Theoretical analysis of 2n accidentals rates

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Introduction

A given photon pulse may cause multiple neutron-producing reactions, ranging from zero to "infinity" reactions. The number of neutron-producing reactions actually occurring in a given pulse is denoted by the random variable [math]n[/math], and is assumed to follow the Poissonian distribution. Each neutron-producing interaction is said to produce [math]v_{i}[/math] correlated neutrons, where the random variable [math]v_{i}[/math] is the distribution of the number of neutrons produced in a single neutron-producing reaction. Each of the [math]v_i\text{'}s[/math] are independent and identically distributed random variables, so the purpose of the subscript is to distinguish between several distinct neutron-producing interactions which may occur in a single pulse.

The beam has a Bremsstrahlung end point of 10.5 MeV, which energetically allows for only two possible neutron-producing interactions, 1n-knochout and photofission. Thus, [math]v_{i}[/math] is equal to the photofission neutron multiplicity, plus a contribution at [math]v_{i}=1[/math] from 1n-knockout events. The analysis that follows does not need to distinguish between 1n-knockout events and photofission events that emit a single neutron, since in both cases, a single neutron is emitted that is uncorrelated with all other neutrons.

Variable reference

variable Description
[math]n , n_1 , n_2[/math] A random variable for the number of neutron producing reactions occurring in a single pulse. [math]n_1[/math] and [math]n_2[/math] are used to distinguish between the two different [math]n's[/math] of a two pulse pair.
[math]v[/math], [math]v_i[/math] A random variable for the number of correlated neutrons produced by a single neutron-producing reaction in a given pulse. The index only distinguishes between distinct and independent instances of [math]v[/math].
[math]p(d^2)[/math] is as the probability of detecting a pair of two uncorrelated neutrons, provided that we know nothing else about them except that they are uncorrelated..
[math]\lambda[/math] Poissonian mean for the number of neutron-producing interactions per pulse.
[math]M_{SP}[/math] Represents the series [math]\prod_{i\neq j}v_iv_j[/math], which is the number of possible accidentals that exist in a given pulse.
[math]M_{DP}[/math] Represents the series [math](\sum_{i}^{n_1}v_i)\times(\sum_{j}^{n_2}v_j)[/math], which is the total number of two neutron pairs that can be formed across two separate pulses.
[math]T_{SP}[/math] The number of terms of the form [math]v_iv_j[/math] in [math] M_{SP}[/math].
[math]T_{DP}[/math] The number of terms of the form [math]v_iv_j[/math] in [math] M_{DP}[/math].
[math]SP_{acc}[/math] represents to event that an accidental is detected in a single pulse.
[math] DP_{acc}[/math] represents the event that a two-neutron pair is detected when examining events across two separate pulses.

Accidental estimation

The rate of detected accidentals can be estimated from the two-neutron "coincidence" rate of different pulse (DP) events. Under suitable conditions, the rate of accidentals is very close to 1/2 of two-neutron rate in DP events.

During the derivation of this result, a pulse is considered in which there are [math]n[/math] neutron-producing interactions, where each individual reaction produces [math]v_i[/math] correlated neutrons, where [math]i[/math] ranges from 1 to [math]n[/math]. This event is denoted as, [math]v_1...v_n[/math], and its probability by the expression, [math]p(v_1...v_n)[/math].


Assumptions

There are two approximations which are required to make this result valid.

1. Poissonian statistics

The number of neutron-producing reactions occurring per pulse, [math]n[/math], follows the Poissonian distribution. This can be seen as an approximation to the binomial distribution, where the number of independent trials is much greater than the typical number of successes. Each beam electron constitutes an independent binomial trial, where a success is the event that an electron produces a Bremsstrahlung photon, which then makes it's may to the target and induces a neutron-producing reaction. With trillions of electrons per pulse, there were on the order of one photo-nuclear reactions per pulse.

2. Small detection probability

The probability of detecting two specific uncorrelated neutrons in a given pulse, denoted by [math]p(d^2)[/math], is independent of how many other neutrons are emitted in that same pulse. These two neutrons could be among, say, 8 other neutrons which are emitted in the same pulse, and the assumption is that [math]p(d^2)[/math] has the same value as it would if no other neutrons are emitted.

This is only an approximation, however, since each detector can register at most one hit per pulse, and so the effective efficiency of the entire array drops as the number of particles emitted increases. In other words, it is being assumed that neutrons do not "compete" against one another for a chance to be detected. This approximation is justified because each detector covers only 0.5% out of [math]4\pi[/math] sr, and the rate of detected two-neutron coincidences per pulse is on the order of 10E-5, and no triple neutron events were recorded. I am making a point to address this, because it accounts for the fact that the SP/DP accidental ratio is significantly greater than 1/2 for photons, which are strictly accidentals.

This assumption can be expressed mathematically as:

[math]p(d^2|v_1...v_n) = p(d^2)[/math]

As a result, to find the probability of detecting an accidental in a given pulse, simply count all possible accidental pairs in that pulse and multiply the result by [math]p(d^2)[/math]:

[math]p(SP_{acc}) = M_{SP}*p(d^2)[/math]
where,
[math]SP_{acc}[/math] is the event that an accidental is detected.
[math]M_{SP}[/math] is the number of accidentals in a given pulse, in other words, the number of ways to form pairs of two uncorrelated neutrons.

When examining events in pulse pairs, if a neutron event occurs in the same detector in both pulses, the event with the larger time of flight is thrown out. This way, [math]p(d^2)[/math] is the same for any pair of uncorrelated neutrons, whether they were produced during one, or two separate pulses. Thus, the same line of reasoning used for finding the probability of detecting two uncorrelated neutron in single pulse, can be applied to the case of pulse pairs:

[math]p(DP_{acc}) = M_{DP}*p(d^2)[/math]
where,
[math]DP_{acc}[/math] is the event that a neutron is detected in both pulses
[math]M_{DP}[/math] is the total number of two neutron pairs that can be made across two separate pulses.

Combinatorics

Same pulses

The total number of accidentals in a given pulse, [math]M_{SP}[/math] can be calculated directly:

[math]M_{SP} = \sum_{i\neq j}v_iv_j[/math]

For example, consider a pulse with three fission events producing sets of 2, 4, and 3 correlated neutrons. Counting every possible uncorrelated neutron pairs can be done as follows:

[math]2\times 4 + 2\times 3 + 3\times 2 = 20 \text{ pairs}[/math]

If there were 5 fissions producing 2, 5, 2, 1 and 2 neutrons, then we have:

[math] 2\times 5 + 2\times 2 + 2\times 1 + 5\times 2 + 5\times 1 + 2\times 1 = 33 \text{ pairs}[/math]

The most important point to note here, is that the number of terms in such a sequence depends only on the number of reactions, [math]n[/math]:

[math]T_{SP} = \frac{1}{2}n(n-1)[/math]
where [math]T_{SP}[/math] is the number of terms.

The importance of [math]T_{SP}[/math], is that we will ultimately seek the expectation value of the number of accidentals detected per pulse, and as it turns out, this is expressible in terms of [math]E[T_{SP}][/math], [math]E[v][/math], and [math]p(d^2)[/math], where the E denotes the expected value.

The expected number of accidentals existing per pulse (but not necessarily detected), is a sum of products of independent random variables, and by the linearity of expectation, its expected value becomes the sum of the product of the individual expectation values:

\begin{array}{lcl} E[M_{SP}|n] &=& E[\sum_{i\neq j}v_iv_j|n]\\ & = & \overbrace{\bar{v}^2 + \bar{v}^2 + ... +\bar{v}^2}^{T_{SP} \text{ terms}}\\ & = &T_{SP}\bar{v}^2 \\ &=&\frac{1}{2}n(n-1)\bar{v}^2 \end{array}

The expression, [math]E[M_{SP}|n][/math], is known as a conditional expectation, which means that the expectation is computed for some fixed [math]n[/math]. The steps above are using the fact that [math]M_{SP}[/math] is a sum of terms with the form [math]v_iv_j[/math]. In taking the expectation of the sum, each term reduces identically to [math]\bar{v}^2[/math].

Next, the total expectation of [math]M_{SP}[/math] is computed by taking the expectation again, but this time with respect to n:

\begin{array}{lcl} E[M_{SP}] & = & E \left [\frac{1}{2}n(n-1)\bar{v}^2 \right ]\\ & = & \frac{1}{2} \left (E[n^2]- E[n]\right ) *\bar{v}^2 \end{array}

[math]n[/math] is Poissonian, so the variance is equal to the mean, or in other words, [math]E[n] = E[n^2] - E[n]^2[/math]. Substituting this in for [math]E[n][/math] gives:

\begin{array}{lcl} E[M_{SP}] &=& \frac{1}{2}E[n]^2\bar{v}^2 \\ & = & \frac{1}{2} \lambda^2\bar{v}^2 \end{array} where [math] \lambda[/math] is the mean number of neutron producing reactions occurring per pulse. In the section above, it was shown that the probability of actually detecting an accidental in some pulse, denoted by [math]P_{acc}[/math], is proportional to the number of accidentals available times the two-neutron efficiency. To arrive at the final result, we take the expectation of [math]P_{acc}[/math]:

\begin{array}{lcl} E[P_{acc}] &=& E[M_{SP} \times p(d^2)] \end{array}

and since [math]p(d^2)[/math] and [math]M_{SP}[/math] are independent: \begin{array}{lcl} E[P_{acc}] &=& E[M_{SP}] \times E[p(d^2)] \\ & = & \frac{1}{2} \lambda^2 \bar{v}^2 p(d^2) \end{array}

[math]E[p(d^2)][/math] becomes [math]p(d^2)[/math] since it's a constant value. This same variable will appear in the result for DP events, allowing in to cancel in the ratio, SP/DP.

Different pulses

When forming pairs of uncorrelated neutrons across separate pulses, the total number of pairs is given by:

[math]M_{DP} = (\sum_{i=1}^{n_1}v_i)(\sum_{j=1}^{n_2}v_j)[/math]

Here, [math] T_{DP}[/math], or the total number of terms of the form [math]v_iv_j[/math] after expanding, is the product of the number of reactions in each pulse:

[math] T_{DP} = n_1 * n_2 [/math]

So the expected number of two-neutron pairs available in two-pulse events is:

[math] E[M_{DP}] = E[T_{DP}] *\bar{v}^2 [/math]

Since the values of [math]n[/math] from two separate pulse are independent, [math]E[T_{DP}] = \lambda^2[/math]. Finally, the expected number of two-neutron pairs available when looking at DP events is:

[math] E[M_{DP}] = \lambda^2 \bar{v}^2[/math]

Following the same reasoning as for same pulse accidentals, the probability of a two-neutron event in a two pulse pair is:

\begin{array}{lcl} E[DP_{acc}] &=& E[M_{DP} \times p(d^2)] \\ &=& \lambda^2 \bar{v}^2 \times p(d^2) \end{array}