Re: [Phys-L] intermediate axis theorem

[John Denker <jsd@av8n.com>]

On 12/26/2012 07:26 AM, Carl Mungan wrote:
> I'm looking for an intuitive argument for the intermediate-axis
> theorem (IAT),  rather than a formal derivation via Euler's
> equations.

Gack!  In my previous answer, I violated an important principle:
When talking about stability, it pays to say explicitly "stable
with respect to WHAT?"

Web searches indicate that I am not the first person to make 
this mistake in the context of this problem.

So, there are two separate versions of this problem to be
considered ... plus an interesting interaction between the
two versions.

A) The easy version applies to an isolated _almost_ rigid
 system.  There is a tiny bit of internal dissipation.  The
 dissipation arises from the fact that the system is not 100% 
 rigid.

 The assumptions guarantee that the system has constant
 angular momentum ... but not necessarily constant rotational
 energy.  Some of the energy might be dissipated as heat.

 This can be analyzed in terms of basic 3rd-grade physics.
 As surely as balls roll downhill and are stable at the
 bottom of the hill, this system is stable (in the
 thermodynamic sense) when rotating around the axis with
 the largest angular momentum.

B) The more challenging case, which was probably the one CM
 intended, assumes constant rotational energy as well as 
 constant angular momentum.

 Here we are interested in stability in the Lyapunov sense.
 That is, given nearby initial conditions, do we get nearby
 trajectories in the long term?

 The intuitive way to handle this is to think about it in
 angular momentum space.  The contours of constant energy
 are ellipsoids in this space.  Let's choose one such
 contour, i.e. let's choose a definite energy.

 If the norm of the angular momentum is as small as possible 
 (consistent with the chosen energy) the angular momentum must 
 be at the short axis of the ellipsoid.  If the norm is merely 
 /near/ the smallest possible value, the angular momentum must 
 be near the short axis.  The geometry of spheres intersecting
 with an ellipsoid guarantees it.

 Similarly, if the norm of the angular momentum is at or near
 the largest possible value, the angular momentum vector must
 be at or near the long axis of the ellipsoid.

 Now (!) consider the intermediate case.  There are lots of
 places the angular momentum vector could wander relative to
 the ellipsoid.  (In any inertial frame the angular momentum
 vector doesn't wander at all, but the ellipsoid does, as
 we re-orient the object.  Alternatively, in a frame comoving
 with the body, the ellipsoid stays fixed but the angular
 momentum wanders.)

 So, in the intermediate case I haven't yet proved that it
 will tumble, but I have explained why it could tumble (and 
 could not tumble in the other two cases).  Proving that it 
 *will* tumble might require a more detailed analysis.

 I tried googling for images of the intersection of spheres
 with an ellipsoid.  Here is the crucial intermediate case:
   http://api.ning.com/files/50GrcSOUEsM8m6GLfnIFLQe5jrT5bi-VvI51otk7sHFl*xzmqT8NEvugFn2MEdwAMjqW9eFeom3F0KQu161tQaDyJoqt*Ero/CirclesOfEllipsoid.png

 It should be trivial to cobble up more such images using x3d 
 aka VRML.  I can easily visualize such things in my mind's 
 eye, but when explaining it to somebody else it helps to 
 have the picture.

C) In the real world, the object always has some internal
 friction.  If the object is spinning on the "big" axis, it 
 is unconditionally stable, so the friction doesn't matter.
 Interestingly, if it is spinning smoothly on the "small" 
 axis, the friction might act only slowly.  This axis is 
 therefore in some sense metastable.  In  the intermediate 
 case, the system is wobbling like crazy, causing lots of 
 internal friction, so it will rather quickly decay to the 
 lowest-energy state.

 This case (C) is a weird combination of the previous two
 cases.