introduction: absorption, hydrostatic balance, heating...

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Introduction: absorption, hydrostatic balance, heating rate I Coverage: pp. 173-202 on scattering, absorption emission I Main points: I Scattering direction and efficience depends on wavelength λ and particile size (r) via the size parameter I Three distinct scattering regimes: Rayleigh, Mie, Geometric I 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 1/34

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Page 1: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

1/34

Page 2: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

Note that in this case z ispositive downward!

2/34

Page 3: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

3/34

Page 4: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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)

4/34

Page 5: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

5/34

Page 6: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

6/34

Page 7: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

7/34

Page 8: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

8/34

Page 9: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

9/34

Page 10: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

10/34

Page 11: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

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

11/34

Page 12: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

)I and so the vertical optical depth is:

dτλ = kλρgdz = rgρ0 exp

(− z

)dz

12/34

Page 13: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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 ′

)dz ′

0− τλ = Hρkλrgρ0

[0− exp

(− z

)]τλ = Hρkλrgρ0 exp

(− z

)= 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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Page 14: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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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Page 15: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

dzexp(−τ) = F0

C

Hexp(−z/H) exp(−τ)

d2F

dz2= −FC

H2exp(−z/H) exp(−τ) +

FC

Hexp(−z/H) exp(−τ)

dz= 0

1

H+

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.

15/34

Page 16: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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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Page 17: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

Beers law for ozone

17/34

Page 18: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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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Page 19: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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.

19/34

Page 20: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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.

20/34

Page 21: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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.

21/34

Page 22: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

22/34

Page 23: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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.

23/34

Page 24: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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)

24/34

Page 25: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

25/34

Page 26: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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)

26/34

Page 27: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

27/34

Page 28: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

28/34

Page 29: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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.

29/34

Page 30: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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

Page 31: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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 )

µ(13)

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Page 32: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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)

32/34

Page 33: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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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Page 34: Introduction: absorption, hydrostatic balance, heating ratephil/courses/eosc582_2012/textfiles/e582... · Size parameter (scattering) How light scatters depends on the relative size

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 .

34/34