implies
Tg = 330 K ! No Way
Again - - - it is clear that at the surface
convection dominates
Before Finishing
If τ g ≈ 2 ≈ σ abs fabs N
Using N = 2 × 1025 mol/cm2 (see early lecture)
fabs = 1% (H2O, CO2, O3 ....)
σ abs ≈ 10-23 cm2
IR absorbers have small cross sections relative to UV
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VENUS (Problem for set 2)
Te = 230 Tg = 750
Therefore: g* = ?
Therefore: Use cross section from previous slide, pure CO2
N = ?
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TopWhy isn’t Te = T() at the top?
=emissivety
Te4(1-) T4
T4
€
Te4 = 2 ε T4
T = 1
21/4 Te = asymptotic value of T (skin T)
€
T4 (z) = Te
4
2(1+ τ*)
τ * → 0 at top of atmosphere
Earlier we calculated that Te ≈ 250 K at Earth
T(∞) = 1
21/4Te ≈ 210 K
This, of course, ignores the direct thermosphere heating
Direct means of obtaining T(∞)
Goody + Walker (Skin T)
How Thin Layer at Top is Heated
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Thermal Structure Tropopause to Mesopause
small iscontent heat but the
hot'' is reThermosphe :also Note
radiativepurely not isit Suggests
2) (K 330 (calc.)T :T Surface
)T(bracket they heating ozone theIgnoring
km 12 215 se)T(Tropopau
km 80 200 e)T(Mesopaus
K 210 )T( Find
K 255 T Start with
g
e
≈≈
∞≈≈
≈∞≈
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TIME CONSTANT FOR RADIATIVE EQUILIBRIUM
€
Initially : Equilibrium
ρ cp ∂T
∂t = − OUT + IN = 0
Imbalance : suddenly add heat can write as
σ T4 → σ [T + ΔT]4 in a layer of atmosphere Δz
ρ cp dT
dt = -
1
Δz[2σT4 − 2σ (T + ΔT)4 ]
dT
dt ≈
8 σ T3
Δz ρ cp
ΔT
time ≈ T/[dT/dt] ≈ Δz ρ cp
8 σ T3
cp ≈ 1000 J/kg K ρ = 1.3 kg/m3 σ = 5.7 × 10-8 J
m2
1
K4 s Troposphere Δz → H
t rad ≈ 9 days
Therefore, thermal changes are slow in the radiative region
vs what we are familiar with in the troposphere
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Carbon concentrationvs. time
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Carbon Concentration Long Term
Later we will look at the carbon cycle
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GREEN HOUSE EFFECT
.
Constants Time andcarbon for Reservoirs :discuss to
evolution catmospheri discuss weback when come willWe
OCEAN ATMOSPHERE !K 2 - 1
ONLY ATMOSPHERE !K 5 - 2.5 T
1800) since increase 30% ~ ( ?CO Double
! T decrease25K z
T increase30K z O Increase :Note
K 288 K 33 K 255
T Te
K 33
K 3 ),CH O,Other(N
K 2 O
K 7 CO
K 21 OH
2
3
42
3
2
2
+==Δ
<>
⇒
=+Δ+
++++
L
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However,
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#4 Summary Things you should know
Planck Radiation LawLocal Thermodynamic Equilibrium:
LETRadiative TransportGreenhouse Effect Surface temperatureSkin TemperatureRadiative Time Constant