Derivation Notes on Particle Correlation in Particle Distribution Function

Derivation Notes on 3 Particle Correlation using Particle Distribution Function

Mixed event subtraction
Conventions: we use particle number $$N$$, instead of density $$\rho$$,  and  $$N_{ev}$$  is the total number of events, events are differentiated by  $$(i,j)$$ ,  $$N_t^T$$  is the total number of trigger of all the events. $$N_a^{i}(\phi)\equiv N^{i}(\phi)$$ is the number of associated particles of event $$ i $$ emitted at angle  $$\phi$$. $$N_{ta}^{i}(\phi,\phi+\Delta\phi_1)\equiv N^{i}(\phi,\phi+\Delta\phi_1)$$ is the number of particles pairs of event  $$i$$, a pair is to have the a trigger particle emitted at angle  $$\phi$$  and an associated particle at angle  $$\phi+\Delta\phi_1$$ $$N_{taa}^{i}(\phi,\phi+\Delta\phi_1,\phi+\Delta\phi_2)\equiv N^{i}(\phi,\phi+\Delta\phi_1,\phi+\Delta\phi_2)$$ is the number of particle triples of event $$i$$, a triple is to have a trigger particle emitted at angle  $$\phi$$  and two associated particles at angle  $$\phi+\Delta\phi_1$$ and $$\phi+\Delta\phi_2$$.

Self-correlation is not explicitly written down, only mixed event subtractions, please check

Terms like


 * $$\begin{align}

\frac{1}{N_{ev}N_t^T}\int d\phi\sum_{i,j}N^{i}(\phi,\phi+\Delta\phi_1)N^{j}(\phi+\Delta\phi_2) \end{align}$$

where $$N^{i}(\phi,\phi+\Delta\phi_1)$$  is the number of particles pairs of event  $$i$$, a pair is to have the triggers emitted at angle  $$\phi$$  and associated particles at angle  $$\phi+\Delta\phi_1$$ ,  $$N^{j}(\phi+\Delta\phi_2)$$  is the number of associated particles of event $$ j $$ emitted at angle  $$\phi+\Delta\phi_2$$. Since the summation of $$i, j $$ indiscriminatingly runs over all events, the above expression has  $$N_{ev}^2$$  pairs of events and  $$N_{ev}$$  of them are proper-event-pairs. The advantage of above expression is that the two summation procedures can be factorized, therefore calculated when one treats every single event, instead of processing all the events simultaneously. The problem is that summation should be carried out only for different event paris ( $$i\ne j$$ ), contribution from the same event ($$i=j$$)  should not be taken into account. Fortunately, this can be done as follows without increasing numerical labor


 * $$\begin{align}

&\frac{1}{N_{ev}N_t^T}\\ &\rightarrow\frac{1}{(N_{ev}-1)N_t^T} \int d\phi\sum_{i,j}N^{i}(\phi,\phi+\Delta\phi_1)N^{j}(\phi+\Delta\phi_2)  \\ &\rightarrow\int d\phi(\sum_{i,j}N^{i}(\phi,\phi+\Delta\phi_1)N^{j}(\phi+\Delta\phi_2)-\sum_{i}N^{i}(\phi,\phi+\Delta\phi_1,\phi+\Delta\phi_2)) \end{align}$$

Another term is little bit more complicated


 * $$\begin{align}

\frac{1}{N_{ev}^2N_t^T}\int d\phi\sum_{i,j,k}N^{i}(\phi)N^{j}(\phi+\Delta\phi_1)N^{k}(\phi+\Delta\phi_2) \end{align}$$

The conventions are exactly the same as above, one needs to


 * $$\begin{align}

&\frac{1}{N_{ev}^2N_t^T}\\ &\rightarrow\frac{1}{(N_{ev}^2-3N_{ev}+2)N_t^T} \int d\phi\sum_{i,j,k}N^{i}(\phi)N^{j}(\phi+\Delta\phi_1)N^{k}(\phi+\Delta\phi_2)\\ &\rightarrow \int d\phi\left(\sum_{i,j,k}N^{i}(\phi)N^{j}(\phi+\Delta\phi_1)N^{k}(\phi+\Delta\phi_2)-\sum_{ij}N^{i}_{TA}N^{j}_A-\sum_{ij}N^{i}_{TA}N^{j}_A-\sum_{ij}N^{i}_{AA}N^{j}_T+2\sum_iN^{i}_{TAA}\right) \end{align}$$

Self-correlation

 * One should consider self-correlation effect even one uses density distribution instead of real particles.
 * When the momentum thresholds of trigger and associated particle overlap, it is very complicated... imagine 3$$\times$$ or  2 $$\times$$  (trigger-associated, associated-associated) cases to discuss for two-particle terms in the code, and more cases for the (fortunately unique) three particle term. Obviously there is no easy way to get around?
 * Consider the simple case when trigger and associated particle have exactly the same momentum interval. Some thought shows that self-correlation actually depends on the fluctuation, besides the particle distribution function. For example


 * $$\begin{align}

\bar{f}=1.6 \end{align}$$

Let assume one has 40%  chance to emit 1 particle and  60%  chance to emit 2 particles. An intuitive approach give the correlation


 * $$\begin{align}

C= 2 \times 1\times 60%+1\times 0\times 40%=1.2 \end{align}$$

If the distribution number is


 * $$\begin{align}

\bar{f}=n.m\equiv(n+m) with  n\in N, 0(n.m)(n.m-1) \end{align}$$

Therefore we overestimated the correlation comparing to the result obtained by natively subtracting 1. However, everything is based on how the probability distributed around the mean value $$\bar{f}$$


 * If one simply does not consider it, sometime self-correlation cancels out little bit in 3 cumulant cross terms. In the above formulae, self-correlation has not been considered. However, for instance, in the second factor of $$c3$$ term $$N^i(\phi)N^j(\phi+\Delta\phi_1,\phi+\Delta\phi_2)$$, one should have considered self-correlation, and in the above term  $$N^{i}_{AA}N^{j}_TN^{i}_{AA}N^{j}_T$$, one should not, since it was not considered in  $$\sum_{i,j,k}N^{i}(\phi)N^{j}(\phi+\Delta\phi_1)N^{k}(\phi+\Delta\phi_2)$$. It implies another 2 case discussion for any two-particle term... really too much to handle.

The follow results are for NEXUS+SPheRIO sj1-sja, the code was implemented considering 2(3) and 2(3) was avoided by considering non-overlapping trigger and associated particles. It seems that the result is not smooth when using the above self-correlation subtraction. Big question marks...
 * Implementation