Derivation Notes on Random Eccentricity Rotations

Derivation Notes on Random Eccentricity Rotations


 * This note is to study the effect of eccentricity angle $$\Phi_n$$  on two particle correlation and flow fluctuation. It is based on discussions in our group meetings.

Eccentricity Plane Rotation
Initial geometrical eccentricities are defined according to D. Teaney[cite] as


 * $$\begin{align}

\varepsilon_n \equiv \sqrt{^2+^2} \end{align}$$

where the average $$$$  of a quantity  $$f(r)$$  is defined by
 * $$=\int r drd\theta \epsilon(r,\phi) f(r) $$

and $$ \epsilon(r,\theta)$$ is the energy density distribution on the transverse plane


 * $$\begin{align}

\epsilon(r,\phi)=\epsilon_0+\sum_{n=1}\left\{2\epsilon_n^c(r) \cos(n\phi)+2\epsilon_n^c(s) \sin(n\phi) \right\} \end{align}$$

It is worth noting here that we will not use the definition of eccentricities introduced previously by B. Alver[cite]


 * $$\begin{align}

\varepsilon_n^{Alver} \equiv \sqrt{^2+^2} \end{align}$$

though most of the results derived in following can be applied to this definition without major modification

Now if we introduce eccentricity angles


 * $$\begin{align}

\Phi_n \equiv \frac{1}{n} \text{atan}2(,) \end{align}$$ one may rewrite eccentricities as the following


 * $$\begin{align}

\varepsilon_n=  \end{align}$$

This can be shown explicitly since


 * $$\begin{align}

&\\ &= \\ &=\cos(n\Phi_n)+\sin(n\Phi_n) \\ &=\sqrt{^2+^2} \left(\frac{\cos(n\Phi_n)}{\sqrt{<r^n\cos(n\phi)>^2+<r^n\sin(n\phi)>^2}}+\frac{\sin(n\Phi_n)<r^n\cos(n\phi)>}{\sqrt{<r^n\sin(n\phi)>^2+<r^n\sin(n\phi)>^2}}\right) \\ &=\varepsilon_n(\cos(n\Phi_n)^2+\sin(n\Phi_n)^2) \\ &=\varepsilon_n \end{align}$$

and


 * $$\begin{align}

&<r^n\sin(n(\phi-\Phi_n))>\\ &=<r^n[\sin(n\phi)\cos(n\Phi_n)-\cos(n\phi)\sin(n\Phi_n)]> \\ &=<r^n(\sin(n\phi)>\cos(n\Phi_n)-<r^n\cos(n\phi)>\sin(n\Phi_n) \\ &=0 \end{align}$$

Eccentricity has been discussed extensively and is understood to be connected to flow components. Mathematically, flow components contribute to two particle correlation due to Fourier expansion


 * $$\begin{align}

<\frac{dN_{pair}}{d\Delta\phi}>\equiv \int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi) =(\frac{N}{2\pi})^2(1+ \sum _{n=1}^{\infty} 2v_n^2 \cos(n\Delta\phi)) \end{align}$$

If one takes into consideration of mixed event subtraction and notice the fluctuation between events in terms of multiplicity and flow, for proper events, one actually gets


 * $$\begin{align}

<\frac{dN_{pair}}{d\Delta\phi}>_{proper}\equiv <\int \frac{d\phi}{2\pi} \frac{dN}{d\phi}(\phi) \frac{dN}{d\phi}(\phi+\Delta\phi)> =\frac{<N^2>}{(2\pi)^2}+ \sum _{n=1}^{\infty} \frac{1}{2\pi^2}<(Nv_n)^2> \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}

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

From above expressions, one see that event planes, $$\Psi_n$$, do not affect two particle correlation. Indeed only multiplicities, flow components and its fluctuations are the quantities in questions here. This argument also gives us an impression that eccentricity angles $$\Phi_n $$ are not important, if they are only correlated to event planes. In fact, $$\Phi_n$$  has not been quite studied in literature.

In the following, we will first show (numerically) the effect of eccentricity angle \Phi_n  on flow fluctuation therefore two particle correlation throw a simple model. Then we will argue that, a major difference between one-tube approach of two particle correlation and flow decomposition approach, is that, in one-tube IC, $$\Phi_n$$  are automatically correlated owing to the position of the tube, while in the later case,  $$\Phi_n$$  has not been taken into consideration. But before anything, we will write down the derivations.

To show explicitly, we first apply clockwire rotations of each eccentricity component by an angle $$\Phi_n'$$  (or contra-clockwire ratations of the axes by  $$\Phi_n'$$ ), one finds


 * $$\begin{align}

&\Phi_n \Rightarrow \frac{1}{n} \text{atan}2(<r^n\sin(n(\phi-\Phi_n'))>,<r^n\cos(n(\phi-\Phi_n'))>) \\ &= \frac{1}{n} \text{atan}2(<r^n\sin(n\phi)>\cos(n\Phi_n')-<r^n\cos(n\phi)>\sin(n\Phi_n'),<r^n\cos(n\phi)>\cos(n\Phi_n')+<r^n\sin(n\phi)>\sin(n\Phi_n')) \\ &= \frac{1}{n} \text{atan}2(\frac{<r^n\sin(n\phi)>}{<r^n\cos(n\phi)>}-\frac{\sin(n\Phi_n')}{\cos(n\Phi_n')},1+\frac{\sin(n\Phi_n')}{\cos(n\Phi_n')}\frac{<r^n\sin(n\phi)>}{<r^n\cos(n\phi)>}) \\ &= \frac{1}{n} \text{atan}2(\tan(n\Phi_n)-\tan(n\Phi_n'),1+\tan(n\Phi_n')\tan(n\Phi_n)) \\ &= \frac{1}{n} \text{atan}2(\tan(n(\Phi_n-\Phi_n')) ) \\ &=\Phi_n-\Phi_n' \end{align}$$ This implies an trivial effect, that the eccentricity angles consequently rotate by the same amounts. Therefore eccentricities do not change their values when one rotates individually each $$ \Phi_n$$.

Now we show above assertion the other way around. If one rotates the axes for angles $$\Phi_n' $$, by carrying out the definition of eccentricities, one has


 * $$\begin{align}

&\varepsilon_n \Rightarrow \sqrt{<r^n\cos(n(\phi-\Phi_n'))>^2+<r^n\sin(n(\phi-\Phi_n'))>^2} \\ &= \sqrt{(<r^n\cos(n\phi)>\cos(n\Phi_n')+<r^n\sin(n\phi)>\sin(n\Phi_n'))^2+(<r^n\sin(n\phi)>\cos(n\Phi_n')-<r^n\cos(n\phi)>\sin(n\Phi_n'))^2}\\ &=\sqrt{(<r^n\cos(n\phi)>^2+<r^n\sin(n\phi)>^2+2<r^n\sin(n\phi)><r^n\sin(n\phi)>\cos(n\Phi_n')\sin(n\Phi_n')-2<r^n\sin(n\phi)><r^n\sin(n\phi)>\cos(n\Phi_n')\sin(n\Phi_n')} \\ &= \sqrt{<r^n\cos(n\phi)>^2+<r^n\sin(n\phi)>^2} \\ &= \varepsilon_n \end{align}$$

It is worth noting that above derivation does not depend on explicit form of the expansion of $$\epsilon(r,\phi) $$, neither the definition of average, therefore can be easily applied to B. Alver's definition on eccentricity (as we promised). Now we write out explicitly the integrations contained in the average, and make use of the fact that rotation of axes will not modify the forms of  $$\epsilon_n^c(r),  \epsilon_n^s(r)$$  since they are not functions of the angles. For completeness we write down here explicitly the derivation. We note that the following derivations rely on the detailed form of definition of the average, therefore it is less general.


 * $$\begin{align}

&<r^n\cos(n(\phi-\Phi_n'))>\\ &=\int rdrd\phi\epsilon(r,\phi)r^n\cos(n(\phi-\Phi_n') \\ &=\int rdrd\phi \left\{\epsilon_0+\sum_{m=1}[2\epsilon_m^c(r) \cos(m\phi)+2\epsilon_m^s(r) \sin(m\phi) ]\right\} r^n\cos(n(\phi-\Phi_n') \\ &=\int rdrd\phi \left\{2\epsilon_n^c(r) \cos(n\phi)+2\epsilon_n^s(r) \sin(n\phi) \right\} r^n\cos(n(\phi-\Phi_n') \\ &=\int r^{n+1}drd\phi \left\{2\epsilon_n^c(r) \cos(n\phi)\cos(n\phi)\cos(n\Phi_n')+2\epsilon_n^s(r) \sin(n\phi)\sin(n\phi)\sin(n\Phi_n') \right\} \\ &=\int r^{n+1}dr \left\{\epsilon_n^c(r) \cos(n\Phi_n')+\epsilon_n^s(r) \sin(n\Phi_n') \right\} \end{align}$$

similarly


 * $$\begin{align}

<r^n\sin(n(\phi-\Phi_n'))> =\int r^{n+1}dr \left\{\epsilon_n^c(r) \cos(n\Phi_n')-\epsilon_n^s(r) \sin(n\Phi_n') \right\} \end{align}$$

Therefore one sees that explicitly


 * $$\begin{align}

&\sqrt{<r^n\cos(n(\phi-\Phi_n'))>^2+<r^n\sin(n(\phi-\Phi_n'))>^2}\\ &=\sqrt{(\int r^{n+1}dr \epsilon_n^c(r))^2+(\int r^{n+1}dr \epsilon_n^s(r))^2}\\ &=\sqrt{<r^n\cos(n\phi)>^2+<r^n\sin(n\phi)>^2}\\ &=\varepsilon_n \end{align}$$

To sum up, if one only cares about $$\varepsilon_n$$  but not  $$\Phi_n$$, one may rotate  $$\Phi_n$$  the way he likes without modifying of $$ \varepsilon_n$$. One point of our one-tube model is that $$\Phi_n$$  as well as  $$\Psi_n$$  are correlated. The correlations lead to non-trivial effect. Topics can be discussed are two particle cumulant, three particle cumulant, in-plane out-of-plane, they will present non-trivial effect due to the fluctuation in flow and correlation among $$\Psi_n$$  produced during the hydrodynamic evolution. Now we turn to $$\Phi_n$$, see how things might be different if one ignores the correlation among  $$\Phi_n$$. In other word, one must treat a tube as a whole, rather than only talking about its $$\varepsilon_n $$ components.

One tube model with eccentricity random rotations
In this simple model, we want to show that correlation between the angles $$\Phi_n$$  may greatly affect the flow fluctuation therefore particle correlations. Though the overall odd index $$\Phi_n$$  are not correlated (as discussed by many authors), a part of their components connected to each tube shall play an important role. To show this, we insert by hand random shifts in each $$\Phi_n \rightarrow \Phi_n-\Phi_n'$$  in the initial conditions of our one-tube model, with  \Phi_n'  being random angles. The eccentricities of the model will not be altered by the random rotations in Fourier components of IC. We want to show that the resulting evolution and two particle correlation can be very different. We ran 500 events each with different set of $$\Phi_n$$  random shifts. Few other cases are also discussed following comments of the discussions.

Results Though the initial energy distribution looks very different, the resulting two particle correlations give us a big surprise!?
 * The energy distribution of one-tube model
 * Energy distribution when its $$\Phi_1 \rightarrow \Phi_{15}$$  were reshuffled, keeping  $$\varepsilon_n$$  unchaged
 * All three cases have the same initial eccentricities except the $$\varepsilon_3 =0$$  case (see below)

Above plot show the two particle correlation of one event, if one does event align and mixed event subtraction, one gets a flat correlation, i.e., no correlation at all. This is because the cumulant actually depends on difference
 * Two particle correlation of one tube model


 * $$\begin{align}

<N^2v_n^2>-<Nv_n>^2 \end{align}$$

In one tube case,


 * $$\begin{align}

&N = const. \\ &<v_n^2>=<v_n>^2 \end{align}$$

This implies the disappearance of two particle correlation if there is no fluctuation at all.

Above numerical calculation shows that
 * Two particle correlation of reshuffled $$\Phi_n$$  case


 * $$\begin{align}

<N^2v_n^2>-<Nv_n>^2 \ne 0 \end{align}$$ In fact, it is always positive, if say, the fluctuation is Guassian. The results were obtained by using 400 events with event-by-event fluctuations.

Flow eccentricity linear dependence check
Above results motivated us to do the following calculations, aiming at check explicitly the linear dependence of flow/eccentricity.


 * Energy distribution when $$\Phi_2$$  is rotated by  $$\frac{\pi}{2} $$
 * Energy distribution when one sets $$\varepsilon_3 = 0 $$, while keeping  $$\Phi_n$$  and other  $$\varepsilon_n$$  unchanged (one sees a triangular shape appears to give a triangular eccentricity contribution to compensate that from the tube)
 * Two particle correlation of $$\Phi_2=\pi/2$$  case
 * Two particle correlation of $$\varepsilon_3=0$$  case


 * The corresponding distribution of triggers ( $$p_T>1.5\text{GeV}$$ ) and associated particles ( $$0.5 \text{GeV} <p_T < 1.5 \text{GeV}$$ ). I also tried different momentum thresholds (including those to increase the momentum thresholds), but results look similar. We do have peaks in the trigger distribution, but I think its splitting angle caused the final two particle correlation. The final correlation can also be understood in terms of flows,  $$v_3, v_4$$  are very small.
 * The following shows a $$p_T>3.5 \text{GeV}, 2.5\text{GeV} < p_T <3.5\text{GeV}$$  two particle correlation of  $$\varepsilon_3=0$$  case.


 * In the following plot we show the flows in different cases, where for the reshuffled case we also show the standard deviation of flows, which is huge! We also see in the $$\Phi_2=\pi/2$$  case,  $$v_2$$  significantly decreased while  $$\varepsilon_2$$  remains unchanged.

NEXUS IC with eccentricity random rotations
Based on the previous results, we see that indeed in those cases v3 is small, however, its deviation is big and two particle correlation depends on deviation not only its average value. In specific, for one event, one has


 * $$\begin{align}

\frac{dN_{pair}}{d\Delta\phi}=\sum_i2v_n^2\cos(n\Delta\phi) \end{align}$$ The observed raw two particle correlation is given by


 * $$\begin{align}

<\frac{dN_{pair}}{d\Delta\phi}>=\sum_i2<v_n^2>\cos(n\Delta\phi) \end{align}$$ where the average should be carried out event-by-event. Therefore $$<v_3> \sim 0$$  does not imply  $$<v_3^2> \sim 0$$. This effect can be better seen when one introduce event by event fluctuations, rather than a single event. In the following, we will run several hundreds of NEXUSSPheRIO events and set $$\varepsilon_3 =0$$  for each event, hopefully we will have small  $$<v_3> \sim 0$$  but big enough  $$<v_3^2>$$  to give double peak structure on away side.

Here I attach few plots on flow and eccentricity of NEXUSSPheRIO events, they are obtained by using 0-5%, around 300 events. From the plots, we see big standard deviations, they will not easily disappear when one rips away triangular eccentricity.