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

Windows and Absorptions in the Solar Spectrum

0.2 0.6 1.0 1.4 1.8 2.2 2.6 3.0m

Radiation: Solar and Earth Surface

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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

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

€

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

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 ?

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

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.

Slab of Atmosphere Absorption

I + dI

dz

I

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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

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

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.

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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)

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

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

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

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

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

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

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

Radiative Transport Solution (cont.)Use Eq. (3) and (4)

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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 ]

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.

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*

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

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

Greenhouse Effect is Complex

PLANETARY ENERGY BALANCE G+W fig 3-5

113 86Convective30

IR Radiation To Space67

GROUND

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Absorbed

Clouds 21

Atmos. 22

Ground 24

67

Back to space

Reflect Ground 7

Clouds 26

Albedo 33

Incoming solar radiation

mesopause

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

=

−=

−=−=

+−=

+−=

=

=

=

↓

↑

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

=

=

=

=−−=

−=−

−=

≈

↑

↑

↓

↓↑

↓↑

↓↓

↑↑

(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

+=

=

+=

=

↓

↑

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

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

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

VENUS (Problem for set 2)

Te = 230 Tg = 750

Therefore: g* = ?

Therefore: Use cross section from previous slide, pure CO2

N = ?

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

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

≈≈

∞≈≈

≈∞≈

TIME CONSTANT FOR RADIATIVE EQUILIBRIUM

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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

Carbon concentrationvs. time

Carbon Concentration Long Term

Later we will look at the carbon cycle

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

However,

#4 Summary Things you should know

Planck Radiation LawLocal Thermodynamic Equilibrium:

LETRadiative TransportGreenhouse Effect Surface temperatureSkin TemperatureRadiative Time Constant