Is Rama stable?

[dasmiller]

No. The angular velocity around the other two axes will oscillate with frequency (A-B)(A-C)/BC. Since they started small, they remain small.

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?

It will not reorient, as that would violate conservation of angular momentum.

I was using 'reorient' because I didn't want to imply that the spin axis wasn't fixed in inertial space. The body changes orientation, but the axis of rotation doesn't. If this is a violation of conservation of angular momentum for the spin-axis-near-min-axis case, why is it not a violation of angular momentum for a body initially spinning about the intermediate axis?

 

[ben m]

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?

The deviation is only as big as the perturbation. Small perturbation, small deviation. That's how all stable systems work.

By contrast, for an object spinning around its "middle" axis, any arbitrarily small perturbation will make it tumble. The size of the deviation grows and becomes extremely large, even if the perturbation was small.

This is how all stable equilibria work. Think of, say, a mass sliding around the bottom of a frictionless bowl. If you give it a small push away from the bottom, it will undergo small oscillations about its rest position. A larger push will give larger oscillations. That's stable. By contrast, consider a mass sitting on the peak of a frictionless hill; a small push leads to a large excursion down the hill. That's unstable.

 

[sol invictus]

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?

As ben says, the definition of "stable" is that small perturbations do not grow but instead lead to small oscillations around the stable point.

I was using 'reorient' because I didn't want to imply that the spin axis wasn't fixed in inertial space. The body changes orientation, but the axis of rotation doesn't.

The body cannot reorient in the way you proposed unless either energy or angular momentum conservation is violated (such as if a torque is applied, or if the body can lose energy through some interactions).

If this is a violation of conservation of angular momentum for the spin-axis-near-min-axis case, why is it not a violation of angular momentum for a body initially spinning about the intermediate axis?

It is. A body initially spinning around the intermediate axis does not reorient to spin around one of the stable axes, at least not so long as its energy and angular momentum are constant. It just tumbles around a lot, going very far from the intermed axis and only occasionally coming back.

What you may be thinking of is that rotations around the min axis require the most energy. So if you have a process that reduces the energy which retaining the angular momentum (which is sometimes a good approximation for certain kinds of dissipation), the body will eventually re-orient itself to rotate around the max axis.

 

[Dorfl]

What you may be thinking of is that rotations around the min axis require the most energy. So if you have a process that reduces the energy which retaining the angular momentum (which is sometimes a good approximation for certain kinds of dissipation), the body will eventually re-orient itself to rotate around the max axis.

Doesn't every non-rigid body have the ability to reduce energy while retaining angular momentum?

 

[sol invictus]

Doesn't every non-rigid body have the ability to reduce energy while retaining angular momentum?

No, I wouldn't say every. But generally it is easier to lose energy than it is to lose angular momentum (that's why galaxies are flat and the planets orbit the sun in a plane).

I suppose a non-rigid body can lose energy by generating heat from internal motions and radiating it away. That process doesn't quite conserve the angular momentum of the body, but it should be close enough. However it could take an extremely long time for this to significantly affect the motion if the body is nearly rigid. Moreover not many bodies are totally isolated in space, so interactions with the surrounding environment are often going to be more important.

 

[dasmiller]

The deviation is only as big as the perturbation. Small perturbation, small deviation. That's how all stable systems work.

<snip>

This is how all stable equilibria work. Think of, say, a mass sliding around the bottom of a frictionless bowl. If you give it a small push away from the bottom, it will undergo small oscillations about its rest position. A larger push will give larger oscillations. That's stable. By contrast, consider a mass sitting on the peak of a frictionless hill; a small push leads to a large excursion down the hill. That's unstable.

Interestingly, we simply have a different use of terminology here. In aircraft dynamics, for example, if a small perturbation isn't damped, then the aircraft isn't considered "stable."

But that being said, I was operating under some confusion (see next post)

 

[dasmiller]

The body cannot reorient in the way you proposed unless either energy or angular momentum conservation is violated (such as if a torque is applied, or if the body can lose energy through some interactions).

So I woke up this morning troubled by the fact that in my view of rotating rigid bodies, the motions wouldn't necessarily be reversible. If one magically and perfectly reversed the rotation, the rigid body wouldn't eventually work its way back to the condition it was in when the perturbation was applied. That's a pretty serious flaw for a rigid body that isn't interacting with its environment.

After more pondering, I think I was led astray by way too many years working with satellites. Most real-world satellites don't act like rigid bodies because of fuel slosh. I believe that fuel motion provides the sort of energy loss (not necessarily angular momentum loss) I'd need to get the nutation to grow. Indeed, as I recall, the dedamping (the term "amplification" was reserved for other effects) constants were always associated with fuel motion.

But I was definitely confused on rigid body rotation. Now I'm not entirely sure that I'm unconfused on the rotation of a body with significant quantities of viscous fluids, but . . . that may be a bit ambitious for a Saturday morning.

 

[sol invictus]

Interestingly, we simply have a different use of terminology here. In aircraft dynamics, for example, if a small perturbation isn't damped, then the aircraft isn't considered "stable."

If there's dissipation, small oscillations will damp out so long as they don't grow due to some dynamical instability. So the two uses are consistent.

So I woke up this morning troubled by the fact that in my view of rotating rigid bodies, the motions wouldn't necessarily be reversible. If one magically and perfectly reversed the rotation, the rigid body wouldn't eventually work its way back to the condition it was in when the perturbation was applied.

Precisely. Damping is an irreversible process that involves an increase in entropy, so it can only happen when the "rigid" body is coupled to many other degrees of freedom (such as molecules of fuel).

But I was definitely confused on rigid body rotation. Now I'm not entirely sure that I'm unconfused on the rotation of a body with significant quantities of viscous fluids, but . . . that may be a bit ambitious for a Saturday morning.

There's a beautiful geometric way to visualize the motion which this thread reminded me of. Attach a coordinate frame to the rigid body, with the origin at the center of mass and the axes aligned with the principal axes of the body. Now use that coordinate system. It's non-inertial, which means the angular momentum vector in those coordinates doesn't remain constant. But because these coordinates are related to inertial coordinates by a rotation, the length of the angular momentum vector is constant. So is the total energy, of course. Therefore

[latex]${\rm total \, angular \, momentum}=\vec L^2 = L_x^2+L_y^2+L_z^2 ={\rm constant}$[/latex]
[latex]${\rm kinetic \, energy}=T={1 \over 2}\left( L_x^2/A+L_y^2/B+L_z^2/C \right) = {\rm constant}$[/latex]

The first equation is a sphere, the second is an ellipsoid. Therefore for some given angular momentum and energy, the motion will be along a line of intersection between those two shapes. That line of intersection is some closed curve on the sphere. It can be a small circle around the min axis or a small circle around the max axis. But it can't be a small circle around the mid axis, as you can see if you visualize the intersection - it must be a very large circle that only occasionally passes close to it.

So you can see the stability purely from that, which is pretty nice. No need to write any differential equations at all.

 

[Dorfl]

No, I wouldn't say every. But generally it is easier to lose energy than it is to lose angular momentum (that's why galaxies are flat and the planets orbit the sun in a plane).

I suppose a non-rigid body can lose energy by generating heat from internal motions and radiating it away. That process doesn't quite conserve the angular momentum of the body, but it should be close enough. However it could take an extremely long time for this to significantly affect the motion if the body is nearly rigid. Moreover not many bodies are totally isolated in space, so interactions with the surrounding environment are often going to be more important.

Sorry, I meant kinetic energy, not just energy. That can be decreased without changing angular momentum, or requiring transporting away, can't it?

Could Rama's atmosphere heating up then absorb some of the kinetic energy, leading to it eventually becoming unstable?

 

[sol invictus]

Sorry, I meant kinetic energy, not just energy. That can be decreased without changing angular momentum, or requiring transporting away, can't it?

Under some circumstances, yes.

Could Rama's atmosphere heating up then absorb some of the kinetic energy, leading to it eventually becoming unstable?

Any system will eventually evolve so as to maximize its free energy subject to whatever constraints are in place. In this case in the medium-long term that would probably correspond to a uniform rotation around the axis with the maximum moment while the excess energy goes into either heating the atmosphere/water or radiation. In the very long term I suspect it would correspond to zero rotation, even if Rama is completely isolated (because it can very slowly radiate away its angular momentum as well).

 

[dasmiller]

If there's dissipation, small oscillations will damp out so long as they don't grow due to some dynamical instability. So the two uses are consistent.

You'd think that I'd learn not to disagree, but the two uses don't sound consistent to me.

An aircraft is not considered to be stable if it simply maintains a perturbed orientation, or if it oscillates indefinitely due to a perturbation. The undamped rotating rigid body is considered to be stable even though the perturbation causes permanent oscillation.

Maybe we'll agree to disagree on that one.

Anyway, back to Rama and my earlier remark about 'tuning' the ocean. Barring any active measures (e.g. pumps), I suspect that the ocean can only dissipate energy. 'Tuning' would mean designing the ocean to get a particular dedamping constant, but since the ocean would always be dissipating energy, you'd never be able to get a negative dedamping constant. The ocean would always cause perturbations to grow.

 

[Myriad]

[obsessive sf fan]The Raman ocean is frozen except when it approaches a star system.[/obsessive sf fan]

Respectfully,
Myriad

 

[sol invictus]

You'd think that I'd learn not to disagree, but the two uses don't sound consistent to me.

An aircraft is not considered to be stable if it simply maintains a perturbed orientation, or if it oscillates indefinitely due to a perturbation. The undamped rotating rigid body is considered to be stable even though the perturbation causes permanent oscillation.

Normally, stable equilibria occur at the bottom of a potential well. Like ben said, think of a marble rolling around near the bottom of a bowl. If the marble is right at the bottom it stays there, but if you move it a little off it oscillates around the stable point. Give it a little friction and the oscillations damp, and eventually it's at rest at the stable point again.

This stable point around the min axis is very unusual, though, because it's not a minimum of the energy - it's a max (with fixed angular momentum at least). Nevertheless it's stable in the sense that small oscillations don't grow in the ideal case with no friction... but it does seem to me that damping actually makes them grow rather shrink. So in this case I think I agree with you that the two uses don't agree. And thanks for pointing it out, because I'm not sure I know another example like that.

The ocean would always cause perturbations to grow.

You mean, around the min axis? Around the max axis it should be stable in all senses of the word.

 

[dasmiller]

You mean, around the min axis? Around the max axis it should be stable in all senses of the word.

Yes, I was thinking specifically of Rama nominally spinning around the min axis.

 

[Dr Adequate]

Since we know that Rama does not obey Newton's Third Law, any analysis of its motion is going to be fraught with difficulties.

 

[Dorfl]

Since we know that Rama does not obey Newton's Third Law, any analysis of its motion is going to be fraught with difficulties.

Point. But I thought it behaved classically when in flight. It seemed like the non-Newtonian stuff only happened when it had just refuelled near a star.

 

[nathan]

Point. But I thought it behaved classically when in flight. It seemed like the non-Newtonian stuff only happened when it had just refuelled near a star.

Hey, this is like a detective story -- some participants have more information than others

 

[dasmiller]

Normally, stable equilibria occur at the bottom of a potential well. Like ben said, think of a marble rolling around near the bottom of a bowl. If the marble is right at the bottom it stays there, but if you move it a little off it oscillates around the stable point. Give it a little friction and the oscillations damp, and eventually it's at rest at the stable point again.

This stable point around the min axis is very unusual, though, because it's not a minimum of the energy - it's a max (with fixed angular momentum at least). Nevertheless it's stable in the sense that small oscillations don't grow in the ideal case with no friction... but it does seem to me that damping actually makes them grow rather shrink. So in this case I think I agree with you that the two uses don't agree. And thanks for pointing it out, because I'm not sure I know another example like that.

It seems to me that the marble-in-the-bowl might actually be another example. If I whip the marble around so that it's quickly circling just below the rim of the bowl, it that trajectory stable? (that's not a rhetorical question. I really don't know, but it seems like it could be). It's certainly not a minimum-energy case.

 

[sol invictus]

It seems to me that the marble-in-the-bowl might actually be another example. If I whip the marble around so that it's quickly circling just below the rim of the bowl, it that trajectory stable? (that's not a rhetorical question. I really don't know, but it seems like it could be). It's certainly not a minimum-energy case.

If it's just below the rim it's not, because it can fly out under a small perturbation. If it's farther down it might be, depending on the shape of the bowl.

But you make a good point. Orbits are examples of this, for example the earth's orbit around the sun. Even subject to no external perturbations, a two-body gravitational system will eventually spiral together, because that configuration has lower energy and the excess can be radiated away in various forms (including gravitational radiation) while increasing the entropy.

 

[ben m]

If it's just below the rim it's not, because it can fly out under a small perturbation. If it's farther down it might be, depending on the shape of the bowl.

But you make a good point. Orbits are examples of this, for example the earth's orbit around the sun. Even subject to no external perturbations, a two-body gravitational system will eventually spiral together, because that configuration has lower energy and the excess can be radiated away in various forms (including gravitational radiation) while increasing the entropy.

A Keplerian orbit isn't the minimum of a global energy function, but it is the minimum of an energy function constrained by angular momentum conservation. I wondered if their might not be a similar "effective energy" function for Rama, and a wobbly-Rama might be shown to be making small oscillations around the effective minimum. But on reflection I'm not sure this makes sense---I can't quite see how this breaks up into a dynamical variable and a constraint equation.

But I haven't done solid-body mechanics in ages.