Finite atmospheres

infinite atmosphere is not a good model for a planetary atmosphere, even for that of Venus. We shall now consider the same radiative heating problem for finite atmospheres with a ground layer, the

properties of which were given in Section II.1. We shall solve the problem in the same way as we did for semi-infinite atmospheres

and then we shall consider the special case of a finite atmosphere with no ground. This has no value in planetary atmosphere studies but is included for completeness by which it emphasises many of the salient features of the radiative heating problem.

The linear radiation field is defined in the same way as before and we shall derive exact expressions for its intensity. There are

two further radiation fields in this problem, namely, the reduced visible ground radiation field and the reduced thermal ground

radiation field. By virtue of the isotropy of the emission from the ground, the radiation scattered from these fields is independent of azimuth, so that there is no need to include any contribution from these in the azimuthally dependent linear radiation field. The scattering of the linear radiation field is one example of the general problem of one-dimensional radiative transfer along a line of finite length and with radiation incident upon both ends of the line. It will be expedient to solve the general situation because other examples of this problem will occur in Chapter III. Let xo be the total optical length of the linear medium and let x be the

115

optical distance measured in terms of the extinction coefficient 4 from one end. Let I + (x) represent the intensity of the radiation flowing in the positive oc-direction at a point x, and let I~ (32)

represent the intensity of the radiation flowing in the opposite : direction at the point X . Let I* be the intensity of the radiation incident upon the medium in the positive X-direction at the origin, X = 0, and let Io be the intensity of the radiation incident upon the » medium in the negative X-direction at the point, X= Xa . These

are defined in each problem and hence create the two boundary

conditions. .

T+(°) = T*

and I (<O * I„ .

(11-47) i

Let o, be the albedo for single scattering and let [3 be the fraction of the scattered radiation that is scattered forwards, the

remaining fraction, (1 - (3), being the scattered backwards.

The subscript, unity, on the albedo serves as a reminder that it is not necessarily the same quantity as the albedo for single scattering in the complete problem of scattering in a finite atmosphere.

It is, in fact, equal to <*> (1 - a) in terms of the parameters defined earlier for the schematic phase function, and represents the albedo for scattering into the delta-function spikes.

The two equations of transfer for the intensities in the positive and negative x-directions are

= -rw and

o! x

respectively, where B" (3c) are the two source functions. The source function, B*J'(x), is the sum of the radiation scattered forwards from

(x) and the radiation scattered backwards from .1” (x) , whilst the source function, B"(x), is the sum of the radiation scattered forwards from I (x) and the radiation scattered backwards from I (x) That is,

= Z, & I*Cx) +

0-/0 !’(*■>. Hence, we obtain

d.1 (*) = -( I - £>,^5) X* (x) +

I (x)

otx

(11-48) and - fr'W • - (>-<3,/S') r(x) + £,<.-/?) x+u)

These two equations combine to give

2 t +

I (x) - cr* I+(x) « o , cl x2

the solution of which is

where

VM = C, e + Ct e

-<rx _(11-49)

(11-50)

and C, and are constants of integration. Equations (11-48) and (11-49) combine to give

rw

w, 0-^)

C

crx r n ~frK r -s?

117

Applying the two boundary conditions given by equations (11-47) to equations (11-50) and (11-51) we obtain

and

C, = W,

(i-/6)T0 -

-or) T„ e.

-cr Xo ( I- to,

p

+ or) e - ( I ~ jg - cr) £ Ca = I* - Ci . crxo _(11-52) <rx

Consider the following special cases.

(i) To"° ; To • This is the case of no incident radiation; and equations (11-49) to (11-52) yield the trivial solution

C, = C-3. - 0 ,

I*(x) - 1'M^ o,

(ii) To ~O . This is the case of incident radiation upon one

end only for which we obtain

and

C, =

- (i- <3.^ - I„

+ e- crXo

( 1 - -V CT ) G - ( 1- <2,^ <r)

- crxo

(11-53)

c, = i: - c,

(iii) To -• 0. This is the case of radiation incident upon the end, X = Xo only, for which we obtain

c, -

+ ~ { I-w, „ cr) & (11-54) (I — yg’) I.

and

Cx = - c,.

(iv) r„+ ~ To - To . We have the same incident radiation upon each end of the medium and therefore, we have

{ &,(i~

}

and

Cv = T( p) “ (t- - or) -crxo

£ / /«» x o“Xo f \ ?

[ {i-co, 6 + <r)e - (i-£o,p-o-)e j

C2 * 10 - Ci.

(11-55)

The linear radiation field in our problem of a finite plane- parallel atmosphere with parallel radiation incident upon its upper

surface corresponds to the special case (ii) with it ~ 7T F. The albedo in question is the albedo for scattering into the delta- function spikes, which is to, = co (1 - a). The geometry of the problem demands that

x-'tf/

and that the radiation field exists for (i = (io only and for or + 7V only. Consequently, the first two moments of the linear radiation field are

I .... b” oo (i-cQ ( + o*" J q _d- A-TV W ( l - ti) ( I j -o"tZp« (11-56)

r

1

r „

1

and « Cl fA0 L

ter J

e + Q Ll~(11-57)

+ d - <TJ £.

It to (i-ot) (i~/0 k-TT (l-ot) (i-p)

where C» and CA are given by equations (11-5 3) with X« « IT F. The reduced incident radiation field can be written down

119

immediately. It is merely the incident intensity exponentially attenuated from the point of entry into the atmosphere to the depth in question. We thus have

££ (v, jz, /) = rrFe

and M - A Fe 4 S (p -1*0} f ~ /o) 9 W « -1 jAo Fe“T/r\ (n-58)

Before proceding with the solution for the scattered radiation we shall consider several special cases. Firstly, in the limit as 'to tends to infinity, the constants

C,

and

Cx

tend to zero and

7T F respectively. Thus, equations (11-49) and (11-51) tend to equations (II-1I) and (11-12), so that the limiting forms of the expressions for finite atmospheres agree with those obtained for

semi-infinite atmospheres In the preceding sub-section. Secondly, in the case of isotropic scattering, cr is unity and the albedo, tu, 9 is zero, This causes equation (11-51) to be indeterminate.

However, when the scattering is isotropic the linear radiation field is identical to the reduced incident radiation field given by

equation (11-58). Another apparent singularity arises in the third special case for which (3 = 1.0. The solution can be found by

appropriately adjusting the equations of transfer, equations (11-48) and then proceeding as before. Hence, we obtain

I+(x) = 1; e 0-

and

T(x) « o ,

so that OS- (vl ~ trx'/po _and ~(rr-/Ho_{, (11-59)} '+ '

where cr = 1 - w, , in this case, These results are obvious from physical reasoning., When the spike scattering is forward only

the situation is the same as though the scattered radiation were not scattered. Thus, the incident radiation is attenuated by the absorption coefficient and isotropic part of the scattering

coefficient rather than the extinction coefficient.

We are now in a suitable position to construct the source function for the scattered radiation and solve the equation of transfer. The scattered radiation is, of course, not the true scattered radiation field because part of this is included in the linear radiation field. The most convenient way to treat the ground radiation is to consider, it an external source of isotropic stellar intensity, G-s , and isotropic thermal radiation, .

The values of G-s and Gp are found by applying boundary conditions . The emission coefficient for the scattered radiation is made up of three terms; the radiation scattered isotropically from the linear radiation field; the radiation scattered anisotropically from the

scattered radiation field; and the radiation scattered anisotropically from the reduced visible ground radiation. Hence

Rtt

+1

f f

js ~ ~~ Ifo (t, p', j£') dp' dfi' 4- J J o -1 +1 + ( Kj

<■ <r,) a j

4- ( + crs O -I -m r

r

p , 0 J

n' , $

' ) G-jr £

c£|/

)/? ~ I li*_tprr <|

The limits of the last integral are p/ = 0 and p/ - 1, rather than p/ = -1 and p/ = +1, because the integrand is zero for all negative values of p' . The source function, , is the ratio of

121

and the phase function used above is the schematic one of equation (1-29). The third integral in the emission coefficient deserves mention. By virtue of the three-part nature of the phase

function, this integral will divide into three separate integrals. The one that arises from the isotropic part of the phase function involves an integral that is known as an exponential integral function. Details of the exponential integral functions, their definitions and their properties are found in the Appendix.

The other two integrals of this third integral of the emission coefficient involve a type of delta function which we shall define as

s _-1

s O ; § O .

It is introduced merely to aid the mathematical expression of the equation of transfer; and it will be cancelled during the solution of the equation. Using these functions, the source function

becomes

+■ (a (l - d.} (i --p ) Jy ("£ } - ja) 4* ~ £*> ol Grs -f*

. (11-60)

Consequently, the equation of transfer for anisotropic scattering in a finite plane-parallel atmosphere with a ground layer is

p 41s (* ■

- ZocZJsC't) - (u (<~o<)^ Ij ( *t, <^) ~

- <S(i-dXi-/Crs(r ,-p)

£s E2.(u-r)~

• . -C'to-'v)/! - Cr»-v)M

-to (i-«d C G-se

- cXG-»0 6y)£ £ye.

. (n-6i)

We shall solve this equation by Eddington1s method in the same way as we did in the preceding subsection for semi-infinite atmospheres o Firstly., we integrate the equation of transfer by applying the Lo -operator, to obtain '

- ( I- u> ) X ft} - (X ot 3E (-t) - p (J-5 Ez(t«,-t) , (11-62)

and secondly, we integrate the equation of transfer by applying the i, -operator, to obtain

XKs(r) = [ 1-Hs(x) - Xw (z^-0 frs r3<r.-x).

ckx. &

Using the Eddington approximation, equation (1-14), and defining

2f * l - & (l- <x.) (2^-{‘) (11-63)

we obtain

Ci.) " 3 $ R s Ci) — S

Eg, C T,-1;) .

(ii --64)

> <%

123

(

t

1-S') ls(t) =

- l[i-»U)]CsE«h.-t),

(11-65)

t

where

3 8* f t- t*t ) .

(11-66)

Equation (11-65) is a second order inhomogeneous total differential equation with constant coefficients and can be solved by normal

analytical techniques. The solution involves further transcendental functions, the F^-functions, which are integrals of products of

exponential and exponential integral functions. They are defined in Section 2 of the Appendix and several of their properties are also listed. To avoid long strings of constants, equation (11-56) for the mean intensity of the linear radiation field will be written

ct-c/ja,,

* /U e

+ /5

t e , (n-67)

where A6 and Av can be found from equations (11-56), (11-58) or

(11-59) which ever is appropriate. The solution of equation (11-65) is

S-V

-ft

-w/pe

e + $xe +

+ he. *

* As G-s | e Fa L-S/r^u ~ e Fa IX

(11-68)

where As and Ax are arbitrary constants, and

A, = 3ZU* fls/C Fpt - ffO ,

and /A 5 - “ 3 £ I - CI - t*> ) 3 / S .

The boundary conditions to be applied to this solution all involve the scattered flux, Ks(t) , which is found by using equations

(11-66) and (11-68). The derivatives of the F*.-functions are given in the Appendix, so that we have

-S'* CTT/^o

Hs(t) = & A e - S A* e + q~/^e

_ <r r /

- Q'/Ih. e + (>-y) Eif'Vo-'t)

sap* **

- SAs f e. Fa t e R IX(%~t')']J

(

ii

-70)

The conditions required to determine Grs , A, and Ax are the two Eddington approximate boundary conditions pertinent to the two surfaces of fhe atmosphere and the equation defining the parameter A. The two Eddington approximate boundary conditions are

= 2Ps(o) and = -2H$Cr„). _(11-71)

The parameter, A, is defined as the ratio of the stellar radiation flux reflected by the ground, to the stellar radiation flux incident on the ground. Now the outward flux from the ground, by

definition of G-s , is

+i

-i

125

and the flux incident upon the ground is - ^f-7r (v,,). Equating the emergent flux with \ times the incident flux, we obtain

Grs =

-4-X [ Hs(ro) + Hb-jTo')].

(11-73)

Equations (11-71) and (11-73) provide three equations for three

unknowns and the solution for these unknowns involves no difficulty. Thus we have completed the solution for the first two moments of the scattered radiation field.

The emission coefficient of the thermal radiation is comprised of five terms. The radiation from the three stellar radiation fields, the linear, the scattered and the reduced visible ground radiation fields, is converted into thermal radiation and emitted isotropically. The first three terms of the emission coefficient are from these three sources. The other two sources consist of the radiation absorbed from the thermal and reduced thermal ground radiation fields and then emitted conservatively and isotropically. We have defined Gq as the intensity of the thermal radiation emitted isotropically from the ground, so that the emission coefficient

for the thermal radiation is

2.TT +1 o -l 4 4ir in- i s' S' 4- ttr t o 0 G-p CL + J o o (i,/*') oU'. (11-74)

The factor, n, in the attenuation coefficient in the fourth term of this equation is due to the fact that the thermal ground

radiation is attenuated according to kp whereas the optical depth scale is measured in terms of (tq+c's) „ The source function, which is the emission coefficient divided by the absorption coefficient, KP , is thus

'BpCx) = 3^ M ~ E’x E 4-

+ Vi(i-ta) L 3s Ot) + (t) + £ 6-5 J , (11-75)

and hence, the equation of transfer for the thermal radiation is

I* 41? (*> r) = a 1.(1:,/*) - a - j_ ck g r<to-r)/n] -

'Jr

n

*»

- (l-fcj) Jsfr) - (i-w) - i £(-u) £s Ha.ffo-'C). (11-76)

1

This equation is solved by applying the two integral operators, Lo and L, , and replacing the resultant moment Kp(i) in the second

equation by CTpCT)/3 according to the Eddington approximation. Hence, we obtain

Jl My (x? = - Ei C (%-X)/v\l - X £s ) E2 -

ysM - G-&)

YgUt), (11-77)

127

Using equations (11-67) and (11-68) for Jurt(*y) and CTs('e)

respectively we integrate equation (11-77) to obtain

ht

(

t

) = t, - i E

j

t (T.-n/ni-

-(l-w)

r i zfa i

L ~ $ J

h-S) [ A,

* fU £

J1

~ ~ 00 ) j*o L (^5 +

) 6

~(Aw* /!•»*)€. I

(11-79)

where details of the integrations of the exponential integral functions and the Fv< -functions are given in the Appendix.

Equation (11-79) involves two unknown constants, 8, and Gp; ‘ and these must be determined by two boundary conditions. However, at this stage in the solution only one boundary condition can be applied successfully, and that is the condition of conservation of energy. So far, our solutions have been valid for atmospheres with and without a ground. However, the application of the

principle of conservation of energy is different in the two cases.

We shall postpone consideration of atmospheres with no ground, and firs' consider atmospheres with a ground layer. For these atmospheres

the principle of conservation of energy expresses itself mathematically as the condition of zero net flux, which is

H* M s ( A ) +■ Up M t HC“ 0 , _(11-80)

fields. It is given by

HyouU (**) - JL &S Ej + 1 £3 C(tP-i:)Ad .

2. 2> (11-81)

The substitution of equations (11-57), (11-70), (11-79) and (11-81) into equation (11-80), followed by a lengthy algebraic reduction yields the condition that the constant, B, , must be zero for energy to be conserved in the atmosphere.

It is interesting to note that this result is independent of £rr . Consider the energy balance at the ground surface. The emergent thermal flux is equal to the incident thermal flux plus the absorbed fraction, (1 - \) of the incident stellar flux. The emergent thermal flux is

1 f f

r (11-82)

O o

the incident thermal flux is flux is - H-ir H HsC'to) + 3

4-tr _{, and the incident stellar}

The energy balance is therefore

J

IT

An-+

J -Mrh

(11-83)

However, this is not an equation for 6rp . The total flux of

ground radiation,

H<younA (To)

is given by 7r ( )

from equations (11-72) and (11-82). Thus an equation for (%) can be written involving equations (11-73) and (11-83).

129

4K * IT ( + 6-S)

- [ UXri+Mt^Gv.)] -4k(H)[ HsCTuU Wh«(n)]-UM/r,). •

A glance at this equation shows it to be nothing more than equation (11-80), the equation of constant net flux, which does not involve the constant, Gp . Thus we see that the greater the flux out of the ground the greater the flux reflected back into the ground by the atmosphere. Nevertheless, it is still quite surprising that this state of affairs should exist. Xt is quite analogous, however, with a similar arbitrariness encountered by Chandraskhar

(1960) in work on the exact solutions of similar problems using the principles of invariance. These principles will be discussed and used in Section II.5.1. Chandrasekhar found that, for

conservative problems, the principles of invariance yielded arbitrary solutions that could not be resolved by appeal to the flux-integral alone, but by appeal to the K-integral as well. The situation here is similar. The problem is conservative and is exact because we are considering fluxes only. This was shown to be so by the second method outlined in Section II.3.1. and is true here also, even though it is not proved directly. Our solutions are arbitrary and we shall see that a unique solution is only possible after using the second moment integral of the equation of transfer, which is the K-integral, and is expressed in an approximate form by equation (11-78). Furthermore, the arbitrariness vanishes if the thermal radiation transfer is considered to be non-conservative.

r.(t)

= ba - 3 G-p Hu. [

1 -

z

— 3 (I - co )

[

■

_{L _ 2 As Si/iv-T)}

n

_{- 3 (i-fi-) L «,e + d5e} r . st

-sti_l~

I_{- 1 5 J}

• -

r ,

crt/f'o L ( As * Ac.) c

n <rx

— 36*-u) /is

F

sc-t.-t) -SOt.-T)

3

■ e GCS/VT)]|.( 11-84)

vS1

The constants, and Crf are found by using the two Eddington approximate boundary conditions

XpCo) = 2HPM and TpCn):: - 2.Hp(To) . (11-85)

This procedure is straightforward and completes the solution of the equations of transfer for the scattered and thermal radiation fields in an anisotropically scattering finite plane-parallel atmosphere with a conservative Lambertian ground layer.

An important special case is that of the finite atmosphere with no ground at its lower surface. The scattering problem is

exactly the same as the scattering problem of the finite atmosphere with a ground layer of albedo, equal to zero. Consequently,

the source function is given by equation (11-60), the equation of transfer by equation (11-61) and the mean intensity and flux of the scattered radiation field by equations (11-68) and (11-70) respectively. In these equations the parameter, A, and hence the intensity, , are zero. The linear radiation field is,are zero.

131

of course, unchanged. However, the two problems are not the same for the thermal radiation field. In the case of the

ground being absent, the intensity, , is zero also, but with this restriction, equations (11-75), (11-76) and (11-79) still represent the source function, the equation of transfer and the thermal flux respectively. The major change in the physics of the problem arises in the expression of the principle of

conservation of energy, which is no longer the equation of zero net flux, equation (11-80). Energy must be conserved in the atmosphere, so we state that the net flux into the atmosphere at its upper surface must equal the net flux out of the atmosphere at

In document The radiative heating of plane parallel and spherical atmosphere (Page 125-159)