I am building a computational model of ellipsoidal cell network formation and I would like to use a particle order parameter to study my model's behavior. I have come across this article Steinhardt, Nelson, Ronchetti, PRB (1983), that was apparently the one that started everything, but the math is quite exotic for me. I have found that the formula:

$$ S = 1/3<cos^2θ - 1> $$

is the order parameter for a 3-D system. This is convenient as S = 0 means no order, while S = 1 means complete alignment. A colleague at the univ has mentioned that for a 2-D particle system the previous equation for S becomes:

$$ S = cos2θ $$

but I don't think this can be right, as in the following example: 2 particles have a 90 degree angle between them, so they are completely unaligned. cos2θ = cos2*90 = -1 which would imply they are completely aligned, which is wrong. Also, the range is [-1:1] instead of 0 < S < 1.

I would really appreciate it, if anyone could point me to the right direction.

  • $\begingroup$ NB 1: $ S = \cos 2\theta = 2 \cos^2 \theta -1 $ (second equality is a trigonometric identity); that is, the suggestion by the colleague of Dima1982 is equivalent to that of @user56071 $\endgroup$ – eldad-a Jun 4 '20 at 6:44
  • $\begingroup$ NB 2: the $\theta$ that goes into the ensemble average in the order parameter $S$ is the angle with respect to the director, denoted below by $\psi_m$ in the answer by @user56071. In the test case of two particles with $90^{\circ}$ between them, one indeed finds that the largest eigenvalue of $\mathbf{Q}$ is $0$. One way to build intuition here is to pick the best candidate for the director, which would be $45^{\circ}$ to each of the two particles; averaging over $ \cos \left( 2 \cdot \frac{\pi}{4} \right) = \cos \frac{\pi}{2} = 0 $ . $\endgroup$ – eldad-a Jun 4 '20 at 7:06
  • $\begingroup$ The way I understood the question, @user56071 should be the accepted answer. $\endgroup$ – eldad-a Jun 4 '20 at 7:13

I met a paper published in PRL using the orietational order parameter and global bond_orietational order parameter. Although the structures in this paper are about the short chain alkanes, i think the definitions of the two kind of order parameters may provide you some insights into your question. enter image description here

  • $\begingroup$ I have looked into the article you posted and it seems that they are also doing this for 3D molecules. So the formula would hold for 3D. I cannot find a 2D formula anywhere and I am thinking that maybe the same could be used for 2D as well. I need to be careful with the assumptions made for 3D in the original article I posted. I am sorry I cannot upvote you. I need more reputation. :) I am still waiting for other answers. If I don't get any today, I will accept yours. $\endgroup$ – Dima1982 May 22 '13 at 14:53
  • $\begingroup$ It does not matter. We all hope the questions can be desirably solved and we can both learn from the questions and answers. $\endgroup$ – Legend_Dyson May 23 '13 at 1:48

I realize that this question was asked some time ago but I looked at this page when I was trying to figure this out in the context of 2D liquid crystals.

Legend_Dyson gave an answer that mentioned the tensor order parameter $\mathbf{Q}$ --- let's talk about that first (and use their notation). Essentially, we are really interested in the tensor $\left\langle\hat{u}_m\hat{u}_m\right\rangle$ (where $\left\langle\cdot\right\rangle$ is the ensemble average over the $m$ nematogens) BUT this is not traceless, which is a property that is generally useful. However, $\mathrm{Tr}\left[\left\langle\hat{u}_m\hat{u}_m\right\rangle\right]=1$ since $\hat{u}_m$ is a unit vector. In 3D, we can make it traceless by subtracting off $\mathbf{I}/3$ but in 2D it should be $\mathbf{I}/2$. The generalization is $$\mathbf{Q}\propto\left\langle\hat{u}_m\hat{u}_m\right\rangle-\frac{\mathbf{I}}{d}.$$ This tensor represents the orientational order. Before we talk about the scalar order parameter $S$ that you are really interested in, let's scaling the tensor by $d-1$ (since I know that it will be needed later to get $0 \leq S \leq 1$). So the tensor order parameter in any dimension is $$\mathbf{Q}=\frac{d\left\langle\hat{u}_m\hat{u}_m\right\rangle-\mathbf{I}}{d-1}.$$ The scalar order parameter $S$ is the largest eigenvalue of $\mathrm{Q}$ and the director $\hat{n}$ is the corresponding eigenvector.

Some math will show you that you could write things like $$\mathbf{Q}=S\frac{d\left\langle\hat{n}\hat{n}\right\rangle-\mathbf{I}}{d-1}$$ and $$S=\frac{\left\langle d \left(\hat{u}\cdot\hat{n}\right)^2-1\right\rangle}{d-1} = \frac{\left\langle d \cdot \cos^2\psi_m-1\right\rangle}{d-1},$$ (where $\psi_m$ is the angle between each orientation $\hat{u}_m$ and the director $\hat{n}$), which are useful if you know the director $\hat{n}$. In 3D, $S=\left\langle 3 \cdot \cos^2\psi_m-1\right\rangle/2$ (like it says in Wikipedia), while in 2D $S=\left\langle 2 \cdot \cos^2\psi_m-1\right\rangle$.

I hope this was helpful. I'm going to leave one last note. If you aren't interested in solving the eigensystem to find $\hat{n}$ and $S$ there is a shortcut in 2D. In 2D, you can write \begin{align} \mathbf{Q} &= \begin{pmatrix} Q_{xx} & Q_{xy} \\ Q_{xy} & -Q_{xx} \end{pmatrix} = S\begin{pmatrix} \cos\left(2\theta\right) & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) \end{pmatrix}. \end{align} From this form, you can see $S=\sqrt{Q_{xx}^2+Q_{xy}^2}$ and it is also known that the director is $\hat{n}=\left[\cos\theta,\sin\theta\right]$, which can be calculated by just solving for the angle $\theta$.

  • $\begingroup$ Thank you for the helpful answer! To clarify why "...traceless, which is a property that is generally useful.": traceless-ness is desired such that in the isotropic case -- uniform angular distribution, $\mathbf{Q} = \mathbf{0}$; so as an order parameter it indeed becomes zero when the associated order is absent. $\endgroup$ – eldad-a Jun 5 '20 at 1:05
  • $\begingroup$ An mathematically equivalent approach is to express the orientations as points on the unit circle in the complex plane. Hence, for the sample $ \{ \theta_m \}$, which are the angles of the $\hat{u}_m$, construct the complex number : $ \rho = \langle e^{i\,2\theta} \rangle $ ; the director is oriented at $ \theta_{\hat{n}} = \arg \{ \rho \} / 2 $ and $S = | \rho | $. NB. The factor $2$ is to rescale $[0,\pi] \to [0,2\pi] $ , and is delibrately not applied when retrieving $S$. $\endgroup$ – eldad-a Jun 5 '20 at 2:00

In this article PhysRevE.67.031608 an order parameter is given for a 2D system. Originally, it was introduced by Halperin and Nelson. It is the absolute value of the sixth Fourier component of the bond angle distribution function, which is constant in the isotropic fluid and consists of six equally spaced peaks in the solid phase. It is given by:

$$ \Phi_6 = \Big\langle \frac{1}{N} \sum_{m=1}^{N}\frac{1}{N_b}\sum_{n=1}^{N_b}e^{6i\theta_{mn}} \Big\rangle $$

The angular brackets indicate the configurational average and $\theta_{mn}$ is the angle between some fixed axis (e.g. x or y) and the bond joining the m-th particle with a neighboring n-th particle. $N_b$ is the number of particle-neighbor bonds.

The local bond-orientational order parameter is:

$$ \phi_6 =\Big\langle \frac{1}{N_b}\sum_{n=1}^{N_b}e^{6i\theta_{mn}} \Big\rangle $$

that is the order parameter of a single particle. $|\Phi_6|$ and $|\phi_6|$ produce a number between 0 and 1 that represents the orientational order.

I guess this answers my question.

The only question I have now is if the input angle $\theta_{mn}$ needs to be in radians or degrees. I have tried some mock examples by hand and I couldn't understand my own results! Any help would be appreciated!

EDIT: I have come to this answer by reading about the "hexatic" phase of fluids, nematic liquid crystals and the two dimensional melting. As to how feasible these are to use in my research, remains to be seen.


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