Transcript Document

Review:
defined radiance as
irradiance (brightness)
/B
Wm-2 Sr-1
L = dF / (dW. ds. cosq) (in W.Sr-1. m-2)
then total radiant flux F = ∫∫ L(f,q dW. cosq
And E (irradiance over all wavelengths) = Ltotal(Q,f) = ∫ Ll (Q,f dl
Spectral Radiance (L at a particular wavelength) defined as: L(l,q,f) - W m-2 sr-1 mm -1
Within dr,
L changes (dL) from…
sources due to
scattering &
emission
losses due to
scattering &
absorption
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
F
C
D
H
Problems:
• absorption by molecules in atmosphere
• attenuation of signal by scattering
• EMR emitted by atmosphere at same freq. as
signal of interest.
A
I
E
G
B
http://rst.gsfc.nasa.gov/Intro/Part2_4.html
dLl/dr = A + B + C + D
A = absorption =
-sa(l) Ll
B = emission =
sa(l) Bl(T)
C = scattering out = -ss(l) Ll
D = scattering in =
ss(l) <Ll’>
where <Ll’> =
1/4p∫∫Ll Q ’ ,f ’ P 
q q’ f ’
P  = scattering phase function
 = scattering angle
(angle between Q ,f
Q ’,f ’
))
dLl/dr = sa(l)[Bl(T) - Ll Q,f ] + ss(l)[<Ll’>- Ll Q,f ]
absorp. coeff.
scatter. coeff.
extinction coeff.
sa
ss
se
m-1
m-1
m-1
In other words:
At position X (x,y,z), and along direction vector r (r,q,f)
dL(l,X,r) = -se(l,X) L(l,X,r)dr + J(l,X,r)dr
Term 1: Represents a loss of photons
se(l,X) = Beam attenuation coefficient
= sa(l,X) + ss(l,X)
sa = Volume absorption coefficient
ss = Volume scattering coefficient
dr
Both have units of 1/length
Term 2: Represents a source of photons
J = Jth + Jscat
emitted
scattered
SOURCES
Jth(l,X) = emittance along path x Planck function for T(X)
= sa(l,X) B(l,T(X))
dr
T(X) = temperature at X
e(l,X) = thermal emittance/distance
but e(l,X) = sa(l,X)
Jscat(l,X) = sum of scattering from all directions
=
(q,)
dr
(q,)
ys
r
gs(r,r,l,X) = volume scattering function
(Probability/distance that a photon
moving in a direction r will be scattered
into the direction r)
Define ys= scattering angle from (q,) to (q,)
cos ys = cos q cos q + sin q sin qcos(-)
may see gs(r,r,l,X) = gs(q,;q,;l,X) = gs(ys,l,X)
also note ss(l,X) =
Inherent Optical Properties
Independent of illuminating radiance
Determined by substance itself
se, sa, ss, e, p(ys )(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
d(l,z) =
’
dl = d0 = TOA
dl d = s
Q
dl
dl = 0
dl ( , ) = ∫s
dl (s,s ) = ∫s /m
where m = cosQ
Single Scattering albedo
Probability of a scatter vs an absorption
when a photon interacts with a particle.
wo =
if wo = 1 ---> no absorbtion
if wo = 0 ---> all absorbtion
Examples:
Bare soil
Sand, desert
Grass
Forest
Snow (clean)
Snow (dirty/wet)
Sea surface (>25°)
Sea surface (low sun)
10-25%
25-40%
15-25%
10-20%
75-95%
25-75%
< 10 %
10-70%
Scattering
Let  = 2pr/l ; r == radius of scatterer( e.g. raindrop, dust)
Mie scattering -   1; l  2pr (wavelength and diameter similar)
examples: radars and raindrops (microwave)
Visual l and aerosals (400-700nm)
IR and cloud droplets (~10mm)
This true-color image acquired May 4,
2001, by the Sea-viewing Wide Field-ofview 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.
http://earthobservatory.nasa.gov/Newsroom/NewImages/Images/S2001124.L1A_HNSG.jpg
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 l 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 1999
www.visibleearth.nasa.gov
Rayleigh scattering -   01 (diameter of scatterer much smaller)
examples:
Visual/IR l 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.
raindrops
Geometric
drizzle
r (m 
dust haze
cloud drops
Negligible
air molecules
radar
0.1 1 10 100 1e3 1e4 1e5 1e6
l (m 
Bragg (resonant) scattering
active 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 Bragg
scattering 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 have
been 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).
mdLl/ddl = -Ll(Q,f)+ alBl(T) + wl/4p ∫0
ddl
2p ∫
-1
1
Ll(m’,f’)P(s)dm’df’
se(l)dz = vertical optical depth
al= sa(l)/ se(l) = absorption number
alBl(T) = emitted energy
wl/4p ∫0
2p ∫
-1
1
Ll(m’,f’)P(s)dm’df’ = scattering term
-Ll(Q,f) = radiance
Now … we want to simplify equation ….
Beer-Bouguer-Lambert Law
Assume that no sources of radiance are possible along a path:
dL(s) = -se(s) L(s)ds + J(s)ds
0
s
dL(s)/ L(s) = -se(s) ds
Integrating …
If we define path optical depth as,
= direct transmittance, td from s to the boundary s1
s1
Schwartzchild’s Equation
no scattering (ss=0) but include a source function from emission: B(l,T)
dL(l,s) = -se(l,s) L(l,s)ds + se(l,s) B(l,T(s))ds
since dd = -se(l,s) ds, then…
multiply by e-d dd, and integrate from s to s1
(prime means along the path)
radiance at s1
=
radiance emitted at s’
radiance at s
x
+
sum
of
x
direct transmittance
direct transmittance
from s’ to s1
from s to s1
normal or vertical path optical depth
d(l,z) =
’
This differs from the path optical depth by cos q
d(l,s) = d (l,z)/m
where m = cos q
From now on d = d (l,z) is our vertical coordinate
Solutions
at d
at d
at 0
at dt
The radiative transfer equation
is then…
radiance at the top of the atmosphere
As an example:
0
m=cos q
q
direct
-d(z)/m
transmittance = e
summing all changes
along the path gives…
Jscat
d(z)
Jth
dd
direct transmittance = e
-dt/m
radiance change at height z
L(dt;m,f)
dt