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function (1.50), compute the integral over the azimuth and level items with
the equal numbers m in the left-hand and right-hand parts of the equation.
Only the items with equal numbers m will be nonzero in the product of the
series (the phase function by the radiance) due to the orthogonality of the
trigonometric functions:
π
2
ϕ cos m 2 ϕ d
ϕ =
0for m 1 =
cos m 1
m 2 ,
0
π
=
for
m 1
m 2 ,
π
2
ϕ cos m 2 ϕ d
ϕ =
sin m 1
0.
0
Finally, obtain:
1
= ω 0 (
τ
)
B m (
p m (
µ ) I m (
µ ) d
µ
τ
µ
µ 0 )
τ
µ
τ
µ
,
,
,
,
,
,
2
(1.66)
−1
+ ω 0 (
τ
)
Sp m (
τ
µ
µ 0 )exp(−
τ|µ 0 ).
,
,
4
Further from (1.51) the following equation is derived:
τ
µ
µ 0 )
dI m (
,
,
µ
=
I m (
τ
µ
µ 0 )+ B m (
τ
µ
µ 0 ) ,
,
,
,
,
(1.67)
τ
d
with boundary conditions:
I m (0,
µ
µ 0 )
=
µ 0 > 0and
,
0, for
(1.68)
I m (
τ 0 ,
µ
µ 0 )
=
µ 0 < 0.
,
0for
The following relations are correct for it:
τ
µ 0 )exp τ
d
τ
1
µ
I m (
τ
µ
µ 0 )
=
B m (
τ ,
µ
τ
µ
,
,
,
> 0,
µ
0
(1.69)
τ 0
µ 0 )exp τ
d
τ
1
µ
I m (
B m (
τ ,
τ
τ
µ
µ 0 )
=
µ
µ
,
,
,
< 0.
µ
τ
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