introduction: absorption, hydrostatic balance, heating...

Introduction: absorption, hydrostatic balance, heating rate

I Coverage: pp. 173-202 on scattering, absorption emissionI Main points:

I Scattering direction and efficience depends on wavelength λand particile size (r) via the size parameter

I Three distinct scattering regimes: Rayleigh, Mie, GeometricI Absorption depends on line strength and line width (function

of temperature, pressure, wavelength)I Vertical optical thickness relates the physical propoerties of

absorbers/scatterers to driect beam transmission

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Beers law for direct beamDefine the “slant path” asds = dz/ cos θ

Note that in this case z ispositive downward!

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Size parameter (scattering) How light scatters depends onthe relative size of the wave-length and the particle as de-termined by the size parameterx = 2πr

λ

I x � 1: little scattering

I x < 1: Rayleigh scatter(blue sky, evenly lit)

I x > 1: Mie scattering(forward, clouds)

I x � 1: geometricscattering (refraction)

I Radar (3 cm) on raindrops (1 mm) andsunlight (0.5 µm) onnitrogen molecules (1nm) are both Rayleighscattering

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Polar scattering plot (also figure 12.2)

I Rayleigh scattering isevenly split betweenforward and backscatteer (which is whythe blue sky is evenly lit

I Mie scattering is mainlyforward (which is whyclouds let so much lightthrough)

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Scattering efficiency (also figure 12.6)

I Note that x canchange either becauseof λ is changing or ris changing (K ∝ λ−4

for small λ

I Use this diagram toexplain why sunsetsare red and moonsare blue

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Absorption: Lorentz & doppler line shapes (figure 9.7)

I The width of thelorentz line dependson collison frequency,which is proportionalto pressure.

I The width of thedoppler line dependson molecular speed,which is proportionalto√

temperature

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Absorption lines are very narrow (figure 9.6)

I Top figure goes fromλ = 5.18 µm toλ = 5.16 µm

I Bottom figure goesfrom λ = 8.65 µm toλ = 8.59 µm

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Slantwise vs. vertical optical depth

I The slant optical thickness (or slant optical depth) is definedas:

dσλ = dTr =dFλFλ

= −kλρgds

(where kλ is the mass absorption coefficient ( m2 kg−1))

I In words, it’s a measure of how “thick” a gas with density ρgis for photons with wavelength λ in a layer of thickness ds.

I The “vertical optical depth” dτλ uses dz instead of ds. Sincedz = ds ∗ cos θ, the vertical optical depth is dτλ = σλ cos θ

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What is ρg? Start with: What is ρ?

I Start with a layer with density ρ

I Assume hydrostatic balance for a layer of thickness ∆z , areaA, volume V = A∆z

I Force down= ma = ρVg = ρA∆zgI pressure force up= dp AI Balance requires dpA = −ρA∆zg orI dp = −ρgdz

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Hydrostatic atmosphere, cont.

I We want to integrate dp′ = −ρgdz ′ from the surface to apressure level p.

I Neglecting water vapor, the equation of state gives p = ρRdT

I So use this:

dp′ = − p

RdTgdz ′∫ p

p0

dp′

p′= −

∫ z

0

g

RdTdz ′ = −

∫ z

0

1

Hdz ′

define the pressure scale height: Hp as:

1

Hp=

1

z

∫ z

0

1

Hdz ′

which leaves:

∫ p

p0

d ln p′ = − z

Hp

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Hydrostatic atmosphere, cont.I Finish the integration:∫ p

p0

d ln p′ = − z

Hp

ln

(p

p0

)= − z

Hp

p = p0exp

(− z

Hp

)I and use the equation of state to get density:

ρRdT = ρ0RdT0exp

(− z

Hp

)ρ = ρ0

T0

Texp

(− z

Hp

)≈ ρ0 exp

(− z

Hρ

)where Hρ is the density scale height. Typical values are Hp =10 km and Hρ=8 km

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Finally, what is ρg?

I If the gas is well mixed, then the mixing ratio:

rg =kg gas

kg air=

ρgρair

= constant

I Which means that

ρg = rgρ0 exp

(− z

Hρ

)I and so the vertical optical depth is:

dτλ = kλρgdz = rgρ0 exp

(− z

Hρ

)dz

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now use ρg to get τλ (p. 192-193

I Integrate the optical thickness between some height z ′ = zand τ ′λ = τλ and the top of the atmosphere where z ′ =∞ andτ ′λ = 0. We need to introduce a minus sign because we aremoving downward (negative z direction) from the top:

∫ 0

τλ

dτ ′λ = −∫ ∞z

kλρgdz′ = −

∫ ∞z

kλrgρ0 exp

(− z ′

Hρ

)dz ′

0− τλ = Hρkλrgρ0

[0− exp

(− z

Hρ

)]τλ = Hρkλrgρ0 exp

(− z

Hρ

)= Hρkλrgρ = Hρβa

where βa = kλρg = kλrgρ is the volume absorption coefficient.

I So the vertical optical depth measured from the top of theatmosphere is proportional to the density for a well-mixedabsorber.

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Level of maximum heating from ozone absorption

I The stratosphere is produced by solar heating due to ozoneabsorption of ultraviolet photons.

I Recall the heating rate equation from day 8:

dT

dt= − 1

ρcp

dFNdz

where ∆FN is the net upward flux.

I Since we know τ , we know the direct beam flux from Beer’slaw (Lecture 8)

Fλ = Fλ0exp(−τλ)

and since Fλ from the sun is downwards, and we are assumingno reflection, Fλ = −FN and dT

dt ∝dFλdz

I So the heating rate will be maximum where d2Fλ

dz2= 0

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heating rate: find τ where d2Fdz2 = 0 (Petty eq. 7.60)

I since F = F0 exp(−τ) and τ = C exp(−z/H):

dF

dz= −F0

dτ

dzexp(−τ) = F0

C

Hexp(−z/H) exp(−τ)

d2F

dz2= −FC

H2exp(−z/H) exp(−τ) +

FC

Hexp(−z/H) exp(−τ)

dτ

dz= 0

1

H+

dτ

dz= 0

1

H− C

Hexp(−z/H) = 0

1

H− τ

H= 0

τ = 1

I So the heating rate is maxium where τ = 1.

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Vertical absorption profile for downward direct solarintensity

I Why does the directbeam change fastestat τλ = 1?

I What height is this?

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Beers law for ozone

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Why are we getting straight lines of different slopes?

I Beer’s law:

Iλ = Iλ0 exp (−τλ/ cos θ)

ln(Iλ) = ln(Iλ0)− τλ/ cos θ

(assuming a plane parallel atmosphere)

I If kλ is constant:

τλ =

∫ ∞z

kλρgdz′ = kλ

∫ ∞z

ρgdz′ = kλ × Constant

I so the slope of the line is proportional to the mass absorptioncoefficient, which is strongest for ozone in the ultraviolet (0.4µm)

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Discovery of the ozone hole

I The ozone hole was discovered using land based sunphotometers in 1985.

I The hole was also detected by the TOMS instrument

I The cause wasn’t fully understood until aircraft measurementswere made in 1987.

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Summary

I Coverage 173-202 on scattering, absorption, emissionI Main points:

I Scattering direction and efficience depends on wavelength λand particile size (r) via the size parameter

I Three distinct scattering regimes: Rayleigh, Mie, GeometricI Absorption depends on line strength and line width (function

of temperature, pressure, wavelength)I Vertical optical thickness relates the physical propoerties of

absorbers/scatterers to direct beam transmissionI Absorption (and it turns out, emission) is maximum whenτ/µ=1.

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Schwartzchild equation

I Reading: Petty pages 205-211I Questions:

I How do we handle an atmosphere that is not only absorbing,but emitting radiation?

I Major problem is that now temperature, absorptivity, emissivityand transmisivity are all changing with height

I Major opportunity that if you know some of these quantitiesyou can use remote sensing to get others (like the temperaturegiven the transmittance).

I The approach: the previous slide we know that absorption ismaximum when τλ = 1. It will turn out that the maximumradiation escaping to space also occurs at τλ = 1. So bymeasuring radiance at several wavelengths we can getinformation about emission at different optical depths.

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Example: High resolution infrared radiation sounder(HIRS): 12 IR channels

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HIRS weighting functions

Using these 7 weighting functionswe can invert 7 radiance measure-ments to find a vertical profile oftemperature in the atmosphere.Below we show that these weight-ing functions peak where the op-tical depth at that wavelength isequal to 1. This is the height oforigin for the majority of photonsreaching the satellite.

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First step: constant temperature

I From last lecture:

I Review Beer’s law, letting µ = cos θ and defining τλ increasingupward:

dIλ = −Iλρair rgaskλds = −Iλρair rgaskλdz/µNow use dτλ/µ = ρair rgasdz/µ and integrate

from (Iλ0, 0) to(Iλ, τλ)∫ Iλ

Iλ0

dI ′λI ′λ

=

∫ Iλ

Iλ0

d ln I ′λ = −∫ τλ

0dτ ′λ/µ

ln I ′λ∣∣IλIλ0

= ln

[IλIλ0

]= −

∫ τλ

0dτ ′λ/µ = −τλ/µ

and taking exp of both sides gives:

Iλ = Iλ0 exp(−τλ/µ)

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absorptivity, emissivity, transmissivity

I Remember the relationships between Trλ, τλ, αλ and ελ:

Trλ =IλIλ0

= exp(−τλ/µ)

and if τλ � 1 then Taylor says :

dTrλ = d exp(−τλ/µ) = − exp(−τλ/µ) dτλ/µ ≈−(1− τλ/µ) dτλ/µ ≈ −dτλ/µ

or do the expansion first then take the differential:

dTrλ = d exp(−τλ/µ) ≈ d(1− τλ/µ) = −dτλ/µconservation of energy without reflection says:

Trλ + αλ = 1

dTrλ = −dαλ

dαλ ≈ dτλ/µ

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Kirchoff’s law and the Schwartzchild equation

I Kirchoff says ελ = αλ so:

dελ = dαλ = dτλ/µ

I So the radiance gain due to emission is:

dI emissionλ = dελBλ(T ) = κλρgBλ(T )dz/µ

I and combine this with the absorption loss to getSchwartzchild’s equation (Petty 8.4)

µdIλdτλ

= −Iλ + Bλ(T ) (1)

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Schwartzchild, cont.

I In class we derived the following: if the temperature T (andhence Bλ(T )) is constant with height∫ Iλ

Iλ0

dIλIλ − Bλ

= −∫ τλ

0dτ ′λ/µ (2)

ln

(Iλ − Bλ

Iλ0 − Bλ

)= −τλ/µ (3)

Iλ − Bλ = (Iλ0 − Bλ) exp(−τλ/µ)) (4)

where µ = cos θ

or rearranging:

Iλ = Iλ0 exp(−τλ/µ) + Bλ(Tatm)(1− exp(−τλ/µ)) (5)

which is Petty 8.33, p. 211

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What the equation is saying:

I If we set µ = 1 (straight up) and define:

Trλ = exp(−τλ/µ) = 1− absorption = 1− αλ

ελ = (1− exp(−τλ/µ))

this gives

Iλ = Iλ0Trλ(τλ, µ) + Bλ(Tatm)ελ

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Integrating dελ for thicker layers:

I On the last slide I said that

ελ = (1− exp(−τλ/µ)) (6)

I But I also said that:

dελ = dαλ = dτλ/µ

which, if I integrate from τ ′λ = 0 to τ ′λ = τλ gives:

ελ = τλ/µ (7)

Both (6) and (7) can’t be right.

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The answerI To see where the difference comes from, go back to

conservation of energy:

Trλ + αλ = 1 (8)

dTrλ = −dαλ = −dελ (9)

dαλ ≈ dτλ/µ (10)

The third equation is only true for very thin layers whereτλ/µ� 1, because in that case:

dTrλ = d exp(−τλ/µ) ≈ d(1− τλ/µ)

So for a thicker layer we need to integrate (9) without thisapproximation to get the right ελ:∫ ελ

0dε′λ = ελ = −

∫ τλ

0d exp(−τ ′λ/µ) = 1−exp(τλ/µ) = 1−Trλ

which is the correct result for either thick or thin layers.30/34

Temperature changing with height

I Here’s the Schwartzchild equation again (τ increasing upwardwith height):

µdIλdτλ

= −Iλ + Bλ (11)

I But with T changing with height, we need to use anintegrating factor to solve this (Petty 8.8, page 206):

d(I exp(τ/µ)) = exp(τ/µ)dI + I exp(τ)dτ/µ (12)

I Now integrate this from 0 up to the level of interest (τ). Firstmultiply both sides by exp(τ/µ) and rearrange:

exp(τ/µ)dIλ + Iλ exp(τ/µ)dτ

µ= exp(τ/µ)Bλ(T )

dτ

µ(13)

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Schwartzchild, cont.I Now use use the chain rule in reverse to combine the two

terms on the left, and integrate:

d (Iλ exp(τ/µ)) = exp(τ/µ)Bλdτ

µ(14)∫ τ

0d(I ′λ exp(τ ′/µ)) =

∫ τ

0exp(τ ′/µ)B ′λ

dτ ′

µ

Now impose the boundary condition that at τ ′ = 0: (15)

I ′λ = Iλ0, Trλ = exp(0) = 1

Iλ exp(τ/µ)− Iλ0 =

∫ τ

0exp(τ ′/µ)Bλ(T ′)

dτ ′

µ

dividing through by exp(τ/µ) we get :

Iλ(τ, µ) = Iλ0(µ) exp(−τ/µ)+∫ τ

0exp

(−(τ − τ ′)

µ

)Bλ(T )

dτ ′

µ(16)

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Here’s a sketch of equation (16)

Figure : Sketch of surface and layer emission terms for an isothermallayer for radiance at zenith angle µ = 0.866 (θ = 30◦)

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Summary:

I Next lecture we’ll show that:

∫ τ

0exp

(−(τ − τ ′)|µ|

)Bλ(T )

dτ ′

|µ|=

∫ z

0B(z ′)W ↑(z ′)dz ′

So the peak of the weighting function can select whichheights make the main contribution to the upward intensity

I Given different weighting functions with different peaks, wecan make a good guess at the vertical profile of B(T ), andtherefore the temperature T .

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