Lecture 5: Radiative transfer theory where light comes from and how it gets to where it’s going Tuesday, 19 January 2010 Ch 1.2 review, 1.3, 1.4

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Lecture 5: Radiative transfer theory where light comes from and how it gets to where it’s going Tuesday, 19 January 2010 Ch 1.2 review, 1.3, (scattering) (refraction) (Snell’s Law) Review On Solid Angles, (class website -- Ancillary folder: Steradian.ppt) Last lecture: color theory data spaces color mixtures absorption Reading

The Electromagnetic Spectrum (review) Units: Micrometer = m Nanometer = m Light emitted by the sun The Sun

Light from Sun – Light Reflected and Emitted by Earth Wavelength, μm W m -2 μm -1 W m -2 μm -1 sr -1 The sun is not an ideal blackbody – the 5800 K figure and graph are simplifications

Atmospheric Constituents Constant Nitrogen (78.1%) Oxygen (21%) Argon (0.94%) Carbon Dioxide (0.033%) Neon Helium Krypton Xenon Hydrogen Methane Nitrous Oxide Variable Water Vapor ( %) Ozone (0 – 12x10 -4 %) Sulfur Dioxide Nitrogen Dioxide Ammonia Nitric Oxide All contribute to scattering For absorption, O 2, O 3, and N 2 are important in the UV CO 2 and H 2 O are important in the IR (NIR, MIR, TIR)

Solar spectra before and after passage through the atmosphere

Atmospheric transmission

Modeling the atmosphere To calculate  we need to know how k in the Beer-Lambert- Bouguer Law (called  here) varies with altitude. Modtran models the atmosphere as thin homogeneous layers. Modtran calculates k or  for each layer using the vertical profile of temperature, pressure, and composition (like water vapor). This profile can be measured made using a balloon, or a standard atmosphere can be assumed.  o is the incoming flux

Radiosonde data Altitude (km) Relative Humidity (%) Temperature ( o C) Mt Everest Mt Rainier

Radiant energy – Q (J) - electromagnetic energy Solar Irradiance – I toa (W m -2 ) - Incoming radiation (quasi directional) from the sun at the top of the atmosphere. Irradiance – I g (W m-2) - Incoming hemispheric radiation at ground. Comes from: 1) direct sunlight and 2) diffuse skylight (scattered by atmosphere). Downwelling sky irradiance – I s↓ (W m -2 ) – hemispheric radiation at ground Path Radiance - L s↑ (W m -2 sr -1 ) (L p in text) - directional radiation scattered into the camera from the atmosphere without touching the ground Transmissivity –  - the % of incident energy that passes through the atmosphere Radiance – L (W m -2 sr -1 ) – directional energy density from an object. Reflectance – r -The % of irradiance reflected by a body in all directions (hemispheric: r·I) or in a given direction (directional: r·I·  -1 ) Note: reflectance is sometimes considered to be the reflected radiance. In this class, its use is restricted to the % energy reflected. IgIg L s↑ I toa 0.5º I s↓ L Terms and units used in radiative transfer calculations

DN = a·I g ·r + b Radiative transfer equation I g is the irradiance on the ground r is the surface reflectance a & b are parameters that relate to instrument and atmospheric characteristics This is what we want Parameters that relate to instrument and atmospheric characteristics

DN = g·(  e ·r ·  i ·I toa ·cos(i)/  +  e · r·I s↓ /  + L s↑ ) + o gamplifier gain  atmospheric transmissivity eemergent angle iincident angle rreflectance I toa solar irradiance at top of atmosphere I g solar irradiance at ground I s↓ down-welling sky irradiance L s↑ up-welling sky (path) radiance oamplifier bias or offset Radiative transfer equation DN = a·I g ·r + b

The factor of  Consider a perfectly reflective (r=100%) diffuse “Lambertian” surface that reflects equally in all directions. Lambert

The factor of  Consider a perfectly reflective (r=100%) diffuse “Lambertian” surface that reflects equally in all directions. If irradiance on the surface is I g, then the irradiance from the surface is r·I g = I g W m -2. The radiance intercepted by a camera would be r·I g /  W m -2 sr -1. The factor  is the ratio between the hemispheric radiance (irradiance) and the directional radiance. The area of the sky hemisphere is 2  sr (for a unit radius). So – why don’t we divide by 2  instead of  ?

∫ ∫ L sin  cos  d  d  L 22  00 Incoming directional radiance L  at elevation angle  is isotropic Reflected directional radiance L  cos  is isotropic Area of a unit hemisphere: ∫ ∫ sin  d  d  22  00 The factor of  Consider a perfectly reflective (r=100%) diffuse “Lambertian” surface that reflects equally in all directions.

i I toa cos(i) I toa  g  i I toa cos(i) ii r reflectance r (  i I toa cos(i) ) /  reflected light “Lambertian” surface ee e L s↑ (L p ) Highlighted terms relate to the surface i

i I toa cos(i) I toa  g  i I toa cos(i) ii r reflectance r (  i I toa cos(i) ) /  reflected light “Lambertian” surface ee e Measured L toa DN(I toa ) = a I toa + b L toa  e r (  i I toa cos(i) ) /  +  e r I s↓ /  + L s↑ L s↑ (L p ) Highlighted terms relate to the surface I s↓ L s↑ =r I s↓ /  i Lambert

Next lecture: Atmospheric scattering and other effects Mauna Loa, Hawaii