--ok, first, my proof was erroneous. To explain the error: Actually, in my proof each planet plainly has not just "at least one," but in fact "an entire cone," of unstable directions you can perturb it in. Indeed if we are at an energy-flatspot these are doublecones. That is, there are 3 degrees of freedom per planet. Now the demands that the angular momentum and center of mass stay fixed, is 6 demands total. So then, we argue that if there are N>=3 bodies, then N*3>6 so that there will be a way to perturb them which decreases energy but keeps ang.mom. and cent.of mass both fixed. BUT, the trouble with that is, the 6D subspace of "meeting demands" might lie entirely outside the "allowed doublecones". So therefore, I think my proof does not quite work. If however the number N of bodies exceeds 6, then just using the central axial lines of the doublecones the proof seems to work. So revised claim is, if N>6 no stable configuration is possible. If 3<=N<=6 then maybe possible and I suppose one should try to work them all out. --Second, somebody was complaining that lagrange configurations are stable. This incidentally is "experimentally proven" by the existence of the "Trojan & Greek" asteroids. My response is "yes, BUT that is NOT a configuration where all the bodies are stationary in some rotating reference frame." The trojan asteroids are not stationary; they perform some sort of oscillations around the Lagrange point. It is (observations indicate, and probably also theorems) possible for N-body configurations to have "attractor trajectories" but those are trajectories, not stationary points. http://en.wikipedia.org/wiki/Trojan_(astronomy) My theorem was only about a scenario where all the bodies are fixed in position in some rotating reference frame, not more general scenarios. III. http://en.wikipedia.org/wiki/Lagrangian_point has some discussion of Lagrange points and their stability. Does this discussion for L4 and L5 contradict my alleged theorem? I'm not sure but think so. On 1/11/13, math-fun-request@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:

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Today's Topics:

1. Re: rounded tetrahedron (Fred lunnon) 2. Re: rounded tetrahedron (Dan Asimov) 3. Re: rounded tetrahedron (Bill Gosper) 4. Re: rounded tetrahedron (Fred lunnon) 5. Re: stable time-invariant configurations of planets (sci-fi myth) (rcs@xmission.com)

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Message: 1 Date: Fri, 11 Jan 2013 03:57:16 +0000 From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] rounded tetrahedron Message-ID: <CAN57Yqv5oONmxnarB-9jJBMTHYE5uX+3c2DJnJv0EMikeRnSbg@mail.gmail.com> Content-Type: text/plain; charset=UTF-8

On 1/11/13, Dan Asimov <dasimov@earthlink.net> wrote:

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Right, assuming the proof holds up.

There does seem to be an lot of detail involved in Roberts' presentation. As far as I can tell however, it does at least attempt to cover all the options.

...

But I'm more inclined to believe the validity of the claim in the Math Intelligencer article that you cite. ... P.S. Fred, what do you mean by "elsewhere" in your posted quoted below?

Kawohl & Weber (Math. Intell., Scott's link) give no justification or reference for the constant-width claim, which I consider on the face of it very surprising and quite possibly erroneous [it takes one to know one ...]; "elsewhere" refers to width along diametral lines away from planes of symmetry.

Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all. So the Minkowski yields a distinct surface.

Fred Lunnon

...

In fact, they write:

----- Incidentally, the Minkowski sum (?)M_V ? (?)M_F, which one obtains half way in the process of morphing M_V into M_F, would render a body with tetrahedal symmetry. It actually has the same constant width as MV and MF .

...

Its volume, however, is larger than that of the Meissner bodies, due to the Brunn-Minkowski inequality. -----

Here M_V is the Meissner tetrahedron where starting from the Reuleaux tetrahedron, 3 edges sharing a vertex have been modified in the same way, to get a body of constant width. Likewise for M_F, but in this case the 3 edges bound one face of the Reuleaux tetrahedron.

I believe the example in the example Scott Huddleston linked to below is identical to the example (?)M_V ? (?)M_F mentioned in the Intelligencer article.

On 2013-01-10, at 4:20 PM, Fred lunnon wrote:

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On 1/10/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:

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See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf

Looks like that probably wraps Dan's question up, albeit at some length!

...

Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.

That was obviously true --- but there seems no reason to expect that the width would be constant elsewhere.

WFL

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Message: 2 Date: Thu, 10 Jan 2013 20:34:00 -0800 From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] rounded tetrahedron Message-ID: <BA5ABB5A-333B-4C31-829A-5A2E24F2BA12@earthlink.net> Content-Type: text/plain; charset=us-ascii

But the Minkowski average in question does not involve the Reuleaux tetrahedron as summand.

It is the sum of two types of Meissner tetrahedra: One (M_V) where 3 Reuleaux arcs that share a vertex are modified, and one (M_F) where 3 Reuleaux arcs that share a face are modified.

--Dan

On 2013-01-10, at 7:57 PM, Fred lunnon wrote:

...

Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all.

...

So the Minkowski yields a distinct surface.

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Message: 3 Date: Thu, 10 Jan 2013 23:12:13 -0800 From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] rounded tetrahedron Message-ID: <CAA-4O0HFfGMr+Vb2prysvjuHEwSV00+6hPFVturf8yetN7KmWg@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1

On 1/8/2013 5:35 PM, James Propp wrote: Does anyone know which sort of rollers the MoMath exhibit "Coaster Rollers" uses? Jim Propp

GeorgeHart>Jim, There are four shapes of rollers in the MoMath ride-on exhibit. Three are acorn-like surfaces of revolution and the fourth is a Meissner tetrahedron. One of them is very similar to this very cool commercially available metal set: http://www.grand-illusions.com/acatalog/Solids_of_Constant_Width.html The second has a point at the apex but is differentiable everywhere else, and the third is based on a regular pentagon (extended with radii and rotated to a surface of revolution). (I also wanted to have one sphere in the mix, with a different color, to emphasize that all the others aren't spheres, but that was vetoed...) George http://georgehart.com/

Too bad. Viewing the rollers through the transparent floor should reveal the ball moving more slowly due to its greater circumference.

It must be a bear keeping that floor descratched. You probably want a Gorilla Glass sandwich. --rwg

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Message: 4 Date: Fri, 11 Jan 2013 15:22:44 +0000 From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] rounded tetrahedron Message-ID: <CAN57Yqu2H3z0N07=maowxdNmb3iVQ5z=9X=fc+1QvFq2bxnZng@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1

My argument concerns just the cross-section of the region around one particular tetrahedral edge of the surface, and cut off by the extended neighbouring faces of the tetrahedron.

The Minkowski average combines a Reuleaux "cusp" with a Meissner cyclide: its cross-section is the mean of a pair of adjacent arcs with a single arc, yielding some kind of non-circular curve with a rounded apex of greater curvature, and only its endpoints in common with either summand.

The Roberts comprises an lower portion shared with the Releaux arcs, capped higher up by a circular arc.

The two curves are evidently incongruent; therefore so are the completed surfaces.

Finally, while I'm prepared to believe that there may be some obvious or well-known reason why this global Minkowski surface should retain the constant curvature of its summands, the matter remains currently beyond my comprehension --- can anybody cast any light here?

Fred Lunnon

On 1/11/13, Dan Asimov <dasimov@earthlink.net> wrote:

...

But the Minkowski average in question does not involve the Reuleaux tetrahedron as summand.

It is the sum of two types of Meissner tetrahedra: One (M_V) where 3 Reuleaux arcs that share a vertex are modified, and one (M_F) where 3 Reuleaux arcs that share a face are modified.

--Dan

On 2013-01-10, at 7:57 PM, Fred lunnon wrote:

...

Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all.

...

So the Minkowski yields a distinct surface.

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Message: 5 Date: Fri, 11 Jan 2013 09:37:24 -0700 From: rcs@xmission.com To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] stable time-invariant configurations of planets (sci-fi myth) Message-ID: <20130111093724.e62zkokn00s8gs8w@webmail.xmission.com> Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; format="flowed"

Warren's proof would seem to prove that the equilateral triangle configuration is unstable. My impression is that it's at least neutrally stable (for some mass ratios). I haven't done the details, so I'll leave this as a question, rather than an assertion of fact. [One possible out: the rotation might be non-uniform?]

Rich

---- Quoting meekerdb <meekerdb@verizon.net>:

...

On 1/10/2013 3:53 PM, Warren Smith wrote:

...

Suppose you have N planets moving under Newton's laws of gravity& motion (treat as point masses). Further, suppose that in an appropriate -- rotating about the center of mass -- reference frame, all the planets are stationary. Finally, let the configuration be STABLE against small perturbations.

MY CLAIM: If N>2 then no such configuration exists.

Why not? I.The pseudo-potential from the centrifugal force from the rotation, is proportional to -r, where r=distance to rotation axis.

II. The (actual) potential cased by gravitational attractions behaves proportionally to -1/r where r is the distance to the attractor.

III. All (pseudo and actual) potentials sum.

IV. The Laplacian LI of the psuedo-potential (I) is LI = -2/r. The Laplacian of the actual potential (II) is LII = 0. Thus upon summing we see that every body experiences a total potential whose Laplacian is NEGATIVE. Since Laplacian is a sum of second derivatives in orthogonal directions, that means at least one such second derivative is NEGATIVE. In other words, each planet can be moved infinitesimally in a way which makes its potential LOWER. Hence that planet was not in a (which would have been stable) potential-minimum.

Escape hatch: if there are only 2 (or 1) planets then you cannot move any planet at all (relative to the center of mass) without violating conservation of angular momentum and without keeping center of mass fixed (except for perturbations which increase energy), so the 2-body and 1-body problems are stable. But with 3 or more planets it is easy to see that some linear combination of the unstable shifts must exist which preserves both total angular momentum and center of mass, since you have>=3 degrees of freedom but only 2 constraints in a (linear, since infinitesimal perturbations) system... that is we have a 3-dimensional (or more) space of perturbations and within it there must exist a 2D (or less) subspace preserving these two invariants.

Right. Perturbing a mass at an L4 or L5 causes it to make a small orbit around the Lagrange point. Of course this also causes a small perturbation in the motion of the two larger bodies.

Brent Meeker

...

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