1. Within dr, L changes (dL) from… sources due to scattering & emission losses due to scattering & absorption Spectral Radiance, L(, ,  ) - W m -2 sr -1  m -1

2.  dL /d  = -L (  )+  B (T) +  /4  ∫ 0 2  ∫ -1 1  L (  ’,  ’)P(  s )d  ’d  ’  =  a ( )/  e ( ) = absorption number d      e ( )dz = vertical optical depth  B (T) = emitted energy  /4  ∫ 0 2  ∫ -1 1  L (  ’,  ’)P(  s )d  ’d  ’ = scattering term -L (  ) = radiance Now … we want to simplify equation ….

3. Beer-Bouguer-Lambert Law Assume that no sources of radiance are possible along a path: dL(s) = -  e (s) L(s)ds + J(s)ds 0 dL(s)/ L(s) = -  e (s) ds s s1s1 Integrating … = direct transmittance,  d from s to the boundary s 1 If we define path optical depth as,

4. no scattering (  s =0) but include a source function from emission: B(,T) Schwartzchild’s Equation dL(,s) = -  e (,s) L(,s)ds +  e (,s) B(,T(s))ds multiply by e -  d , and integrate from s to s 1 radiance at s 1 = radiance at s x direct transmittance from s to s 1 + sum of radiance emitted at s’ x direct transmittance from s’ to s 1 (prime means along the path) since d  = -  e (,s) ds, then…

5. normal or vertical path optical depth  (,z) = ’ This differs from the path optical depth by cos   (,s)  (,z)/  where  = cos  From now on  =  (,z) is our vertical coordinate Solutions The radiative transfer equation is then… at  at  at  t

6. 0 tt  (z)   =cos  dd J scat J th direct transmittance = e -  t /  direct transmittance = e -  (z)/  L(  t ; ,  ) radiance change at height z radiance at the top of the atmosphere As an example: summing all changes along the path gives…

7. Special Solutions #1, #2 and #3 We will develop special solutions of the radiative transfer equation for radiance at the top of the atmosphere. These solutions will apply to the major applications of satellite-based remote sensing. Appropriate assumptions will be employed to simplify the solutions to illustrate conceptual principles. These include… Solution #1 - No path radiance Solution #2 - Path radiance from emission only Solution #3 - Path radiance from single-scattering only (Microwave solutions will be presented later)

8. Special Case #1 No Path Radiance We are at a wavelength where B(,T) ~ 0 and there is no scattering solution:  ~ ? Also known as Beer's Law if  = 0  =0.01 0.11.0 7 e -  =99%90.5%36.8%0.1%

9. Optical Depths Examples The SAGE II (Stratospheric Aerosol and Gas Experiment II) sensor was launched into a 57 degree inclination orbit aboard the Earth Radiation Budget Satellite (ERBS) in October 1984. During each sunrise and sunset encountered by the orbiting spacecraft, the instrument uses the solar occultation technique to measure attenuated solar radiation through the Earth's limb in seven channels centered at wavelengths ranging from 0.385 to 1.02 micrometers. The exo atmospheric solar irradiance is also measured in each channel during each event for use as a reference in determining limb transmittances. Unlike the lower atmosphere (or troposphere, which extends from the surface to roughly 10 km), the stratosphere does not have rain clouds as a mechanism to quickly wash out pollutants. Therefore, a heavy influx of aerosol pollutants, like the plume from Mount Pinatubo, will remain in the stratosphere for years until the processes of chemical reactions and atmospheric circulation can filter them out. In the case of Mount Pinatubo, the result was a measurable cooling of the Earth's surface for a period of almost two years. Because they scatter and absorb incoming sunlight, aerosol particles exert a cooling effect on the Earth's surface. The Pinatubo eruption increased aerosol optical depth in the stratosphere by a factor of 10 to 100 times normal levels measured prior to the eruption. ("Aerosol optical depth" is a measure of how much light airborne particles prevent from passing through a column of atmosphere.) Consequently, over the next 15 months, scientists measured a drop in the average global temperature of about 10F (0.60C)

10. This 10km resolution image of Water Cloud Optical Thickness MODIS. Uses bands 1 (648nm) and band 2(858nm)

11. Special Case #2 Emitted Path Radiance Only Here emission is the only source of photons and there is no scattering, so... J (,z) =  a (,z) B(,T(z)) and  e (,z) =  a (,z) Also, L o (, ,  ) =  s (,  ) B(,T s ) The solution becomes: (known as Schwartzchild’s Equation)  ( )

12. For which wavelengths does this solution apply?12 For which wavelengths below does term 1 dominate? For which wavelengths below does term 2 dominate?

13. Special Case #3 Source due to single-scattered path radiance only (note in general multiple scattering is required) J = J scat only Single scattering implies each photon is scattered only once along the path from the source to the satellite. Therefore, the only source of photons L(r’,,X) at some place (X) in the atmosphere is the radiance from the source: L(r’,,X) = L(  0,  0,,X) = L(  0,  0, ) e -  (,z)/  o The path radiance is then L(  0,  0, ) What conditions are required for single scatter to dominate? 00

14. times the probability of transmitting through the atmosphere with out interacting with scatterers. times the probability of the scattered radiance being directed toward the satellite ( ,  ) times the probability of a scattering interaction (rather than an absorption) times the probability of an interaction with a scatterer (1 - the probability of no interaction) The radiance entering the top of the layer plus The radiance that leaves the surface The radiance that reaches the top of the scattering layer consists of… At the top of the atmosphere the result is… If we apply this to a single homogeneous layer…

15. We need to simplify since it implies that we need to know the radiance coming from all other directions r’(  ’,  ’) in order to calculate the radiance in the direction r( ,  ). Substituting J = J scat … Other Approximations: Some approaches simplify sources as in Special Solution #3 above. Some approaches focus on simplifying the scattering phase function p(  s ).

16. 0.39-0.76  m --> visible window 8.5-12.5  m --> IR window 2-4 & 6cm -->  wave window Radar bands Ku = 2.1cm X = 3.6 cm C = 6 cm