January 2013 - mathfun@mailman.xmission.com

[math-fun] stable time-invariant configurations of planets (sci-fi myth); flaw in my argument
by Warren Smith 12 Jan '13

12 Jan '13

--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(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > 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(a)xmission.com) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Fri, 11 Jan 2013 03:57:16 +0000 > From: Fred lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] rounded tetrahedron > Message-ID: > <CAN57Yqv5oONmxnarB-9jJBMTHYE5uX+3c2DJnJv0EMikeRnSbg(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > On 1/11/13, Dan Asimov <dasimov(a)earthlink.net> wrote: >> 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: >> >>> On 1/10/13, Huddleston, Scott <scott.huddleston(a)intel.com> wrote: >>>> 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 >>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > > ------------------------------ > > Message: 2 > Date: Thu, 10 Jan 2013 20:34:00 -0800 > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] rounded tetrahedron > Message-ID: <BA5ABB5A-333B-4C31-829A-5A2E24F2BA12(a)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. > > > > > ------------------------------ > > Message: 3 > Date: Thu, 10 Jan 2013 23:12:13 -0800 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] rounded tetrahedron > Message-ID: > <CAA-4O0HFfGMr+Vb2prysvjuHEwSV00+6hPFVturf8yetN7KmWg(a)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 > > > ------------------------------ > > Message: 4 > Date: Fri, 11 Jan 2013 15:22:44 +0000 > From: Fred lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] rounded tetrahedron > Message-ID: > <CAN57Yqu2H3z0N07=maowxdNmb3iVQ5z=9X=fc+1QvFq2bxnZng(a)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(a)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. >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > > ------------------------------ > > Message: 5 > Date: Fri, 11 Jan 2013 09:37:24 -0700 > From: rcs(a)xmission.com > To: math-fun(a)mailman.xmission.com > Cc: rcs(a)xmission.com > Subject: Re: [math-fun] stable time-invariant configurations of > planets (sci-fi myth) > Message-ID: <20130111093724.e62zkokn00s8gs8w(a)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(a)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 >> >>> >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > > ------------------------------ > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > End of math-fun Digest, Vol 119, Issue 17 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Is the universe homogeneous at large enough scales?
by Warren Smith 11 Jan '13

11 Jan '13

] > From: Dan Asimov <dasimov(a)earthlink.net> > I'm skeptical of the claim that this minimum spanning-tree > is any less random than the rest of the quasars depicted > in the Picture linked below. (Given that gravity attracts, > at least on relatively small cosmological scales.) > > How could one decide, with reasonable confidence? > > --Dan --I'm also skeptical. You need a reasonable preferably-simple statistic, you need to account for the "cherrypicking effect," and you need to clearly describe your reasoning. They failed on all three counts, but did use a lot of adjectives about how they thought their tests were very "conservative." Without any of that, if one had their software which found the object, one could just run it on randomly generated "quasars" data and see how often it found their-quality or better objects -- if only 1 time per 1000 datasets, then yes, this would be 99.9% significant. This might be an expensive way to proceed but would be pretty foolproof.

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[math-fun] Is the universe homogeneous at large enough scales?
by Warren Smith 11 Jan '13

11 Jan '13

A 73-quasar object called the "large quasar group" has been found that is 4 billion light years long. It was found with a 3D min-spanning-tree algorithm on a dataset of 105783 quasars. This is allegedly the largest thing ever seen. Quasars tend to clump together in organized-looking structures, not "randomly distributed." This seems a good deal larger than "galaxy superclusters." Is this object for real, or merely what would have been expected to happen by examining enough random points? They argue the former, but I find their argument unclear and hence unconvincing. They think this object has 40% greater "local density" than it would have had with uniform universe. Roger G. Clowes, Kathryn A. Harris, Srinivasan Raghunathan, Luis E. Campusano, Ilona K. S"ochting And Matthew J. Graham: A structure in the early Universe at z~1.3 that exceeds the homogeneity scale of the R-W concordance cosmology, Monthly Notices of the Royal Astronomical Society (11 January 2013) http://mnras.oxfordjournals.org/content/early/2013/01/07/mnras.sts497 Picture: http://mnras.oxfordjournals.org/content/early/2013/01/07/mnras.sts497/F1.la…

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[math-fun] stable time-invariant configurations of planets (sci-fi myth); flaw in my argument
by Warren Smith 11 Jan '13

11 Jan '13

And also, there are no "attractor trajectories" either. That is, in a Hamiltonian system, there is no energy (or entropy) loss mechanism, so no trajectory can attract all nearby ones. However, in the solar system there are slight non-Hamiltonian effects (beyond merely Newton laws) such as tidal friction losses. It is only as a result of these, that certain "attractor" configurations, such as trojans, and also such as 2:1 resonances, tidal locking of Earth's moon (does not rotate) have happened. To return to the original problem -- which we now see is more interesting that I thought -- PUZZLE: find every configuration of N point masses, which (1) form a stationary configuration in rotating reference frame, and which (2) is stable. FACT 1: The N bodies must all lie in a single plane perpendicular to the rotation axis since otherwise could decrease energy by "flattening" configuration into said plane. FACT 2: My repaired proof, if no further problems spotted, shows N<=6. FACT 3: Requirement (1) is a set of polynomial equations which can be solved by Grobner basis, elimination, etc methods, but it might get pretty damn difficult especially if you are naive about it. Apparently Lagrange settled the "2 and a half bodies" problem (2 bodies, plus one more of negligible mass) but I do not think even the 3-body case has been done?

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[math-fun] stable time-invariant configurations of planets (sci-fi myth); flaw in my argument
by Warren Smith 11 Jan '13

11 Jan '13

I think (?) Meeker in his cryptic message wondering what "stable" means (huh? I thought everybody knew what stable meant) was trying to worry about Lagrange Trojans being stable. Anyway, I claim no Lagrange points are known to be stable (at least in the math I saw) because the math I saw only works for infinitesimal third masses, not finite nonzero masses. Perhaps that answers Meeker. Of course, with very small third masses the instability (if there is any) would have a very long growth time.

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Re: [math-fun] stable time-invariant configurations of planets (sci-fi myth)
by Warren Smith 11 Jan '13

11 Jan '13

> The fictitious force is proportional to omega-squared r, and omega is > assumed constant. So doesn't that mean the pseudo-potential is > proportional to r^2? --aggghh! Yes, that was dumb. However, my proof remains just as valid as before, and actually gets simpler since the laplacian of this is simpler (namely, now is a negative constant).

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[math-fun] cable-dome
by Warren Smith 11 Jan '13

11 Jan '13

re earlier discussion about optimal domes, and also earlier discussion re "quadrangulations" etc... an interesting -- and it seems highly economical -- way to make a dome is the "cable dome." This has cables going round the "lines of latitude" of a dome, carrying pure-tensile stress, and struts going up the "meridians" of the dome and fastened to the cables where they cross. The struts carry pure compressive stress. The struts and cables can be made from different materials. It also is possible to make the struts not be along meridians, but instead between each two adjacent parallel cable-rings, a zigzag of struts could be employed thus partitioning said annular region into triangles instead of quadrilaterals. The whole thing seems cheap, light, and easy to erect. You can roof it with membranes, e.g. a greenhouse. Such cable-dome structures were built for use as Olympic stadia in Korea and then later in Atlanta Georgia. The triangulated method clearly produces a rigid dome. But it is more expensive than the quadrangulated scheme (more beams). One might naively imagine the quadrangulated dome would be nonrigid and "floppy". I think that is true but I also think it nevertheless will have (if designed right) the property that the desired geometry is an energy-minimum, so it will stably self-restore after any distortion.

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[math-fun] Graham's Number
by Hans Havermann 11 Jan '13

11 Jan '13

John Baez on Google+: ".. where did Graham's number really come from?" ".. I asked Graham. And the answer was interesting. He said he made up Graham's number when talking to Martin Gardner! Why? Because it was simpler to explain than his actual upper bound - and bigger, so it's still an upper bound!"

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[math-fun] stable time-invariant configurations of planets (sci-fi myth)
by Warren Smith 11 Jan '13

11 Jan '13

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. QED -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] rounded tetrahedron
by Bill Gosper 11 Jan '13

11 Jan '13

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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