Lecture9

# Lecture9 - Chapter 8 Atmospheric Emission 1 Perspective:...

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1 1 Chapter 8 Atmospheric Emission 2 Perspective : Radiation traversing a medium can be extinguished by absorption or scattering out of the path, enhanced by scattering into the path, and enhanced by emission into the path. Chapter 7: Extinction (absorption or scattering out of the path) Chapter 8: Emission (into the path) Chapter 11 & 13: Scattering into the path Absorption : dI abs = - β a I ds (differential form of Beer’s law) β a ds = absorptivity (a) = fraction of radiation lost to absorption Emission : dI emit = ε B , where B is the Planck Function (fn of λ and Temp) Kirchhoff : absorptivity = emissivity for LTE conditions dI emit = β a B ds Net change in radiant intensity is dI = dI emit + dI abs = β a ds (B – I) , which gives: () I B = a ds dI β Schwarzschild’s Equation (Eq. 8.4) I=Incident radiance, B=Planck radiance, (W m -2 sr -1 ) β a = Absorption extinction coefficient (m -1 ) Schwarzschild’s Equation describes Radiative Transfer (RT) in a non-scattering LTE atmosphere. (LTE = Local Thermodynamic Equilibrium)

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2 3 How does the observed radiance depend on the (absorption) optical depth of the atmosphere? Consider the following geometry: S is the position of the detector. s is an arbitrary position along the source/detector path. ●τ (s) is the optical path between point s and the detector. () ( ) ' ' ds s s S s a = β τ 8.5 s S This is opposite to the convention used when we originally introduced Beer’s law, where τ increased as the path length through the medium increased. This is done because we are interested not only in an external source (e.g., the sun shining through the atmosphere), but also the atmospheric emission itself as a source. Here we define the geometry such that the optical path, τ , is 0 at the detector. External Source Atmosphere 4 ' ' ds s s S s a = 8.5 From last slide ds d a = Differentiate wrt s: 8.6 Why is there a negative sign? s S Here we define the geometry such that the optical path, τ , is 0 at the detector. We have defined τ to increase with increasing distance from the detector – in other words, we are assuming a stationary (constant position) detector and a movable source. The source might include both an external source and emission from the atmosphere along the path. Mathematically (assuming constant β a for simplicity): External Source () a a a a S s a s ds d S ds d ds s ds d ds d = = = = 0 ' '
3 5 () I B = a ds dI β 8.4 Substitute 8.6 into 8.4 to get: B I = τ d dI Knowing the answer you wish to end up with , multiply both sides by exp(- τ ) and rearrange: ) exp( B ) exp( I ) exp( = d dI 8.7 8.9 Back to: How does the observed radiance depend on the (absorption) optical depth of the atmosphere? ds d a = 8.6 We just showed: Then presciently recognize that: [] ) exp( I ) exp( I ) exp( = d d d dI 8.8 [i.e., d (f × g) = (f × dg) – (g × df) ] 6 Substitute 8.8 into 8.9 to get: ) exp( B ) exp( I = d d Now integrate from the sensor ( τ = 0) to an arbitrary point τ′ : = ' 0 ' 0 ) exp( B ) exp( I d d d d = ' 0 ) exp( B ) 0 ( I ) ' exp( ) ' ( I d + = ' 0 ) exp( B ) ' exp( ) ' ( I ) 0 ( I d 8.13 8.10 8.12 8.11 Rearrange: )

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## Lecture9 - Chapter 8 Atmospheric Emission 1 Perspective:...

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