- Email: [email protected]
Numerical solution of the transfer equation for polarized continuum radiation
J. Quonr. Specwosc. Radiot. Trans/er. Vol. I I, pp. 1675-1680. Pergamon Press 1971. Printed in Great Britain
NUMERICAL SOLUTION OF THE TRANSFER EQUATION FOR POLARIZED CONTINUUM RADIATION S. DUMONT Observatoire de Paris, Meudon, France (Receiued 8 January 197 1)
Abstract-We show how to generalize FEAUTRIER’S’~) numerical method to solve the transfer equation of polarized radiation in the case where axial symmetry of the radiation is not assumed.
1. POSING
OF THE
PROBLEM
only the solution of the transfer equation for polarized radiation in the form given by CHANDRASEKHAR (2) for the case of a plane-parallel diffusing atmosphere obeying Rayleigh’s law :
WE CONSIDER
where p = cos f3and 0 and 9 are the polar angles of the direction n of the incident radiation ; 13’and cp’are those of the direction n’ of the scattered radiation.
There is no difficulty in generalizing our proposed solution to an atmosphere emits radiation. The transfer equation becomes now, 2n
which
1
PI d$ d#, (2) ss 0 -1 where x is the probability of the radiation being scattered. We shall later examine the question of transfer under the influence of a magnetic field. The vector I, which represents the radiation, has the components I,, I, for the intensities in the direction of the meridian plane and in the perpendicular direction, and the two Stokes parameters U and V. The two other Stokes parameters are I = I, + I, and Q = II - I,. The phase matrix P isC2’ 0 (4 ZJ2 (6 Z12 (1,I)@>0 pg
(1,rJ2 3
20, I)(Z,r) 0
I--(1 -x)B-2
=
(6 rJ2
(1,r) (r, 4
2(r, r)(r, r) (1,I)(r, r) +(r, NZ, r) 0
0 1675
0 0 (4 I)(r, I)-(r,
0(Z,r)
1676
S. DUMONT
with (I, I) = sin U sin 0’ + cos 0 cos (I’ cos(cp’ - cp), (r, I) = cos 0 sin(cp’- cp), (1, F) = - cos 0’ sin( cp’- cp), (r? r) = cos(cp’- cp). We shall now generalize
FEAUTRIER'S(')method
2. THE
EQUATIONS
Let us consider first the integral scattering source function S, viz.
(3) may be written
IN THEIR
on the right-hand
277
Equation
by separating
NEW
the terms p > 0 and p < 0.
FORM
side of equation
(l), which is the
1
as
(4) setting
J(T,P> cp,cp’)=
s 0
W’(pL, cp,- $3 cp’)I(z,- p’, cp’) (5)
+ P(pL,cp,I*‘, @MT, $3 cp')l d,u’, From this we obtain .
S(T,I*, cp)=
s
&
[J(T, ,u,cp,cp’)+ J(s, P,
cp,TX+ cp')l W.
(6)
0
Let us consider now the directions n defined with polar angles 8 and cp, -n with n- 0 and 7~+ cp, m with 8 and n + cp and -m with TI- 8 and cp, and the directions (defined in an analogous way) of the scattering radiation n’, -n’, m’, -m’. We write the transfer equations in these four directions by taking equations (5) and (6) into account. The results are II
’
dI(n) = I(n)-& __ dt
dI( - n) -7iF
1
ss 0 0 n
= I( -n)-t
[P31( - m’) + P&m’) + P,I(n’) + P,I( -n’)] d$ d@,
(7)
1
J-s 0 0
[&I( - m’) + P;I(m’) + PkI(n’)+ P;I( - n’)] d$ d#,
(8)
Numerical solution of the transfer equation for polarized continuum radiation
PF=
1
ss II
dVm)
I(m)-;
[PJ( - m’) + P,I(m’) + PJ(n’) + P,I( - n’)] d$ dq’,
0 0 n
dI( - m) -EC -’ dr
1(-m)-&
1677
(9)
1
ss 0 0
[P;I( - m’) + PbI(m’) + P;I(n’) + P;I( - n’)] d$ dq’,
(10)
where we have introduced vectors such as Y(n) = p(n) + I( - n)]/2 and F(n) = [I(n) - I( - n)]/2, which have the components y,(n) = [z(n) + z( - n)]/2 and f,(n) = [z(n) - z( - n)]/2 if z represents one of the components of the vector I(n). We have set P, = PQ.4 cp,$3 VP’)= P(K rc+ cp>$3 n+ CP’X P, = P(K cp, - $3 7c+ 9’) = P(K rr+ cp, - $2 cp’), P, = P(cL, 40, - p’, cp’) = P($4 rc+ cp, - $3 lx+ cp’), P, = P(,4 cp,$3 rc+ 40’)= P(c1,7c+ 40,$9 cp’), P; = P(-/44o,$,(P’) P; = P(-Acp, P; = P(-c1,
= P(-/&71+(P,$,n+cp’),
-$,rc+(P’) cp, -n’,
Pk = P(-p,cp,
= P(-/471+(P,
cp’) = P(-L4a+cp,
-$,cp’), -$,“$.V’),
$9 n+cp’) = P(-p,n+%p’,cp’).
These matrices are such that P,I( - m’) + PJ(m’) = 2P,X(m’) if the vector X has components y,(m’), y,(m’), f”(m’) and j&m’). There are seven other similar relations. Then, combining equations (7) and (8), on the one hand, and equations (9) and (lo), on the other hand, we obtain n
Wn) qy=
1
ss Y(n)-& ss
F(n)-& [(PA-P;)X(m’)+(P,
-Pk)X(n’)l d$d@,
(11)
0 0
1
II
Wn) qy=
0
0
77
1
[(Pa + P’AX(m’) + (P, + Pk)X(n’)] d$ dp’,
(12)
[(Pr - Pk)X(m’) +(P,-
P;)X(n’)l d$ d@,
(13)
f P;)X(n’)] d$ dq’.
(14)
and
’
dY(m)= dz
F(m)--&
ss
b 0 x
Wm)
PT=
1
ss
Y(m)-&[(PI +P&)X(m’) +(P, 0 0
But, P4 + Pi and P, + P:, are matrices
such that the two first rows are, respectively, identical to those of 2P, and 2P, while the two other rows are zero. The same holds for the matrices P, - Pi and P, -Pi, where the two first rows are zero and the latter two are, respectively, identical to those of 2P4 and 2Pi. Writing equations (11) and (12) for each component of the vectors Y and F, we obtain the following eight equations for the direction n if P, = $aij)
S. DUMONT
1678
and P, = ~bijl (the other matrices have the same terms except, possibly, for opposite signs and some of the bij equal to the aij) : *
d!(n)
-$
h(n)
PF=
1
ss
[~l~Y~(m’)+u12y,(m’)-~13f~m’)
(15)
0 0
+al,y,(n’)+a,,y,(n’)+a,,f,(n’)ld~’d~’, = f;(n),
p$$
(16) 1
R
dfi(n)
y,(n)-& IS[a2Mm’) + YW)
I-(dz=
0
0
(17) + az2(y,(m’) + y,(n’)) + a2,(fv(m’)
py
+fdn’)ll W dv’,
= fi(n),
(18)
pdfub) ___ = ydn), dr
(19) [b,,y,(m’)+a,,y,(m’)-b,,f,(m’) 0 0 (20)
+ ~3&(n’) + a32yl(n’) f
~33fvWl dN dq’,
dfv(n) __ dT
= ydn),
(21)
[ - L&W) + a&b’)1 W hf.
(22)
00 The equations (13) and (14) yield eight similar equations among which are those for f,(m) and y,(m) which are identical with equations (17) and (18). As CHANDRASEKHAR~‘) pointed out, the parameter I/ is independent because some terms are zero in the matrix P and therefore it is possible to solve equations (21) and (22), as well as the analogous equations for direction m, independently. 3. SOLUTION OF THE EQUATIONS It is possible to adapt FEAUTRIER’S(‘) method to this problem by replacing equations such as (15) and (16) by second-order equations. We must then solve the following simultaneous second-order equations : 2
P
d2yl(nor 4 dz2
= y,(n or m) - S,(n or m),
(23)
Numerical solution of the transfer equation for polarized continuum radiation
P
2dzyr(n or ml = dz=
y,(n or m) - S,(n or m),
p zd2fdn Or m, = fu(n or m)- Sdn or m) ;
dT=
1679
(24)
(25)
we must also solve the following two simultaneous equations :
p2d2h4n or ml = fv(n dz=
or m) - Sy(n or m),
(26)
which may be written as
pdr2=
1
ss R
,d’Xb)
X(n)-& [P4X(m’) + PIX(n’)] d$ dcp’,
(27)
0 0 A1
2d2Wm) ’
= x(m)-lt;;s
dt=
1 [PIX(m’) + P,X(n’)] d$ dcp’. 0 0
(28)
The integrals in the source function S are discretized and each of the preceding equations is replaced by K equations if the integral is calculated with K values of p’ and cp’.Let us write, for instance, these relations which replace equation (23) in the direction n :
(29)
where i = 1,K and tl is a weighting factor. We obtain without difficulty the equations for the two Stokes parameters I and Q since y, = y,+ y, and yQ = y, -y,. For example, equations (23) and (24) in the direction n yield 7t
2d2Yh) f’
dz=
= yr(4-~
1
ss 0 0
KbIl +azl +a12+a22)yI(m’)
+(b11+u21-u12-~,,)YQ(m’)+(-b,,+u,,)2f,(m’) +bll
+u2l
+ (a 13 + a2
+u12+u22)YIb’)+(ull
+a21-u12-a22)YQb’)
(30)
3Mub’)lW W
and n
2d=YQ(“) pr=
YQ(n)-&
+@,,
1
ss 0 0
[(bll-a21
+a12-a22)YIW
-~2l-~l2+~22)YQ(~‘)-~~,3+~23)2fU(m’)+(~,,-~2l+~l2--22)Yl(n’)
+~~ll-~21-~12+u22~YQ~n’~+~~13-~2~~~fv(~’~~
Similar equations are also obtained for the direction m.
WW.
(31)
1680
S. DUMONT
If we consider a slab of gas with depth r1 illuminated on one side by polarized radiation. then we have the following boundary conditions : (a) at the top r = 0, there is no entering radiation and I( - n) = 0, I( - m) = 0 and, therefore, y,(n) = f,(n). The relations (16) (18). (19) and (21) become now dX(n or m) P (b) at the ground so that
dt
= X(n or m).
r = rI and we know that the entering
dX(n or m) P
dr
radiation
(32) is I(n or m) = I,(n or m)
= Ir(n or m)- X(n or m).
(33)
If the incident polarized radiation I 1 has axial symmetry, U 1 and VI are zero, CJ and I/ remain zero in the entire slab and the problem reduces to one already solved by DUMONT(“) as used by DBBARBATet u/.‘~’ CONCLUSION
Our method may be applied to the problem of transfer in the chromosphere continuum radiation originating from a sun-spot.
REFERENCES 1. 2. 3. 4.
P. S. S. S.
FEAUTRIER, C. R. Acad. Sci. Paris 258, 3 189 (1960). CHANDRASEKHAR, Radiatiw Transfer, 2nd Edn. Dover Pub]., New York (1960) DUMONT, C. R. Acad. Sri. Paris 268, 1678 (1969). DBBARBAT, S. DUMONT and J.-C. PECKER, Astron. Astrophys. 8, 231 (1970).
of polarized
NUMERICAL SOLUTION OF THE TRANSFER EQUATION FOR POLARIZED CONTINUUM RADIATION S. DUMONT Observatoire de Paris, Meudon, France (Receiued 8 January 197 1)
Abstract-We show how to generalize FEAUTRIER’S’~) numerical method to solve the transfer equation of polarized radiation in the case where axial symmetry of the radiation is not assumed.
1. POSING
OF THE
PROBLEM
only the solution of the transfer equation for polarized radiation in the form given by CHANDRASEKHAR (2) for the case of a plane-parallel diffusing atmosphere obeying Rayleigh’s law :
WE CONSIDER
where p = cos f3and 0 and 9 are the polar angles of the direction n of the incident radiation ; 13’and cp’are those of the direction n’ of the scattered radiation.
There is no difficulty in generalizing our proposed solution to an atmosphere emits radiation. The transfer equation becomes now, 2n
which
1
PI d$ d#, (2) ss 0 -1 where x is the probability of the radiation being scattered. We shall later examine the question of transfer under the influence of a magnetic field. The vector I, which represents the radiation, has the components I,, I, for the intensities in the direction of the meridian plane and in the perpendicular direction, and the two Stokes parameters U and V. The two other Stokes parameters are I = I, + I, and Q = II - I,. The phase matrix P isC2’ 0 (4 ZJ2 (6 Z12 (1,I)@>0 pg
(1,rJ2 3
20, I)(Z,r) 0
I--(1 -x)B-2
=
(6 rJ2
(1,r) (r, 4
2(r, r)(r, r) (1,I)(r, r) +(r, NZ, r) 0
0 1675
0 0 (4 I)(r, I)-(r,
0(Z,r)
1676
S. DUMONT
with (I, I) = sin U sin 0’ + cos 0 cos (I’ cos(cp’ - cp), (r, I) = cos 0 sin(cp’- cp), (1, F) = - cos 0’ sin( cp’- cp), (r? r) = cos(cp’- cp). We shall now generalize
FEAUTRIER'S(')method
2. THE
EQUATIONS
Let us consider first the integral scattering source function S, viz.
(3) may be written
IN THEIR
on the right-hand
277
Equation
by separating
NEW
the terms p > 0 and p < 0.
FORM
side of equation
(l), which is the
1
as
(4) setting
J(T,P> cp,cp’)=
s 0
W’(pL, cp,- $3 cp’)I(z,- p’, cp’) (5)
+ P(pL,cp,I*‘, @MT, $3 cp')l d,u’, From this we obtain .
S(T,I*, cp)=
s
&
[J(T, ,u,cp,cp’)+ J(s, P,
cp,TX+ cp')l W.
(6)
0
Let us consider now the directions n defined with polar angles 8 and cp, -n with n- 0 and 7~+ cp, m with 8 and n + cp and -m with TI- 8 and cp, and the directions (defined in an analogous way) of the scattering radiation n’, -n’, m’, -m’. We write the transfer equations in these four directions by taking equations (5) and (6) into account. The results are II
’
dI(n) = I(n)-& __ dt
dI( - n) -7iF
1
ss 0 0 n
= I( -n)-t
[P31( - m’) + P&m’) + P,I(n’) + P,I( -n’)] d$ d@,
(7)
1
J-s 0 0
[&I( - m’) + P;I(m’) + PkI(n’)+ P;I( - n’)] d$ d#,
(8)
Numerical solution of the transfer equation for polarized continuum radiation
PF=
1
ss II
dVm)
I(m)-;
[PJ( - m’) + P,I(m’) + PJ(n’) + P,I( - n’)] d$ dq’,
0 0 n
dI( - m) -EC -’ dr
1(-m)-&
1677
(9)
1
ss 0 0
[P;I( - m’) + PbI(m’) + P;I(n’) + P;I( - n’)] d$ dq’,
(10)
where we have introduced vectors such as Y(n) = p(n) + I( - n)]/2 and F(n) = [I(n) - I( - n)]/2, which have the components y,(n) = [z(n) + z( - n)]/2 and f,(n) = [z(n) - z( - n)]/2 if z represents one of the components of the vector I(n). We have set P, = PQ.4 cp,$3 VP’)= P(K rc+ cp>$3 n+ CP’X P, = P(K cp, - $3 7c+ 9’) = P(K rr+ cp, - $2 cp’), P, = P(cL, 40, - p’, cp’) = P($4 rc+ cp, - $3 lx+ cp’), P, = P(,4 cp,$3 rc+ 40’)= P(c1,7c+ 40,$9 cp’), P; = P(-/44o,$,(P’) P; = P(-Acp, P; = P(-c1,
= P(-/&71+(P,$,n+cp’),
-$,rc+(P’) cp, -n’,
Pk = P(-p,cp,
= P(-/471+(P,
cp’) = P(-L4a+cp,
-$,cp’), -$,“$.V’),
$9 n+cp’) = P(-p,n+%p’,cp’).
These matrices are such that P,I( - m’) + PJ(m’) = 2P,X(m’) if the vector X has components y,(m’), y,(m’), f”(m’) and j&m’). There are seven other similar relations. Then, combining equations (7) and (8), on the one hand, and equations (9) and (lo), on the other hand, we obtain n
Wn) qy=
1
ss Y(n)-& ss
F(n)-& [(PA-P;)X(m’)+(P,
-Pk)X(n’)l d$d@,
(11)
0 0
1
II
Wn) qy=
0
0
77
1
[(Pa + P’AX(m’) + (P, + Pk)X(n’)] d$ dp’,
(12)
[(Pr - Pk)X(m’) +(P,-
P;)X(n’)l d$ d@,
(13)
f P;)X(n’)] d$ dq’.
(14)
and
’
dY(m)= dz
F(m)--&
ss
b 0 x
Wm)
PT=
1
ss
Y(m)-&[(PI +P&)X(m’) +(P, 0 0
But, P4 + Pi and P, + P:, are matrices
such that the two first rows are, respectively, identical to those of 2P, and 2P, while the two other rows are zero. The same holds for the matrices P, - Pi and P, -Pi, where the two first rows are zero and the latter two are, respectively, identical to those of 2P4 and 2Pi. Writing equations (11) and (12) for each component of the vectors Y and F, we obtain the following eight equations for the direction n if P, = $aij)
S. DUMONT
1678
and P, = ~bijl (the other matrices have the same terms except, possibly, for opposite signs and some of the bij equal to the aij) : *
d!(n)
-$
h(n)
PF=
1
ss
[~l~Y~(m’)+u12y,(m’)-~13f~m’)
(15)
0 0
+al,y,(n’)+a,,y,(n’)+a,,f,(n’)ld~’d~’, = f;(n),
p$$
(16) 1
R
dfi(n)
y,(n)-& IS[a2Mm’) + YW)
I-(dz=
0
0
(17) + az2(y,(m’) + y,(n’)) + a2,(fv(m’)
py
+fdn’)ll W dv’,
= fi(n),
(18)
pdfub) ___ = ydn), dr
(19) [b,,y,(m’)+a,,y,(m’)-b,,f,(m’) 0 0 (20)
+ ~3&(n’) + a32yl(n’) f
~33fvWl dN dq’,
dfv(n) __ dT
= ydn),
(21)
[ - L&W) + a&b’)1 W hf.
(22)
00 The equations (13) and (14) yield eight similar equations among which are those for f,(m) and y,(m) which are identical with equations (17) and (18). As CHANDRASEKHAR~‘) pointed out, the parameter I/ is independent because some terms are zero in the matrix P and therefore it is possible to solve equations (21) and (22), as well as the analogous equations for direction m, independently. 3. SOLUTION OF THE EQUATIONS It is possible to adapt FEAUTRIER’S(‘) method to this problem by replacing equations such as (15) and (16) by second-order equations. We must then solve the following simultaneous second-order equations : 2
P
d2yl(nor 4 dz2
= y,(n or m) - S,(n or m),
(23)
Numerical solution of the transfer equation for polarized continuum radiation
P
2dzyr(n or ml = dz=
y,(n or m) - S,(n or m),
p zd2fdn Or m, = fu(n or m)- Sdn or m) ;
dT=
1679
(24)
(25)
we must also solve the following two simultaneous equations :
p2d2h4n or ml = fv(n dz=
or m) - Sy(n or m),
(26)
which may be written as
pdr2=
1
ss R
,d’Xb)
X(n)-& [P4X(m’) + PIX(n’)] d$ dcp’,
(27)
0 0 A1
2d2Wm) ’
= x(m)-lt;;s
dt=
1 [PIX(m’) + P,X(n’)] d$ dcp’. 0 0
(28)
The integrals in the source function S are discretized and each of the preceding equations is replaced by K equations if the integral is calculated with K values of p’ and cp’.Let us write, for instance, these relations which replace equation (23) in the direction n :
(29)
where i = 1,K and tl is a weighting factor. We obtain without difficulty the equations for the two Stokes parameters I and Q since y, = y,+ y, and yQ = y, -y,. For example, equations (23) and (24) in the direction n yield 7t
2d2Yh) f’
dz=
= yr(4-~
1
ss 0 0
KbIl +azl +a12+a22)yI(m’)
+(b11+u21-u12-~,,)YQ(m’)+(-b,,+u,,)2f,(m’) +bll
+u2l
+ (a 13 + a2
+u12+u22)YIb’)+(ull
+a21-u12-a22)YQb’)
(30)
3Mub’)lW W
and n
2d=YQ(“) pr=
YQ(n)-&
+@,,
1
ss 0 0
[(bll-a21
+a12-a22)YIW
-~2l-~l2+~22)YQ(~‘)-~~,3+~23)2fU(m’)+(~,,-~2l+~l2--22)Yl(n’)
+~~ll-~21-~12+u22~YQ~n’~+~~13-~2~~~fv(~’~~
Similar equations are also obtained for the direction m.
WW.
(31)
1680
S. DUMONT
If we consider a slab of gas with depth r1 illuminated on one side by polarized radiation. then we have the following boundary conditions : (a) at the top r = 0, there is no entering radiation and I( - n) = 0, I( - m) = 0 and, therefore, y,(n) = f,(n). The relations (16) (18). (19) and (21) become now dX(n or m) P (b) at the ground so that
dt
= X(n or m).
r = rI and we know that the entering
dX(n or m) P
dr
radiation
(32) is I(n or m) = I,(n or m)
= Ir(n or m)- X(n or m).
(33)
If the incident polarized radiation I 1 has axial symmetry, U 1 and VI are zero, CJ and I/ remain zero in the entire slab and the problem reduces to one already solved by DUMONT(“) as used by DBBARBATet u/.‘~’ CONCLUSION
Our method may be applied to the problem of transfer in the chromosphere continuum radiation originating from a sun-spot.
REFERENCES 1. 2. 3. 4.
P. S. S. S.
FEAUTRIER, C. R. Acad. Sci. Paris 258, 3 189 (1960). CHANDRASEKHAR, Radiatiw Transfer, 2nd Edn. Dover Pub]., New York (1960) DUMONT, C. R. Acad. Sri. Paris 268, 1678 (1969). DBBARBAT, S. DUMONT and J.-C. PECKER, Astron. Astrophys. 8, 231 (1970).
of polarized