introduction: absorption, hydrostatic balance, heating...
TRANSCRIPT
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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