Idiots Guide to Statistics Fission Events

8/6/2019 Idiots Guide to Statistics Fission Events

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LLNL-TR-414245

The Idiot's Guide to the StatisticalTheory of Fission Chains

S. Walston

June 26, 2009


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The Idiots Guide to the

Statistical Theory of Fission Chains

S. Walston trying to understand R. P. Feynman,

F. de Hoffmann, R. Serber, G. I. Bell, W. Hage, D. M. Cifarelli,

N. Snyderman, and M. Prasad

May 14, 2009

1 THE BINOMIAL AND POISSONDISTRIBUTIONS[1]

The number of arrangements, or permutations, ofn objects is n!, since the firstposition can be occupied by any one of the n objects, the second by any of the(n 1) remaining objects, and so on. The number of ordered subsets containingm objects out ofn is, by similar reasoning,

n(n 1) (n m 1) =n!

(n m)!(1)

If we simply ask for the number of subsets containing m objects out ofn with-

out regard to the order in which they appear (number of combinations of nobjects taken m at a time), we must divide the above result by m! since eachcombination may be arranged in m! ways. Thus the number of combinations ofn things taken m at a time is the binomial coefficient

nm

=

n!

m!(n m)!(2)

The result ofn successive flips of a fair coin can be represented by a seriesof letters, each either h or t. For example, hhth t. The probability of anysuch outcome is

12

n; the number of such arrangements with exactly m heads

is again

n!

m!(n m)! = n

m

(3)

so that the probability of getting exactly m heads is

Pn(m) =

nm

1

2

n(4)

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Generalizing slightly, we may ask for the probability of exactly m successesand n m failures in n repetitions of an experiment, if the probability of a

success is p and the probability of a failure is (1p). The answer is easily foundby similar reasoning to be

Pn(m) =

nm

pm(1 p)nm (5)

which is the binomial distribution. Generalizing further, we can inquire aboutan experiment with three possible outcomes rather than simply success andfailure. Let p1 be the probability for the first outcome, p2 the probability forthe second, and 1 p1 p2 the probability for the third; let m1 be the numberof instances of the first outcome, m2 the number of instances of the second, andn m1 m2 the number of instances of the third. The probability would thenbe

Pn(m1,m2) =n!

m1!m2!(n m1 m2)!pm11 p

m22 (1 p1 p2)

nm1m2 (6)

This is the most elementary example of the multinomial distribution.A limiting case of the binomial distribution which is of interest results when

n and p 0 in such a way that the product np = remains finite. Underthe present conditions, with m n

n!

(n m)! nm (7)

(1 p)nm

1

n

n e (8)

Therefore,

P(m; ) =me

m!(9)

which is the Poisson distribution. This is shown in Fig. 1.Using similar reasoning, we can examine the limiting case of the multinomial

distribution which results when n and p1, p2 0 in such a way thatnp1 = 1 and np2 = 2 where both remain finite. Under the present conditions,with m1,m2 n

n!

(n m1 m2)! nm1nm2 (10)

(1 p1 p2)nm1m2

1

1

n

2

nn

e(1+2)

(11)

Therefore,

P(m1,m2; 1,2) =m11

m22 e

(1+2)

m1!m2!(12)

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0 2 4 6 8 10 12 14 16 18 200

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Poisson Distribution with =5

m

Probability

Figure 1: The Poisson distribution gives the probability to count m randomevents if the expected number is . In this example, = 5.

2 INTRODUCTION TOTIME DEPENDENCE

The Poisson distribution gives the probability of counting n events over a givenamount of time t when the events occur independently of one another (i.e.randomly) at an average rate :

Pn(t) = (t)netn!

(13)

Now consider the probability to count n events over time t + t:

Pn(t + t) =[(t + t)]

n e(t+t)

n!(14)

By using the series expansions for binomials and exponentials, this can be rewrit-ten as

Pn(t + t) =et

n!

(t)n + n(t)n1t +

n(n 1)

2!(t)n2(t)2 +

1 t + (t)22!

+

(15)

Rearranging in powers of t and writing only terms up to first order gives

Pn(t + t) =(t)net

n!

(t)net

n!t +

(t)n1et

(n 1)!t + (16)

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= Pn(t) Pn(t)t + Pn1(t)t + (17)

= Pn

(t)(1 t) + Pn1

(t)t + (18)

The first term on the right may be interpreted as the probability of countingn events over time t and zero events over time t; the second term is theprobability of counting n 1 events over time t and one event over time t,and so on.

By rewriting Eq. 18 in a particular way, one can see the definition of thederivative:

limt0

Pn(t + t) Pn(t)

t=

tPn(t) = Pn(t) + Pn1(t) (19)

The probability generating function can be constructed by multiplying Pn(t)by xn and summing over n:

(t, x) =

n=0

Pn(t)xn =

n=0

(t)

n

e

t

n! xn (20)

Multiplying Eq. 19 by xn and summing gives

t

n=0

Pn(t)xn =

n=0

Pn(t)xn +

n=1

Pn1(t)xn1x (21)

t= (x 1) (22)

The initial condition that (0, x) = 1 is obvious since at t = 0, no events wouldhave been counted. The solution to this differential equation is the familiar

(t, x) = e(x1)t (23)

The factorial moments of the count distribution are obtained by differenti-ating this generating function with respect to x and then setting x = 1. Thefirst factorial moment is

x

x=1

= t =

n=1

nPn(t)xn1

x=1

=n=1

nPn(t) (24)

and the second factorial moment is

2

x2 x=1 = (t)2 =

n=2

n(n 1)Pn(t)xn2

x=1

=n=1

n2Pn(t) n=1

nPn(t)

=

n=1

n2Pn(t) t (25)

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0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Probability of Counting the (n+1)th Random Event as a Function of Total Wait Time

t

Probability

n=0

n=1

n=2

n=3

Figure 2: Probability to count the (n + 1)th (i.e. next) random event as afunction of total waiting time t in units of1.

(Note that when n = 1, the two terms cancel, so summing from n = 1 andsumming from n = 2 give the same result.)

The variance is also easy to obtain from Eqs. 24 and 25:

2 =

n=2n2Pn(t)

n=1nPn(t)

2=

(t)2 + t

[t]2 = t

(26)

as expected.The Poisson distribution, Eq. 13, gives the probability of counting n random

events occurring at a rate of per unit time during a measurement interval t.The probability of counting the (n+1)th random event after waiting an amountof time t is just the cumulative distribution function (CDF) of Eq. 13,

CDF(t) =

t0

(x)nex

n!dx

= 1

t

(x)nex

n!dx

= 1 (n + 1, t)n!

(27)

This is shown in Fig. 2. For a process that generates events as a function of timeaccording to a Poisson distribution, the probability distribution of waiting timesfrom an arbitrary starting point (which may be some particular event) to the

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0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Probability of Countingn Random Events as a Function of Measurement Interval

t

Probability

n=0

n=1

n=2

n=3

Figure 3: Probability to count n = k 1 events as a function of measurementinterval t in units of1.

kth event, where we have defined k = n + 1, is then obtained by differentiatingthe CDF (Eq. 27) with respect to t,

d

dt

1

(k,t)

(k 1)!

=

ktk1et

(k 1)!

=ktk1et

(k)

(28)

This is the gamma distribution and is shown in Fig. 3.

3 COUNTING DISTRIBUTIONSFROM FISSION CHAINS

Suppose 1 is the probability to count one neutron from a fission chain, and2 is the probability to count two neutrons coming from the same fission chain.We could ask for the probability of detecting three neutrons each coming fromthree different fission chains and a single instance of two neutrons coming fromthe same fission chain, for a total of five neutrons. By Eq. 12, the answer would

be

P(3, 1) =312e

(1+2)

3!(29)

To get the total probability of detecting five neutrons, there are several pos-sibilities which must each be considered: The five neutrons could come from

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And because independent probabilities multiply, the probability of detecting kneutrons from a single fission chain during the time interval T is simply the

product of these probabilities.Consider a fission chain which produces n neutrons. Of these n neutrons,

suppose a total ofm actually get detected. Of the m total detected neutrons,only k are detected during the finite time interval T (i.e. n m k).

The fission chain neutron multiplicity Pn is the probability that a fissionchain generates n neutrons. The fission chain is assumed to create the neu-trons instantaneously (i.e. the neutrons being detected are not fast neutrons).Because there is no upper bound on the length of the fission chain, we mustconsider all possibilities by summing over n. The details of this quantity willbe dealt with in Section 7.

The probability of detecting exactly m neutrons out of a possible n neutrons,if the probability of detection is , is just the binomial distribution,

Pn(m) =

nm

m(1 )nm (35)

Consider now only the m neutrons which ultimately get detected (neutronswhich one way or another avoid detection were dealt with in Eq. 35). Ifat time t = 0, one starts with m0 neutrons in the system, and after some timeinterval dt, m neutrons remain undetected in the system, then m0 m = dmneutrons were detected during dt. Thus, the probability for a neutron to getdetected in a time interval dt is

dm

m0= dt (36)

where is the neutron detection rate (or alternately 1 is the neutron lifetime

against detection). This differential equation of course has the familiar solutionm(t) = m0et, according to which a neutron created at time s has a proba-bility e(ts) of surviving undetected until time t. The probability of gettingdetected during the infinitesimal interval dt on the other hand is the product ofthe probability that the neutron survives from s until t and the probability thatit is detected during the interval dt, thus e(ts)dt, remembering that inde-pendent probabilities multiply. The probability of detecting k out of the totalm detected neutrons within the infinitesimal time interval dt is just a binomialdistribution in this probability, e(ts)dt (assuming for the moment that allof the n neutrons were created at the same time s),

Pm(k) = mk e

(ts)dtk

1 e(ts)dt

mk

If all of the m neutrons were not created at the same time s, but rather at anytime before dt, then we must integrate over s to get the correct probability. LetR be the rate at which fission chains are initiated. There can be contributionsfrom both spontaneous fission as well as from induced fission. I would preferto avoid the details ofR for now as it ultimately enters into the combinatorial

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moments of the count distribution in a non-trivial way. The probability ofinitiating a fission chain within the time ds is then R ds. The probability of

detecting k out of the total m detected neutrons within the infinitesimal timeinterval dt is then

Pm(k) =

mk

0

e(ts)dt

k 1 e(ts)dt

mkR ds

If the time interval for detection is some finite value T rather than infinitesimal,a further modification becomes necessary:

Pm(k) =

mk

0

T0

e(ts) dt

k 1

T0

e(ts)dt

mkR ds

This, however, only takes account of the neutrons which got created before the

time interval T. The probability of detection for neutrons which got createdduring the time interval T is, by similar reasoning

Pm(k) =

mk

T0

Ts

e(ts)dt

k 1

Ts

e(ts)dt

mkR ds

The total probability to detect k neutrons during a finite time interval T froma total ofm detected neutrons, allowing the neutrons to be created at any timebefore the end of the detection interval, is thus

Pm(k) =

mk

0

T0

e(ts)dt

k 1

T0

e(ts)dt

mkR ds

+

T0

Ts

e(ts)dt

k 1

Ts

e(ts)dt

mkR ds

(37)

There is in principle no limit to the number n of neutrons produced by thefission chain, nor indeed the number m which are ultimately detected exceptthat n m k. The probability of detecting k neutrons from a single fissionchain within the time interval T is thus the product of Pn, Pn(m), and Pm(k)summed over the possible combinations ofn and m:

k(T) =

n=kPn

n

m=kPn(m)Pm(k)

=n=k

Pnn

m=k

nm

m(1 )nm

mk

0

T0

e(ts)dt

k 1

T0

e(ts)dt

mkR ds

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+T

0

T

s

e(ts)dtk

1 T

s

e(ts)dtmk

R ds(38)

For the sake of brevity, let us define the following:

=

T0

e(ts)dt = es

1 eT

(39)

=

Ts

e(ts)dt = 1 e(Ts) (40)

k(T) = R

n=k

Pn

n

m=k

nm m(1 )nm m

k

0

k (1 )mk ds +

T0

k (1 )mk ds

= Rn=k

Pn

0

nm=k

nm

m(1 )nm

mk

k (1 )mk ds

+

T0

nm=k

nm

m(1 )nm

mk

k (1 )

mk ds

(41)

The identity for the product of binomial distributions allows the simplification

nm=k

nm

m(1 )nm

mk

k (1 )mk =

nk

()k(1 )nk

(42)

and similarly for . The equation for k(T) becomes

k(T) = R

0

n=k

Pn

nk

()k(1 )nk ds

+RT0

n=k

Pn

nk

()k(1 )nk ds (43)

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5 COMBINATORIAL MOMENTS OF

THE COUNT DISTRIBUTIONThe qth combinatorial moment Mq of the probability distribution bn(T) is justthe qth factorial moment of bn(T) divided by q!, or more simply

Mq =n=q

nq

bn(T) (44)

It will be useful to note that M1 = c, where c denotes the average number ofcounts recorded per unit time. The factorial moments of bn(T) can be computedfrom the probability generating function, which can be constructed in the usualway by multiplying bn(T) by zn and summing over n, as in Eq. 20. Thefollowing algebraic steps show how to write the probability generating function

in terms of the j in a very concise form:

(z) =n=0

znbn(T) (45)

= e(P

k=1k)

1 + z1 + z2

212!

+ 2

+ z3

313!

+ 12 + 3

+z4

414!

+212

2!+ 13 +

222!

+ 4

+

= e(P

k=1k)

1 + z1 + z2

21

2!+ z22 + z

3 31

3!+ z312 + z

33

+z441

4!+ z4

212

2!+ z413 + z

4 22

2!+ z44 +

= e(P

k=1k)

1 +z1 + z

22 + z33 + z

44 +

+

z2

212!

+ z312 + z413 + z

4 22

2!+

+

z3

313!

+ z4212

2!+

+

z4

414!

+

+

= e(P

k=1k)

1 +z1 + z

22 + z33 + z

44 +

+1

2!

z1 + z

22 + z33 + z

44 + 2

+

= e(P

k=1k)1 +

j=1

zjj+ 1

2!

j=1

zjj

2

+

= e(P

k=1k)e(

P

j=1zjj)

= e(P

k=1(zk1)k) (46)

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The factorial moments are then computed by differentiating Eq. 46 with respectto z. The qth combinatorial moment Mq is thus

Mq =1

q!

dq

dzq

z=1

(47)

This enables one to express Mq in terms of the k, as follows:

M1 =d

dz

z=1

=

k=1

kkzk1

e(P

k=1(zk1)k)

z=1

=k=1

kk =k=1

k1

k (48)

M2 =

k=2

k2

k +

1

2! k=1

k1

k2

(49)

M3 =k=3

k3

k +

k=2

k2

k

k=1

k1

k

+1

3!

k=1

k1

k

3(50)

We can now define the combinatorial moments of the k(T) as

Yq(T) =k=q

kq

k(T) (51)

and Eqs. 48, 49, and 50 can be written as

M1 = Y1 = c (52)

M2 = Y2 +c2

2!(53)

M3 = Y3 + Y2c +c3

3!(54)

The analysis of Eqs. 24, 25, and 26 can be extended to derive the varianceof the fission chain probability distribution bn. The variance can be written as

2b =

n=0

n2bn(T)

n=0

nbn(T)2

(55)

and, analogously to Eqs. 24 and 25,

z

z=1

=

n=1

nbn(T)zn1

z=1

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=

n=1

nbn(T)

2

z2

z=1

=

n=2

n(n 1)bn(T)zn2

z=1

=

n=1

n2bn(T) n=1

nbn(T)

The variance may thus be expressed in terms of derivatives of the probabilitygenerating function as

2b =2

z2

z=1

+

z

z=1

z

2z=1

(56)

Using the definition of the combinatorial moments from Eq. 47,

dq

dzq

z=1

= q!Mq (57)

the variance may be expressed as

2b = 2M2 + M1 M21 (58)

= 2Y2 + c2 + c c2

= 2Y2 + c (59)

Consider again the combinatorial moments of the k(T) as defined in Eq.51, and recall the expression for k(T) from Eq. 43:

Yq(T) =

k=q k

qR

0

n=k

Pn nk ()k(1 )nk ds

+R

T0

n=k

Pn

nk

()k(1 )nk ds

= R

0

n=k

Pn

nk=q

nk

kq

()k(1 )nk ds

+R

T0

n=k

Pn

nk=q

nk

kq

()k(1 )nk ds (60)

This can be simplified further with the identity

nk=q

nk k

q

()k

(1 )nk

= n

q

()q

(61)

thus

Yq(T) = Rn=q

Pn

nq

q

0

q ds +

T0

q ds

(62)

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by considering all the possible configurations the system can be in at time t.For example, there could be n neutrons at time t, and nothing happens during

the time t; there could be n + 1 neutrons at time t and one could withprobability 1 p get absorbed or leak out during t; there could be n+ 1 neutrons at time t and during t one neutron could with probability p induce a fission with probability C which produces neutrons; there could ben+ 2 neutrons at time t and two could get absorbed or leak out during t; andso on. The probability that there are n neutrons in a multiplying medium attime t + t would then just be the sum of each of these probabilities,

Pn(t + t) = Pn(t)

1

t

n

+ (1 p)Pn+1(t) (n + 1)t

+ p

Pn+1(t)C (n + 1 )t

+ (1 p)2Pn+2(t)

n + 2

2

t

2+ (66)

where is the mean neutron lifetime against leakage and absorption (eitherthrough gamma conversion or fission). Thus, in the first term, each of the nneutrons has a probability of (1 t/) of not interacting during the timeinterval t. In the second and third terms, one of the neutrons has a proba-bility t/ of interacting during the interval t, and so on. By expanding thefirst term with the binomial expansion and multiplying through by /t, thederivative can be constructed

limt0

Pn(t + t) Pn(t)

t=

tPn(t)

and the rate equation is found to be

tPn(t) = Pn(t)n

+ (1 p)Pn+1(t) (n + 1)

+ p

Pn+1(t)C (n + 1 ) (67)

As in Eq. 20, let the probability generating function and its first derivativewith respect to x be

f(t, x) =n=0

Pn(t)xn (68)

f

x=

n=1

Pn(t)nxn1 (69)

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And as was done with Eq. 21, we can multiply Eq. 67 by xn and sum to give

t

n=0

Pn(t)xn =

n=1

Pn(t)nxn1x

+n=0

(1 p)Pn+1(t) (n + 1) xn

+ p

n=

Pn+1(t)C (n + 1 )xnx (70)

f

t=

x + (1 p) + p

Cx

f

x(71)

Let us define the coefficient in brackets as

g(x) = x + (1 p) + p

Cx (72)

Its first derivative with respect to x is then

g(x) = 1 + p

Cx1 (73)

Note that g(1) = 0 (because C is assumed to be properly normalized), g(1) =

1 + pwhere

=

C (74)

keff = p (75)The first moment of Eq. 71 is, by the product rule,

2f

tx

x=1

= g(x)f

x

x=1

+ g(x)2f

x2

x=1

(76)

t

n=1

Pn(t)n = (1 + p)n=1

Pn(t)n (77)

This differential equation has the familiar solution

n=1Pn(t)n = e

t(1keff) (78)

This is the probability that a neutron which was created at time t = 0 sur-vives until time t. In the absence of fission, it would just be et/ with thedecay probability being determined only by the neutron lifetime. With fission,the neutron population is replenished with the consequence that the survivalprobability decays more slowly.

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The probability that a neutron induces a fission between times tf and tf+dt isdt/f where f is the neutron lifetime against fission and is longer than the total

neutron lifetime by a factor of 1/p, i.e. p = /f. And because independentprobabilities multiply, the probability that a neutron created at time t = 0survives until time tf, and then induces a fission between times tf and tf + dt is

just et(1keff) dt/f. The total number of induced fissions is on average is then

0

et(1keff) dt/f =

f (1 keff)

=p

(1 keff)(79)

The system multiplication can be defined as

M =1

(1 keff)(80)

and using the definition ofkeff from Eq. 75

p

(1 keff)=

M 1

(81)

7 RATE EQUATION FOR THE FISSIONCHAIN NEUTRON POPULATION

The analysis of Section 6 can be extended by considering the probability that,at time t+ t, there are m neutrons in the multiplying medium and n neutronswhich have left the medium (either through non-fission absorption or leakage).Eq. 66 becomes

Pm,n(t + t) = Pm,n(t)

1

t

m

+ (1 p)Pm+1,n1(t) (m + 1)t

+ p

Pm+1,n(t)C (m + 1 )t

+ (1 p)2Pm+2,n2(t)

m + 2

2

t

2+ (82)

The corresponding rate equation is found to be

tPm,n(t) = Pm,n(t)m

+ (1 p)Pm+1,n1(t) (m + 1)

+ p

Pm+1,n(t)C (m + 1 ) (83)

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The probability generating function then becomes

f(t,x,y) =

m=0

n=0

Pm,n(t)xmyn (84)

Analogously to Eq. 71, we can multiply Eq. 83 by xmyn and sum over m andn,

t

m=0

n=0

Pm,n(t)xmyn =

m=1

n=0

Pm,n(t)mxm1xyn

+

m=0

n=1

(1 p)Pm+1,n1(t) (m + 1) xmyn1y

+ p

m=

n=0

Pm+1,n(t)C (m + 1 )xmxyn (85)

which can be written more simply as

f

t=

x + (1 p)y + p

Cx

f

x(86)

Define the coefficient in brackets as

g(x, y) = x + qy + pC(x) (87)

where

C(x) =

Cx (88)

q = 1 p (89)

Now consider the quantity G(x, y) such that

f

G= g(x, y)

f

xG

x

f

G= g(x, y)

G

x

f

xf

x= g(x, y)

G

x

f

xG

x=

1

g(x, y)

G(x, y) =

dxg(x, y)

(90)

Eq. 86 can now be written as

f

(t/)=

f

G(91)

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8 COMBINATORIAL MOMENTS OF

THE FISSION CHAIN NEUTRONMULTIPLICITY DISTRIBUTION

The qth combinatorial moment of the fission chain neutron multiplicity distri-bution Pn is just the qth factorial moment of Pn divided by q!. The factorialmoments are in turn computed by differentiating Eq. 101 with respect to y. InEq. 93, we assumed there was one neutron in the system at t = 0. This impliesthat, for the probability generating function of Eq. 101, the first fission is aninduced fission. For fission chains initiated by induced fission, the first threefactorial moments are thus

h(y) = (1 p) + pC[h(y)]h(y)

=1 p

1 pC[h(y)](102)

h(y) = pC[h(y)] [h(y)]2

+ pC[h(y)]h(y)

=pC[h(y)] [h(y)]

2

1 pC[h(y)](103)

h(y) = pC[h(y)] [h(y)]3

+ 3pC[h(y)]h(y)h(y) + pC[h(y)]h(y)

=pC[h(y)] [h(y)]

3+ 3pC[h(y)]h(y)h(y)

1 pC[h(y)](104)

From Eqs. 84 and 99, it can be seen that

h(y) =

n=0Pn(t)y

n (105)

and because Pn is assumed to be properly normalized, h(1) = 1. From Eq. 88,

C(x) =

Cx1

C(1) =

C = (1) = (106)

C(x) =

( 1)Cx2

C(1) =

( 1)C = (2) (107)

C(x) =

( 1)( 2)Cx

3

C(1) =

( 1)( 2)C = (3) (108)

Applying Eq. 106 to Eq. 102, we have

h(1) =1 p

1 p= Me (109)

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where Me is variously known as the escape multiplication or leakage multipli-cation of the system. It is useful to note that

p

1 p=

Me 1

1=

M 1

(110)

Setting y = 1 in Eq. 103 and applying Eqs. 106, 107, 109, and 110, we have

h(1) =p(2)M

2e

1 p

= M2eMe 1

1(2) (111)

Setting y = 1 in Eq. 104 and applying Eqs. 106, 107, 108, 109, and 110, wehave

h(1) =p(3)M3e1 p

+3p(2)Me

1 pp(2)M2e

1 p(112)

= M3eMe 1

1(3) + 3M

3e

Me 1

1

22(2) (113)

The corresponding combinatorial moments of Pn for chains initiated by inducedfission are

n=1

Pn

n1

=

h(1)

1!= Me (114)

n=2

Pnn2 =

h(1)

2!= M2e

Me 1

1

(2)

2!= M2e

Me 1

12 (115)

n=3

Pn

n3

=

h(1)

3!

= M3eMe 1

1

(3)3!

+ M3e3

3!

Me 1

1

2(2!)2

(2)2!

2

= M3eMe 1

13 + 2M

3e

Me 1

1

22

2 (116)

= M3eMe 1

1

3 + 2

Me 1

12

2

(117)

where

==

C =

()!

(118)

are the combinatorial moments of the neutron multiplicity distribution for in-duced fission. The two terms in Eq. 116 correspond to the three diagrams in

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Figure 4: Distinct chains which produce three neutrons starting from an inducedfission.

Fig. 4 (the two lower diagrams contain the same factors which accounts for the2 multiplying the second term).

The diagrams in Fig. 4 suggest something analogous to Feynman diagrams,and they can be drawn more stylistically as in Fig. 7. We define torepresent a neutron which does not induce another fission and is thus availablefor detection and

to represent an induced fission. As with Feynman diagrams,

there are rules which translate the diagrams into mathematics. By comparingEq. 116 with Fig. 7, it can be seen that the rules are as follows:

= Me (119)

=Me 1

1 (120)

where is the number of lines coming out of the fission. And as with Feynmandiagrams, the final result is the sum over all diagrams. Figs. 5, 6 and 7 can beseen to correspond to Eqs. 114, 115, and 116 respectively.

Including chains initiated by a spontaneous fission complicates the situationsomewhat, however the rules in Eqs. 119 and 120 still apply. To these must beadded one additional rule: Let

represent a spontaneous fission. Spontaneous

fission is the same as induced fission except that there is no factor Me11 sincethere is no incoming neutron, and the combinatorial moments of the neutronmultiplicity distribution for spontaneous fission,

S =

=

CS (121)

must be used instead of those for induced fission from Eq. 118. Thus,= S (122)

Applying these rules to the diagrams in Figs. 8, 9, and 10, we can determine the

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Figure 5: An induced fission which produces one neutron is indistinguishablefrom a single neutron doing nothing.

Figure 6: Distinct chain which produces two neutrons starting from an inducedfission.

Figure 7: Distinct chains which produce three neutrons starting from an inducedfission.

Figure 8: Distinct chain which produces one neutron starting from a sponta-neous fission.

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Figure 9: Distinct chains which produce two neutrons starting from a sponta-neous fission.

Figure 10: Distinct chains which produce three neutrons starting from a spon-taneous fission.

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combinatorial moments of the fission chain neutron multiplicity distribution PSnfor chains initiated by a spontaneous fission. The corresponding combinatorial

moments of PSn are

n=1

PSn

n1

= MeS1 (123)

n=2

PSn

n2

= M2e S2 + M

2e S1

Me 1

12 (124)

= M2e

S2 +

Me 1

1S1 2

(125)

n=3

PSn

n3

= M3e S3 + M

3e S1

Me 1

13 + 2M

3e S2

Me 1

12

+2M3e S1Me 1 1

22

2 (126)

= M3e

S3 +

Me 1

1(S1 3 + 2S2 2)

+2

Me 1

1

2S1 2

2

(127)

Let us now reconsider the factor R from, most notably, Eq. 62 among others.Let

R =

FI Induced fission rateFS Spontaneous fission rate

(128)

where FI is a function of the neutron flux either from external sources or fromwithin the object, e.g. (, n). We can define the leading term on the right handside of Eq. 62 as

Rq = FI

n=q

Pn

nq

+ FS

n=q

PSn

nq

(129)

Summing the contributions from chains initiated by both induced and sponta-

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neous fission, we find that

Rq =

FIMe + FSMeS1 q = 1

FIM2e

Me 1

12

+FSM2e

S2 +

Me 1

1S1 2

q = 2

FIM3e

Me 1

1

3 + 2

Me 1

12

2

+FSM3e

S3 +

Me 1

1(S1 3 + 2S2 2)

+2Me 1

12

S1 22

q = 3

(130)

9 EXTRACTING USEFUL INFORMATION

FROM THE COUNT DISTRIBUTIONS

Let Nbe the number of time intervals of duration T which are examined, and letBn(T) be the number of those time intervals in which n neutrons were detected.So in other words, suppose that during the first time interval, six neutrons werecounted; B6 would be incremented by one. During the next time interval, sayeight neutrons were counted; B8 would be incremented by one, and so on forall N time intervals. In this way, the count distribution Bn(T) is built up.

The probability distribution bn Bn/N is just the probability of counting nneutrons during a time interval of duration T. The total number of neutronscounted during all N time intervals is

nTotal =

n=0

nBn (131)

As a practical matter, the combinatorial moments Mq of the distributionbn(T) are easy to compute. From Eqs. 52, 53, and 54, the quantities Yq can beexpressed in terms of Mq as

Y1 = M1 = c =nTotalN

(132)

Y2 = M2 c2

2! (133)

Y3 = M3 Y2c c3

3!

= M3 M2c +c3

3(134)

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By applying Eqs. 65 and 129 to Eq. 62 we also know that

Y1 = R1T (135)

Y2 = R22

T

1 eT

(136)

Y3 = R33

T

3 4eT + e2T

2

(137)

It is convenient to define the following:

Y2F =Y2Y1

(138)

R2F =R2R1

(139)

Y3F =Y3Y1

(140)

R3F =R3R1

(141)

Applying these definitions to Eqs. 136 and 137, the quantities normally used inthe analysis are found to be

Y2F = R2F

1

1 eT

T

(142)

Y3F = R3F2

1

3 4eT + e2T

2T

(143)

or, as expressed in terms of the combinatorial moments of the count distribu-tions,

Y2F =M2

c

c

2!(144)

Y3F =M3

c M2 +

c2

3(145)

It is worth noting that

limT0

Y2F = 0 (146)

limT0

Y3F = 0 (147)

limT

Y2F = R2F (148)

limT

Y3F = R3F2 (149)

and the dependence of Y2F and Y3F on the duration of the time interval T is

shown in Fig. 11.As an example, consider a 4.48 kg ball of-phase Pu comprised of 94% 239Pu

and 6% 240Pu with keff = 0.77 and Me = 3.23. The induced fission neutronmultiplicity distribution for 239Pu is shown in Fig. 12 and the spontaneousfission neutron multiplicity distribution for 240Pu is shown in Fig. 13. Theapproximation is made that 239Pu does not undergo spontaneous fission and

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-2 -1 1 2 3

0.2

0.4

0.6

0.8

1

log10(T)

Y2F

R2F

Y3F

R3F2

Figure 11: Dependence ofY2F (purple) and Y3F (blue) on the duration of thetime interval T.

0 1 2 3 4 5 6 7 80

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Induced Fission Neutron Multiplicity Distribution for Pu239

Number of Neutrons

Probability

Figure 12: Induced fission neutron multiplicity distribution for 239Pu.

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0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Spontaneous Fission Neutron Multiplicity Distribution for Pu240

Number of Neutrons

Probability

Figure 13: Spontaneous fission neutron multiplicity distribution for 240Pu.

240Pu does not undergo induced fission. Thus, no fission chains are initiated byinduced fission and FI = 0.

240Pu produces spontaneous fission neutrons at therate of 1020 g1s1, making FS = 1.26 10

5 s1 for this example.Suppose the bare Pu ball is placed near a neutron detector along with an

amount of moderator such that 1 = 40 s, and the detector is allowed tocount for 100 ms. The 100 ms total counting time is then divided into 100ms/T time gates with T = 1, 2, 3 512 s and the probability distributionsbn

(T) are determined for each T. The probability distribution for T = 10 sis shown in Fig. 14. For each bn(T), Y2F(T) and Y3F(T) are computed. Thetime dependence of these values follow the functional forms in Eq. 142 and Eq.143, respectively, and can be fit to extract , R2F and R3F

2. This is shown inFigs. 15 and 16 respectively. Also shown on these plots are the true values ofY2F(T) and Y3F(T) calculated using Eq. 130 from the moments of the neutronmultiplicity distributions in Figs. 12 and 13, and the values listed above for FI,FS, and Me.

If the efficiency of the detector is known a priori, and if reasonable as-sumptions about FI, FS, and the neutron multiplicity distributions (e.g. Fig.12 and/or 13) can be made, then the fits to Y2F(T) and Y3F(T) in Figs. 15 and16 and consequent values for R2F and R3F

2 can be used to estimate the mul-tiplication by solving Eq. 130 for Me. The fitted value for also affords some

insight as to the amount of moderation present; larger values for the neutronlifetime, 1, tend to indicate more moderation.

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0 10 20 30 40 50 6010

5

104

103

102

101

100

Probability to Count n Neutrons in a 10 s Time Period

Number of Neutrons Counted 10 s

Probability

BeRP BallPoisson

Figure 14: The probability distribution bn with T = 10 s for a 100% efficientneutron detector (i.e. = 1) near the 4.48 kg Pu Ball. There is moderatorpresent such that 1 = 40 s. The detector was allowed to count for 100 ms.A Poisson distribution with the same count rate is shown for comparison.

0 50 100 150 200 250 300 350 400 450 5000

2

4

6

8

10

12

14

16

18

Y2F

vs. T

T (s)

Y2F

/

Data

Fit

Theory

Figure 15: Y2F/ as a function ofT from 1 s to 512 s. For each point, the 100ms total counting time was divided into 100 ms/T time gates and the probabilitydistribution bn(T) was determined for each T. Y2F was then computed fromeach bn(T) (Data). The data was fit to determine and R2F (Fit). The truevalue (Theory) is shown for comparison.

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0 50 100 150 200 250 300 350 400 450 5000

100

200

300

400

500

600

Y3F

vs. T

T (s)

Y3F

/

2

Data

FitTheory

Figure 16: Y3F/2 as a function ofT from 1 s to 512 s. For each point, the 100

ms total counting time was divided into 100 ms/T time gates and the probabilitydistribution bn(T) was determined for each T. Y3F was then computed fromeach bn(T) (Data). The data was fit to determine and R3F

2 (Fit). The truevalue (Theory) is shown for comparison.

10 FURTHER READING

The following references are useful in developing an understanding of the sta-tistical theory of fission chains.

References

[1] Matthews, J. and Walker, R. L. (1970). Mathematical Methods of Physics,Second Edition. Menlo Park: Addison-Wesley Publishing Company, Inc.

[2] Feynman, R. P. (1946). Statistical Behavior of Neutron Chains, LosAlamos Report LA-591 (classified).

[3] Feynman, R. P., de Hoffmann, F., and Serber, R. (1956). Dispersion ofthe Neutron Emission in U-235 Fission, Journal of Nuclear Energy 3, 64.

[4] Bell, G. I. (1963). Probability Distribution of Neutrons and Precursors in

a Multiplying Assembly. Annals Phys. 21, 243.

[5] Hage, W. and Cifarelli, D. M. (1985). Correlation Analysis with NeutronCount Distributions in Randomly or Signal Triggered Time Intervals forAssay of Special Nuclear Materials, Nucl. Sci. Eng. 89, 159.

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[6] Bohnel, K. (1985). The Effect of Multiplication on the Quantitative De-termination of Spontaneously Fissioning Isotopes by Neutron Correlation

Analysis, Nucl. Sci. Eng. 90, 75.

[7] Hage, W. and Cifarelli, D. M. (1985). On the Factorial Moments of theNeutron Multiplicity Distribution of Fission Cascades, Nucl. Instr. andMeth. A236, 165.

[8] Hage, W. and Cifarelli, D. M. (1986). Models for a Three-Parameter Anal-ysis of Neutron Signal Correlation Measurements for Fissile Material As-say, Nucl. Instr. and Meth. A251, 550.

[9] Snyderman, N. and Prasad, M. (2005). Appendix A: Theory of FissionChains and Count Distributions, Lawrence Livermore National LaboratoryTechnical Report UCRL-TR-218042.

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