meto 621
DESCRIPTION
METO 621. Lesson 9. Solution for Zero Scattering. If there is no scattering, e.g. in the thermal infrared, then the equation becomes. This equation can be easily integrated using an integrating factor e t. Solution for Zero Scattering. Solution for Zero Scattering. - PowerPoint PPT PresentationTRANSCRIPT
METO 621
Lesson 9
Solution for Zero Scattering
• If there is no scattering, e.g. in the thermal infrared, then the equation becomes
dI
d S
I B (T)
• This equation can be easily integrated using an integrating factor e
dI
de I e
d
d(I e ) Be
Solution for Zero Scattering
Solution for Zero Scattering• Consider a straight path between point P1 and P2 . The optical path from P1 to an intermediate point P is given by
(P1,P) ds ds0
P
P1
P
ds (P) (P10
P1
)
• Integrating along the path from P1 to P2
dtd
dt (P1 )
(P2 )
(Ie t ) I (P2) e (P2 ) I (P1) e (P1 )
dt B(t)e t
(P1 )
(P2 )
Solution for Zero Scattering
• Dividing through by eP2) we get
I (P2) I (P1) e (P2 ) (P1 ) dt B (P1 )
(P2 )
(t)e t (P2 )
I (P1) e (P1 ,P2 ) dt B (P1 )
(P2 )
(t)e t(P ,P2 )
• But what does this equation tell us about the physics of the problem?
Physical Description
Solution for Zero Scattering
• Break up the path from P1 to P2 into small elements s with optical depths
• When n is zero then n is equal to 1• Hence is the blackbody emission from
the element ds• The intensity at P1 is I[t(P1)]• This intensity will be absorbed as it moves
from P1 to P2 , and the intensity at P2 will be I[t(P1)]exp[-(t(P2)-t(P1)]
Solution for Zero Scattering
• Now consider each small element P with a with an optical depth
• Emission from each element is B• The amount of this radiation that reaches P2 is
texpP,P2)] where is the optical depth between P and P2
• Hence the total amount of radiation reaching P2
from all elements is
t)()()(
)(
),(),(
0
2
1
22 P
P
PPtPPnetBeB
Isothermal Medium – Arbitrary Geometry
22
2
22
10
then,0P such thatorigin theRedefine
0
12
1
PP
PtPP
eBeI
dteBePIPI
If the medium is optically thin, i.e. τ(P2) <<1 then the second term becomes B τ(P2).
If there is no absorption or scattering then τ=0 and the intensity in any direction is a constant, i.e. I[τ(P2)]=I[τ(P1)]
Isothermal Medium – Arbitrary Geometry
If we consider the case when τ>>1 then the total intensity is equal to B(T). In this case the medium acts like a blackbody in all frequencies, i.e. is in a state of thermodynamic equilibrium.
If ones looks toward the horizon then in a homogeneous atmosphere the atmosphere has a constant temperature. Hence the observed intensity is also blackbody
Zero Scattering in Slab Geometry• Most common geometry in the theory of
radiative transfer is a plane-parallel medium or a slab
• The vertical optical path (optical depth) is given the symbol as distinct from the slant optical path s
• Using z as altitude (z) = s |coss • The optical depth is measured along the
vertical downward direction, i.e. from the ‘top’ of the medium
Half-range Intensities
Half-Range Quantities in Slab Geometry
• The half-range intensities are defined by:
I(,,) I (, /2,)
I (,,) I (, /2,)
• Note that the negative direction is for the downward flux,
Half-Range Quantities in Slab Geometry
• The radiative flux is also defined in terms of half-range quantities.
F d cos I
( ˆ ) d d sin cos I
0
/ 2
0
2
(,,)
d d I
0
/ 2
0
2
(,,)
F d cos I
( ˆ ) d d sin cos I
0
/ 2
0
2
(,,)
d d I
0
/ 2
0
2
(,,)
Half Range Quantities
• In the limit of no scattering the radiative transfer equations for the half-range intensities become
dI(,,)
dI
(,,) B()
dI (,,)
dI
(,,) B( )
Formal Solution in Slab Geometry
• Choose the integrating factor e for the first equation, then
d
dI e / dI
d
1
I
e /
B ()
e /
• This represents a downward beam so we integrate from the “top” of the atmosphere (=0) to the bottom (
Slab geometry
or
)'('
),,0(),*,('
'
/'*
0
/*/'*
0
Bed
IeIeId
dd
/)'*(
*
0
/ )'('
),,0(),*,(* eB
deII
Slab Geometry
I (,,) I
(0,,)e / d '
0
B ( ')e ( ' ) /
• For an interior point, * , we integrate from 0 to . The solution is easily found by replacing * with