1 3 µ2 p ν (τ ν ) = I ν. s ν (τ ν ) + (1) µ dq ν. dτ ν. and. = Q ν {( 1 µ 2) p ν (τ ν ) }, (2)

Transcription

1 1. Plane-parallel atmospheres with polarized scattering. In the treatment by Chandrasehar (196, pp 38-5), the transfer equations are expressed in terms of I l and I r, the intensities in directions parallel and perpendicular to the meridian plane. In terms of Stoes parameters, this is I = I l +I r and Q = I l I r, where the meridian plane is the reference plane for Q. Because of the symmetry of the problem, the polarization can only be parallel or perpendicular to the meridian plane (Note: In an atmosphere where the temperature increases with depth, the emerging radiation will be most intense in the vertical direction. Then, if we are looing near the limb of the star, scattering will tend to polarize light perpendicular to the z-axis. This will mae I r > I l, so that we will expect negative values of Q.) We then find the radiation can be described by the following two transfer equations (Harrington, 197): µ di ν dτ ν = I ν { s ν (τ ν ) + ( ) } 1 3 µ2 p ν (τ ν ) (1) and µ dq ν dτ ν = Q ν {( 1 µ 2) p ν (τ ν ) }, (2) where we have defined two auxiliary functions, s ν (τ ν ) and p ν (τ ν ): s ν (τ ν ) = (1 λ ν ) J ν + λ ν B ν (τ ν ) (3) p ν (τ ν ) = 3 8 (1 λ ν) { (J ν 3K ν ) + 3 ( J Q ν K Q ν )} (4) Here, J ν and K ν have their usual definitions: J ν = I ν dµ K ν = I ν µ 2 dµ, (5) while we define J Q ν = Q ν dµ K Q ν = Q ν µ 2 dµ. (6) 1

2 The quantity λ ν is the ratio of pure absorption to extinction, and (1 λ ν ) the ratio of scattering to extinction. We then see that s ν (τ ν ) is just the usual source function for unpolarized radiation. We can regard p ν (τ ν ) as a sort of source function for the polarization. It will tend toward zero as τ ν becomes large. Recalling Eddington s approximation, 3K ν J ν, and we see that the first term of the r.h.s. of eq. (4) will vanish at depth. Since the second term depends only on Q ν, and Q ν = I l I r will decrease as the radiation field becomes more isotropic, it will also vanish at depth. We may use the formal solutions of the transfer equations to write the corresponding integral equations for s ν (τ ν ) and p ν (τ ν ) (Harrington, 197): s ν (τ ν ) = (1 λ ν ) [ Λ τν (s ν ) M τ ν (p ν ) + λ ν B ν (τ ν ) (7) Here, Λ τ is the familiar Λ-operator, p ν (τ ν ) = 3 8 (1 λ ν) [M τν (s ν ) + N τν (p ν ) (8) Λ τ {f(t)} = 1 2 f(t) E 1 ( t τ ) dt. (9) If we ignore the term involving M τ (p), we see that eq. (7) is just the familiar Schwarzschild- Milne equation for unpolarized radiation in a plane-parallel atmosphere. The new M τ and N τ operators (which have no relation to the M n (τ) and N n (τ) used in Kourganoff (1963) p. 259) are defined as N τ {f(t)} = M τ {f(t)} = [ 1 f(t) 2 E 1( t τ ) 3 2 E 3( t τ ) [ 5 f(t) 3 E 1( t τ ) 4E 3 ( t τ ) + 3E 5 ( t τ ) dt (1) dt (11) The ernels of all three operators have a logarithmic singularity at t = τ. The ernel of the M-operator has a zero at τ M = ; it is positive for t τ < τ M and negative for t τ > τ M. Fig. 1 shows the behavior of this ernel. Thus the M τν (s ν ) term in eq. (8) is a measure of the anisotropy of the radiation at τ ν. Radiation from layers with t τ < τ M will be traveling horizontally, while radiation from layers with t τ > τ M will be traveling vertically and will give a negative contribution to p ν. The ernel of the N-operator is always positive. 2

3 Fig. 1. The ernel of the M-transform over.25 < τ <

4 From the formal solutions of the transfer equations we then obtain the emergent radiation field: I ν (,µ) = { ( ) } 1 s ν (τ ν ) + 3 µ2 p ν (τ ν ) e τν/µ dτ ν µ (12) Q ν (,µ) = {( 1 µ 2 ) p ν (τ ν ) } e τν/µ dτ ν µ (13) 2. The equations of the grey atmosphere problem. In the grey atmosphere problem, where absorption and scattering are assumed not to vary with frequency, we can replace all the forgoing variables with those integrated over frequency, writing the integrated quantities without the ν subscript. In addition, in this case, the condition of radiative equilibrium is j ν dν = κ B ν dν = κ J ν dν, (14) where j ν is the thermal emission coefficient. Kirchhoff s law provides the first equality. κ is the (frequency independent) coefficient of pure absorption (as opposed to scattering), so the second equality implies that B = J, and thus by eq. (3), J = s (But this does not imply that J ν = s ν!) This last result then reduces the frequency integrated form form of eq. (7) to s(τ) = Λ τ (s) M τ(p) (15) which we must solve along with the frequency integrated eq. (8): p(τ) = 3 8 (1 λ) [M τ(s) + N τ (p). (16) Now if s(τ) and p(τ) are solutions of equations (15) and (16), then for any constant c, c s(τ) and c p(τ) are also solutions. This follows from the linearity of the operators (see, Kourganoff, 1963, p 41). Thus any solution of eqns. (15) and (16) will have an arbitrary scale; to set the the scale, we may specify the emergent flux from the atmosphere. The operator that yields the flux for unpolarized radiation is the Φ-operator, defined as Φ τ {f(t)} = 2 τ τ f(t) E 2 (t τ) dt 2 f(t) E 2 (τ t) dt. (17) 4

5 If we were dealing with the unpolarized problem, and the source function were s(τ), then Milne s second integral equation would hold: Φ τ {s(t)} = F, where F is the net flux. Φ operating on any function s(τ) which is a solution of the grey problem, Λ τ {s(t)} = s(τ), must yield a value F which is the same for all values of τ. Furthermore, we see that the Φ-operator can set the scale of the solution since Φ τ {c s(t)} = c F for any constant c. In our case with polarization, if we integrate eqn. (1) over frequency and then integrate over µ, noting that in radiative equilibrium J s =, and further that we have the result 1 1 ( ) 1 3 µ2 dµ =, (18) d dτ { 1 } 2 I µ dµ 1 1 = 2 I µ dµ = constant = F (19) 1 Using equation (12) for I and reversing the order of integration results in the polarization analog of Milne s second equation: where we introduce another new operator Φ (4) τ {f(t)} = 2 Φ τ {s(t)} + Φ (4) τ {p(t)} = F, (2) τ 2 f(t) τ f(t) { } 1 3 E 2(t τ) E 4 (t τ) { } 1 3 E 2(τ t) E 4 (τ t) dt dt. (21) The ernel of Φ (4) τ is zero at τ = and is otherwise negative. It reaches a minimum of at τ M.3, the zero of the M-transform. In figure 2 we show the ernel of Φ τ with the ernel of Φ (4) τ scaled up by a factor of Numerical solution of the grey atmosphere problem with polarization. One method of solution is to lay down a grid of discrete points in τ: τ 1,τ 2,...,τ N. Our solution will be represented by two vectors, s = s 1,s 2,...,s N and p = p 1,p 2,...,p N We will represent the Λ-, M-, N-, Φ- and Φ (4) - operators of equations (15), (16) and (2) by N N matrices. E.g., we will find a matrix Λ ij such that for a vector f = f 1,f 2,...,f N of the values of some function f(τ) at our τ-points, the matrix product Λ ij f gives the Λ transform of the function f at the τ points. We will show below how to compute such matrix representations based on spline approximations to the function. 5

6 Fig. 2. The ernel of Φ and of 1 Φ (4) for < τ < 4. 6

7 4. Spline approximation of the functions. Following Press et al. Numerical Recipes (1992), consider a function y(x) for which we have tabulated values y i = y(x i ) for some x i,i = 1, 2,,N. The linear interpolation between any two points x i and x i+1 is just y(x) = A y i + B y i+1 where A = (x i+1 x)/h i and B = 1 A = (x x i )/h i (22) and h i is the interval h i = x i+1 x i. They show that the cubic spline interpolating polynomial can be written y(x) = A y i + B y i+1 + C y i + D y i+1 (23) where y i and y i+1 are the second derivatives of the tabulated function y(x) and C and D are C = h2 ( i A 3 A ) and D = h2 ( i B 3 B ) (24) 6 6 Thus, since A and B are linear in x, equation (23) gives y as a cubic polynomial in x. But what are the 2nd derivatives y i? They are not the actual 2nd derivatives of y(x), which are of course unnown, but rather the 2nd derivatives of our cubic spline. It can be shown that to mae the 1st derivative of the spline smooth across the interval boundaries x i (and have the 2nd derivative continuous there), the y i must satisfy the following condition: h i 1 y i 1 + 2(h i 1 + h i )y i + h i y i+1 = 6 [ 6 y i y i + 6 y i+1 (25) h i 1 h i 1 h i h i for i = 2, 3,,N 1. This gives, however, only N 2 equations for the N 2nd derivatives. A natural cubic spline is one with y 1 = y N =. With this choice we have enough equations to find the remaining y i for i = 2,,N 1. These equations form a tridiagonal system which can be solved efficiently. However, for our purposes, we do not wish to solve the equations directly; rather, we want to express the y i formally in terms of the y i. Let y be the N-element column vector y 1,y 2,,y N, and let y be the (N 2)-element column vector y 2,y 3,,y N 1. Then equations (25) can be written as A y = B y (26) where A is an (N 2) (N 2) matrix and B is an (N 2) N matrix. We thus see that we can express the 2nd derivatives simply as y = C y where C = A 1 B, (27) A 1 being the inverse of A. The product C is an (N 2) N matrix. With the row index i = 2,, (N 1) and column index j = 1,,N, the non-zero elements of B are B i,i 1 = 6 [ 6, B i,i = + 6 and B i,i+1 = 6. (28) h i 1 h i h i 1 h i 7

8 Lewise, the elements of A are A i,i 1 = h i 1, A i,i = 2(h i 1 + h i ) and A i,i+1 = h i, (29) where the index i = 2,, (N 1), so that the elements A 2,1 and A N 1,N do not exist. Thus, given any array of points x i, we can compute the matrix C. Then the 2nd derivatives for the spline fit of any function are given by the matrix multiplication (27). Things are a bit more complex if, instead of natural splines, we want to impose a specific slope on the fit at one or both boundaries. The first derivative of expression (23) is dy dx = y i+1 y i h i h i 6 {( 3A 2 1 ) y i ( 3B 2 1 ) y i+1} At the lower boundary, x = x 1, A =, B = 1 and we have [ dy = y 2 y 1 h 1 dx x 1 h 1 6 {2 y 1 + y 2} (31) and at the upper boundary, x = x N, A = 1, B = and [ dy = y N y N 1 + h N 1 { y dx h N 1 6 N y N} x N which we write in the form of equations (25) as ) 2h 1 y 1 + h 1 y 2 = ( 6h1 y 1 + and h N 1 y N 1 + 2h N 1 y N = Our system of equations then become ( 6 h N 1 ) y N 1 + (3) (32) ( ) [ 6 dy y 2 6 (33) h 1 dx x 1 ( 6 ) [ dy y N + 6 (34) h N 1 dx x N A y = B y + z (35) where z is a column vector [z,z 1,,z N 1,z N with only two non-zero elements: z = 6[dy/dx x1 and z N = 6[dy/dx xn. Now, however, A is an N N matrix, as is B. The elements of A are still given by equation (29) for rows 2 through N 1, but now the elements A 2,1 and A N 1,N do exist. Further, the first row of A is given by [2h 1,h 1,,, and the last (Nth) row is [,,h N 1, 2h N 1. The elements of B are as given in equation (28) for rows 2 through N 1, while the first and last rows of B are [( 6/h 1 ), (6/h 1 ),,, and [,, (6/h N 1 ), ( 6/h N 1 ). Then, multiplying through by A 1 we obtain the desired result y = C y + d, where C = A 1 B and d = A 1 z. (36) 8

9 Our ultimate aim is to integrate the product of a ernel function and the spline fit to an arbitrary function y(x). To do this we have to expand equation (23) to isolate the powers of x. After some algebra we see that, over the interval x i x x i+1, where C = C + C 1 x + C 2 x 2 + C 3 x 3 (37) C = x i+1(x 2 i+1 h 2 i) 6h i, C 1 = 3x2 i+1 h 2 i 6h i, C 2 = x i+1 2h i, C 3 = 1 6h i (38) while for the coefficient D D = x i(x 2 i h 2 i) 6h i, D 1 = 3x2 i h 2 i 6h i, D 2 = x i 2h i, D 3 = 1 6h i (39) We thus can write the cubic spline fit as y(x) = (a i + α i ) + (b i + β i ) x + γ i x 2 + δ i x 3 for x i x x i+1 (4) where the coefficients are a i = 1 [x i+1 y i x i y i+1, α i = 1 [ xi+1 (x 2 i+1 h 2 h i 6h i)y i x i (x 2 i h 2 i)y i+1, (41) i b i = 1 [ y i + y i+1, β i = 1 [ (3x 2 h i 6h i+1 h 2 i)y i + (3x 2 i h 2 i)y i+1, (42) i γ i = 1 [ xi+1 y i x i y i+1 2h i and δ i = 1 [ y 6h i + y i+1. (43) i At this point, recall from equation (27) or (36) that each of the y i can be obtained as a weighted sum over all the y : y i = C i, y or y i = C i, y + d i. (44) This means that each of the factors α,β,γ and δ of equation (4) can be expressed as a sum over the y i multiplied by coefficients whose value can be pre-computed based only on the choice of the x i grid. Suppose we wish to evaluate the integral of y(x) times some ernel K(x) over the interval x i to x i+1. We see this becomes T i = where xi+1 x i y(x) K(x) dx = (a i + α i ) I () i + (b i + β i ) I (1) i + γ i I (2) i + δ i I (3) i, (45) I (n) i = xi+1 x i x n K(x) dx. (46) 9

10 Note that for a given x-grid the I (n) i are just numbers that need be evaluated only once. As the simplest possible example, consider just the integral of y(x), which corresponds to K(x) = 1. Then I () i = x i+1 x i, I (1) i = 1 2 (x2 i+1 x 2 i), I (2) i = 1 3 (x3 i+1 x 3 i), I (3) i = 1 4 (x4 i+1 x 4 i) (47) and hence a i I () i = [x i+1 y i x i y i+1, α i I () i = 1 6 [ xi+1 (x 2 i+1 h 2 i)y i x i (x 2 i h 2 i)y i+1, (48) b i I (1) i = 1 2 (x i+1+x i ) [ y i + y i+1, β i I (1) i = 1 12 (x i+1+x i ) [ (3x 2 i+1 h 2 i)y i + (3x 2 i h 2 i)y i+1 γ i I (2) i (49) = x3 i+1 x 3 [ i xi+1 y i x i y i+1, (5) 6h i δ i I (3) i = x4 i+1 x 4 [ i y 24h i + y i+1 i We now add these and regroup to obtain where U i = 1 6 and V i = 1 6 (51) T i = 1 2 h i (y i + y i+1 ) + U i y i + V i y i+1 (52) { x i+1 (x 2 i+1 h 2 i) 1 2 (x i+1 + x i )(3x 2 i+1 h 2 i) + x i+1 (x 3 i+1 x 3 h i) 1 } (x 4 i+1 x 4 i 4h i) i { x i (x 2 i h 2 i) 1 2 (x i+1 + x i )(3x 2 i h 2 i) + x i (x 3 i+1 x 3 h i) 1 } (x 4 i+1 x 4 i 4h i) i After tedious manipulation it turns out that (53) (54) U i = V i = 1 24 (x i+1 x i ) 3 = 1 24 h3 i!! (55), So, recalling equation (44), we finally obtain T i = h i 2 (y i + y i+1 ) 1 [ 24 h3 i y i + y i+1 h i = 2 (y i + y i+1 ) h3 i 24 [C i, + C i+1, y (56) The first term is just the result of integrating a linear fit to points y i and y i+1 ; the remaining part is the integral of the spline correction to the linear fit. The bottom line is that we have expressed the integral T i in the form T i = W y, where W = h i 2 (δ ii + δ i,i+1 ) h3 i 24 [C i, + C i+1, (57) and the range of i is i = 1,, (N 1). Here, δ ij stands for the Kronecer delta. The elements W are independent of the y and can be computed from the x alone. 1

11 Finally, the total integral of y(x) over the range [x 1,x N is just T = N 1 i=1 T i = W y, where the vector W = N 1 i=1 W. (58) For example, on the interval [,1 with x =,.1,.2,, 1., we get W = , ,.96488,.1967, ,.1138,.1138, ,.1967,.96488, , In case, instead of natural splines, we wish to specify the slope of the spline fit at the boundary, equations (57) and (58) become and T = D + T i = D i + W y W y, where D i = h3 i 24 (d i + d i+1 ) (59) N 1, where D = i=1 D i = h3 i 24 [( 2 ) d i (d 1 + d N ) i=1. (6) If we have information on y at x 1 and/or x N, then this will improve the fit. Consider the following cubic polynomial: y(x) = 4 3x + 2x 2 x 3 with derivative y (x) = 3 + 4x 3x 2 (61) Let s integrate it over the interval [ 1, 3. The integral of y(x) over this range is exactly Let us choose an x-grid of 5 evenly spaced values: [ 1,, 1, 2, 3. If we assume a natural spline fit with this grid we obtain W i = [ , , , , We then see that T = W y = , an error of 3.6% (62) We might expect an exact fit to a cubic polynomial, but note that y = 4 6x which is at odds with y = at x = 1 and x = 3. If instead we force y ( 1) = 1 and y (3) = 18, then we obtain W i = [.5, 1, 1, 1,.5 and D = 2/3. This leads to T = D + W y = = 22 3, (63) the exact result, as expected. If we were to consider a higher order polynomial for y(x), neither would give the exact result, but the weights based use of y (x 1 ) and y (x N ) would be significantly more accurate. 11

12 5. Matrix representation of the integral operators. To compute the matrix representations of the Λ, M, N, Φ, and Φ (4) operators where the function to be transformed is represented by a cubic spline, we must have the integrals of 1,x,x 2 and x 3 against the exponential integral functions E 1,E 2,E 3,E 4, and E 5. These are: E1 (x)dx = E 2 (x) x E1 (x)dx = E 3 (x) e x x 2 E 1 (x)dx = 2E 4 (x) x e x x 3 E 1 (x)dx = 6E 5 (x) e x (3 + x + x 2 ) E2 (x)dx = E 3 (x) x E2 (x)dx = 2E 4 (x) e x x 2 E 2 (x)dx = 6E 5 (x) + e x (1 x) x 3 E 2 (x)dx = 24E 6 (x) e x (6 + x 2 ) E3 (x)dx = E 4 (x) x E3 (x)dx = 3E 5 (x) e x x 2 E 3 (x)dx = 12E 6 (x) + e x (2 x) x 3 E 3 (x)dx = 6E 7 (x) e x (11 x + x 2 ) E4 (x)dx = E 5 (x) x E4 (x)dx = 4E 6 (x) e x x 2 E 4 (x)dx = 2E 7 (x) + e x (3 x) x 3 E 4 (x)dx = 12E 8 (x) e x (18 2x + x 2 ) E5 (x)dx = E 6 (x) x E5 (x)dx = 5E 7 (x) e x x 2 E 5 (x)dx = 3E 8 (x) + e x (4 x) x 3 E 5 (x)dx = 21E 9 (x) e x (27 3x + x 2 ) If we then consider M(x) = 1 2 E 1(x) 3 2 E 3(x), we find that M(x)dx = 1 2 E 2(x) E 4(x) x M(x)dx = 1 2 E 3(x) 9 2 E 5(x) + e x x 2 M(x)dx = E 4 (x) + 18E 6 (x) e x (3 x) x 3 M(x)dx = 3E 5 (x) 9E 7 (x) + e x (15 2x + x 2 ) 12

13 while with N(x) = 5 3 E 1(x) 4E 3 (x) + 3E 5 (x) we obtain N(x)dx = 5 3 E 2(x) + 4E 4 (x) 3E 6 (x) x N(x)dx = 5 3 E 3(x) 12E 5 (x) + 15E 7 (x) 2 3 e x x 2 N(x)dx = 1 3 E 4(x) + 48E 6 (x) 9E 8 (x) e x (6 x) x 3 N(x)dx = 1E 5 (x) 24E 7 (x) + 63E 9 (x) 2 3 e x (63 5x + x 2 ) and for Φ (4) (x) = 2 3 E 2(x) 2E 4 (x) we have Φ (4) (x) dx = 2 3 E 3(x) + 2E 5 (x) x Φ (4) (x) dx = 4 3 E 4(x) 8E 6 (x) e x x 2 Φ (4) (x) dx = 4E 5 (x) + 4E 7 (x) 4 3 e x (4 x) x 3 Φ (4) (x) dx = 16E 6 (x) 24E 8 (x) e x (24 3x + x 2 ) We assume a grid of optical depths [τ,τ 1,,τ N 1, at which the functions are nown and at which we want to evaluate the transforms. Usually, τ =, while τ N 1 may be the central plane of a slab, or the far surface (so that τ N 1 = also), or perhaps τ N 1 >> 1 for a semi-infinite atmosphere. To form the matrix representation of the Λ-operator we will thus need to evaluate the integrals Λ which give the contribution at τ i due to the emission from the material in the layers between τ and τ +1 : Λ = 1 2 τ+1 τ f(τ)e 1 ( τ τ i )dτ = 1 2 for the case where τ τ i, while if τ +1 τ i we have Λ = 1 2 τ+1 Now from equation (4) we see that τ f(τ)e 1 ( τ τ i )dτ = 1 2 τ+1 τ i τ τ i f(τ i + x)e 1 (x)dx, (64) τi τ τ i τ +1 f(τ i x)e 1 (x)dx, (65) f(τ i + x) = (a + α ) + (b + β )(τ i + x) + γ (τ i + x) 2 + δ (τ i + x) 3 (66) so that equation (64) can be written as Λ = (a + α ) I () + (b + β ) I (1) + γ I (2) + δ I (3). (67) where we have indicated the integrals by I (n) = ± 1 2 τ+1 τ i τ τ i x n E 1 (x)dx where the (+) sign is for this (τ τ i ) case, (68) 13

14 and, after expanding eqn (66), we see that the primed coefficients are given by Lewise, equation (65) can be written as a = a + τ i b, α = α + τ i β + τ i 2 γ + τ i 3 δ, β = β + 2τ i γ + 3τ i 2 δ, γ = γ + 3τ i δ. (69) Λ = (a + α ) I () + (b + β ) I (1) + γ I (2) + δ I (3), (7) but for this case, where τ +1 τ i, equation (68) requires the lower (minus) sign: 1 2 τi τ τ i τ +1 x n E 1 (x)dx = 1 2 τ+1 τ i τ τ i x n E 1 (x)dx = I (n), (71) which is why we used ± in eqn (68). (We cannot just tae the absolute value of the integral because some ernels, le M(x), may tae negative values.) Now the coefficients are a = a, α = α, b = b, β = β, γ = γ, δ = δ. (72) From equations (41), (42) and (43), and defining = τ +1 τ, we see that a = 1 (τ +1 τ i )y 1 (τ τ i )y +1, α = 1 { τ+1 (τ ) τ i (3τ ) + 3τi 2 τ +1 τi 3 1 { τ (τ ) + τ i (3τ 2 2 ) 3τi 2 τ + τi 3 } y+1 } y + (73) b = ( 1 ) y + ( 1 ) y +1, β = 1 { } (3τ ) + 6τ i τ +1 3τi 2 y + 1 { } (3τ ) 6τ i τ + 3τi 2 y+1 (74) γ = 1 {τ +1 τ i } y + 1 { τ + τ i } y δ = 1 y + 1 y (75) (76) Now we want to arrange this in the form Λ = X y + Y y +1 + U y + V y +1 (77) From equations (67), (7) and (73)-(76) we see that X,Y,U and V are given by 14

15 [ X = 1 (τ +1 τ i )I () I (1) [, Y = 1 (τ τ i )I () I (1) (78) (6 ) U = [ τ +1 (τ ) τ i (3τ ) + 3τi 2 τ +1 τi 3 () I [ (3τ ) 6τ i τ τi 2 (1) I + 3 [τ +1 τ i I (2) I(3) (79) (6 ) V = [ τ (τ 2 2 ) τ i (3τ 2 2 ) + 3τ 2 i τ τ 3 i ± [ (3τ 2 2 ) 6τ i τ + 3τ 2 i I () I (1) 3 [τ τ i I (2) ± I(3) (8) Here, the upper sign of the ± and pair applies when τ τ i and the lower when τ +1 τ i. The integral Λ can thus be expressed as Λ = n=1 L () n y n where L () n = X δ n + Y δ n,+1 + U C,n + V C +1,n (81) But this is only the contribution to the transform from the material between τ and τ +1. To find the Λ-transform at τ i, we need the sum over all from = 1 to = N 1: ( N 1 N 1 N 1 ) Λ i = Λ = L () n y n = L () n y n = L (i) n y n (82) n=1 where, carrying out the sum over we have L (i) n = X i,n + Y i,n 1 + N 1 n=1 U i, C,n + N 1 n=1 V i, C +1,n (83) where for n = 1, Y i,n 1 is absent, and for n = N, X i,n is absent. The X and Y terms alone give the transform which results from a linear approximation to the function f(τ). We thus see that if we construct the matrix Λ i,n = L (i) n, where each row is evaluates the transform at the i = 1,,N values of τ i, then Λ i,n is the matrix approximation to the Λ-transform that we require. Further, this result extends immediately to the M- and N-transforms by use of the appropriate functions for the I (n) integrals. The Φ and Φ (4) -transforms are slightly different. Consider the partial contribution to the Φ-transform at point τ i by radiation from the layers between τ and τ +1. If τ τ i, then we see from equation (17) that Φ = 2 τ+1 τ f(τ)e 2 (τ τ i )dτ = 2 τ+1 τ i τ τ i f(τ i + x)e 2 (x)dx (84) 15

16 On the other hand, when τ +1 τ i, we have Φ = 2 τ+1 τi τ f(τ)e 2 (τ i τ)dτ = 2 f(τ i x)e 2 (x)dx (85) τ τ i τ +1 If we now define the integrals I (n) by the expression I (n) = 2 τ+1 τ i τ τ i x n E 2 (x)dx (86) we see that I (n) > for τ τ i but I (n) < for τ +1 τ i, as required. Note that this differs from the form of the earlier equation (68) only in that we do not reverse the sign for τ +1 τ i. With this definition of the I (n), equations (78) through (83) also can be applied to the Φ-transform, and the Φ (4) -transform is analogous. We now have prescriptions for evaluating the matrix approximations of the transforms for a slab extending from τ 1 to τ N. However, in the treatment of a semi-infinite atmosphere, we must include the contributions to the integrals of material below the last point, τ N. To do this we must mae some assumption about the behavior of the function f(τ) over the interval [τ N,. One simple possibility would to be to assume the function constant over this interval: f(τ) = f(τ N ). However, we now that in the unpolarized grey case, the asymptotic behavior of the source function is linear in τ. Thus, let us suppose that we can write [ df f(τ) = f(τ N ) + (τ τ N ) for τ > τ N. (87) dτ τ N Now, from eqn (32), the derivative of the spline fit to f at τ N is [ df = f ( ) N f N 1 N 1 [f + N 1 + 2f dτ N 1 6 N. (88) τ N in terms of f and the 2nd derivatives f. The first term is just the extrapolated linear fit to the last two points. Let us define the integrals and I () in = 1 2 We then see that 1 2 τ N E 1 ( τ τ i ) dτ = 1 2 τ N τ i E 1 (x) dx = 1 2 E 2(τ N τ i ) (89) I (1) in = 1 x E 1 (x) dx = 1 [ e (τ N τ i ) E 3 (τ N τ i ). (9) 2 τ N τ i 2 τ N (τ τ N ) E 1 ( τ τ i ) dτ = [ τ i I () in + I(1) in τ N I () in In this case, since 2E 3 (x) = e x xe 2 (x), this simplifies further: = I(1) in (τ N τ i )I () in (91) I (1) in (τ N τ i )I () in = I( ) in = 1 2 E 3(τ N τ i ). (92) 16

17 We thus see that we can write the contribution to the Λ-transform at τ i due to emission from τ > τ N, using eqns (87)-(92), as ( ) ( ) ( ) N 1 [f Λ in = f N 1 + f N + I ( ) in N 1 + 2f N. (93) I( ) in N 1 I () in + I( ) in N 1 Thus, as in eqn (81) we can write the contribution from beyond τ N as Λ in = n=1 L (in) n = + 6 L (in) n y n where now the elements of L (in) n are (94) ( ( N 1 6 I( ) in N 1 I ( ) in ) δ n,n 1 + ) C N 1,n + ( I () in + I( ) in N 1 ( N 1 I ( ) in 3 ) δ n,n For natural splines, y N = so the C N,n = and the last term vanishes. So we then must add the vector L n (in) from below τ N. to each L (i) n ) C N,n (95) of eqn (83) to account for the radiation 6. Boundary conditions for a finite slab. In addition to the semi-infinite atmosphere, another important problem is that of a finite slab. For this case, we may simply tae a grid of points through the whole slab and then use the natural y = conditions at both boundaries. But this problem may be symmetric about the central plane (but not always: e.g., if the slab is illuminated from one side). If we do have symmetry, we will only need half half as many points if our grid is defined to cover τ τ N, where the optical thicness of the whole slab is 2τ N. In this case, we will want to specify the first derivative at τ N, since τ N is the mid-plane of a slab and thus y N = by symmetry. (This does not mean that y N =, however.) In the case where we set y N =, z N of eqn (35) is also zero and as a result d of eqn (36) vanishes and hence does not complicate the expression for y. Now the function f(τ)) (e.g., s(τ) and p(τ)) will be symmetric about τ N so that f(τ) = f(2τ N τ). The transform at some τ i will now have contributions from both the layer between τ and τ +1 and the corresponding layer on the far side between (2τ N τ ) and (2τ N τ +1 ) where the function f(τ) is described by the same spline fit. So for the Λ-transform, we have Λ = 1 2 τ+1 τ f(τ)e 1 ( τ τ i )dτ τN τ 2τ N τ +1 f(2τ N τ)e 1 (τ τ i )dτ. (96) Note that for the second term, (τ τ i ) will always be positive. We let x = τ τ i and define τ i = 2τ N τ i. We then see that the far-side contribution can be written as 17

18 Λ (f) = 1 τ i τ f(τ 2 τi τ i x)e 1 (x)dx. (97) +1 By analogy with eqn (66), the spline approximation to f(τ) over this interval can be written as f(τ i x) = (a + α ) + (b + β )(τ i x) + γ (τ i x) 2 + δ (τ i x) 3 (98) and after expanding and collecting coefficients of powers of x, as with eqn (67) and eqn (7), we can write Λ (f) = (a (f) + α (f) ) I(f,) + (b (f) + β (f) ) I(f,1) + γ (f) I (f,2) + δ (f) I (f,3) (99) where now the coefficients are a (f) = a + τi b, b (f) = b, α (f) = α + τi β + τi 2 γ + τi 3 δ, β (f) and the integrals are = β 2τ i γ 3τ i I (f,n) = 1 τ i τ 2 τi τ +1 2 δ, γ (f) = γ + 3τ i δ, δ (f) = δ. (1) x n E 1 (x)dx. (11) Once again, we go bac to eqns (41)-(43), collecting the coefficients of y, y +1, y, and y +1 to express the far-side contribution in the form Λ (f) = X (f) y + Y (f) y +1 + U (f) y + V (f) y +1 (12) where now X (f) = 1 [ (τ +1 τ i )I (f,) + I (f,1), Y (f) [ = 1 (τ τi )I (f,) + I (f,1) (13) (6 ) U (f) = [ τ +1 (τ ) τi (3τ ) + 3τi 2 τ +1 τi + [ (3τ ) 6τ i τ τ i 2 I (f,1) + 3 [τ +1 τ i I (f,2) 3 I (f,) + I (f,3) (14) (6 ) V (f) = [ τ (τ 2 2 ) τi (3τ 2 2 ) + 3τi 2 τ τi 3 I (f,) [ (3τ 2 2 ) 6τ i τ + 3τ i 2 I (f,1) 3 [τ τ i I (f,2) I (f,3) (15) Finally, eqn (44) is used for the y, and the contribution from the far half of the slab has exactly the same form as eqn (83). We then add the near and far sides to get the total Λ i,n matrix transform. The Φ i,n and Φ (4) i,n-transforms are not special cases here since the radiation from the far side is traveling upwards (positive) for all near-side layers. 18

19 7. The emergent radiation. From equations (12) and (13), we have the emergent radiation in terms of the source terms s(τ) and p(τ) : I(,µ) = We can define a transform E µ as s(τ) e τ/µ dτ µ + ( 1 3 µ2 ) Q(,µ) = ( 1 µ 2) E µ {f(t)} = p(τ) e τ/µ dτ µ t/µ dt f(t) e µ p(τ) e τ/µ dτ µ (16) (17) (18) so that the emergent radiation is given by ( ) 1 I(,µ) = E µ (s) + 3 µ2 E µ (p) ; Q(,µ) = ( 1 µ 2) E µ (p) (19) To evaluate E µ based on the values of s and p at the grid points τ i, we will need to find E µ, for = 1,, (N 1): E µ, = τ+1 τ f(τ) e τ/µ dτ µ = x+1 x f(µx) e x dx (11) where x i = τ i /µ and f(µx i ) = f(τ i ) = f i. As before, we assume the function is represented by a cubic spline. Then, going bac to eqn (4), so we can write it terms of the integrals E (n) µ, f(µx) = (a + α ) + (b + β )µx + γ µ 2 x 2 + δ µ 3 x 3 (111) E µ, = (a + α ) E () µ, + (b + β ) E (1) µ, + γ E (2) µ, + δ E (3) µ, (112) = µn x+1 and the integrals against e x e x dx = e x x x n e x dx where x = τ /µ, x +1 = τ +1 /µ, (113) are: x e x dx = e x (1 + x) x 2 e x dx = e x (2 + 2x + x 2 ) x 3 e x dx = e x (6 + 6x + 3x 2 + x 3 ) We next write E µ, as E µ, = X µ, f + Y µ, f +1 + U µ, f + V µ, f +1 (114) 19

20 Referring to eqn (41)-(43), we see that ( ) X µ, = 1 τ +1 E () µ, E(1) µ, ( ), Y µ, = 1 τ E () µ, E(1) µ, (115) (6 )U µ, = τ +1 (τ ) E () µ, (3τ ) E (1) µ, + 3τ +1 E (2) µ, E(3) µ, (116) and thus, using eqn (44), (6 )V µ, = τ (τ 2 2 ) E () µ, + (3τ2 2 ) E (1) µ, 3τ E (2) µ, + E(3) µ, (117) E µ, = {X µ, δ n + Y µ, δ n,+1 + U µ, C,n + V µ, C +1,n } f n (118) n=1 Finally, just as with eqns (81)-(83), our transformation is approximated by where E µ (f) = N 1 E (µ) n = X µ,n + Y µ,n 1 + E µ, = N 1 n=1 U µ, C,n + E (µ) n f n (119) N 1 and, once again, for n = 1, Y µ,n 1 is absent, while for n = N, X µ,n is absent. Thus, having chosen a set of M angles µ m, we can compute the E n (µm) Then, the emergent intensity and polarization are obtained from V µ, C +1,n (12) for each of them. I(,µ m ) = n=1 E (µm) n s n + ( ) 1 3 N µ2 E n (µm) p n (121) n=1 Q(,µ m ) = ( 1 µ 2) N n=1 E (µm) n p n (122) In the case of a slab with symmetric sources, where the last grid point τ N is the midpoint of the slab, we must add the contributions from the far side. From each slab we have a contribution E (f) µ, = τ τ τ τ +1 f(τ) e τ/µ dτ µ, where τ = 2τ N (123) By the symmetry of the function f, f(τ τ) = f(τ). Let x = τ/µ. Then we can write E (f) µ, = µx µx µx µx +1 f(µx µx) e x dx. (124) 2

21 The cubic spline representation of f is of the form f(µx µx) = (a + α ) + (b + β )µ(x x) + γ µ 2 (x x) 2 + δ µ 3 (x x) 3 (125) The point of this is that at the edges of the interval, f(µx µx ) = f(τ ) = f. The moment integrals we then need are just E (f,n) µ, and we then write = µ n x x x x +1 x n e x dx, where x = 2τ N /µ, x = τ /µ, etc. (126) E (f) µ, = (a(f) + α (f) ) E(f,) µ, + (b (f) + β (f) ) E(f,1) µ, + γ (f) E (f,2) µ, + δ (f) E (f,3) µ, (127) where now (with τ = 2τ N ) the coefficients are a (f) = a + τ b, b (f) = b, α (f) = α + τ β + τ 2 γ + τ 3 δ, β (f) = β 2τ γ 3τ 2 δ, γ (f) = γ + 3τ δ, δ (f) = δ. (128) Proceeding as in section 6, we then write the partial contribution from the far side as E (f) µ, = X(f) µ, y + Y (f) µ, y +1 + U (f) µ, y + V (f) µ, y +1 (129) where now X (f) µ, = 1 [ (τ +1 τ )E (f,) µ, + E (f,1) µ, [, Y (f) µ, = 1 (τ τ )E (f,) µ, + E (f,1) µ, (13) (6 ) U (f) µ,i = [ τ +1 (τ ) τ (3τ ) + 3τ 2 τ +1 τ 3 E (f,) µ, + [ (3τ ) 6τ τ τ 2 E (f,1) µ, + 3 [τ +1 τ E (f,2) µ, + E (f,3) µ, (131) (6 ) V (f) µ, = [ τ (τ 2 2 ) τ (3τ 2 2 ) + 3τ 2 τ τ 3 E (f,) µ, [ (3τ 2 2 ) 6τ τ + 3τ 2 E (f,1) µ, 3 [τ τ E (f,2) µ, So finally, for the symmetric slab, equation (12) acquires additional terms: E (f,3) µ, (132) E (µ) n = ( ) ) X µ,n + X µ,n (f) + (Y µ,n 1 + Y (f) µ,n 1 + N 1 ( V µ, + V (f) µ, ) + N 1 ( ) U µ, + U (f) µ, C,n C +1,n (133) 21

22 8. Solution of some specific cases by the matrix transform method. We now return to the grey problem that we abandoned in section 3. Let us consider this problem for the semi-infinite atmosphere. Combine the vectors s and p into one vector sp = s 1,s 2,...,s N,p 1,p 2,...,p N of length 2N. We then can express equations (15) and (16) as the following 2N 2N system: [ Λ ij I ij (1 λ)m ij 3 M ij 3 8 (1 λ)n ij I ij [ si p i = [ (134) where I ij is an N N identity matrix. As it stands, we cannot solve this system since the r.h.s. is identically zero. But we can use the matrix form of the flux equation (2) Φ ij s + Φ (4) ij p = F (135) This provides N equations. We can replace the first N equations with these and solve the system [ [ Φ ij Φ (4) [ ij si F 3 (1 λ)m 3 8 ij (1 λ)n = (136) 8 ij I ij p i where the F on the right hand side represents an N-element column vector with each element equal to F. This wors, but another method seems to give better values just near the surface. We only need to add any one of the flux equations to the first N equations of (134) to set the scale. A satisfactory approach seems to be to tae the equation for the flux at the surface (the first equation, i = 1) Φ 1n s n + n=1 n=1 and add it to each of the first N equations (134): [ 1 Λ ij I ij + Φ 1j M 3 ij + Φ (4) 1j 3 8 (1 λ)m ij 3 8 (1 λ)n ij I ij Φ (4) 1n p n = F (137) [ si p i = [ F The resulting linear system can be solved by any standard numerical method. (138) All the foregoing have been implemented in J. The verb tau grid constructs a logarithmic series of points τ =,τ 1,,τ N when provided with the arguments tau grid τ 1, n, τ N, where n is the number of points per decade. For example tau grid We provide some results for a fine grid: τ 1 =.1,n = 2,τ N = 25, which yields 11 grid points. As opposed to the semi-infinite atmosphere, there is no grey solution for integrated radiation from a slab, since the equations have no source for the radiation (in the semiinfinite case, it streams in from beyond the lower boundary). Slabs must have an internal or external source added to the equations. 22

23 Turning to the monochromatic case described by equations (7) and (8), we see that they lead to the following set of equations: [ Iij (1 λ i )Λ ij 1 3 (1 λ i)m ij 3 8 (1 λ i)m ij I ij 3 8 (1 λ i)n ij [ si p i = [ λi B i where B i = B ν [T(τ i ) and we allow that λ i, the fraction of opacity due to pure absorption, may vary with optical depth zone τ i. 23

24 9. References. Bohren, C.F., and Huffman, D.R. 1983, Absorption and Scattering of Light by Small Particles, (John Wiley & Sons, New Yor). Chandrasehar, S., 196, Radiative Transfer, (Dover Pub., NY) a reprint of the 195 edition published by Oxford. Harrington, J.P., 197, Polarization of Radiation from Stellar Atmospheres. The Grey Case, Astrophysics and Space Science, 8, Hovenier, J.W., van der Mee, C. and Dome, H. 24, Transfer of Polarized Light in Planetary Atmospheres: Basic Concepts and Practical Methods, (Kluwer, Dordrecht). Kourganoff, V., 1963, Basic Methods in Transfer Problems, (Dover Pub., NY) a reprint of the 1952 edition published by Oxford. Press, W.H., Teuolsy, S.A., Vetterling, W.T., and Flannery, B.P., 1992, Numerical Recipes in Fortran 77, Vol. 1 (Cambridge University Press). van de Hulst, H.C. 1981, Light Scattering by Small Particles (Dover Pub., NY), a reprint of the original 1957 edition. 24