1
$\begingroup$

When you spin a top, while it first wobbles, it will eventually reach a period when it spin 'smoothly', then finally falls to the ground. I'm curious about whether its motion (or maybe the path that a particular point/axis traces?) is chaotic. I read that the motion of an object rotated about its second axis is quasi-period while not chaotic – how does this apply to spinning tops? Would the presence of friction in real-life have any significance? (because without friction/resistant forces, a top would spin in the same manner forever, so that motion would be periodic.)

(I have no background in this topic – I am just interested because I recently read James Gleick's Chaos and became curious about other chaotic systems in our world. Let me know if my question doesn't make sense.)

$\endgroup$
0
1
$\begingroup$

Let me first go over the property that makes a system satisfy the definition of 'chaotic':

If you start multiple simulation runs, with minimally different starting conditions, the states of the respective simulation runs will diverge, and eventually the states will be anywhere within the available state space.

One of the usual examples: the solar system. Our solar system is long term stable, so you can predict that the Earth will see solar eclipses as long as the solar system exists, but in multiple simulation runs the exact time of the eclipses will diverge.

Every celestial body of the solar system exerts some gravitational pull on every other celestial body. Because of that any difference in initial conditions, no matter how small, is continuously multiplied and the differenes accumulate.

A system can be a chaotic system if it has multiple members that all interact. It seems to me the minimum amount of members is three.

In the case of a spinning top:
In that case the system consists of two members: the spinning top, and the surface that the spinning top is standing on.

When you model that case in idealized form (no friction) then there is no room for chaotic effect.

In terms of celestial mechanics the counterpart would be as follows: as primary, say: a white dwarf, and as secondary a small celestial body that is small enough to have lost internal heat to the extent that it is completely solid. Let the secondary have a spin, and let its spin axis have an angle with respect to the plane of its orbit. Then that secondary will have a precession of the equinoxes (precession of the equionoxes is gyroscopice precession). That precession will remain the same forever. The secondary is solid, so there is no way for it to shed rotational kinetic energy. As long as the rotation rate remains the same, and the distance to the primary remains the same, the rate of precession will remain the same. No room for chaotic effect.

Now, in the case of a real world spinning top the tip has a radius, so the tip can slide a bit or walk around a bit, and that introduces a degree of unpredictability. But that type of unpredictability is different from chaotic effect.

Part of the the definition of chaotic effect is that it is not just occuring in the messy real world, chaotic effect also manifests in the idealized case.

If your simulation is the idealized case, and the chaotic effect is still there, that is when you have actual chaotic effect.

Conversely, if you can make the effect go away by moving to an idealized case, then the physics is not actually chaotic effect.

$\endgroup$
1
  • $\begingroup$ A system can be a chaotic system if it has multiple members that all interact. It seems to me the minimum amount of members is three. – This is clearly wrong. The double pendulum (without friction) is clearly chaotic, but has only two interacting members. What is correct is that you need to have at least three independent dynamical variables for chaos (which means four degrees of freedom in a conservative system, since you cannot access one degree due to energy conservation) but that is given here: You have the inclination as well as three angular momenta. $\endgroup$ – Wrzlprmft May 14 '20 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.