The Order Parameter

Module 2: Synchronization and Collective Behaviour

To visualize the Kuramoto model, picture each phase $\theta_j$ as a point $e^{i\theta_j}$ running around the unit circle in the complex plane (Strogatz 2000).

Figure 1: Oscillators (blue dots) on the unit circle. The order parameter (red) points towards the centroid of the phases.

Definition

The complex order parameter is the centroid of these points:

\[ r e^{i\psi} = \frac{1}{N} \sum_{j=1}^N e^{i\theta_j}. \tag{1}\]

$r \in [0, 1]$ is the coherence. $r = 1$ when all phases coincide. $r \approx 0$ when the phases are spread around the circle.
$\psi$ is the mean phase, the phase of the collective rhythm.

In Figure 1 the initial phases cover only part of the circle, so the red vector has a clear length and direction. To see the vector move as the phases evolve, run the script amlab/odes_coupled/kuramoto.py. You build the same animation yourself in Animation.

Mean-field form

Multiply Equation 1 by $e^{-i\theta_i}$ and take the imaginary part:

\[ r\sin(\psi - \theta_i) = \frac{1}{N}\sum_{j=1}^N \sin(\theta_j - \theta_i). \]

The Kuramoto model therefore becomes

\[ \dot\theta_i = \omega_i + K r \sin(\psi - \theta_i),\quad i = 1, \dots, N. \tag{2}\]

This rewriting is exact for every $N$ (Strogatz 2000). Each oscillator no longer sees the others one by one. It sees only the mean field: it is pulled toward the mean phase $\psi$ with effective strength $Kr$. Computing $r$ and $\psi$ costs $O(N)$ operations per step, while the double sum in the pairwise form costs $O(N^2)$.

Feedback between coherence and coupling

Equation 2 shows why synchronization can appear spontaneously (Strogatz 2000):

Oscillators are pulled toward $\psi$ with strength $Kr$.
If more of them cluster, $r$ grows.
A larger $r$ increases the pull $Kr$, which recruits more oscillators.

The loop is opposed by the spread of natural frequencies. Oscillators whose $|\omega_i|$ exceeds $Kr$ cannot follow the mean field and keep drifting. The balance sets a threshold coupling $K_c$, derived in Critical Coupling.

Exercise

Suppose the $N$ phases are independent and uniform on $[0, 2\pi)$. Show that $\mathbb{E}[r^2] = 1/N$. What does this imply for the value of $r$ you measure in an incoherent simulation?

Solution

\[ r^2 = \left|\frac{1}{N}\sum_j e^{i\theta_j}\right|^2 = \frac{1}{N^2}\sum_{j,k} e^{i(\theta_j - \theta_k)}. \]

The $N$ terms with $j = k$ equal $1$. For $j \ne k$ the phases are independent, so $\mathbb{E}[e^{i(\theta_j-\theta_k)}] = \mathbb{E}[e^{i\theta_j}]\,\mathbb{E}[e^{-i\theta_k}] = 0$. Hence $\mathbb{E}[r^2] = N/N^2 = 1/N$. An incoherent population with $N = 100$ shows $r$ of order $0.1$, not $0$. You will see this floor in Finite-Size Effects.

Exercise

Compute $r$ for three oscillators at phases $0$, $2\pi/3$ and $4\pi/3$, and for three oscillators at $0$, $0$ and $\pi$.

Solution

The first three points are the cube roots of unity, which sum to zero, so $r = 0$. For the second set, $e^{0} + e^{0} + e^{i\pi} = 1$, so $r = 1/3$ and $\psi = 0$.

References

Strogatz, Steven H. 2000. “From Kuramoto to Crawford: Exploring the Onset of Synchronization in Populations of Coupled Oscillators.” Physica D: Nonlinear Phenomena 143: 1–20.

--- title: "The Order Parameter" subtitle: "Module 2: Synchronization and Collective Behaviour" format: html --- To visualize the Kuramoto model, picture each phase $\theta_j$ as a point $e^{i\theta_j}$ running around the unit circle in the complex plane [@strogatz2000kuramoto]. ```{python} #| label: fig-kuramoto-circle #| fig-cap: 'Oscillators (blue dots) on the unit circle. The order parameter (red) points towards the centroid of the phases.' #| fig-width: 5 #| fig-height: 5 #| echo: false import numpy as np import matplotlib.pyplot as plt from amlab.odes_coupled.kuramoto import initialize_oscillators, kuramoto_order_parameter num_oscillators = 100 theta, omega = initialize_oscillators( num_oscillators, distribution="normal", scale_omega=0.5, scale_phase=0.6, seed=1) x = np.cos(theta) y = np.sin(theta) r, phi, rcosphi, rsinphi = kuramoto_order_parameter(theta) fig, ax = plt.subplots(figsize=(5, 5)) ax.set_xlabel(r"$\cos\theta$") ax.set_ylabel(r"$\sin\theta$") ax.set_xlim(-1.1, 1.1) ax.set_ylim(-1.1, 1.1) ax.set_aspect("equal") ax.grid(True) circle = plt.Circle((0, 0), 1, color="lightgray", fill=False) ax.add_artist(circle) ax.scatter(x, y, s=50, color="blue", alpha=0.5) ax.plot([0, rcosphi], [0, rsinphi], color="red", linewidth=2) ax.plot([rcosphi], [rsinphi], "ro", markersize=6) plt.show() ``` ## Definition The complex **order parameter** is the centroid of these points: $$ r e^{i\psi} = \frac{1}{N} \sum_{j=1}^N e^{i\theta_j}. $$ {#eq-r-exp} - $r \in [0, 1]$ is the **coherence**. $r = 1$ when all phases coincide. $r \approx 0$ when the phases are spread around the circle. - $\psi$ is the **mean phase**, the phase of the collective rhythm. In @fig-kuramoto-circle the initial phases cover only part of the circle, so the red vector has a clear length and direction. To see the vector move as the phases evolve, run the script [amlab/odes_coupled/kuramoto.py](../../amlab/odes_coupled/kuramoto.py). You build the same animation yourself in [Animation](kuramoto-animation.qmd). ## Mean-field form Multiply @eq-r-exp by $e^{-i\theta_i}$ and take the imaginary part: $$ r\sin(\psi - \theta_i) = \frac{1}{N}\sum_{j=1}^N \sin(\theta_j - \theta_i). $$ The Kuramoto model therefore becomes $$ \dot\theta_i = \omega_i + K r \sin(\psi - \theta_i),\quad i = 1, \dots, N. $$ {#eq-ode-mean-field} This rewriting is exact for every $N$ [@strogatz2000kuramoto]. Each oscillator no longer sees the others one by one. It sees only the mean field: it is pulled toward the mean phase $\psi$ with effective strength $Kr$. Computing $r$ and $\psi$ costs $O(N)$ operations per step, while the double sum in the pairwise form costs $O(N^2)$. ## Feedback between coherence and coupling @eq-ode-mean-field shows why synchronization can appear spontaneously [@strogatz2000kuramoto]: - Oscillators are pulled toward $\psi$ with strength $Kr$. - If more of them cluster, $r$ grows. - A larger $r$ increases the pull $Kr$, which recruits more oscillators. The loop is opposed by the spread of natural frequencies. Oscillators whose $|\omega_i|$ exceeds $Kr$ cannot follow the mean field and keep drifting. The balance sets a threshold coupling $K_c$, derived in [Critical Coupling](critical-coupling.qmd). ::: {.callout-note} ## Exercise Suppose the $N$ phases are independent and uniform on $[0, 2\pi)$. Show that $\mathbb{E}[r^2] = 1/N$. What does this imply for the value of $r$ you measure in an incoherent simulation? ::: ::: {.callout-tip collapse="true"} ## Solution $$ r^2 = \left|\frac{1}{N}\sum_j e^{i\theta_j}\right|^2 = \frac{1}{N^2}\sum_{j,k} e^{i(\theta_j - \theta_k)}. $$ The $N$ terms with $j = k$ equal $1$. For $j \ne k$ the phases are independent, so $\mathbb{E}[e^{i(\theta_j-\theta_k)}] = \mathbb{E}[e^{i\theta_j}]\,\mathbb{E}[e^{-i\theta_k}] = 0$. Hence $\mathbb{E}[r^2] = N/N^2 = 1/N$. An incoherent population with $N = 100$ shows $r$ of order $0.1$, not $0$. You will see this floor in [Finite-Size Effects](kuramoto-finite-size.qmd). ::: ::: {.callout-note} ## Exercise Compute $r$ for three oscillators at phases $0$, $2\pi/3$ and $4\pi/3$, and for three oscillators at $0$, $0$ and $\pi$. ::: ::: {.callout-tip collapse="true"} ## Solution The first three points are the cube roots of unity, which sum to zero, so $r = 0$. For the second set, $e^{0} + e^{0} + e^{i\pi} = 1$, so $r = 1/3$ and $\psi = 0$. :::