METO 621

Lesson 10

Upper half-range intensity

• For the upper half-range intensity we use the integrating factor e-

( ) νν

νν

/// )(1 −−++

−+ −=⎟⎟⎠

⎞⎜⎜⎝

⎛−= e

BeI

d

dIeI

d

d

• In this case we are dealing with upgoing beams and we integrate from the bottom to the top.

Upper half-range intensity

€

Iν+(0,μ,φ) = Iν

+(τ *,μ,φ)e−τ */ μ +dτ '

μ0

τ *

∫ Bν (τ ')e−τ ' / μ

Upper half-range intensity

• To find the intensity at an interior point , integrate from * to and obtain

€

Iν+(τ ,μ,φ) = Iν

+(τ *,μ,φ)e−(τ *−τ ) / μ +dτ '

μτ

τ *

∫ Bν (τ ')e−(τ '−τ ) / μ

• What happens when approaches zero. This is where the line of sight traverses an infinite distance parallel to the slab. Here

€

Iτ±(τ ,μ = 0,φ) = Bν (τ )

Formal solution including Scattering and Emission

• Note that the source is now due to thermal emission and multiple scattering

€

S(τ , ˆ Ω ) = 1− a(τ )[ ]B(τ ) +a(τ )

4πdω' p(τ , ˆ Ω ', ˆ Ω )I(τ , ˆ Ω ')

4π

∫

• The independent variable is the extinction optical depth, the sum of the absorption and scattering optical depths. We can write

)ˆ,()ˆ,()ˆ,(

Ω+Ω−=Ω

SId

dI

Formal solution including scattering and emission

• The method of using an integrating factor can be applied as before

€

I τ (P2), ˆ Ω [ ] = I τ (P1), ˆ Ω [ ]e−τ (P1,P 2) + dt Sτ (P1 )

τ (P2 )

∫ (t, ˆ Ω )e−τ (P P2 )

• In slab geometry the solutions become

€

I−(τ ,μ,φ) = I−(0,μ,φ)e−τ / μ +dτ '

μ0

τ

∫ S(τ ',μ,φ)e−(τ −τ ' ) / μ

Formal solution including scattering and emission

€

I+(τ ,μ,φ) = I+(τ *,μ,φ)e−(τ *−τ ) / μ +dτ '

μτ

τ *

∫ S (τ ')e−(τ '−τ ) / μ

• where

€

I±(τ ,μ = 0,φ) = S (τ ,μ = 0,φ)

€

S(τ ,μ,φ) = 1− a[ ]B(τ )

+a

4πdφ' dμ ' p(μ ',φ';μ,φ)I+(τ ,μ ',φ')

0

1

∫0

2π

∫

+a

4πdφ' dμ ' p(−μ ',φ';μ,φ)I−(τ ,μ ',φ')

0

1

∫0

2π

∫

• and

Radiative Heating Rate

• The differential change of energy over the distance ds along a beam is

€

∂(d4 E) = dIν dA dt dν dω

• If we divide this expression by dsdA, (the unit volume, dV), and also dνdt then we get the time rate of change in radiative energy per unit volume per unit frequency, due to a change in intensity for beams within the solid angle d.

Radiative Heating Rate

• Since there is (generally) incoming radiation from all directions, the total change in energy per unit frequency per unit time per unit volume is

∫∫ ∇⋅Ω=π

νπ

ν 44

)ˆ( Idds

dId

• The spectral heating rate H is

)ˆ(4∫ ∇⋅Ω−=π

νν Η Id

Radiative Heating Rate

• The net radiative heating rate H is

€

Η =− dν0

∞

∫ dω( ˆ Ω ⋅∇Iν )4π

∫

• In a slab geometry the radiative heating rate is written

€

H = − dνδFν

δz0

∞

∫ = −2π dν cosθd(cosθ)∂Iν

∂z−1

+1

∫0

∞

∫

where Fν = Fν+ − Fν

− is the radiative flux in the

z direction

Separation into diffuse and direct(Solar) components

• Two distinctly different components of the shortwave radiation field. The solar component:

€

IS−(τ ,θ,φ) = F Se−τ / μ 0δ( ˆ Ω − ˆ Ω 0)

= F Se−τ / μ 0δ(μ − μ0)δ(φ − φ0)• We have two sources to consider, the Sun and the rest of the medium

€

I−(τ ,μ,φ) = Id− (τ ,μ,φ) + IS

−(τ ,μ,φ)

Diffuse and direct components

• Assume (1) the lower surface is black, (2) no thermal radiation from the surface, the we can write the half range intensities as

€

− dI−(τ , ˆ Ω )

dτ= I−(τ , ˆ Ω ) − (1− a)B

−a

4πdω' p(+ ˆ Ω ',− ˆ Ω )I+(τ , ˆ Ω ')

+

∫

−a

4πdω' p(− ˆ Ω ',− ˆ Ω )I−(τ , ˆ Ω ')

+

∫

Diffuse and direct components

• And for the +ve direction

€

− dI+(τ , ˆ Ω )

dτ= I+(τ , ˆ Ω ) − (1− a)B

−a

4πdω' p(+ ˆ Ω ',+ ˆ Ω )I+(τ , ˆ Ω ')

+

∫

−a

4πdω' p(− ˆ Ω ',+ ˆ Ω )I−(τ , ˆ Ω ')

+

∫

Diffuse and direct components

∫

∫∫

−

−

−

+

−

−

−−−−

ΩΩ−Ω−−

ΩΩ−Ω+−ΩΩ−Ω−−

−−Ω+Ω=Ω

−Ω

−

)'ˆ,()ˆ,'ˆ('4

)'ˆ,()ˆ,'ˆ('4

)'ˆ,()ˆ,'ˆ('4

)1()ˆ,()ˆ,()ˆ,()ˆ,(

τωπ

τωπ

τωπ

τττ

τμ

τ

τμ

d

dS

SdSd

Ipda

Ipda

Ipda

BaIId

dI

d

dI

Now substitute the sum of the direct and diffuse components

Diffuse and direct components

gets one hence

/ and beamsolar direct theis But dIdII SSS−−− −=

€

− dId− (τ , ˆ Ω )

dτ= Id

− (τ , ˆ Ω ) − (1− a)B − S*(τ ,− ˆ Ω )

−a

4πdω' p(+ ˆ Ω ',− ˆ Ω )Id

+(τ , ˆ Ω ')+

∫

−a

4πdω' p(− ˆ Ω ',− ˆ Ω )Id

− (τ , ˆ Ω ')−

∫

Diffuse and direct components

• where

€

S*(τ ,− ˆ Ω ) =a

4πdω' p(− ˆ Ω ',− ˆ Ω )F S

−

∫ e−τ / μ 0δ( ˆ Ω − ˆ Ω 0)

=a

4πp(− ˆ Ω 0,− ˆ Ω )F Se−τ / μ 0

• One can repeat this procedure for the upward component

Diffuse and direct components

€

− dId+(τ , ˆ Ω )

dτ= Id

+(τ , ˆ Ω ) − (1− a)B − S*(τ ,+ ˆ Ω )

−a

4πdω' p(+ ˆ Ω ',+ ˆ Ω )Id

+(τ , ˆ Ω ')+

∫

−a

4πdω' p(− ˆ Ω ',+ ˆ Ω )Id

− (τ , ˆ Ω ')−

∫

Diffuse and direct components

€

S*(τ ,+ ˆ Ω ) =a

4πdω' p(− ˆ Ω ',+ ˆ Ω )F S

−

∫ e−τ / μ 0δ( ˆ Ω − ˆ Ω 0)

=a

4πp(− ˆ Ω 0,+ ˆ Ω )F Se−τ / μ 0

Diffuse and direct components

• If we combine the half-range intensities we get

€

udI(τ ,u,φ)

dτ= I(τ ,u,φ) −

a

4πdφ'

0

2π

∫ du', p(u' .φ';u,φ)I(τ ,u',φ')−1

+1

∫

− (1− a)B − S*(τ ,u,φ)

• Where u is cos and not |cos|