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The milne problem for a semi-infinite medium with anisotropic scattering and absorption of neutrons
Reactor
Saence
and Teclmology
(Journal
of Nuclear
Energy
Parts A/B)
1963, Vol.
17, pp. 559 to 569.
Pergamon
Press
Ltd.
Printed
in Northern
Ireland
THE MILNE PROBLEM FOR A SEMI-INFINITE MEDIUM WITH ANISOTROPIC SCATTERING AND ABSORPTION OF NEUTRONS”? L. M. ROMANOVA 1. THE MILNE PROBLEM INFINITE MEDIUM WITH SCATTERING
FOR A SEMIANISOTROPIC
THE following problem is investigated : a semi-infinite medium z > 0 is uniformly filled with a non-absorbing material which scatters neutrons without changing their velocity, according to the following law u(cos Y) = 2
[l + gPkos
p1) + hP,(cos y)l
(1)
(where p is the scattering angle; g, h are constants; and P,, Pz are Legendre polynomials). It is assumed that there are no sources in the medium and that no neutrons enter the region z > 0 from the outside. A neutron current is assumed to exist in the scattering medium in the -z direction. The problem consists of determining the steady-state neutron flux distribution in the medium and the angular distribution of neutrons leaving the plane z = 0. For the case of isotropic scattering, the problem can be solved by the Wiener-Hopf (WIENERand HOPF, 1931) method. Their method can also be applied to problems where anisotropic scattering is considered; in our work we will use a variation of the Wiener-Hopf method, given by PLACZEKand SEIDEL(1947), which is more amenable to generalized cases. Let us further say that the mathematical algorism used in this work is not dependent on the actual physics of the radiation and can therefore be applied to other problems in transport theory. The transport equation and some relationships derived from it The transport equation for the problem being considered is as follows:
0
y(z,
B’)o(cos
p)
sin 0’de’ da.
q) =
P,(COS
+2
(1.1)
i
e/j
(n-mm)!pycosejpm
m=L (n + m)! (COS
n
n
et) cos m(p - /?‘),
equation (1.1) becomes
sly
/-J=&
Y(Z9 Pu)=
+
+ hPdP)YdZ)l. (1.2) fl
Here
i-4dp i PICUMZ,
w(z) =
J-1
where yO(z) denotes the scalar neutron flux, and yl(z) the scalar neutron current at point z. After integration of equation (1.2) over d,u between the limits -1 to +l, one obtains dz
0
ey,(cos
x
dy,(z)
vI
=II
P,(COS
-=
aY cmes + y(s, e> 2n
Here, 0 and 8’ are the angles between the neutron velocity vectors and the positive direction of the z-axis, a is the corresponding azimuthal angle, pl is the angle between the initial and scattered neutron velocity vectors, and y(z, 0) is the neutron distribution function. The spatial units are normalized to the neutron mean free path, and the velocity is normalized to the neutron velocity. The density of scattering nuclei is considered to be unity. The expression o(cos p) is given by equation (l), g = 3 cos a, = 3, where cos v is the average cosine of the scattering angle; and 0 G h G 2, since a(cos 9) > 0. Expressing cos 0 as p, and using the relationship for Legendre polynomial multiplication
o
’
yl(z) = F = const.
The scalar neutron current is thus reduced to a constant. Multiplying equation (1.2) by ,Uand integrating as above, one obtains dK = -( 1 - &g)F, K(z) dz
-
* The results of this work were obtained in 1951. 7 Translated by A. LORENZ from the Russian work Some Mathematical Problems in Neutron Physics (1960).
= - (1 - &g> Fz + K(O). 559
(1.3)
L. M. ROMANOVA
560
s +1
K(z) represents
,~?y(z, ,u) d,u. Since K(z) > 0 for
-1 all z, then -(l - &g)I; must be positive; also since (1 - &) > 0, then -F > 0. We obtain the following integral equation for the neutron distribution function in the semi-infinite medium z > 0:
medium is thus reduced to the finding of the Laplace transform of the neutron density. The integral equation for the Laplace transform of the neutron density and the solution of this equation
In order to find C&,(S),we shall use the Placzek and Seidel method. We shall start with the transport equation (1.2) given above, and shall substitute for yr(z) and y*(z) the following relationships YAZ) = F
and Y&) = QNO) obtaining
I+
hP&4wdEN
dE, -1
(1.4)
If we substitute for yr(z) and y&r) = $Y(z) - iyO(z) and integrate equation (1.4) over dp, we obtain a variation of Milne’s integral equation:
p ;
+
(1 - &WI - &w,,(z),
y(z, PI = ; [ 1 -
+ SFgp + f
t
4
2P2W] wow
- ig)Fz + W)IP,cu).
Let us furthermore substitute the Laplace transform and divide the relationship thus obtained by (1 + s,u), where s is a compIex number. The whole expression is then integrated over dp from - 1 to + 1. The resulting relationship is an integral equation in Q,,(s):
{sf(l+;)s2 - $Do(s) - Fgs2 - $FgE,(z)
-
Jwl
- 4
(1 - f g) F[3E,(z)
+ ; jw> [3&(z) - -wN
+ ;
(1.5)
where E,(z) =
Lo $
dt,
[(
z > 0.
u-7) Applying the following boundary conditions
The neutron distribution function y(z, ,u) is obtained from equation (1.5) with the aid of equation (1.4). We will limit our investigations to the determination of the scalar neutron flux yO(z), and the angular distribution of neutrons leaving the plane z = 0. For the latter we obtain from equation (1.4) :
+
- fg)F-SK(O)])
=s2[(l+;)s2-;]G(s)+;Fs2-;K(O)s’.
s1
Y(O,PI = -
X
(9 - 3) x [(I
l [l -gJ2Lu+o(-~)
G
i$Fgp+ ; PA,4 [Cl- 58) FP + W31,
y(0, EC)= 0,
O
we have: G(s) =
’ puw(O,/J) d,u * s -1 1fsCL
The expression (1.6) for ~(0, ,u) is given in terms of @o(- l//J). Let us now introduce the following notation:
(1.6) + T (s2 - 3)[(1 - +g)F - S(O)],
where m e+y,,(z) dz. @‘o(s)= s0 The problem of determining the angular distribution of neutrons leaving the boundary of the scattering
f(s)=
~(+s2[(1
(1 +$-$)(I
-klnz)
+;)s2-;]G(s)+;s’[F-sK(O)l.
-$,
The Milne problem for a semi-infinite medium
561
Equation (1.7) can now be written as :
Let us analyse the properties of this function. The scalar neutron flux can be expressed in the form: y,,(z) = O(e”‘),
k -=c1;
then the function O&r), as well as k. For Is1+ co in this halfplane O(s) = 0((s13). The domain of the analytic function H(s) corresponds with the domain of the analytic function G(s). It can be seen from the expression for G(s) that it is analytic in the half-plane Re(s) < 1. For IsI-+ cc in this half-plane H(s) = 0(js13). The functionf(.s) is analytic in the region - 1 < Re(s) < +l,*ands = +I ands = -1 areits branch points. We shall choose that branch off(s) which is real for - 1 < s < + 1. For IsI + co in this region the functionf(s) approaches unity. At the points = 0 this function has a second-order pole; f(s) has no other poles in the region - 1 < Re (s) < 1. Let us prove this by using the principle of the argument.? The number of poles the function f(s) can have within a closed contour C is
FIG. 1
For large values of z it is possible to determine the asymptotic behaviour of yO(z)without solving equation (1.7). This can be expressed in the form of a sum of residues of the function Q,(s) es2 relative to the poles of the function CD,(s). From equation (1.7) we find:
l f’ods = -!var arg f(s) ’ 2%- c
N = 5Z sJ-(s)
where var, argf(s) designates a full increment of arg f(s) after integration along the closed contour; each pole is counted once. Let us consider a square contour C whose sides are parallel to the co-ordinate axis, which lies within the region - 1 < Re (s) < 1, and which includes within its contour the point s = 0 (see Fig. 1). Within this contour,f(s) is analytic, continuous and does not have any poles on the contour itself. (See WIENERand HOPF,Lemma 3 ; within the region - 1 < Re (s) < 1J(s) has only a finite number of poles, it is therefore possible to choose a contour which will satisfy the above conditions.) As s moves along the contour from point A to B in a counter-clockwise direction, the point w = f(s) moves along a closed contour, for even powers off(s). When s moves from B to A in the same direction, the point w = f(s) also moves along a closed contour which corresponds to the former. The point w = 0 is situated within the contour C, therefore var, arg f(s) = 2.2~
and N = 2.
;[F-sK(O)][l
+
-
(1-;)(I -&I;)] S2f(4 (1.9)
This equality shows that (D,,(s)has one second-order pole s = 0. The expansion of QO(s) into a series around s = 0 has the form Q,(s) = -31; and we obtain: w,&)
= -3F(l
- &)[z + 4, zo = -
mu (1 - +g)F ’
It can be seen that K, which indicates the rate of growth of vo(z) to infinity is equal to zero. In order to solve equation (1.7) let us define s2 I(S) = Tf
1
(4.
* A wider region of analyticity will be assigned to the function (s) in the text below. t Translator’s note: The author Wiener-Hopf technique.
makes reference
to the
This function, analytic in the domain - 1 < Re(s) < 1, does not have any poles and approaches unity as Is1--+
L. M. ROMANOVA
562
co in that domain. It is therefore possible (1) to expand the function T(S):
(l.lOa)
Equating this expression for Q,,(s) with that given in equation (1.9) we see that the values s = +1/3/z/(4 + h) are not poles of the function a,,(s). Under these conditions the first term in the numerator of equation (1.13) is therefore equal to zero, and the constants C and K(0) are determined as a function of the current -F: 6
1
U)T_(-U)'
(l.lOb)
(1 + k(u)
T+(S)is analytic in the half-plane Re(s) < ,!I T_(S) is analytic on the half-plane Re(s) > -8. T+(S) and T_(S) do not have poles and approach unity as Is] 3 co in those domains. Equation (1.8) can now be expressed as S-l H(s). CD(s)= S + 1 T_(S) T+(s) 1
s2
(1.14)
’= -F(l+U)T_(U) + (1 -
. ,
(1.11)
K(o)
=
-%[(I
where
+
U)T_(U)
a=
- (1 - a>T-(-a) +
(1
-
(,
.15)
U)T_(-U)]’
3h -. 4+h
J
The real integral T_(U) which is obtained from equation (1.106) by transformation of the contour integral onto the real axis has the form of
The left-hand side of this equasin2 x tion is analytic in the half-plane In Re(s) > 0, and the right-hand 1 + a + ; COPx (1 - x cot x) - $ a IQ’* 1 side is analytic in the half-plane T_(U)= eXp dx . 7T u2 + (1 a”) sin2 x Re (s) < 1. Since these two planes [ UO I intersect, each side is the analytical continuation of the other one. The left-hand side, as From the expressions for T_(S) and T+(S) it can be well as the right-hand side of equation (1.11) established that T_( -s) = 1/[T_+(S)] which leads to the approaches infinity in these half-planes as ]s14. Both equality sides can be expressed in fourth-order polynomials 1 T-(-U) = (1.16) using Liouville’s theorem : T(U)7_(U) ., ’
1
where s-l =
-
(1.12)
H(S).
T+(S) Expanding the function [(s series, we find that
1 -u2h T(U) = - . u2 4
in a Laurent
l)/~.+(s)]H(s)
c, = c, = c2 = c, = 0,
c4
=
c
#
0.
Angular distribution of leakage neutrons From expressions (1.6), (1.13), (1.14), (1.15) and (1.16) one obtains
And we can write equation (1.12) as: cS2(S
@o(s) =
+
l)T_(S)
s2[(l
-
;
(S2 -
3)[F - SK(O)]
+ $),. _ 3$]
T_(_~)
=exp,_:jrl+‘+l~~~~,:“tx)-~.,.
(1.17)
The Milne problem for a semi-infinitemedium Note that the angular distribution of neutrons at the interface is not dependent on the coefficient g. It is therefore seen that the angular distribution of neutrons at the interface will be the same for isotropic and linear scattering. The scalar flux y,,(z)
The scalar neutron flux can be found with the inverse Laplace transform
(1.18) Let us write: H(s) - ; Q,(s) =
(9 - 3)[F - SK(O)]
Fg
+$)s2_?!]
0
s
n
n/2
exP !E [S
K(O)
1
1--Uln2
&_($)P
0 < P < 1,
0
2 sin2 x In
3(1 + COPX)(1 - x cot x) - 1 dx 1 - (1 - p2) sin2 x
1.
The expression In r(l/,~) is solved for in the same manner as it was done for the case of isotropic scattering treated in PLACZEK (1947). Noting that
P
s
dx = arc tan 01 tan x), 1 - (1 - p2)sin2x
the expression for In ~-(1/p) is integrated
e-ziu dp )(
$fJ12)](1 +
=
4a2 (1 + a)e[r(a>12 4 h
0
where
1 i-_ P
(1 + a>r(a)
1
X
Substituting p for -,u in equation (1.17) and assuming -F = 1 and h = 2, an expression for the neutron angular distribution at boundaries is obtained :
*
The solution of the integral analogous to equation (1.18), is given in MARK (1947). The solution for this case is :
- Jg)(z + z0) -
- Sg)(z + %),
where z0 is the extrapolation length. Calculations done for the angular distribution of neutrons at boundaries can be used to solve the integral for cases other than the asymptotic one.
+2.
The function H(s) has a branch point at s = 1 and is analytic in the plane between 1 and co. The function f(s) has branch points at s = 1 and s = -1 and is analytic in the plane between - co and -1 and 1 and co. The function [H(s)/f(s)] and therefore O,,(s) will in all cases by analytic in the half-plane Re(s) < 1, between -co and - 1. The analytic behaviour of @,,(s) in the positive half-plane follows from the determination of $(s) and the asymptotic behaviour of yO(z). The @,(s) is therefore analytic in all of the s plane between - co and - 1, except for the secondorder pole at s = 0.
ye(z) = -3F{(l
%a&) = -3F(l
YK4 I4 =
f(s) $[(I
00 the integral term drops out and one
Calculation of the neutron angular distribution at boundaries and of the extrapolation distance for thecaseofh=2andg=O
eSZ@&)dS, CJ> 0.
Y&) = Aj
When z finds
563
13-P 1-p
(1 + a)“[T_(a)]” - 7
” = - (1 - Qg)F = (1 - *g)a ’(I + a>2[T_(a)12+ 42 ’ h
by parts,
L. M. ROMANOVA
564
yielding : ln,(-l-)
=i
rarctanbtanx) K(0) = 2[T-(l)12 - 1 = o.71486 2[T-(l)12 + 1
JO
X,
6(1 - x cot x) 3x(1 + COPx) - cot x 1 + sin2 x ( dx 3(1 + COPx)(1 - x cot x) - 1 --
3 nl2 cot x . arc tan @ tan x) dx. Vr s0
~~(0)=
TABLE l.-ANGULAR
cot x . arc tan (p tan x) dx
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
is given in LE CAINE(1950). The result has the following form 1 n/2 cot x . arc tan (,u tan x) dx = &In (1 + p). ?r 1 YO The expression for the angular distribution can now be written in the form
2. THE
X
PROBLEM
MEDIUM
6(1 - x cot x) sin2 x
3(1 + cot2 x)(1 - x cot x) - 1
1
dx > 1
O
WITH AND
0~50000 0.62363 O-72518 O-82126 0.91464 I.00639 I.0971 1.1870 1.2794 1.3653 1.4539
FOR
A SEMI-
ANISOTROPIC
ABSORPTION
The problem solved in this section is analogous to the one treated above except for the fact that the medium is assumed to be a neutron absorber; that 1,/l= l/(1 - a), where 1, and 1 are the scattering and total neutron mean-free path in the medium, respectively, and that 0 < a < 1.
0
s-m
MILNE
Y@, P)
h = 0, y,(O) = 1
= 1
0.43532 057690 0.69150 0.79950 0.90450 0.98840 1.1095 1.2105 1.3440 1.4108 1.5098
SCATTERING
aI2 arc tan (p tan x)
3x( 1 + COPx) - cot x 1 +
Ylm
0.74170 0.98295 1.1782 1.3622 1.5410 1.6835 1.8903 2.0624 2.2336 2.4036 2.0724
INFINITE
37_(l)
[,
YK4 P)
Y(O,PI
Jo
1 x exp?7
DISTRIBUTION CALCULATION
-F=l
P
(1 + 2[7-(~)12)d1 + p
1.7038.
Due to the condition that g must equal zero when h = 2 (which follows from the condition o(cos v) > 0), we obtain an extrapolation distance of: z, = 0.71486.
This last operation changes the form of the function to one that is more amenable to numerical integration than the previous one. The solution of the integral
w(O,P) =
67_( 1)
2[r(1)12 + 1 = 1.4834,
C=
The transport equation for the given condition is:
aY
P z
+
Yh Pu)=
9
[we(z)
+ gP,(/J)w2(~) +
@2W
w2(z)l.
(2.1)
The unit of linear dimension is taken to be the neutron mean-free path, while that of the velocity is the neutron velocity. The nomenclature of Section 1 will be maintained. If one integrates equation (2.1) over d,u, one obtains (2.2) Multiplying equation (2.1) by ,u first, and then integrating over dp, one finds that dK -=dz
l-
(2.3)
The Milne problem for a semi-infinite medium
Expressions (2.2) and (2.3) show that in the presence of absorption the scalar neutron current yl(z) is not constant, and that K(z) is a linear function of z. Using the methods of Section 1, the following expression for the neutron angular distribution at a boundary is derived :
565
H(s) =
s2 +:a,,--
l-a
(
3
g
)I
G(s)
*-!$2g +
( )I.
hP,(pP’z -;
(2.4)
x
[F(O)- SK(O)]
Here, @,,(s), @ds),%(s 1are the Laplace transforms of the scalar flux y,Jz), the scalar current yl(z) and of yZ(z) respectively. The relationship between CD,(s)and %(s), and @&) and %( s ) can be found from expressions (2.2) and (2.3). Substitution of the expressions for Q1(-l/~) and @,(-l/p) thus obtained, in equation (2.4), yields: l--a
Y(Q p) = - -
2ru i
1+j
h
(
+
T a(1
ag-
-
Fg)p4)@0(E) [(
x ~Wp + Jm
1
1 -+g
>
- w.3 gp2,
(2.5)
where -F(O) is the value of the current and K(0) the magnitude of K(z) at the boundary z = 0. One can also derive, by methods used in Section 1, using the transport equation (2.1), an integral equation in the form @W(s) = H(s),
(2.6)
where
1+ts4
)
+
s2 9h
+-i-
a
l-
X N
*+w
J-l
f(s)
=
x
a(1
+
(
l--
t) - i + p a(1 l-a 3
gs2,
’
a> *+s
1 g 2+(1--l 1
(ag-
:-:a(1
i
2 ln2s l-s
1 -
l-a -Fg
1
1)
!$-:a(* -yg))pz
4
[i
aF(0)
OPY(O9P)dP
G(s) =
x [l+$+
-$(3+1)/J
@(s) =
-
( --g )I l-a
l-
?
(Do(s) - f
(9 - 3)
g) F(0) - sly(O)] - F(O)g.P,
The solution for y,,(z) must be sought in the form O(e”“),with k < 1, (Do(s)are therefore analytic in the half-plane Re(s) > k; when IsI -+ co in this halfplane CD(s)= 0(ls13). The function H(s) is analytic in the half-plane Re(s) < 1; when IsJ + co in this half-plane H(s) is also equal to O(l~1~). The function ,r(s) is analytic in the region - 1 < Re(s) < 1, and s = 1 and s = - 1 are its branch points. Choosing the branch off(s) which is real for -1 < s < + 1, one sees that in that region f(s) -+ 1 when IsI -+ co, and that it has two real zeros, namely s = +Y and s = -_y. In reality, f(O) > 0; for a real j3 close to one &I) = A-a) < 0, therefore, the function f(s) has two zeros on the real axis: s = f~, 1~1< 1. It can be shown, by means of the argument principle* that no other zeros exist for the function f(s) in the region - 1 < Re(s) < 1. Assume that T(S) = [(s2 - l)/(s2 - 2)]f(s), that the function T(S)is analytic, that it does not have zeros and approaches 1 as IsI -+ cc in the region - 1 < Re(s) < 1; it can therefore be assumed that T(S) = T+(s)/T_(s). The expressions for T+(S) and T_(S) are analogous to the ones given by equations (1 .lOa) and (1. lob). * Translator’s note: Wiener-Hopf technique.
The author
makes reference
to the
566
L. M. ROMANOVA
Equation (2.6) can now be written 3 - Y2 1 S-l a>(s) = H(s) . s + 1 T_(S) T+(S)
(2.7)
The left-hand side of this equation is analytic in the half-plane Re(s) > k, the right-hand side in the half-plane Re(s) < 1. Since these two planes overlap, they form an analytic continuum. Both sides of equation (2.7) approach infinity with a 1~1~dependence as Is] + co, and can be expressed in the form of a fourth-order polynomial with the aid of Liouville’s theorem: Y2 1
s2 -
f-D(s)-
S-l = H(s) = c, + c,s + c,s2 + css3 + 1 7_(s) T+(S)
s +
c4.94.
(2.8)
Expanding the function [H(s)(s - l)]/ T+(s) in t o a Laurent series, one finds C,, = C, = 0, so that equation (2.8) can be written as: (Do(s)= s2(C, + c!$ +
C43)(1
+
S)T_(S)
+
(s2 -
Y”)
(2’9)
x(sLy2)((1 +$4+ [ag-:-;a(1 -+gi]s2+~a(l-~~)) If one equates this expression for @&) with one obtained from equation (2.6): 3h
+ F(O)g 1 - k In z ( -+
neutron angular distribution at z = 0 can be construtted from equations (2.5) and (2.9):
3
s2fis) one finds that the roots of equation
(1 +;gs4+ (ag+;-;a(l
Y(O>P) =
_!$g))s”
--Pb-(-;),
-a&v”-b---)(1
_$
2(1 - /W) +:a
(
l-
l-a 3g
1
=o. lnT_(-_!)
s = &! and s = f5 are not poles of the function
k a),
C4 = -W)c(g,
k a),
C3 = --I;(O)& k a), K(0) = --F(O)k,(g, h, a).
Because of their sizes, the equations for a, b, c and k,, are not reproduced here. The expression for the
= -fj~~~“I;~~~~~~d~,
3h pl(X) = x 1 + h _ _h -qa(l 41 4 ( x
(
l-
X [I
l-a
-3
g cot2 x + (1 - a)(1 - x cot X) )
+t--
X cot2 x +
-a)
(ag-y--Ta(l9h 4
a 1(
l-a 3
qg))
1
g cot4 X . )
(2.10)
The Milne problem for a semi-infinite medium
567
The asymptotic solution for the scalar neutron flux y,,,(z) is obtained from the sum of residues at the poles Q&s = v and s = -v : yo’oas(4= --FPMg,
where z0 is the extrapolation z,(g, h, x) = & ln
of (DO(s) es*
(2.11)
4 4 sh v(z + 4;
distance.
(1 + v)T_(v)(~ + bv + CT+) 1 = %ln (1 - v)T_(-v)(u - bv + cvz)
(1 + ~)(a + bv + CV~))~_(V)(~T(V) (1 - v)(a - bv + cv2) ’
(2.12)
Calculation of the neutron angular distribution at boundaries and of the neutron density forg=O,h=2anda=0.5 If one substitutes ,u for -p in equation (2.10) and sets -F(O) = 1, g = 0, h = 2 and a = 0.5, one obtains an expression for the neutron angular distribution at the boundary: (w” + bp + c)(l + ,+W(O?P) =
In1 -
P/U
lnr
L =1u_ 0P ?I 1
; 0
4(1 - Y2@)
,
O<#u
4(sin2 x + v2 cos2 x) : COPx + 3(1 - X cot x)[l + Q cot2 X(1 + cot2 x)] dx 1 - (1 - p2) sin2 x
Substitution of --F(O) = 1, g = 0, h = 2 and u = 0.5 in equations (2.1 l), (2.12) and (2.13), an expression for the asymptotic neutron flux is obtained: Yo&)
=
4v X 3(2v* - 3v2 + 3)
(1 - v2)[a + CV~)~ - b
sh v(z + z,,).
T(V)
where z0 is the extrapolation length defined in equation (2.12). The actual flux at the boundary is ~~(0) = lim s@,,(s) = SC + k,. s-cc In all of these equations v = 0.91089 and is the root of the following equation: Qs4 +
~2 +
(2s* - 3s2 + 3) 1 - k In E i
1
= 0.
The constants a, b, c, k, are determined from the following relationship: [t2(l +
t>r(Ola + Ml + &_(5‘)]b + [r(l + k(&lc - Q . [l(P - v2)(3 - t2)]k0 = $ . [(P - v2)(3 - [2)], [P(l - 81~ - [t3(1 - Dlb + K4(1- t)]c + $[Ht2 - v2)(3 - E2)@))r(E)]k0
= Q[(% - v2)(3 - E”)~(5)r(~)I. Here [ = i(3 + iz/lS) which is one of the roots of 2s4 - 3s2 + 3 = 0, and 52-I T(Q = m
3 * gi
$54 + E2 + (2E4 - 3t2 + 3)
=
0.48608 - iO.087341.
568
L. M. ROMANOVA
The expression for 7_(t) can be obtained by substituting L$for l/p in the equation for r(l/,u). Using the same method as for the case of pure scattering, namely integration by parts, one obtains ln
(1 s L = !_
7_
l”
where
42
arc tan @ tan x) x
[
=o
N(X) -
1+v2 COP x1 ’
3(1 - VJCOf x
dx
N(x) = 2(1 - v2)cot x x 1 + &[3(1 - xcotx)(l + COPX) - l] + $cot”x[(l - xcotx)(l + 3coPx)(1 + Y2cot2 x)[l - Q cot2 x + 3(1 - x cot x)(1 + + cot2 x(1 + cot x”))]
l]
sin x cos x)[l + -!$cot2 x(1 + cot2 x)] - cot x[3( 1 - x cot x)(1 + 2 cot2 x) - l] sin2 x[l - + cot2 x + 3(1 - x cot x)(1 + $ cot2 x)(1 + cot2 x)]
+3(xand
here 6 = Re (l/E), /? = Im (l/C). The values of the integrals in both the real and imaginary parts of In T_(E) are determined by numerical integration. The result is
1 - v2 n’2cot x arc tan (,Mtan x) dx Tr 1 + v2 COPX s0 1 * arc tan @Y) (1 - v2)Y2 =dY 7r (1 + Y”)(V”+ Y”) Y
s
In r_(t) = O-2462 + iO.03342.
0
1
=-
O”arc tan by)
37
s”
coarc tan by)
1 rr
=-
The following set of equations can now be used to calculate the values of the constants:
Y
Y(1 + Y2)
s0
= g In pp
dy - !_ ?r
1.523~ 0.1843~ 4.804~ -1.778~
s0
(PLACZEK, 1947).
Values for the integral 1 n/2 arc tan @Atan x)N(x) dx 7T -1 0 determined by numerical integration. Setting l/,u = v in the expression for In r(l/p), one obtains the value T_(Y)= l-2942. The expression for In T_(@ is equal to: are
lnr(&
= i 1:
arctan([tanx) x
42
1
ZZZ2rr
0
N(x)
-
C
1
1
fl” ln (1 + B tan x)” + d2 tan2 x (1 - /3 tan x)” + d2 tan2 x s [
”
-ii-
1 4rr
x
N(x) -
TABLE 2.-ANGULAR
-
3(1 - v”) cot x dx I + v2 COPx
3~‘~v;2~~;txx 1dx;
1
1.137 1.018 3.384 2.012
which yields a = 1.419; b = O-1185; c = O-5132 and k, = 0.9135, this leads immediately to the value ~~(0) = l-2557. The results obtained for the neutron angular distribution are given in Table 2. The angular distribution function, normalized to a unit current is given in column 2; in column 3 the same distribution is normalized to a unit neutron flux at the boundary. The angular distribution normalized to a unit neutron flux at the boundary for the case of isotropic scattering is given in column 4 for comparison. The latter distribution is taken from LE CAINE (1950).
3(1 - vqcot x dx 1 + v2 COPX I
[ 26tanx arc tan 1 - (d2 + /3”)tan2x
J^ [
x
N(x) -
- 1.711b - 3.509c + l.O72k, = 29170b + 1.859~ + 0*06572k, = + 11.41b + 5*098c - 8*100k, = + 3*468b - 1.155~ + 5.161ko =
DISTRIBUTION CALCULATION
/J
VW, Lc) --F(O) = 1
Y(O, PI y,(O) = 1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0.1283 0.1546 0.2098 0.2787 0.3813 0.5353 0.7731 1.3653 1.8410 3.2700 7.8130
0.1021 0.1231 0.1670 0.2219 0.3036 0.4262 0.6155 1.0870 1.4660 2.6010 6.2210
-
Y@, Fc) h = 0, a = 0.5, W”(O)= 1
-_--__ 0.2500 0.2964 0.3442 0.4012 0.4732 0.5696 0.7076 0.9239 1.3151 2.2456 7.3616
The Milne problem for a semi-infimte medium
The value for the extrapolation distance was found to be z, = O-7892. For the case of isotropic scattering with the same absorption effect (a = 0*5), z,, = l-4408 (in general z0 --f cc as a ---f 1). It can therefore be seen that the value for the extrapolation distance decreases with the presence of anisotropic scattering. This can be explained by the following argument; each effect, the anisotropic scattering and the absorption, considered independently, increases the extrapolation length; however, the effect of anisotropy in the presence of absorption reduces the effective absorption, as a result, the increase of z,, (as a approaches 1) is retarded in the case of anisotropic scattering. Note
569
also that the same effect is seen in the magnitude of the characteristic root Y, which approaches 1 as a -+ 1; for instance for a = 05, van = O-91089, is seen to be smaller than for the isotropic case where vi*,, = 0.9575. The value obtained for the asymptotic neutron flux is v&z)
= I.505 sh 0*9109(z + 0.7892). REFERENCES
LE CAINEJ. (1950) Canad. J. Res. A28, 3. MARK C. (1947) Phys. Rev. 72, 558. PLACZEKG. (19471 Phvs. Rev. 72.7. PLACZEKG. and &D~L W. (194;) Phys. Rev. 72,550. WIENERN. and HOPF E. (1931) Sitz. Ber. Prems. Akad., Phys.M&h. Klasse 30-32, 696.
Saence
and Teclmology
(Journal
of Nuclear
Energy
Parts A/B)
1963, Vol.
17, pp. 559 to 569.
Pergamon
Press
Ltd.
Printed
in Northern
Ireland
THE MILNE PROBLEM FOR A SEMI-INFINITE MEDIUM WITH ANISOTROPIC SCATTERING AND ABSORPTION OF NEUTRONS”? L. M. ROMANOVA 1. THE MILNE PROBLEM INFINITE MEDIUM WITH SCATTERING
FOR A SEMIANISOTROPIC
THE following problem is investigated : a semi-infinite medium z > 0 is uniformly filled with a non-absorbing material which scatters neutrons without changing their velocity, according to the following law u(cos Y) = 2
[l + gPkos
p1) + hP,(cos y)l
(1)
(where p is the scattering angle; g, h are constants; and P,, Pz are Legendre polynomials). It is assumed that there are no sources in the medium and that no neutrons enter the region z > 0 from the outside. A neutron current is assumed to exist in the scattering medium in the -z direction. The problem consists of determining the steady-state neutron flux distribution in the medium and the angular distribution of neutrons leaving the plane z = 0. For the case of isotropic scattering, the problem can be solved by the Wiener-Hopf (WIENERand HOPF, 1931) method. Their method can also be applied to problems where anisotropic scattering is considered; in our work we will use a variation of the Wiener-Hopf method, given by PLACZEKand SEIDEL(1947), which is more amenable to generalized cases. Let us further say that the mathematical algorism used in this work is not dependent on the actual physics of the radiation and can therefore be applied to other problems in transport theory. The transport equation and some relationships derived from it The transport equation for the problem being considered is as follows:
0
y(z,
B’)o(cos
p)
sin 0’de’ da.
q) =
P,(COS
+2
(1.1)
i
e/j
(n-mm)!pycosejpm
m=L (n + m)! (COS
n
n
et) cos m(p - /?‘),
equation (1.1) becomes
sly
/-J=&
Y(Z9 Pu)=
+
+ hPdP)YdZ)l. (1.2) fl
Here
i-4dp i PICUMZ,
w(z) =
J-1
where yO(z) denotes the scalar neutron flux, and yl(z) the scalar neutron current at point z. After integration of equation (1.2) over d,u between the limits -1 to +l, one obtains dz
0
ey,(cos
x
dy,(z)
vI
=II
P,(COS
-=
aY cmes + y(s, e> 2n
Here, 0 and 8’ are the angles between the neutron velocity vectors and the positive direction of the z-axis, a is the corresponding azimuthal angle, pl is the angle between the initial and scattered neutron velocity vectors, and y(z, 0) is the neutron distribution function. The spatial units are normalized to the neutron mean free path, and the velocity is normalized to the neutron velocity. The density of scattering nuclei is considered to be unity. The expression o(cos p) is given by equation (l), g = 3 cos a, = 3, where cos v is the average cosine of the scattering angle; and 0 G h G 2, since a(cos 9) > 0. Expressing cos 0 as p, and using the relationship for Legendre polynomial multiplication
o
’
yl(z) = F = const.
The scalar neutron current is thus reduced to a constant. Multiplying equation (1.2) by ,Uand integrating as above, one obtains dK = -( 1 - &g)F, K(z) dz
-
* The results of this work were obtained in 1951. 7 Translated by A. LORENZ from the Russian work Some Mathematical Problems in Neutron Physics (1960).
= - (1 - &g> Fz + K(O). 559
(1.3)
L. M. ROMANOVA
560
s +1
K(z) represents
,~?y(z, ,u) d,u. Since K(z) > 0 for
-1 all z, then -(l - &g)I; must be positive; also since (1 - &) > 0, then -F > 0. We obtain the following integral equation for the neutron distribution function in the semi-infinite medium z > 0:
medium is thus reduced to the finding of the Laplace transform of the neutron density. The integral equation for the Laplace transform of the neutron density and the solution of this equation
In order to find C&,(S),we shall use the Placzek and Seidel method. We shall start with the transport equation (1.2) given above, and shall substitute for yr(z) and y*(z) the following relationships YAZ) = F
and Y&) = QNO) obtaining
I+
hP&4wdEN
dE, -1
(1.4)
If we substitute for yr(z) and y&r) = $Y(z) - iyO(z) and integrate equation (1.4) over dp, we obtain a variation of Milne’s integral equation:
p ;
+
(1 - &WI - &w,,(z),
y(z, PI = ; [ 1 -
+ SFgp + f
t
4
2P2W] wow
- ig)Fz + W)IP,cu).
Let us furthermore substitute the Laplace transform and divide the relationship thus obtained by (1 + s,u), where s is a compIex number. The whole expression is then integrated over dp from - 1 to + 1. The resulting relationship is an integral equation in Q,,(s):
{sf(l+;)s2 - $Do(s) - Fgs2 - $FgE,(z)
-
Jwl
- 4
(1 - f g) F[3E,(z)
+ ; jw> [3&(z) - -wN
+ ;
(1.5)
where E,(z) =
Lo $
dt,
[(
z > 0.
u-7) Applying the following boundary conditions
The neutron distribution function y(z, ,u) is obtained from equation (1.5) with the aid of equation (1.4). We will limit our investigations to the determination of the scalar neutron flux yO(z), and the angular distribution of neutrons leaving the plane z = 0. For the latter we obtain from equation (1.4) :
+
- fg)F-SK(O)])
=s2[(l+;)s2-;]G(s)+;Fs2-;K(O)s’.
s1
Y(O,PI = -
X
(9 - 3) x [(I
l [l -gJ2Lu+o(-~)
G
i$Fgp+ ; PA,4 [Cl- 58) FP + W31,
y(0, EC)= 0,
O
we have: G(s) =
’ puw(O,/J) d,u * s -1 1fsCL
The expression (1.6) for ~(0, ,u) is given in terms of @o(- l//J). Let us now introduce the following notation:
(1.6) + T (s2 - 3)[(1 - +g)F - S(O)],
where m e+y,,(z) dz. @‘o(s)= s0 The problem of determining the angular distribution of neutrons leaving the boundary of the scattering
f(s)=
~(+s2[(1
(1 +$-$)(I
-klnz)
+;)s2-;]G(s)+;s’[F-sK(O)l.
-$,
The Milne problem for a semi-infinite medium
561
Equation (1.7) can now be written as :
Let us analyse the properties of this function. The scalar neutron flux can be expressed in the form: y,,(z) = O(e”‘),
k -=c1;
then the function O&r), as well as
FIG. 1
For large values of z it is possible to determine the asymptotic behaviour of yO(z)without solving equation (1.7). This can be expressed in the form of a sum of residues of the function Q,(s) es2 relative to the poles of the function CD,(s). From equation (1.7) we find:
l f’ods = -!var arg f(s) ’ 2%- c
N = 5Z sJ-(s)
where var, argf(s) designates a full increment of arg f(s) after integration along the closed contour; each pole is counted once. Let us consider a square contour C whose sides are parallel to the co-ordinate axis, which lies within the region - 1 < Re (s) < 1, and which includes within its contour the point s = 0 (see Fig. 1). Within this contour,f(s) is analytic, continuous and does not have any poles on the contour itself. (See WIENERand HOPF,Lemma 3 ; within the region - 1 < Re (s) < 1J(s) has only a finite number of poles, it is therefore possible to choose a contour which will satisfy the above conditions.) As s moves along the contour from point A to B in a counter-clockwise direction, the point w = f(s) moves along a closed contour, for even powers off(s). When s moves from B to A in the same direction, the point w = f(s) also moves along a closed contour which corresponds to the former. The point w = 0 is situated within the contour C, therefore var, arg f(s) = 2.2~
and N = 2.
;[F-sK(O)][l
+
-
(1-;)(I -&I;)] S2f(4 (1.9)
This equality shows that (D,,(s)has one second-order pole s = 0. The expansion of QO(s) into a series around s = 0 has the form Q,(s) = -31; and we obtain: w,&)
= -3F(l
- &)[z + 4, zo = -
mu (1 - +g)F ’
It can be seen that K, which indicates the rate of growth of vo(z) to infinity is equal to zero. In order to solve equation (1.7) let us define s2 I(S) = Tf
1
(4.
* A wider region of analyticity will be assigned to the function (s) in the text below. t Translator’s note: The author Wiener-Hopf technique.
makes reference
to the
This function, analytic in the domain - 1 < Re(s) < 1, does not have any poles and approaches unity as Is1--+
L. M. ROMANOVA
562
co in that domain. It is therefore possible (1) to expand the function T(S):
(l.lOa)
Equating this expression for Q,,(s) with that given in equation (1.9) we see that the values s = +1/3/z/(4 + h) are not poles of the function a,,(s). Under these conditions the first term in the numerator of equation (1.13) is therefore equal to zero, and the constants C and K(0) are determined as a function of the current -F: 6
1
U)T_(-U)'
(l.lOb)
(1 + k(u)
T+(S)is analytic in the half-plane Re(s) < ,!I T_(S) is analytic on the half-plane Re(s) > -8. T+(S) and T_(S) do not have poles and approach unity as Is] 3 co in those domains. Equation (1.8) can now be expressed as S-l H(s). CD(s)= S + 1 T_(S) T+(s) 1
s2
(1.14)
’= -F(l+U)T_(U) + (1 -
. ,
(1.11)
K(o)
=
-%[(I
where
+
U)T_(U)
a=
- (1 - a>T-(-a) +
(1
-
(,
.15)
U)T_(-U)]’
3h -. 4+h
J
The real integral T_(U) which is obtained from equation (1.106) by transformation of the contour integral onto the real axis has the form of
The left-hand side of this equasin2 x tion is analytic in the half-plane In Re(s) > 0, and the right-hand 1 + a + ; COPx (1 - x cot x) - $ a IQ’* 1 side is analytic in the half-plane T_(U)= eXp dx . 7T u2 + (1 a”) sin2 x Re (s) < 1. Since these two planes [ UO I intersect, each side is the analytical continuation of the other one. The left-hand side, as From the expressions for T_(S) and T+(S) it can be well as the right-hand side of equation (1.11) established that T_( -s) = 1/[T_+(S)] which leads to the approaches infinity in these half-planes as ]s14. Both equality sides can be expressed in fourth-order polynomials 1 T-(-U) = (1.16) using Liouville’s theorem : T(U)7_(U) ., ’
1
where s-l =
-
(1.12)
H(S).
T+(S) Expanding the function [(s series, we find that
1 -u2h T(U) = - . u2 4
in a Laurent
l)/~.+(s)]H(s)
c, = c, = c2 = c, = 0,
c4
=
c
#
0.
Angular distribution of leakage neutrons From expressions (1.6), (1.13), (1.14), (1.15) and (1.16) one obtains
And we can write equation (1.12) as: cS2(S
@o(s) =
+
l)T_(S)
s2[(l
-
;
(S2 -
3)[F - SK(O)]
+ $),. _ 3$]
T_(_~)
=exp,_:jrl+‘+l~~~~,:“tx)-~.,.
(1.17)
The Milne problem for a semi-infinitemedium Note that the angular distribution of neutrons at the interface is not dependent on the coefficient g. It is therefore seen that the angular distribution of neutrons at the interface will be the same for isotropic and linear scattering. The scalar flux y,,(z)
The scalar neutron flux can be found with the inverse Laplace transform
(1.18) Let us write: H(s) - ; Q,(s) =
(9 - 3)[F - SK(O)]
Fg
+$)s2_?!]
0
s
n
n/2
exP !E [S
K(O)
1
1--Uln2
&_($)P
0 < P < 1,
0
2 sin2 x In
3(1 + COPX)(1 - x cot x) - 1 dx 1 - (1 - p2) sin2 x
1.
The expression In r(l/,~) is solved for in the same manner as it was done for the case of isotropic scattering treated in PLACZEK (1947). Noting that
P
s
dx = arc tan 01 tan x), 1 - (1 - p2)sin2x
the expression for In ~-(1/p) is integrated
e-ziu dp )(
$fJ12)](1 +
=
4a2 (1 + a)e[r(a>12 4 h
0
where
1 i-_ P
(1 + a>r(a)
1
X
Substituting p for -,u in equation (1.17) and assuming -F = 1 and h = 2, an expression for the neutron angular distribution at boundaries is obtained :
*
The solution of the integral analogous to equation (1.18), is given in MARK (1947). The solution for this case is :
- Jg)(z + z0) -
- Sg)(z + %),
where z0 is the extrapolation length. Calculations done for the angular distribution of neutrons at boundaries can be used to solve the integral for cases other than the asymptotic one.
+2.
The function H(s) has a branch point at s = 1 and is analytic in the plane between 1 and co. The function f(s) has branch points at s = 1 and s = -1 and is analytic in the plane between - co and -1 and 1 and co. The function [H(s)/f(s)] and therefore O,,(s) will in all cases by analytic in the half-plane Re(s) < 1, between -co and - 1. The analytic behaviour of @,,(s) in the positive half-plane follows from the determination of $(s) and the asymptotic behaviour of yO(z). The @,(s) is therefore analytic in all of the s plane between - co and - 1, except for the secondorder pole at s = 0.
ye(z) = -3F{(l
%a&) = -3F(l
YK4 I4 =
f(s) $[(I
00 the integral term drops out and one
Calculation of the neutron angular distribution at boundaries and of the extrapolation distance for thecaseofh=2andg=O
eSZ@&)dS, CJ> 0.
Y&) = Aj
When z finds
563
13-P 1-p
(1 + a)“[T_(a)]” - 7
” = - (1 - Qg)F = (1 - *g)a ’(I + a>2[T_(a)12+ 42 ’ h
by parts,
L. M. ROMANOVA
564
yielding : ln,(-l-)
=i
rarctanbtanx) K(0) = 2[T-(l)12 - 1 = o.71486 2[T-(l)12 + 1
JO
X,
6(1 - x cot x) 3x(1 + COPx) - cot x 1 + sin2 x ( dx 3(1 + COPx)(1 - x cot x) - 1 --
3 nl2 cot x . arc tan @ tan x) dx. Vr s0
~~(0)=
TABLE l.-ANGULAR
cot x . arc tan (p tan x) dx
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
is given in LE CAINE(1950). The result has the following form 1 n/2 cot x . arc tan (,u tan x) dx = &In (1 + p). ?r 1 YO The expression for the angular distribution can now be written in the form
2. THE
X
PROBLEM
MEDIUM
6(1 - x cot x) sin2 x
3(1 + cot2 x)(1 - x cot x) - 1
1
dx > 1
O
WITH AND
0~50000 0.62363 O-72518 O-82126 0.91464 I.00639 I.0971 1.1870 1.2794 1.3653 1.4539
FOR
A SEMI-
ANISOTROPIC
ABSORPTION
The problem solved in this section is analogous to the one treated above except for the fact that the medium is assumed to be a neutron absorber; that 1,/l= l/(1 - a), where 1, and 1 are the scattering and total neutron mean-free path in the medium, respectively, and that 0 < a < 1.
0
s-m
MILNE
Y@, P)
h = 0, y,(O) = 1
= 1
0.43532 057690 0.69150 0.79950 0.90450 0.98840 1.1095 1.2105 1.3440 1.4108 1.5098
SCATTERING
aI2 arc tan (p tan x)
3x( 1 + COPx) - cot x 1 +
Ylm
0.74170 0.98295 1.1782 1.3622 1.5410 1.6835 1.8903 2.0624 2.2336 2.4036 2.0724
INFINITE
37_(l)
[,
YK4 P)
Y(O,PI
Jo
1 x exp?7
DISTRIBUTION CALCULATION
-F=l
P
(1 + 2[7-(~)12)d1 + p
1.7038.
Due to the condition that g must equal zero when h = 2 (which follows from the condition o(cos v) > 0), we obtain an extrapolation distance of: z, = 0.71486.
This last operation changes the form of the function to one that is more amenable to numerical integration than the previous one. The solution of the integral
w(O,P) =
67_( 1)
2[r(1)12 + 1 = 1.4834,
C=
The transport equation for the given condition is:
aY
P z
+
Yh Pu)=
9
[we(z)
+ gP,(/J)w2(~) +
@2W
w2(z)l.
(2.1)
The unit of linear dimension is taken to be the neutron mean-free path, while that of the velocity is the neutron velocity. The nomenclature of Section 1 will be maintained. If one integrates equation (2.1) over d,u, one obtains (2.2) Multiplying equation (2.1) by ,u first, and then integrating over dp, one finds that dK -=dz
l-
(2.3)
The Milne problem for a semi-infinite medium
Expressions (2.2) and (2.3) show that in the presence of absorption the scalar neutron current yl(z) is not constant, and that K(z) is a linear function of z. Using the methods of Section 1, the following expression for the neutron angular distribution at a boundary is derived :
565
H(s) =
s2 +:a,,--
l-a
(
3
g
)I
G(s)
*-!$2g +
( )I.
hP,(pP’z -;
(2.4)
x
[F(O)- SK(O)]
Here, @,,(s), @ds),%(s 1are the Laplace transforms of the scalar flux y,Jz), the scalar current yl(z) and of yZ(z) respectively. The relationship between CD,(s)and %(s), and @&) and %( s ) can be found from expressions (2.2) and (2.3). Substitution of the expressions for Q1(-l/~) and @,(-l/p) thus obtained, in equation (2.4), yields: l--a
Y(Q p) = - -
2ru i
1+j
h
(
+
T a(1
ag-
-
Fg)p4)@0(E) [(
x ~Wp + Jm
1
1 -+g
>
- w.3 gp2,
(2.5)
where -F(O) is the value of the current and K(0) the magnitude of K(z) at the boundary z = 0. One can also derive, by methods used in Section 1, using the transport equation (2.1), an integral equation in the form @W(s) = H(s),
(2.6)
where
1+ts4
)
+
s2 9h
+-i-
a
l-
X N
*+w
J-l
f(s)
=
x
a(1
+
(
l--
t) - i + p a(1 l-a 3
gs2,
’
a> *+s
1 g 2+(1--l 1
(ag-
:-:a(1
i
2 ln2s l-s
1 -
l-a -Fg
1
1)
!$-:a(* -yg))pz
4
[i
aF(0)
OPY(O9P)dP
G(s) =
x [l+$+
-$(3+1)/J
@(s) =
-
( --g )I l-a
l-
?
(Do(s) - f
(9 - 3)
g) F(0) - sly(O)] - F(O)g.P,
The solution for y,,(z) must be sought in the form O(e”“),with k < 1, (Do(s)are therefore analytic in the half-plane Re(s) > k; when IsI -+ co in this halfplane CD(s)= 0(ls13). The function H(s) is analytic in the half-plane Re(s) < 1; when IsJ + co in this half-plane H(s) is also equal to O(l~1~). The function ,r(s) is analytic in the region - 1 < Re(s) < 1, and s = 1 and s = - 1 are its branch points. Choosing the branch off(s) which is real for -1 < s < + 1, one sees that in that region f(s) -+ 1 when IsI -+ co, and that it has two real zeros, namely s = +Y and s = -_y. In reality, f(O) > 0; for a real j3 close to one &I) = A-a) < 0, therefore, the function f(s) has two zeros on the real axis: s = f~, 1~1< 1. It can be shown, by means of the argument principle* that no other zeros exist for the function f(s) in the region - 1 < Re(s) < 1. Assume that T(S) = [(s2 - l)/(s2 - 2)]f(s), that the function T(S)is analytic, that it does not have zeros and approaches 1 as IsI -+ cc in the region - 1 < Re(s) < 1; it can therefore be assumed that T(S) = T+(s)/T_(s). The expressions for T+(S) and T_(S) are analogous to the ones given by equations (1 .lOa) and (1. lob). * Translator’s note: Wiener-Hopf technique.
The author
makes reference
to the
566
L. M. ROMANOVA
Equation (2.6) can now be written 3 - Y2 1 S-l a>(s) = H(s) . s + 1 T_(S) T+(S)
(2.7)
The left-hand side of this equation is analytic in the half-plane Re(s) > k, the right-hand side in the half-plane Re(s) < 1. Since these two planes overlap, they form an analytic continuum. Both sides of equation (2.7) approach infinity with a 1~1~dependence as Is] + co, and can be expressed in the form of a fourth-order polynomial with the aid of Liouville’s theorem: Y2 1
s2 -
f-D(s)-
S-l = H(s) = c, + c,s + c,s2 + css3 + 1 7_(s) T+(S)
s +
c4.94.
(2.8)
Expanding the function [H(s)(s - l)]/ T+(s) in t o a Laurent series, one finds C,, = C, = 0, so that equation (2.8) can be written as: (Do(s)= s2(C, + c!$ +
C43)(1
+
S)T_(S)
+
(s2 -
Y”)
(2’9)
x(sLy2)((1 +$4+ [ag-:-;a(1 -+gi]s2+~a(l-~~)) If one equates this expression for @&) with one obtained from equation (2.6): 3h
+ F(O)g 1 - k In z ( -+
neutron angular distribution at z = 0 can be construtted from equations (2.5) and (2.9):
3
s2fis) one finds that the roots of equation
(1 +;gs4+ (ag+;-;a(l
Y(O>P) =
_!$g))s”
--Pb-(-;),
-a&v”-b---)(1
_$
2(1 - /W) +:a
(
l-
l-a 3g
1
=o. lnT_(-_!)
s = &! and s = f5 are not poles of the function
k a),
C4 = -W)c(g,
k a),
C3 = --I;(O)& k a), K(0) = --F(O)k,(g, h, a).
Because of their sizes, the equations for a, b, c and k,, are not reproduced here. The expression for the
= -fj~~~“I;~~~~~~d~,
3h pl(X) = x 1 + h _ _h -qa(l 41 4 ( x
(
l-
X [I
l-a
-3
g cot2 x + (1 - a)(1 - x cot X) )
+t--
X cot2 x +
-a)
(ag-y--Ta(l9h 4
a 1(
l-a 3
qg))
1
g cot4 X . )
(2.10)
The Milne problem for a semi-infinite medium
567
The asymptotic solution for the scalar neutron flux y,,,(z) is obtained from the sum of residues at the poles Q&s = v and s = -v : yo’oas(4= --FPMg,
where z0 is the extrapolation z,(g, h, x) = & ln
of (DO(s) es*
(2.11)
4 4 sh v(z + 4;
distance.
(1 + v)T_(v)(~ + bv + CT+) 1 = %ln (1 - v)T_(-v)(u - bv + cvz)
(1 + ~)(a + bv + CV~))~_(V)(~T(V) (1 - v)(a - bv + cv2) ’
(2.12)
Calculation of the neutron angular distribution at boundaries and of the neutron density forg=O,h=2anda=0.5 If one substitutes ,u for -p in equation (2.10) and sets -F(O) = 1, g = 0, h = 2 and a = 0.5, one obtains an expression for the neutron angular distribution at the boundary: (w” + bp + c)(l + ,+W(O?P) =
In1 -
P/U
lnr
L =1u_ 0P ?I 1
; 0
4(1 - Y2@)
,
O<#u
4(sin2 x + v2 cos2 x) : COPx + 3(1 - X cot x)[l + Q cot2 X(1 + cot2 x)] dx 1 - (1 - p2) sin2 x
Substitution of --F(O) = 1, g = 0, h = 2 and u = 0.5 in equations (2.1 l), (2.12) and (2.13), an expression for the asymptotic neutron flux is obtained: Yo&)
=
4v X 3(2v* - 3v2 + 3)
(1 - v2)[a + CV~)~ - b
sh v(z + z,,).
T(V)
where z0 is the extrapolation length defined in equation (2.12). The actual flux at the boundary is ~~(0) = lim s@,,(s) = SC + k,. s-cc In all of these equations v = 0.91089 and is the root of the following equation: Qs4 +
~2 +
(2s* - 3s2 + 3) 1 - k In E i
1
= 0.
The constants a, b, c, k, are determined from the following relationship: [t2(l +
t>r(Ola + Ml + &_(5‘)]b + [r(l + k(&lc - Q . [l(P - v2)(3 - t2)]k0 = $ . [(P - v2)(3 - [2)], [P(l - 81~ - [t3(1 - Dlb + K4(1- t)]c + $[Ht2 - v2)(3 - E2)@))r(E)]k0
= Q[(% - v2)(3 - E”)~(5)r(~)I. Here [ = i(3 + iz/lS) which is one of the roots of 2s4 - 3s2 + 3 = 0, and 52-I T(Q = m
3 * gi
$54 + E2 + (2E4 - 3t2 + 3)
=
0.48608 - iO.087341.
568
L. M. ROMANOVA
The expression for 7_(t) can be obtained by substituting L$for l/p in the equation for r(l/,u). Using the same method as for the case of pure scattering, namely integration by parts, one obtains ln
(1 s L = !_
7_
l”
where
42
arc tan @ tan x) x
[
=o
N(X) -
1+v2 COP x1 ’
3(1 - VJCOf x
dx
N(x) = 2(1 - v2)cot x x 1 + &[3(1 - xcotx)(l + COPX) - l] + $cot”x[(l - xcotx)(l + 3coPx)(1 + Y2cot2 x)[l - Q cot2 x + 3(1 - x cot x)(1 + + cot2 x(1 + cot x”))]
l]
sin x cos x)[l + -!$cot2 x(1 + cot2 x)] - cot x[3( 1 - x cot x)(1 + 2 cot2 x) - l] sin2 x[l - + cot2 x + 3(1 - x cot x)(1 + $ cot2 x)(1 + cot2 x)]
+3(xand
here 6 = Re (l/E), /? = Im (l/C). The values of the integrals in both the real and imaginary parts of In T_(E) are determined by numerical integration. The result is
1 - v2 n’2cot x arc tan (,Mtan x) dx Tr 1 + v2 COPX s0 1 * arc tan @Y) (1 - v2)Y2 =dY 7r (1 + Y”)(V”+ Y”) Y
s
In r_(t) = O-2462 + iO.03342.
0
1
=-
O”arc tan by)
37
s”
coarc tan by)
1 rr
=-
The following set of equations can now be used to calculate the values of the constants:
Y
Y(1 + Y2)
s0
= g In pp
dy - !_ ?r
1.523~ 0.1843~ 4.804~ -1.778~
s0
(PLACZEK, 1947).
Values for the integral 1 n/2 arc tan @Atan x)N(x) dx 7T -1 0 determined by numerical integration. Setting l/,u = v in the expression for In r(l/p), one obtains the value T_(Y)= l-2942. The expression for In T_(@ is equal to: are
lnr(&
= i 1:
arctan([tanx) x
42
1
ZZZ2rr
0
N(x)
-
C
1
1
fl” ln (1 + B tan x)” + d2 tan2 x (1 - /3 tan x)” + d2 tan2 x s [
”
-ii-
1 4rr
x
N(x) -
TABLE 2.-ANGULAR
-
3(1 - v”) cot x dx I + v2 COPx
3~‘~v;2~~;txx 1dx;
1
1.137 1.018 3.384 2.012
which yields a = 1.419; b = O-1185; c = O-5132 and k, = 0.9135, this leads immediately to the value ~~(0) = l-2557. The results obtained for the neutron angular distribution are given in Table 2. The angular distribution function, normalized to a unit current is given in column 2; in column 3 the same distribution is normalized to a unit neutron flux at the boundary. The angular distribution normalized to a unit neutron flux at the boundary for the case of isotropic scattering is given in column 4 for comparison. The latter distribution is taken from LE CAINE (1950).
3(1 - vqcot x dx 1 + v2 COPX I
[ 26tanx arc tan 1 - (d2 + /3”)tan2x
J^ [
x
N(x) -
- 1.711b - 3.509c + l.O72k, = 29170b + 1.859~ + 0*06572k, = + 11.41b + 5*098c - 8*100k, = + 3*468b - 1.155~ + 5.161ko =
DISTRIBUTION CALCULATION
/J
VW, Lc) --F(O) = 1
Y(O, PI y,(O) = 1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0.1283 0.1546 0.2098 0.2787 0.3813 0.5353 0.7731 1.3653 1.8410 3.2700 7.8130
0.1021 0.1231 0.1670 0.2219 0.3036 0.4262 0.6155 1.0870 1.4660 2.6010 6.2210
-
Y@, Fc) h = 0, a = 0.5, W”(O)= 1
-_--__ 0.2500 0.2964 0.3442 0.4012 0.4732 0.5696 0.7076 0.9239 1.3151 2.2456 7.3616
The Milne problem for a semi-infimte medium
The value for the extrapolation distance was found to be z, = O-7892. For the case of isotropic scattering with the same absorption effect (a = 0*5), z,, = l-4408 (in general z0 --f cc as a ---f 1). It can therefore be seen that the value for the extrapolation distance decreases with the presence of anisotropic scattering. This can be explained by the following argument; each effect, the anisotropic scattering and the absorption, considered independently, increases the extrapolation length; however, the effect of anisotropy in the presence of absorption reduces the effective absorption, as a result, the increase of z,, (as a approaches 1) is retarded in the case of anisotropic scattering. Note
569
also that the same effect is seen in the magnitude of the characteristic root Y, which approaches 1 as a -+ 1; for instance for a = 05, van = O-91089, is seen to be smaller than for the isotropic case where vi*,, = 0.9575. The value obtained for the asymptotic neutron flux is v&z)
= I.505 sh 0*9109(z + 0.7892). REFERENCES
LE CAINEJ. (1950) Canad. J. Res. A28, 3. MARK C. (1947) Phys. Rev. 72, 558. PLACZEKG. (19471 Phvs. Rev. 72.7. PLACZEKG. and &D~L W. (194;) Phys. Rev. 72,550. WIENERN. and HOPF E. (1931) Sitz. Ber. Prems. Akad., Phys.M&h. Klasse 30-32, 696.