---
title: "Measure common flow through particle correlations"
description: "Derive distinct-particle pair correlations and four-particle cumulants, with explicit finite-sample and nonflow limits."
canonical_url: "https://sandboxphysics.com/stories/qgp-hydrodynamics/multiparticle-correlations/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/qgp-hydrodynamics/multiparticle-correlations/"
last_modified: "2026-10-27"
---

[Heavy-ion matter and hydrodynamics](https://sandboxphysics.com/stories/qgp-hydrodynamics/)

FR06 · L3 · Chapter 8 of 9

# Measure common flow through particle correlations

The preferred direction changes from collision to collision. If we average all particle angles in the laboratory, the flow pattern can disappear. Differences between angles within the same event preserve the common directional preference, and allow us to estimate it without knowing each event’s orientation.

Before you beginChapter 7 introduces angular harmonics. You need conditional independence and event averages. Complex numbers below are a compact way to add rotating unit arrows; the real part gives a cosine.

Sandbox Physics · 2026-10-27

## Eliminate the unknown event direction

In one event with a fixed harmonic amplitude, let two particles be independent draws conditional on the common angle $\Psi_n$. Their harmonic means are $\langle e^{in\phi}\mid\Psi_n\rangle=v_ne^{in\Psi_n}$. Multiplying a mean by its complex conjugate removes the common angle.

$$
\left\langle\cos n(\phi_1-\phi_2)\mid v_n,\Psi_n\right\rangle=v_n^2
$$

The equality needs conditional independence. A decay pair or another correlated pair can add a contribution that does not factor into these one-particle means.

Across events, write $c_n\{2\}=\langle v_n^2\rangle+\delta_n$. The extra term $\delta_n$ is called nonflow: angular correlation outside the common single-event modulation. Its effect can be important because the flow contribution itself is a small squared number.

## Remove self-pairs before calculating the mean

For $M$ tracks in an event, add the arrows $Q_n=\sum_{j=1}^M e^{in\phi_j}$. Squaring its magnitude sums every ordered pair. It also includes $M$ self-pairs, each with cosine one. Subtract them before dividing by the number of distinct ordered pairs.

$$
\langle2\rangle=\frac{|Q_n|^2-M}{M(M-1)}
$$

This identity is exact for the saved track sample with uniform weights. It does not require large multiplicity or a known event plane.

When multiplicities vary, the event weighting must be specified. The frozen teaching benchmark keeps multiplicity equal in every event, so arithmetic means and pair-count-weighted means coincide. Detector acceptance is uniform by construction; real analyses need acceptance and efficiency corrections.

## Calculate quadruplets without enumerating them

The saved $Q_{2n}$ arrow sum lets repeated-index contributions be removed from the four-particle expression. For $M>3$, the exact numerator and normalisation are:

$$
\begin{aligned}N_4={}&|Q_n|^4+|Q_{2n}|^2-2\operatorname{Re}\!\left[Q_{2n}(Q_n^*)^2\right]\\&-4(M-2)|Q_n|^2+2M(M-3),\\\langle4\rangle={}&\frac{N_4}{M(M-1)(M-2)(M-3)}.\end{aligned}
$$

The conjugate reverses an arrow’s phase. This algebraic identity excludes every repeated track index; it changes the computational cost, not which quadruplets are being averaged.

The implementation is checked against direct enumeration on small track samples. Such a check is independent of the fast Q-vector calculation and catches missing self-correlation subtractions.

## Four particles distinguish a different moment

Define $\langle\!\langle4\rangle\!\rangle$ by averaging $\cos n(\phi_1+\phi_2-\phi_3-\phi_4)$ over distinct quadruplets and events. The cumulant subtracts two products of the pair average.

$$
c_n\{4\}=\langle\!\langle4\rangle\!\rangle-2\langle\!\langle2\rangle\!\rangle^2,\qquad v_n\{4\}=(-c_n\{4\})^{1/4}\ \ (c_n\{4\}<0)
$$

For a fixed flow amplitude and independent emission, the four-particle moment is the fourth power of that amplitude, so the cumulant is its negative fourth power.

With event-by-event fluctuations, the fourth moment is not the square of the second moment. Four-particle and two-particle estimates therefore need not agree even for pure collective flow. Cumulants suppress particular lower-order correlations; finite multiplicity, correlated clusters and acceptance can leave further contributions.

A positive estimated four-particle cumulant has no real fourth-root flow value under this convention. Report that sign and the sample uncertainty. Taking an absolute value would silently change the statistical quantity.

Worked example

## A small population of doublets shifts the pair signal

The synthetic benchmark contains $M=128$ tracks and four pairs whose two tracks have exactly the same angle. Of $128\cdot127$ ordered distinct pairs, eight belong to these doublets. Their fraction is $d=8/(128\cdot127)\simeq0.00049213$.

$$
\mathbb E[\langle2\rangle]=v_n^2+d(1-v_n^2)
$$

Each doublet orientation is drawn from the same single-particle distribution. A within-doublet pair has cosine one; every pair from separate independent emissions has expectation equal to the squared flow.

For $v_n=0.06$, the pure-flow pair value is $0.0036$, but the doublets raise it to about $0.004090$. Ignoring them gives an apparent pair-flow amplitude of about $0.06395$. The nonflow fraction is tiny compared with one, but appreciable compared with the squared signal.

## Predict, then try the model

If the particles stay fixed and only the assumed nonflow is set to zero, do the saved correlators change?

Open transport inference and compare Planted assumptions with Ignore doublets. Expand Saved events and four-particle diagnostics. Observe that the saved pair and fourth-cumulant estimates stay fixed as the assumed background changes.

[Transport parameter inference ↗](https://sandboxphysics.com/labs/qgp-transport-inference/)

Compare your explanation

The sample always includes the same doublets. Ignoring them changes the inference assumption, not the particles. The saved fourth cumulant retains finite-sample fluctuations and cluster contributions.

## Exercises with explanations

1\. For $M=2$, what does the Q-vector pair formula return?

Read the reasoning

$|Q_n|^2=2+2\cos n(\phi_1-\phi_2)$. Subtracting two and dividing by two returns exactly the one distinct-pair cosine. This is a useful small-sample check.

2\. Why does averaging a single-particle cosine over events with uniformly random event planes erase the pattern?

Read the reasoning

Every preferred direction is equally likely in the laboratory frame, so its complex phase averages to zero. In a within-event pair difference the two phases cancel before the event average is taken.

Correlation estimates require a definition of distinct particles, event weights and backgrounds. Multi-particle correlations also depend on how particles are emitted together.

## Sources and reading scope

1. [Gardim et al., Mapping the hydrodynamic response to the initial geometry in heavy-ion collisions, arXiv:1111.6538v2](https://arxiv.org/html/1111.6538v2)

   Version 2 (5 March 2012), §§II–IV: eccentricity estimators, linear response and nonlinear higher harmonics. The teaching exponential damping law is separately defined, not fitted from its event simulations.

2. [Bilandzic, Snellings & Voloshin, Flow analysis with cumulants: direct calculations, arXiv:1010.0233v2](https://arxiv.org/html/1010.0233v2)

   Version 2 (17 October 2011), Eqs. (3)–(4), (11)–(18): nonflow, Q vectors and exact distinct-particle correlators for uniform acceptance.

[← Connect an initial shape to a final angular pattern](https://sandboxphysics.com/stories/qgp-hydrodynamics/eccentricity-flow-response/) [Infer transport while changing the assumed initial state →](https://sandboxphysics.com/stories/qgp-hydrodynamics/conditional-transport-inference/)