# Synchronization phenomenon: A simple explanation?

Being from a mathematical background, physicists' intuitive arguments always seemed challenging for me to follow. I am currently reading a book called "Synchronization: A Universal Concept in Nonlinear Sciences". The book is well-written but from time to time, I have hard time following the intuitive arguments. In its most simple case the authors discuss the effect of a periodic external force on a self-sustained quasilinear oscillator. I have to emphasize that these are arguments in chapter 3 of the book. Things get more formal in chapter 7. The only problem is that in chapter 7 there is no argument about potentials anymore. In chapter 3 the argument mainly relies on a few parameters, $\overline\phi_e$ being the initial phase of the forcing oscillator which has the following phase relation $\phi_e=\omega t+\overline\phi_e$ and the driven oscillator is based on $x(t)=A\sin(\omega_0 t+\phi_0)$ and the external force varies based on $\epsilon \cos(\omega t+\overline\phi_e)$. The following points are the main things that I don't understand:

1. What does potential mean?(I know that potential function defines an invariance most of the times, but how it is defnied for oscillators case?
2. What are big parallel vectors and the small ones in figure 3.3(a) representing?
3. Why they phase locking take place in $\phi_0$
4. Why point 1 is stable and point 2 is not? I imagine stability means the point with minimum level of potential. I hope the answer to this question becomes clear if I get an answer for the first question.

The following figures might be helpful for this clarification:

• Which part you dont' understand? Jul 7 '14 at 1:27
• I just made some of the main questions more clear. Thanks for asking. Jul 7 '14 at 11:08
• You may need a mathematical physics introductory book. Potential, stability and force fields are absolutely needed in order to understand the topic Jul 7 '14 at 11:57
• @Lelesquiz, At least in terms of definition, one might be able to define the terms clearly. Right? I think even that would be enough for me. for example: I know what is a potential in gereral sense. The question is how it is defined here. I edited my questions. Jul 7 '14 at 14:08
• Watch 32 out-of-sync metronomes magically synchronize. 4 minute video. Jul 14 '16 at 5:13

1 What does potential mean?(I know that potential function defines an invariance most of the times, but how it is defnied for oscillators case?

Here it should be defined as the potential energy of the system. This page should clarify a little bit wikipedia entry on potential energy. In general potentials are quite a complex topic, but here it should suffice to know that minus the gradient of the potential gives you the force acting on the body.

2 What are big parallel vectors and the small ones in figure 3.3(a) representing?

The big parallel vectors should represent forces. The small ones should represent those forces projected along the tangent lines to the circle.

3 Why they phase locking take place in $\phi_0$

Because $\phi_0$ is a point of minimum potential energy. Then the gradient in that point is zero. Hence the force on the body is zero (in that point).

4 Why point 1 is stable and point 2 is not? I imagine stability means the point with minimum level of potential. I hope the answer to this question becomes clear if I get an answer for the first question.

Stability is another super-complex topic: you can get an idea here: wikipedia page on stability theory. Simplifying a lot, a particular "situation" (actually a point in the phase space) is stable if after an arbitrary long time interval the system has evolved into a "situation" that is very close to the initial one. In a limit case, initial and final situations can be the same. (Here I'm not referring just to position, since velocity is also needed to specify the "situation". The key concept here is phase space)

What do you need to know here? That an energy minimum is a stable position, while a maximum is an unstable position. Why that if in both position the force acting on the body is zero? Because a small variation in position gives you a force driving you back to the minimum in the first case, while in case of a maximum that force drives you in a direction opposite of the minimum.

That's all. But really consider studying a little bit of mathematical physics before reading your book. In order to appreciate it you need to know those basic concepts; it think that otherwise you could have some real issues understanding the topic.

• Thank you very much. But I have a lot of problems: 1-How do you define this energy on an abstract level.Since in the end we can write the problem in sole dynamical systems framework and there is then no possibility of relying on physical energies. 2-Are you sure about the small ones? Since if that would be the case it should have been zero when the bigger vector gets orthogonal to circle Jul 8 '14 at 9:54
• 3-Oh that I actually knew. I was trying to see if there is justfiication using dynamics rather than potential argument. 4-Oh! Yes I know what stability means in mathematical terms for a dynamical system. I don't know why in mathematical terms point 1 is stable and point 2 is not. So reading through your argument it seems that the force acting in these points is zero which means $\dot x$(being the state) is zero and therefore the system stops at these points. Right? Jul 8 '14 at 9:58
• Once again thank you very much indeed for the trouble of writing this long answer. In any case I would be very glad if you can introduce me a CONCISE mathematical physics book that could cover most of the topics here. Jul 8 '14 at 10:00
• 1) You define potential as that scalar function (if it exists) whose gradient gives you the force on the particle/body. Since $- \nabla U = f = m a$ you just need to specify that $\ddot{x} \propto - \nabla U$ Jul 8 '14 at 12:49
• 2) It isn't actually the projection along the circle, but something that looks similar. Otherwise plot b) would be a cosine, but it isn't. Anyway it looks like the resulting force acting on the body. It is zero at minimum and maximum, while in the other cases it drives your particle to the minimum. Jul 8 '14 at 12:54

The argument here is to relate the phase space of self-sustained oscillator with the system of bead in a ring in viscous fluid.

Mathematically, the force field in this case come from the driving force $\epsilon \cos(\omega t + \bar{\phi_e})$. In the rotating frame of $\omega$, the driving force will become $\epsilon(e^{i \bar{\phi_e}}+e^{-i(2\omega t+ \bar{\phi_e})})$, the second term can be dropped if the detunning is small $2\omega \gg\Delta=|\omega_0 - \omega|$. Also, the original dynamic $\phi(t)$ will become $A e^{i((\omega_0 -\omega)t +\phi_0)}=A e^{i(\Delta t +\phi_0)}$. With zero detuning $\Delta=0$, the phase point doesn't change with time, which is what you expect when you travel along the same rotating frame as the particle.

What does potential mean?(I know that potential function defines an invariance most of the times, but how it is defnied for oscillators case?

If you don't like how the figure draw, you can just rotate the figure until the force vector pointing downward. Then it is exact a bead in a ring in viscous fluid under gravity and you should be able to understand most concept immediately.

For any conservative force field $F(r)$, potential energy $V(r)$ is always well defined by the integral $V(r)=-\int F(r)\cdot dr$. The force here is constant (so it is conservative). Its magnitude is given by $\epsilon$, and the direction depends on how it drive the system.

What are big parallel vectors and the small ones in figure 3.3(a) representing?

In Fig. 3.3, the big vectors is the "force vector" in phase space, small one is its parallel component to the ring.

For non-zero detuning, say $\Delta=|\omega_0 -\omega|>0$, you would expect the particle move counter-clockwise because it rotates faster than your rotating frame $\omega$. This create effective force as shown in Fig. 3.4. Its magnitude is constant, and its direction is always parallel to the ring as expected. Fig. 3.4 explains how these two force can balance out and create two points of equilibrium, or no point of equilibrium.

Why they phase locking take place in ϕ0

For a bead in a ring in viscous fluid under gravity, you will expect that the bead will move to the lowest point, which is $\phi_0$.

Why point 1 is stable and point 2 is not? I imagine stability means the point with minimum level of potential. I hope the answer to this question becomes clear if I get an answer for the first question.

The definition of stability at a point $r(t)$ is that the system will eventual go back to $r(t)$ for any small perturbation $\delta r$ as $t \to \infty$. The definition of equilibrium is that there is no net force.

It is obvious that both point 1 and 2 are equilibrium because there is no force along the ring. However, a small perturbation on the highest point will drive the bead to the lowest point 1, while perturbation on 1 will only drive it back to point 1. That is why point 1 is called stable equilibrium, but point 2 is called unstable equilibrium.

This picture can be somehow understood by the dynamics equation: $$\ddot{x}+\omega_0^2x = f(x,\dot{x}) + \epsilon p(t)$$ The $f(x,\dot{x})$ is the internal dynamic and $\epsilon p(t)$ is the external driving force. To understand the basic idea of limit cycle, you need to know the simplest example: damped and driven harmonic oscillator.

I guess here $f(x,\dot{x}) = -\gamma \dot{x} + \tilde{A} \cos(\omega_0 t + \tilde{\phi}_0)$ in author's mind for the self-sustained oscillator right at the resonance frequency $\omega_0$ of the oscillator. Also external force should be $p(x)=\cos(\omega t + \bar{\phi_e})$.

• Oh, the comments in bounty were for your answer. I am not able to give it until tomorrow. Jul 9 '14 at 22:55
• @Cupitor oops, a bounty. It is not the typical way to use bounty except you want to attract attention to the question. If you think the answer (partially) solve your problem, you should accept it. Jul 9 '14 at 23:20
• I know how the webstie functions my friend. You can also give bounty to thank poeple :) Jul 10 '14 at 0:42
• @Cupitor Thank you :) But you are still doing the things in the wrong order. Jul 10 '14 at 4:31
• Not indeed. The last option for bounty is "Rewarding existing answer" ;) Jul 10 '14 at 10:37