Derivation Notes on Event Plane Correlation

Derivation Notes on Event Plane Correlation

This note is an attempt to carry out an analytic calculation of the observed event plane correlation using peripheral one tube model.

The observed data on the centrality dependence of event plane correlation has an important feature, the strength of the correlation increases from central to peripheral collisions. This feature can be easily understand in terms of the one tube model. The part of the correlation connected to the fluctuation (of flow harmonics and of multiplicity) is effectively described by the one tube model, event planes are correlated as a part of background flow is deflected by the tube. The strength of the correlation correlation remains the same from central to peripheral collisions. On the other hand, the average collective phenomena are parameterized by the background flow coefficients. In this model, we simply assume that there is no event plane correlation in the background flow. As a result, the hadrons emitted from the background dilute the overall event plane correlation due to its multiplicity, namely, $$N_b$$. Since by definition $$N_b$$ will still contributes to the denominator (see below), but the flow, $$v_n^b$$, mostly does not contribute to the event plane correlation. However, as the multiplicity decreases as one goes from central to peripheral collsions, the overall event plane correlation increases.

Elliptic background with isotropic tube distribution
Assuming the following one particle azimuthal distribution in terms of background and flow components generated by one tube model


 * $$\begin{align}

&\frac{dN}{d\phi}(\phi,\phi_t) \equiv \frac{dN}{d\phi}(\phi)=\frac{dN_{background}}{d\phi}(\phi)+\frac{dN_{tube}}{d\phi}(\phi,\phi_t) \\ &\frac{dN_{tube}}{d\phi}(\phi,\phi_t) \equiv \frac{dN_{tube}}{d\phi}(\phi) \end{align}$$

where


 * $$\begin{align}

&\frac{dN_{background}}{d\phi}(\phi)=\frac{N_b}{2\pi}(1+\sum_{n}2v_n^b\cos(n(\phi-\Psi'_n))) \\ & \frac{dN_{tube}}{d\phi}(\phi)=\frac{N_t}{2\pi}\sum_{n}2v_n^t\cos(n(\phi-\phi_t-\delta_n)) \end{align}$$

where $$\phi_t $$ is the azimuthal location of the tube, the azimuthal angle of the hadrons emitted from the tube is measured in respect to the azimuthal location of the tube as well as to $$ \delta_n$$. Here, $$ \delta_n$$ is the same for any tube from any event, and it does not depend on $$\phi_t $$. As shown below, $$\delta_n $$ is important for the sign of the event plane correlation. We note that the it is not necessary to write the distribution as


 * $$\begin{align}

\frac{dN_{tube}}{d\phi}(\phi)=\frac{N_t}{2\pi}(1+\sum_{n=2,3}2v_n^t\cos(n(\phi-\phi_t-\delta_n))) \end{align}$$

due to the fact that the normalization term is isotropic therefore can always be absorbed into the background.

Event plane correlation
The definition of event plane correlation is


 * $$\begin{align}

\left\langle\cos(jk(\Psi_n - \Psi_m))\right\rangle\equiv \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) \cos(jk(\Psi_n - \Psi_m)) \end{align}$$ where $$j$$ is an integer and $$k$$ is the least common multiple (LCM) of $$n$$ and $$m$$. The integral $$\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right)$$ enumerates all possible event plane distributions, and the integral $$\int \frac{d\phi_t}{2\pi}$$ is corresponding to all possible tube orientations. We note that $$f(\phi_t)$$ should be normalized so that $$\int f(\phi_t) \frac{d\phi_t}{2\pi} =1$$ is the tube distribution function, in our first treatment, we take $$f(\phi_t)=1$$  for simplicity.

For a given event (with a specific tube orientation $$\phi_t$$), the event planes $$\Psi_m$$ are determined by the event planes $$\Psi'_n$$ of the background and the tube orientation as follows
 * $$\begin{align}

v_m e^{in\Psi_m}= \left\langle e^{in\phi}\right\rangle = \frac{\int d\phi \frac{dN}{d\phi}e^{im\phi}}{\int d\phi \frac{dN}{d\phi}} = \frac{1}{N_b}\int d\phi e^{in\phi}\left[\frac{N_b}{2\pi}(1+\sum_{n=1}^\infty 2v_n^b\cos(n(\phi-\Psi'_n)))+\frac{N_t}{2\pi}\sum_{n=1}^\infty 2v_n^t\cos(n(\phi-\phi_t-\delta_n)) \right] \end{align}$$

Straightforward calculations give


 * $$\begin{align}

v_m\cos(m\Psi_m)= \left\langle \cos(m\phi)\right\rangle = \frac{1}{N_b}\int {d\phi}\cos(n\phi)\left[\frac{N_b}{2\pi}(1+\sum_{n=1}^\infty 2v_n^b\cos(n(\phi-\Psi'_n)))+\frac{N_t}{2\pi}\sum_{n=1}^\infty 2v_n^t\cos(n(\phi-\phi_t-\delta_n)) \right] = v_m^b \cos(m\Psi'_m)+\frac{N_t}{N_b} v_m^t \cos(m(\phi_t+\delta_m)) \end{align}$$

and


 * $$\begin{align}

v_m\sin(m\Psi_m)= \left\langle \sin(m\phi)\right\rangle = \frac{1}{N_b}\int {d\phi}\sin(n\phi)\left[\frac{N_b}{2\pi}(1+\sum_{n=1}^\infty 2v_n^b\cos(n(\phi-\Psi'_n)))+\frac{N_t}{2\pi}\sum_{n=1}^\infty 2v_n^t\cos(n(\phi-\phi_t-\delta_n)) \right] = v_m^b \sin(m\Psi'_m)+\frac{N_t}{N_b} v_m^t \sin(m(\phi_t+\delta_m)) \end{align}$$

where


 * $$\begin{align}

&(v_m)^2 = (v_m^b)^2 + \left(\frac{N_t}{N_b} v_m^t\right)^2+ 2 \frac{N_t}{N_b} v_m^b v_m^t \cos(m(\Psi'_m-\phi_t-\delta_m))\\ &\langle(v_m)^2\rangle = (v_m^b)^2 + \left(\frac{N_t}{N_b} v_m^t\right)^2 \end{align}$$

As an approximation one may use the average $$\langle(v_m)^2\rangle$$ in the place of $$(v_m)^2$$, so that it does not depend on event plane angles.

Case study
We first discuss the case of $$\left\langle\cos(6(\Psi_2 - \Psi_3))\right\rangle$$. One has


 * $$\begin{align}

&\left\langle\cos(6(\Psi_2 - \Psi_3))\right\rangle \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) \cos(6(\Psi_2 - \Psi_3)) \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) \cos(3(2\Psi_2) - 2(3\Psi_3)) \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) (\cos^3(2\Psi_2) \cos^2(3\Psi_3)-\cos^3(2\Psi_2) \sin^2(3\Psi_3)-3 \sin^2(2\Psi_2) \cos(2\Psi_2) \cos^2(3\Psi_3)\\ &+6 \sin(2\Psi_2) \cos^2(2\Psi_2) \sin(3\Psi_3) \cos(3\Psi_3) -2 \sin^3(2\Psi_2) \sin(3\Psi_3) \cos(3\Psi_3)+3 \sin^2(2\Psi_2) \cos(2\Psi_2) \sin^2(3\Psi_3))\\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{3}}{2\pi} (\cos^3(2\Psi_2) \cos^2(3\Psi_3)-\cos^3(2\Psi_2) \sin^2(3\Psi_3)-3 \sin^2(2\Psi_2) \cos(2\Psi_2) \cos^2(3\Psi_3)\\ &+6 \sin(2\Psi_2) \cos^2(2\Psi_2) \sin(3\Psi_3) \cos(3\Psi_3) -2 \sin^3(2\Psi_2) \sin(3\Psi_3) \cos(3\Psi_3)+3 \sin^2(2\Psi_2) \cos(2\Psi_2) \sin^2(3\Psi_3)) \end{align}$$

We will adopt the approximation that $$v_m$$ is almost a constant, and when substituting the expression for $$\cos(2\Psi_2)$$ and $$\cos(2\Psi_2)$$, a lot of terms will merge. However, the following fact helps to simplify the calculation, namely, the event planes $$\Psi'_2$$ and $$\Psi'_3$$ from the background flow are independent, so any term involving an odd number of factor of $$\sin,\cos\Psi'_2$$ or $$\sin,\cos\Psi'_3$$ will not contribute due the integral with respect to $$\Psi'_2$$ and $$\Psi'_3$$. As a result, only those terms purely involving $$\phi_t$$ may survive. One may then manually count these terms, and the result turns out to be quite simple as follows
 * $$\begin{align}

&\left\langle\cos(6(\Psi_2 - \Psi_3))\right\rangle \\ &= \left(\frac{N_t}{N_b} v_m^t\right)^5 \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{3}}{2\pi} (\cos^3(2(\phi_t+\delta_2)) \cos^2(3(\phi_t+\delta_3))-\cos^3(2(\phi_t+\delta_2)) \sin^2(3(\phi_t+\delta_3))-3 \sin^2(2(\phi_t+\delta_2)) \cos(2(\phi_t+\delta_2)) \cos^2(3(\phi_t+\delta_3))\\ &+6 \sin(2(\phi_t+\delta_2)) \cos^2(2(\phi_t+\delta_2)) \sin(3(\phi_t+\delta_3)) \cos(3(\phi_t+\delta_3)) -2 \sin^3(2(\phi_t+\delta_2)) \sin(3(\phi_t+\delta_3)) \cos(3(\phi_t+\delta_3))+3 \sin^2(2(\phi_t+\delta_2)) \cos(2(\phi_t+\delta_2)) \sin^2(3(\phi_t+\delta_3))\\ &= \left(\frac{N_t}{N_b} v_m^t\right)^5 \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{3}}{2\pi}\cos(3(2(\phi_t+\delta_2)) - 2(3(\phi_t+\delta_3)))\\ &= \left(\frac{N_t}{N_b} v_m^t\right)^5 \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{3}}{2\pi}\cos(6(\delta_2-\delta_3))\\ &= \left(\frac{N_t}{N_b} v_m^t\right)^5\cos(6(\delta_2-\delta_3)) \end{align}$$

One notice that the contribution is quite small, as confirmed by the data, comparing to correlations such as $$\left\langle\cos(4(\Psi_2 - \Psi_4))\right\rangle$$ as discussed below.


 * $$\begin{align}

&\left\langle\cos(4(\Psi_2 - \Psi_4))\right\rangle \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) \cos(4(\Psi_2 - \Psi_4)) \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) \cos(2(2\Psi_2) - \Psi_4) \\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \Pi_i\left(\frac{d\Psi'_{i}}{2\pi}\right) (\cos^2(2\Psi_2) \cos(4\Psi_4)-\sin^2(2\Psi_2) \cos(4\Psi_4)+2 \sin(2\Psi_2) \cos(2\Psi_2) \sin(4\Psi_4))\\ &= \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{4}}{2\pi} (\cos^2(2\Psi_2) \cos(4\Psi_4)-\sin^2(2\Psi_2) \cos(4\Psi_4)+2 \sin(2\Psi_2) \cos(2\Psi_2) \sin(4\Psi_4))\\ &= \left(\frac{N_t}{N_b} v_m^t\right)^3 \int f(\phi_t) \frac{d\phi_t}{2\pi}\int \frac{d\Psi'_{2}}{2\pi}\frac{d\Psi'_{3}}{2\pi}\cos(2(2(\phi_t+\delta_2) - 4(\phi_t+\delta_4)))\\ &= \left(\frac{N_t}{N_b} v_m^t\right)^3 \cos(4(\delta_2-\delta_4)) \end{align}$$

This time, the argument is quite similar, any term involving $$\Psi'_2, \Psi'_4$$ will not contribute. The effect is much bigger due to the fact that its order is 3 instead of 5.