Review: defined radiance as

irradiance (brightness) /BWm-2 Sr-1

L = d/ (d. ds. cos) (in W.Sr-1. m-2)

then total radiant flux = ∫∫ Ld. cos

And E (irradiance over all wavelengths) = Ltotal() = ∫ L d

Within dr, L changes (dL) from…

sources due to scattering & emission

losses due to scattering & absorption

Spectral Radiance (L at a particular wavelength) defined as: L(,,) - W m-2 sr-1 m -1

Sources of radiation received by a satellite

• emitted from the surface (land/water/ice) A • emissions from subsurface layers of the ocean B• direct atmospheric emissions C• direct cloud emissions D

• reflected cloud emissions E• reflected atmospheric emissions F• reflected solar emissions G

• scattered solar H• scattered atmospheric emissions I

A

B

CD

G

F

E

H

IProblems:• absorption by molecules in atmosphere• attenuation of signal by scattering• EMR emitted by atmosphere at same freq. as signal of interest.

http://rst.gsfc.nasa.gov/Intro/Part2_4.html

dL/dr = A + B + C + D

A = absorption = -a() L

B = emission = a() B(T)

C = scattering out = -s() L

where <L’> = 1/4∫∫L’’P’’ P = scattering phase function = scattering angle (angle between ’’))dL/dr = a()[B(T) - L] + s()[<L’>- L]

absorp. coeff. am-1 scatter. coeff. sm-1 extinction coeff. em-1

D = scattering in = s() <L’>

In other words:

At position X (x,y,z), and along direction vector r (r,,)

dL(,X,r) = -e(,X) L(,X,r)dr + J(,X,r)dr

Term 1: Represents a loss of photons

e(,X) = Beam attenuation coefficient

= a(,X) + s(,X)

a = Volume absorption coefficient

s = Volume scattering coefficient

Both have units of 1/length

Term 2: Represents a source of photons

J = Jth + Jscat

emitted scattered

dr

Jth(,X) = emittance along path x Planck function for T(X)

= a(,X) B(,T(X))

T(X) = temperature at X(,X) = thermal emittance/distance

but (,X) = a(,X)

dr

dr

Jscat(,X) = sum of scattering from all directions

=

s(r,r,,X) = volume scattering function(Probability/distance that a photon moving in a direction r will be

scattered into the direction r)

Define s= scattering angle from (,) to (,)

cos s = cos cos + sin sin cos(-) may see s(r,r,,X) = s(,;,;,X) = s(s,,X)

also note s(,X) =

(,)

(,)

s

SOURCES

r

Inherent Optical PropertiesInherent Optical Properties

Independent of illuminating radiance

Determined by substance itself

e, a, s, , p(s )(scattering phase function) refractive index, m= n-in’ (ratio of c in a vacuum:speed with which EMR travels in that substance). At sea level m(air) = 1.0003.

n scattering ; n’ absorption

normal or vertical path optical depth

(,z) =

’

= TOA

) = ∫

ss) = ∫/where = cos

Single Scattering albedo

Probability of a scatter vs an absorption when a photon interacts with a particle.

o =

if o = 1 ---> no absorbtion

if o = 0 ---> all absorbtion

Examples:Bare soil 10-25%Sand, desert 25-40%Grass 15-25%Forest 10-20%Snow (clean) 75-95%Snow (dirty/wet) 25-75%Sea surface (>25°) < 10 %Sea surface (low sun) 10-70%

Scattering

Let = 2r/; r == radius of scatterer( e.g. raindrop, dust)

Mie scattering - 1; r (wavelength and diameter similar)examples: radars and raindrops (microwave)

Visual and aerosals (400-700nm) IR and cloud droplets (~10m)

http://earthobservatory.nasa.gov/Newsroom/NewImages/Images/S2001124.L1A_HNSG.jpg

This true-color image acquired May 4, 2001, by the Sea-viewing Wide Field-of-view Sensor (SeaWiFS) reveals a large, plume of aerosols blowing eastward over the North Atlantic Ocean. The aerosol plume is the regional haze produced by the industrial northeastern United States that you typically see during the summer months. The haze is composed of sulfates and organics that originate from power plants and automotive sources. It is a little surprising to see this much haze so early in the season.

Also, looking closely, beneath the haze you can see a large bloom of phytoplankton in the ocean extending northeastward from the coast of North Carolina. Click on the full image (above) to see another, brighter phytoplankton bloom located about 2,000 km (1,250 miles) due east of Cape Hatteras, North Carolina

geometric scattering - 50 (diameter of scatterer much greater)examples:

Visual and rainbow; halo IR and precipitation

A very well-defined spiral eddy is visible through the haze off the east coast of Japan in this SeaWiFS image. Mar 22 1999www.visibleearth.nasa.gov

Rayleigh scattering - 1 (diameter of scatterer much smaller)examples:

Visual/IR and air molecules IR and cloud droplets

The blue color of the sky is caused by the scattering of sunlight off the molecules of the atmosphere. This scattering, called Rayleigh scattering, is more effective at short wavelengths. Therefore the light scattered down to the earth at a large angle with respect to the direction of the sun's light is predominantly in the blue end of the spectrum.

0.1 1 10 100 1e3 1e4 1e5 1e6

1e-4

1e-3

0

.01

0.1

1

1

0

10

0 1

e3

1

e4

1

e5

1e6

Geometric

Mie

Rayle

igh

Negligible

sola

r

Eart

h/IR

Pass

ive

wave

radar

raindrops

drizzle

cloud drops

dust haze

air molecules

r

Bragg (resonant) scatteringactive radars and

spectrum of sea surface waves

In the incidence angle range between 20° and 70° the main mechanism for the backscattering of microwaves from the ocean surface is described by Braggscattering theory [20]. The power of the backscattered radar signal is therefore dependent on the spectral power density of water surface waves which have the wavelength lB = l0 / 2sin(J) (Bragg wavelength), which depends on the radar wavelength l0 and the incidence angle J. The radar wavelength of the ERS-1/2 SAR is l0 = 5.7 cm and the (mean) incidence angle J = 23 °, the corresponding Bragg wavelength, thus, is lB = 7.2 cm.In Figure 1 a composite of two ERS-1 SAR images acquired on April 16, 1994, at 21:04 UTC

over the southern part of the Baltic Sea is shown (image dimensions 100 km by 100 km). The dark, spiral-like signatures in the bottom half (Pomeranian Bay) are very likely caused by natural surface films which havebeen formed on the water surface due to high biological activity in that particular coastal region in April (spring plankton bloom). The shape of biogenic slicks mostly occurring in coastal waters is caused by interactions with surface currents and eddies. The large, completely dark areas, e.g., north off the island of Rügen, could be caused by surface films or by low wind speed (below the threshold value for wave generation). Note the dark elongated line in the upper left part (south off Sweden) which is very likely caused by mineral oil freshly spilled out from a ship (the bright spot on the right edge of the spill, see the arrow).

dL/d = -L()+ B(T) + /4∫0 2∫ -1

1L(’,’)P(s)d’d’

= a()/ e() = absorption number

de()dz = vertical optical depth

B(T) = emitted energy

/4∫0 2∫ -1

1L(’,’)P(s)d’d’ = scattering term

-L() = radiance

Now … we want to simplify equation ….

Beer-Bouguer-Lambert Law

Assume that no sources of radiance are possible along a path:

dL(s) = -e(s) L(s)ds + J(s)ds0

dL(s)/ L(s) = -e(s) ds

s

s1

Integrating …

= direct transmittance, d from s to the boundary s1

If we define path optical depth as,

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 s1

radiance at s1= radiance at s

x

direct transmittancefrom s to s1

+ sum of

radiance emitted at s’x

direct transmittancefrom s’ to s1

(prime means along the path)

since d = -e(,s) ds, then…

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 equationis then…

at

at

at

at t

0

t

(z)

=cos

dJscat Jth

direct transmittance = e-t/

direct

transmittance = e-(z)/

L(t;,)

radiance change at height z

radiance at the top of the atmosphereAs an example:

summing all changes along the path gives…