Derivation Notes on Flow Decomposition of Three Particle Correlations

Derivation Notes on Flow Decomposition of Particle Correlations - The possible correlation between even and odd event planes

Two particle correlation and cumulant with number fluctuations
Assuming the following one particle azimuthal distribution in terms of Fourier series


 * $$\begin{align}

\frac{dN}{d\phi}(\phi)=\frac{N}{2\pi}(1+\sum_{n=1}^{\infty}2v_n\cos(n(\phi-\Psi_n))) \end{align}$$

Two particle correlation for one event is defined as


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle\equiv \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi) \\ &=\int \frac{d\phi}{2\pi}\frac{N}{2\pi}(1+\sum_{m=1}^{\infty}2v_m\cos(m(\phi-\Psi_m)))\frac{N}{2\pi}(1+\sum_{n=1}^{\infty}2v_n\cos(n(\phi+\Delta\phi-\Psi_n))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(1+2\sum_{m=1}^{\infty}2v_m\cos(m(\phi-\Psi_m))+4\sum_{m,n=1}^{\infty} v_m v_n \cos(m(\phi-\Psi_m))\cos(n(\phi+\Delta\phi-\Psi_n))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(1+4 \sum_{m,n=1}^{\infty} v_m v_n \cos(m(\phi-\Psi_m))\cos(n(\phi+\Delta\phi-\Psi_n))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(1+ 4 \sum _{m,n=1}^{\infty} v_m v_n \frac{1}{2} [\cos((m+n)\phi+n\Delta\phi -(m\Psi_m+n\Psi_n)) + \cos((m-n)\phi-n\Delta\phi -(m\Psi_m-n\Psi_n))])) \\ &=\frac{1}{2\pi}  \int d\phi(\frac{N}{2\pi})^2(1+ 4 \sum _{m,n=1}^{\infty} v_m v_n \frac{1}{2} [\delta_{m+n}\cos((m+n)\phi+n\Delta\phi -(m\Psi_m+n\Psi_n)) + \delta_{m-n} \cos((m-n)\phi-n\Delta\phi -(m\Psi_m-n\Psi_n))]))  \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 ( 1 + 4 \sum_{m,n=1}^{\infty} v_m v_n \frac{1}{2} \delta_{m-n} \cos((m-n) \phi-n\Delta\phi -(m\Psi_m-n\Psi_n)))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(1+ 4 \sum _{n=1}^{\infty} \frac{1}{2}v_n^2 \cos(-n\Delta\phi))  \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(1+ 4 \sum _{n=1}^{\infty} \frac{1}{2}v_n^2 \cos(-n\Delta\phi))  \\ &=(\frac{N}{2\pi})^2(1+  \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi)) \end{align}$$

This is a famous result shown in many papers. If one further carries out an average on $$\Delta\phi$$, one may obtain another observable  $$\left\langle\cos n\Delta\phi\right\rangle$$  introduced by many authors


 * $$\begin{align}

&\left\langle\cos n\Delta\phi\right\rangle\equiv \int \frac{d\Delta\phi}{2\pi}\cos(n\Delta\phi)\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle/\int \frac{d\Delta\phi}{2\pi}\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle  \\ &=\int \frac{d\Delta\phi}{2\pi}\cos(n\Delta\phi)\int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi) /\int \frac{d\Delta\phi}{2\pi}\int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi) \\ &=\int \frac{d\Delta\phi}{2\pi}\cos(n\Delta\phi)(\frac{N}{2\pi})^2(1+ \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi))  /\int \frac{d\Delta\phi}{2\pi}(\frac{N}{2\pi})^2(1+  \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi)) \\ &=(\frac{N}{2\pi})^2v_n^2/(\frac{N}{2\pi})^2 \\ &=v_n^2 \end{align}$$

Now we introduce event-by-event fluctuation. For the proper event, one has


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle_{proper}\equiv \left\langle\int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi)\right\rangle \\ &=\left\langle(\frac{N}{2\pi})^2(1+ \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi))\right\rangle  \\ &=\frac{\left\langle N^2\right\rangle}{(2\pi)^2}+ \sum _{n=1}^{\infty} \frac{1}{2\pi^2}\left\langle(Nv_n)^2\right\rangle \cos(n\Delta\phi) \end{align}$$

For mixed events, one aligns the event plane $$\Psi_2$$  of the two selected events. Since $$\Psi_{2n+1}^A (n\ge 1)$$  of event $$A$$ is generally not correlated to  $$\Psi_{2n+1}^B$$  (the same $$n$$) neither to  $$\Psi_{2n'}^B$$  (any $$n'$$) of another event $$B$$ (In other words, event plane of $$v_1, v_2, v_4, v_6, v_8 \cdots$$ might be correlated, but $$v_5, v_7, v_9$$ are probably not correlated to others.)


 * $$\begin{align}

&\left\langle v_{2n+1}^{A}v_{2n+1}^{B} \cos(-(2n+1)\Delta\phi -(2n+1)(\Psi_{2n+1}^A-\Psi_{2n+1}^B))\right\rangle_{A,B}=0 (n\ge 1) \\ &\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle_{mixed}\equiv \int \frac{d\phi}{2\pi} \left\langle\frac{dN}{d\phi}(\phi)\right\rangle_{}\left\langle \frac{dN}{d\phi}(\phi+\Delta\phi)\right\rangle \\ &=\frac{\left\langle N\right\rangle^2}{(2\pi)^2}+ \frac{1}{2\pi^2}\left\langle Nv_{1}\right\rangle^2 \cos(\Delta\phi)+ \sum_{n=1}^{\infty} \frac{1}{2\pi^2} \left\langle Nv_{2n}\right\rangle^2 \cos(2n\Delta\phi) \end{align}$$

Therefore, cumulant subtraction will only automatically remove the effects of $$v_2$$  and the majority of  $$v_4, v_6\cdots$$

Three particle correlation without number fluctuations
Now we move to three particle correlation. Similarly, one defines


 * $$\begin{align}

\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle\equiv \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi_1)\frac{dN}{d\phi}(\phi+\Delta\phi_2) \end{align}$$ Similar but lengthy derivation shows that


 * $$\begin{align}

\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle=A+B+C \end{align}$$ where


 * $$\begin{align}

&A=(\frac{N}{2\pi})^3 \\ &B=(\frac{N}{2\pi})^3[ \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{\infty} 2v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \\ &C=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{\infty}v_mv_nv_l\cos(n(\phi-\Psi_n))\cos(m(\phi+\Delta\phi_1-\Psi_m))\cos(l(\phi+\Delta\phi_2-\Psi_l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi}\sum_{m,n,l=1}^{\infty}v_mv_nv_l \frac{1}{4}(F(n,m,l)+F(n,m,-l)+F(n,-m,l)+F(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{\infty}v_mv_nv_l \frac{1}{4}(\delta_{n+m+l}F(n,m,l)+\delta_{n+m-l}F(n,m,-l)+\delta_{n-m+l}F(n,-m,l)+\delta_{n-m-l}F(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{\infty}v_mv_nv_l \frac{1}{4}(\delta_{n+m-l}F(n,m,-l)+\delta_{n-m+l}F(n,-m,l)+\delta_{n-m-l}F(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n=1}^{\infty}\frac{1}{4}(T_2+T_3+T_4) \\ &=(\frac{N}{2\pi})^3 \sum_{m,n=1}^{\infty}2(T_2+T_3+T_4) \end{align}$$

where


 * $$\begin{align}

F(n,m,l)=\cos[(n+m+l)\phi+m\Delta\phi_1+l\Delta\phi_2-(n\Psi_n+m\Psi_m+l\Psi_l)] \end{align}$$

and


 * $$\begin{align}

&T_2=v_mv_nv_{m+n} \cos(m\Delta\phi_1-(m+n)\Delta\phi_2-(n\Psi_n+m\Psi_m-(m+n)\Phi_{m+n})) \\ &T_3=v_mv_nv_{m+n} \cos(-(m+n)\Delta\phi_1+m\Delta\phi_2-(n\Psi_n-(m+n)\Psi_{m+n}+m\Phi_m)) \\ &T_4=v_mv_nv_{m+n} \cos(-m\Delta\phi_1-n\Delta\phi_2-((m+n)\Psi_{m+n}-m\Psi_m-n\Phi_n)) \end{align}$$

It is worth giving several comments. 1) The physical interpretation of $$B$$ term is the three particle correlation due to the presence of two particle correlation. 2) For $$C$$ term to be non-zero, one needs the proper combination number among flows, eg, $$T_2$$ terms implies $$n+m-l=0$$. 3) However, if again, one assumes that $$\Psi_{2n+1}^A (n\ge 1)$$  of event $$A$$ is not correlated to $$ \Psi_{2n+1}^B $$ (the same $$n$$) neither to  $$\Psi_{2n'}^B$$  (any $$n'$$) of another event $$B$$. The contribution of the above $$C$$ term is eventually zero when any odd order flow gets involved (even when one gets the correct combination numbers). This is simply due to the fact that odd+odd=even odd+even=odd, consequently the event planes of odd and even order flows eventually present together in the argument of cosine, therefore the contribution goes to zero when carrying out average (on correlation of the same event or among mixed events).

The first simple model: three particle cumulant solely due to triangular flow
Before stepping forward and introducing event-by-event fluctuations, we want to show two results: 1) We will show a simple model, when one only considers $$v_3$$. We argue that the three particle cumulant is zero. This is because, in this case, the three particle correlation is purely due to two particle correlation, in other words, the mixed event contribution cancels exactly the three particle correlation, as we will show explicitly below. Therefore it is a trivial case. 2) In our one tube (shadowing) model, very different from what has been discussed so far as by many other authors, the event plane of $$v_3$$ actually correlation with those of $$v_1$$ and $$v_2$$. The magnitude of correlation between $$v_3$$ and $$v_2$$ is actually very small, so it does not contradict with the existing numerical results from Glauber MC models. As an approach, one may expand the azimuthal flow as follows.


 * $$\begin{align}

\frac{dN}{d\phi}=\frac{N}{2\pi}(1+2\sum_{n=1}^{\infty}v_n\cos(n(\phi-\Psi_n))+2v_1^r\cos(\phi-\Psi_t)+2v_2^r\cos(2(\phi+\frac{\pi}{2}-\Psi_t))+2v_3^r\cos(3(\phi-\Psi_t))) \end{align}$$

where $$v_1^r v_2^r v_3^r$$  measures the amount of residual directed flow, elliptic flow and triangular flow correlated to the same angle $$ \Psi_t $$ due to the presence of tube. The presence of $$\frac{\pi}{2}$$  angle for the convenience of the one-tube (shadowing) model that we will discuss below. It will be shown that these extra terms are realistic and they give interesting non-trivial effect. Though for two particle correlation, it gives very little impact, three particle correlation may present observable difference.

Now the simple model. For a distribution purely consists of $$v_3$$ (together with $$v_0$$ of course), one has


 * $$\begin{align}

\frac{dN}{d\phi}=\frac{N}{2\pi}(1+2v_3\cos(3\phi)) \end{align}$$

For simplicity, again we will not introduce particle number fluctuation in the following calculation. The event plane of $$v_3$$ is taken to be 0 for the first event, and to be a random number for any successive event, whenever mixed event contribution is involved. To calculate the cumulants, one will not align the event plane of $$v_3$$, since the event plane of $$v_3$$ is presumably uncorrelated to \Psi_2 /reaction plane/participant plane which is usually aligned in the same direction for mixed event subtraction. As a result, in the present case, the event planes of mixed events are pointing at random azimuthal angles. First, let us calculate three particle correlation.


 * $$\begin{align}

&\rho_{123}\equiv\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}=A+B+C \\ &A=\frac{N^3}{(2\pi)^3} \\ &B=\frac{N^3}{(2\pi)^3}[ 2v_3^2 \cos(3\Delta\phi_1)+2v_3^2 \cos(3\Delta\phi_2)+ 2v_3^2 \cos(3(\Delta\phi_1-\Delta\phi_2))] \\ &C=0 \end{align}$$

Since there is no particle number fluctuations, averaging over events and then dividing the number of events will give the same result as a single event. In the second place, let us calculate the mixed event contributions. There are four terms


 * $$\begin{align}

&\rho_{12}\rho_3\equiv\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}|_{12,3} \\ &= \int \frac{d\phi_R}{2\pi} \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi_1)\frac{dN}{d\phi}(\phi+\phi_R+\Delta\phi_2) \\ &= \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi_1)\frac{N}{2\pi} \\ &=\frac{N^2}{(2\pi)^2}(1+ 2v_3^2 \cos(3\Delta\phi_1))\frac{N}{2\pi} \\ &=\frac{N^3}{(2\pi)^3}(1+ 2v_3^2 \cos(3\Delta\phi_1)) \end{align}$$

where $$\phi^R$$  indicates the random phase, which is therefore integrated over any angle and then normalized. The fact that $$\phi^R$$  can be integrated first greatly simplifies the calculation, and one only need to calculate effectively a 2 particle correlation. Similarly


 * $$\begin{align}

&\rho_{23}\rho_1 =\frac{N^3}{(2\pi)^3}(1+ 2v_3^2 \cos(3(\Delta\phi_2-\Delta\phi_1))) \\ & \rho_{31}\rho_2 =\frac{N^3}{(2\pi)^3}(1+ 2v_3^2 \cos(3\Delta\phi_2)) \\ & \rho_1\rho_2\rho_3\equiv\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}|_{1,2,3} \\ &= \int \frac{d\phi_{R1}}{2\pi}\int \frac{d\phi_{R2}}{2\pi} \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\phi_{R1}\Delta\phi_1)\frac{dN}{d\phi}(\phi+\phi_{R2}+\Delta\phi_2) =\frac{N^3}{(2\pi)^3} \\ &C_3=\rho_{123}-\rho_{12}\rho_3-\rho_{23}\rho_1-\rho_{31}\rho_2+2\rho_1\rho_2\rho_3=0 \end{align}$$

This concludes our first point, pure $$v_3$$ does give zero three particle cumulants. This conclusion can be easily generated to the case that if the event planes of all flows are not correlated (which is not true for real world where even order flows are correlated with one another), three particle cumulants can only been trivially zero. Mathematically, one only need to realize that, under the above assumption, $$C$$ terms of 3p correlation is trivially zero, and again B term corresponds to 2p correlation, which is exactly what will be generated by mixed event contributions.

We note that this result holds true for any pure $$v_n$$  flow or composition of flows without any correlation in their event planes.

The second simple model: three particle cumulant due to idealized one tube (shadowing) model
Now it is a good moment to show how 3p cumulants will be affected if one has correlation among even and odd flows. A realistic model is the one tube (shadowing) model proposed by us. By Fourier expansion, the IC with only one energetic tube possesses components of $$v_1^r, v_2^r, v_3^r$$, their event planes are correlated such that  $$\Psi_1^r=\Psi_3^r=\Psi_t,  \Psi_2^r=\Psi_t+\pi/2$$. These event planes are not correlated to the overall elliptic flow. This can be understood as an assumption that the orientation of the tube is irrelevant to participant plane. Interesting enough, correlation among $$\Psi_1  \Psi_3$$  was also pointed out by other authors recently but from somewhat different point of view (eg. see arXiv:1010.1876). As we have shown explicitly, the two particle correlation part ($$B$$ term) will be cleanly subtracted by two particle correlation of mixed event. The non-trivial part is actually the $$C$$ term. And still, to illustrate the physics, we will adopt a simplified model at this point. While in a later time, we will show numerical results of more realistic cases. We assume here


 * $$\begin{align}

\frac{dN}{d\phi}=\frac{N}{2\pi}(1+2\sum_{n=1}^{\infty}v_n\cos(n(\phi-\Psi_n))+2v_1^r\cos(\phi-\Psi_t)+2v_2^r\cos(2(\phi+\frac{\pi}{2}-\Psi_t))+2v_3^r\cos(3(\phi-\Psi_t))) \end{align}$$

where we have neglected $$v_2^r$$  only for simplicity, it will not affect the underlying physics. As it will be shown below, the extra $$v_1^r, v_3^r$$  terms will not affect at all two particle correlation, in the sense that: there is no cross terms involving  $$v_1v_1^r$$, neither  $$v_1^rv_3^r $$. The result 2p correlation reads


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle\equiv \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (1+4 \sum_{m,n=1}^{\infty} v_m v_n \cos(m(\phi-\Psi_m))\cos(n(\phi+\Delta\phi-\Psi_n)))  \\ &+\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(4 \sum_{m=1}^{\infty}\sum_{n=1}^3 v_m v_n^r \cos(m(\phi-\Psi_m))\cos(n(\phi+\Delta\phi+\frac{\pi}{2}\delta_{n,2}-\Psi_t))) \\ &+\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (4 \sum_{m,n=1}^3 v_m^r v_n^r \cos(m(\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))\cos(n(\phi+\Delta\phi+\frac{\pi}{2}\delta_{n,2}-\Psi_t))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (1+4 \sum_{m=1}^{\infty} v_m v_m \cos(m(\phi-\Psi_m))\cos(m(\phi+\Delta\phi-\Psi_m))) \\ &+\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2(4 \sum_{m=1}^{3} v_m v_m^r \cos(m(\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_m))\cos(m(\phi+\Delta\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))) \\ &+\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (4 \sum_{m=1}^3 v_m^r v_m^r \cos(m(\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))\cos(m(\phi+\Delta\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))) \\ &=\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (1+4 \sum_{m=1}^{\infty} v_m v_m \cos(m(\phi-\Psi_m))\cos(m(\phi+\Delta\phi-\Psi_m))) \\ &+\int \frac{d\phi}{2\pi}(\frac{N}{2\pi})^2 (4 \sum_{m=1}^3 v_m^r v_m^r \cos(m(\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))\cos(m(\phi+\Delta\phi+\frac{\pi}{2}\delta_{m,2}-\Psi_t))) \\ &=(\frac{N}{2\pi})^2(1+ \sum _{n=1}^{\infty} 2(v_n)^2 \cos(n\Delta\phi)+\sum_{m=1}^3 2(v_m^r)^2 \cos(\Delta\phi)) \end{align}$$

In the above calculation, since $$\Psi_t$$  is not correlated to participant plane, one has to assign to it a random angle, or equivalently, one has to bear in mind that


 * $$\begin{align}

f(\Psi_t\pm\Psi_2) \rightarrow \int \frac{d\Psi_R}{2\pi} f(\Psi_t+\Psi_R\pm\Psi_2) \end{align}$$

which has not been written out explicitly.

A comment is now ready to hand out. If the $$v_3$$  discussed extensively by other author is actually  $$v_3^r$$, or in terms of the parameterization of our model,  $$v_3=0$$  (or it is enough small comparing to  $$v_3^r $$), this difference can never be observed by purely using two particle correlation. The difference we are talking about here is that correlation among event planes.

Now will show explicitly the above statement. We will go further, we will show the difference will eventually affect three particle cumulants. Now we introduce two one-particle distributions, namely,


 * $$\begin{align}

\frac{dN^{(1)}}{d\phi}=\frac{N}{2\pi}(1+2v_1\cos(\phi-\Psi_1)+2\sqrt{v_2^2+v_2^{r2}}\cos(2(\phi-\Psi_2)+2v_3\cos(3(\phi-\Psi_3))  (v_2 \gg v_2^r) \end{align}$$ where $$\Psi_1, \Psi_2, \Psi_3$$  are not correlated, and


 * $$\begin{align}

\frac{dN^{(2)}}{d\phi}=\frac{N}{2\pi}(1+2v_1^r\cos(\phi-\Psi_t)+2v_2\cos(2(\phi-\Psi_2)+2v_2^r\cos(2(\phi-\Psi_t)+2v_3^r\cos(3(\phi-\Psi_t)) (v_1=v_1^r\ \ v_2\gg v_2^r, v_3=v_3^r) \end{align}$$

where $$\Psi_t$$  is determined by the position of the tube therefore not correlated to any event plane, it fluctuates event by event. The second distribution resembles the case where one tube is immersed in a background dominated by elliptic flow, $$v_1^r, v_2^r, v_3^r$$  are coefficients of Flourier expansion of the flow due to the tube. On the other hand, the first distribution only contains directed flow, elliptic flow and triangular flow with their event planes uncorrelated, the term elliptic flow coefficient $$\sqrt{v_2^2+v_2^{r2}}$$  is to guarantee that the two distribution will produce exactly the same two particle correlation.


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(1)}=\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(2)}\equiv\rho_{12} \\ &=(\frac{N}{2\pi})^2(1+ 2(v_1)^2\cos(\Delta\phi) +2(v_2^2)\cos(2\Delta\phi)+2(v_2^{r2})\cos(2\Delta\phi)+2(v_3^2)\cos(3\Delta\phi)) \end{align}$$ (using above result)


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(1)}_{1,2}=\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(2)}_{1,2}\equiv\rho_1\rho_2 =(\frac{N}{2\pi})^2 \\ &C_2^{(1)}=C_2^{(2)}\equiv\rho_{12}-\rho_1\rho_2 =(\frac{N}{2\pi})^2(2(v_1)^2\cos(\Delta\phi) +2(v_2^2)\cos(2\Delta\phi)+2(v_2^{r2})\cos(2\Delta\phi)+2(v_3^2)\cos(3\Delta\phi)) \end{align}$$

Now we want to show it make a big different for 3p correlation. Still we will not introduce particle number fluctuation at this moment. For the first case, it is very similar to the foregoing example, that three particle cumulant is zero.


 * $$\begin{align}

&\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle\equiv \rho_{123} \\ &C_3^{(1)}=\rho_{123}-\rho_{12}\rho_3-\rho_{23}\rho_1-\rho_{31}\rho_2+2\rho_1\rho_2\rho_3=0 \end{align}$$

For the second distribution.


 * $$\begin{align}

\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle\equiv \rho_{123}=A+B+C \end{align}$$ where


 * $$\begin{align}

&A=(\frac{N}{2\pi})^3 \\ &B=(\frac{N}{2\pi})^3[ \sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \end{align}$$

where


 * $$\begin{align}

&\bar v_1 = v_1 \\ &\bar v_2 = \sqrt{v_2^2+v_2^{r2}} \\ &\bar v_3 = v_3 \end{align}$$

and


 * $$\begin{align}

&C=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{3} v_m^rv_n^rv_l^r\cos(n(\phi-\Psi_t+\frac{\pi}{2}\delta_{n,2}))\cos(m(\phi+\Delta\phi_1-\Psi_t+\frac{\pi}{2}\delta_{m,2}))\cos(l(\phi+\Delta\phi_2-\Psi_t+\frac{\pi}{2}\delta_{l,2}))  \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi}\sum_{m,n,l=1}^{3}v_m^rv_n^rv_l^r \frac{1}{4}(F^r(n,m,l)+F^r(n,m,-l)+F^r(n,-m,l)+F^r(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{3}v_m^rv_n^rv_l^r \frac{1}{4}(\delta_{n+m+l}F^r(n,m,l)+\delta_{n+m-l}F^r(n,m,-l)+\delta_{n-m+l}F^r(n,-m,l)+\delta_{n-m-l}F^r(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n,l=1}^{3}v_m^rv_n^rv_l^r \frac{1}{4}(\delta_{n+m-l}F^r(n,m,-l)+\delta_{n-m+l}F^r(n,-m,l)+\delta_{n-m-l}F^r(n,-m,-l)) \\ &=8(\frac{N}{2\pi})^3\int \frac{d\phi}{2\pi} \sum_{m,n=1}^{3}\frac{1}{4}(T_2^r+T_3^r+T_4^r) \\ &=(\frac{N}{2\pi})^3 \sum_{m,n=1}^{3}2(T_2^r+T_3^r+T_4^r) \\ &=(\frac{N}{2\pi})^3 v_1^rv_2^rv_3^r [\cos(-\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-\Delta\phi_2)+\cos(-2\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-2\Delta\phi_2)] \\ &+(\frac{N}{2\pi})^3 v_1^rv_2^rv_3^r[\cos(\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1+\Delta\phi_2)] +2(\frac{N}{2\pi})^3v_1^rv_1^rv_2^r[\cos(-\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1-\Delta\phi_2)] \end{align}$$

where


 * $$\begin{align}

F^r(n,m,l)=\cos[(n+m+l)\phi+m\Delta\phi_1+l\Delta\phi_2-(n(\Psi_t+\frac{\pi}{2}\delta_{n,2})+m(\Psi_t+\frac{\pi}{2}\delta_{m,2})+l(\Psi_t+\frac{\pi}{2}\delta_{l,2}))] \end{align}$$

and


 * $$\begin{align}

&T_2^r=v_m^rv_n^rv_{m+n}^r \cos(m\Delta\phi_1-(m+n)\Delta\phi_2-(n(\Psi_t+\frac{\pi}{2}\delta_{n,2})+m(\Psi_t+\frac{\pi}{2}\delta_{m,2})-(m+n)(\Phi_t+\frac{\pi}{2}\delta_{m+n,2}))) \\ &=v_1^rv_2^rv_3^r [\delta_{m1}\delta_{n2}\cos(-\Delta\phi_1+3\Delta\phi_2)+\delta_{m2}\delta_{n1}\cos(-2\Delta\phi_1+3\Delta\phi_2)]+2v_1^rv_1^rv_2^r\delta_{m1}\delta_{n1}[\cos(-\Delta\phi_1+2\Delta\phi_2)] \\ &T_3^r=v_m^rv_n^rv_{m+n}^r \cos(-(m+n)\Delta\phi_1+m\Delta\phi_2-(n(\Psi_t+\frac{\pi}{2}\delta_{n,2})-(m+n)(\Psi_t+\frac{\pi}{2}\delta_{m+n,2})+m(\Phi_t+\frac{\pi}{2}\delta_{m,2}))) \\ &=v_1^rv_2^rv_3^r [\delta_{m1}\delta_{n2}\cos(3\Delta\phi_1-\Delta\phi_2)+\delta_{m2}\delta_{n1}\cos(3\Delta\phi_1-2\Delta\phi_2)]+2v_1^rv_1^rv_2^r[\delta_{m1}\delta_{n1}\cos(2\Delta\phi_1-\Delta\phi_2) \\ &T_4^r=v_m^rv_n^rv_{m+n}^r \cos(-m\Delta\phi_1-n\Delta\phi_2-((m+n)(\Psi_t+\frac{\pi}{2}\delta_{m+n,2})-m(\Psi_t+\frac{\pi}{2}\delta_{m,2})-n(\Phi_t+\frac{\pi}{2}\delta_{n,2}))) \\ &=v_1^rv_2^rv_3^r [\delta_{m1}\delta_{n2}\cos(\Delta\phi_1+2\Delta\phi_2)+\delta_{m2}\delta_{n1}\cos(2\Delta\phi_1+\Delta\phi_2)]+2v_1^rv_1^rv_2^r [\delta_{m1}\delta_{n1}\cos(\Delta\phi_1+\Delta\phi_2)] \end{align}$$

Now we move on to calculate mixed event contributions


 * $$\begin{align}

&\rho_{12}\rho_3\equiv\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}|_{12,3} \\ &= \int \frac{d\phi_R}{2\pi} \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi_1)\frac{dN}{d\phi}(\phi+\phi_R+\Delta\phi_2) \\ &= \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi_1)\frac{N}{2\pi} \\ &=(\frac{N}{2\pi})^2(1+ 2(\bar v_1)^2 \cos(\Delta\phi_1)+2(\bar v_2^2)\cos(2\Delta\phi_1)+2(\bar v_3^2)\cos(3\Delta\phi_1))\frac{N}{2\pi} \\ &=\frac{N^3}{(2\pi)^3}(1+ 2(\bar v_1)^2 +2(\bar v_2^2)\cos(2\Delta\phi_1)+2(\bar v_3^2)\cos(3\Delta\phi_1)) \\ &\rho_{23}\rho_1 =\frac{N^3}{(2\pi)^3}  (1+  2(\bar v_1)^2 \cos(\Delta\phi_2-\Delta\phi_1)+2(\bar v_2^2)\cos(2\Delta\phi_1)+2(\bar v_3^2)\cos(3(\Delta\phi_2-\Delta\phi_1))) \\ &\rho_{31}\rho_2 =\frac{N^3}{(2\pi)^3}(1+  2(\bar v_2)^2 +2(\bar v_2^2)\cos(2\Delta\phi_2)+2(\bar v_3^2)\cos(3\Delta\phi_2)) \\ &\rho_1\rho_2\rho_3\equiv\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}|_{1,2,3} \\ &= \int \frac{d\phi_{R1}}{2\pi}\int \frac{d\phi_{R2}}{2\pi} \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\phi_{R1}\Delta\phi_1)\frac{dN}{d\phi}(\phi+\phi_{R2}+\Delta\phi_2) =\frac{N^3}{(2\pi)^3} \\ &C_3^{(2)}=\rho_{123}-\rho_{12}\rho_3-\rho_{23}\rho_1-\rho_{31}\rho_2+2\rho_1\rho_2\rho_3=C \\ &=(\frac{N}{2\pi})^3 v_1^rv_2^rv_3^r [\cos(-\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-\Delta\phi_2)+\cos(-2\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-2\Delta\phi_2)] \\ &+(\frac{N}{2\pi})^3 v_1^rv_2^rv_3^r[\cos(\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1+\Delta\phi_2)] +2(\frac{N}{2\pi})^3v_1^rv_1^rv_2^r[\cos(-\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1-\Delta\phi_2)] \end{align}$$

Three particle cumulant is nonzero for the second case.

The second simple model with particle number fluctuation taken into consideration
At last, we want to incorporate the effect of particle number fluctuation. Not spoiling the physics under discussion, we would like to introduce an assumption to greatly simplify our calculations, we assume that flow coefficients do not fluctuate (together with multiplicities). Due to particle number fluctuation C_3^{(1)}  is no longer vanishing, therefore, the above effect we address in the calculation to some extent will be suppressed in real collisions. However, if one restrict himself inside a small centrality window, the effect of particle number fluctuation may be smaller enough comparing to the that of event plane correlations. Two particle cumulants of both distributions have the same form when taking into account particle number fluctuations


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(1f)}=\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(2f)}\equiv\rho_{12}^f \\ &=\frac{\left\langle N^2\right\rangle}{(2\pi)^2}(1+ 2(v_1)^2\cos(\Delta\phi) +2(v_2^2)\cos(2\Delta\phi)+2(v_2^{r2})\cos(2\Delta\phi)+2(v_3^2)\cos(3\Delta\phi)) \end{align}$$ (using above result)


 * $$\begin{align}

&\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(1f)}_{1,2}=\left\langle\frac{dN_{pair}}{d\Delta\phi}\right\rangle^{(2f)}_{1,2}\equiv\rho_1^f\rho_2^f =\frac{\left\langle N\right\rangle^2}{(2\pi)^2} \\ & C_2^{(1f)}=C_2^{(2f)}\equiv\rho_{12}^f-\rho_1^f\rho_2^f \\ &=(\frac{\left\langle N^2\right\rangle}{(2\pi)^2}-\frac{\left\langle N\right\rangle^2}{(2\pi)^2})+\frac{\left\langle N^2\right\rangle}{(2\pi)^2}(2(v_1)^2\cos(\Delta\phi) +2(v_2^2)\cos(2\Delta\phi)+2(v_2^{r2})\cos(2\Delta\phi)+2(v_3^2)\cos(3\Delta\phi)) \end{align}$$

Three particle cumulants possess more complicated forms, but if one notices that the angular parts of the correlation always maintain the same form, attentions will be drawn more to the modifications of multiplicity factors.

Finally one gets for the first distribution,


 * $$\begin{align}

\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle\equiv \rho_{123}^f=A+B+C \end{align}$$ where


 * $$\begin{align}

&A=\frac{\left\langle N^3\right\rangle}{(2\pi)^3} \\ &B=\frac{\left\langle N^3\right\rangle}{(2\pi)^3}[ \sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \\ &C=0 \end{align}$$


 * $$\begin{align}

&C_3^{(1f)}=\rho_{123}^f-\rho_{12}^f\rho_3^f-\rho_{23}^f\rho_1^f-\rho_{31}^f\rho_2^f+2\rho_1^f\rho_2^f\rho_3^f \\ &=(\frac{\left\langle N^3\right\rangle}{(2\pi)^3}+2\frac{\left\langle N\right\rangle^3}{(2\pi)^3}-3\frac{\left\langle N\right\rangle^2\left\langle N\right\rangle}{(2\pi)^3})+(\frac{\left\langle N^3\right\rangle}{(2\pi)^3}-\frac{\left\langle N\right\rangle^2\left\langle N\right\rangle}{(2\pi)^3}) \\ &[ \sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \end{align}$$

For the second distribution.


 * $$\begin{align}

\left\langle\frac{d^2N_{tripple}}{d\Delta\phi_1d\Delta\phi_2}\right\rangle\equiv \rho_{123}^f=A+B+C \end{align}$$ where


 * $$\begin{align}

&A=\frac{\left\langle N^3\right\rangle}{(2\pi)^3} \\ &B=\frac{\left\langle N^3\right\rangle}{(2\pi)^3}[ \sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \\ &C=\frac{\left\langle N^3\right\rangle}{(2\pi)^3} v_1^rv_2^rv_3^r [\cos(-\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-\Delta\phi_2)+\cos(-2\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-2\Delta\phi_2)] \\ &+\frac{\left\langle N^3\right\rangle}{(2\pi)^3} v_1^rv_2^rv_3^r[\cos(\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1+\Delta\phi_2)] +2\frac{\left\langle N^3\right\rangle}{(2\pi)^3}v_1^rv_1^rv_2^r[\cos(-\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1-\Delta\phi_2)] \\ &C_3^{(2f)}=\rho_{123}^f-\rho_{12}^f\rho_3^f-\rho_{23}^f\rho_1^f-\rho_{31}^f\rho_2^f+2\rho_1^f\rho_2^f\rho_3^f \\ &=(\frac{\left\langle N^3\right\rangle}{(2\pi)^3}+2\frac{\left\langle N\right\rangle^3}{(2\pi)^3}-3\frac{\left\langle N\right\rangle^2\left\langle N\right\rangle}{(2\pi)^3})+(\frac{\left\langle N^3\right\rangle}{(2\pi)^3}-\frac{\left\langle N\right\rangle^2\left\langle N\right\rangle}{(2\pi)^3}) \\ &[ \sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_1)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n\Delta\phi_2)+\sum _{n=1}^{3} 2\bar v_n^2 \cos(n(\Delta\phi_1-\Delta\phi_2))] \\ &+\frac{\left\langle N^3\right\rangle}{(2\pi)^3} v_1^rv_2^rv_3^r [\cos(-\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-\Delta\phi_2)+\cos(-2\Delta\phi_1+3\Delta\phi_2)+\cos(3\Delta\phi_1-2\Delta\phi_2)] \\ &+\frac{\left\langle N^3\right\rangle}{(2\pi)^3} v_1^rv_2^rv_3^r[\cos(\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1+\Delta\phi_2)] +2\frac{\left\langle N^3\right\rangle}{(2\pi)^3}v_1^rv_1^rv_2^r[\cos(-\Delta\phi_1+2\Delta\phi_2)+\cos(2\Delta\phi_1-\Delta\phi_2)] \\ &=C_3^{1f}+C \end{align}$$

As has been pointed out, when one considers particle number fluctuation, one gets extra terms. Fortunately, these new contributions affect independently the cumulants, when one concentrates in a small centrality window, their effects can be reasonably small comparing to those due to event plane correlations.