July 2015 - mathfun@mailman.xmission.com

[math-fun] highest mountains in solar system
by Warren D Smith 22 Jul '15

22 Jul '15

The correct metric for comparing mountain amazingness is the mountain's potential energy per unit mass, i.e. height (above mean surface) times g where g is the acceleration of local gravity. NAME.................HEIGHT...........g..................PRODUCT Maxwell Montes(Venus)..11000...8.87..................97570 Mt.Everest............8848m........9.798 m/sec^2.....86693 (m/sec)^2 Olympus Mons(Mars)..21171........3.71.....................78544 Caloris Montes(Merc)...1-3km.....3.7.....................3700-11100 Pluto 11kft mtn..........3353.........0.58....................1945 Given that Jupiter, Saturn, Uranus, Neptune do not have a solid surface (at least not that we can perceive) the champion is on Venus among the 9 planets they used to call planets, Earth is the runner-up. Looking at some nonplanets Mons Huygens(Moon)....4700-5400m......1.62.......7614-8748 Rheasilvia(Vesta).....22km?........0.25....................5500 5km Mtn(Ceres)........5km............0.29.....................1450 18km Mtn(Io)............18km...........1.796...................32000 20km Mtn(Iapetus)......20km.........0.224..................5000 we see that they cannot compete with Mars, Venus, Earth. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Some power-sum inequalities
by Warren D Smith 21 Jul '15

21 Jul '15

Some power-sum inequalities ===Warren D. Smith====July 2015==== Given a vector x of N positive reals, define the E-power-sum of x to be F_E(x) = SUM(0<j<=N) (x_j)^E where E is any real number. As a trivial example, F_0(x)=N. THEOREM: Let x be an N-vector of positive reals. If B<0<A or A<0<B then F_A(x) * F_B(x) >= N * F_{A+B}(x). If 0<=min(A,B) then F_A(x) * F_B(x) <= N * F_{A+B}(x). All these inequalities become equalities if all the x_j are equal, for example if all x_j=1. WHO SHOULD GET THE CREDIT? Normally I would doubt that this result could be new. However, I then found Bruce Reznick: Some inequalities for products of power sums, Pacific J. Maths 104,2 (1983) 443-463 http://projecteuclid.org/euclid.pjm/1102723674 which appears to prove a weaker result via a much lengthier argument. So maybe this actually is a new result (?). Perhaps by return mail I will receive news this is Foobuznik's inequality proved in 1785. EXAMPLE: if A=B=1 then the following inequality holds for all vectors x of nonnegative reals: F_1(x) * F_1(x) <= N * F_2(x) since this is a form of the well known "root mean square inequality." And that will imply the more general inequality F_A(x) * F_A(x) <= N * F_{2A}(x) is valid for any real number A. PROOF: Our first QUESTION is: for which A and B does an inequality between F_A(x) * F_B(x) and N * F_{A+B}(x) hold? 1. We can wlog demand A+B=1 because if A+B=C is not 1, then just redefine x_j to be (x_j)^C, and then (A, B, C) are replaced by (A/C, B/C, 1) whereupon the validity (or non-validity) of the inequality is unaffected. 2. We also may wlog then demand that the sum F_1(x)=1, since if it equalled any other value (say S) we could rescale x by replacing x_j by x_j / S, and again the validity (or non-validity) of the inequality would be unaffected. 3. Observe that the function F_A(x) is a concave-U function of the vector x if either A>1 or A<0, but concave the other way if 0<A<1. That is since the power function mapping z to z^A is concave-U if A>1 or if A<0, for all z>0; and since any sum of concave-U functions is concave-U. 4. The global min of such a concave-U function necessarily occurs when all the x_j are equal. And this includes the global min within the hyperplane F_1(x)=1, since restricting a concave function to a hyperplane, still is concave. 5. And as we remarked before, the inequality we are trying to prove, obviously is an equality under this "x_j all equal" circumstance. 6. Therefore: if A>1 or A<0, then among vectors x with F_1(x)=1, the unique vector minimizing F_A(x) has all entries equal. Similarly for F_B(x). Therefore: F_A(x) * F_B(x) >= N * F_1(x) if A+B=1 with A>1 or B>1, and hence F_A(x) * F_B(x) >= N * F_{A+B}(x) if B<0<A or A<0<B. And similarly from the concavity in the other direction (and maxxing rather than minning) F_A(x) * F_B(x) <= N * F_1(x) if A+B=1 with 0<=A and 0<=B, and hence F_A(x) * F_B(x) <= N * F_{A+B}(x) if 0<=min(A,B). Q.E.D. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] New challenge: Offset Sonsky Gears
by William Somsky 21 Jul '15

21 Jul '15

Numerically, it just comes too close to feel like an accident. It's very similar to the six-planet systems that symmetry arguments say are exact. And I *think* I'm starting to see hints of a pattern to this, but I'm not sure. I'm going to try to examine the specific numbers of this one and see if it can be shown to be exact. (I'm hoping maybe it has a lot of serendipitous things like 'cos alpha = 1/2' in it, which we can put nice, clean values to, rather that stuff like 'cos alpha = 137/293') When I get a bit more time, I'll see about posting my algorithm for generating these numbers. On 07/13/15 23:06, Tom Rokicki wrote: > Can you share any support for your belief, or is this a > puzzle for us to resolve? > > -tom > > On Mon, Jul 13, 2015 at 10:50 PM, William Somsky <wrsomsky(a)gmail.com > <mailto:wrsomsky@gmail.com>> wrote: > > I believe the following eight-planet system to be exact: > > ring = 53, sun = 27, offset = 20.00, planets = 3, 5, 5, 13, 13, > 21, 21, 23 > >

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[math-fun] Hackenbush with double moves
by James Propp 21 Jul '15

21 Jul '15

The group's earlier discussion about double moves dealt mostly with impartial games. What about partizan games, like (Blue-Red) Hackenbush? I conjecture that a Hackenbush position P that under single-move play has value v still has value v under double-move play but in a weakened sense. For instance, if P is a beanstalk of value 1/2 and Q is a beanstalk of value -1 (under single-move play), then I conjecture that under double-move play mP+nQ is m,n-eventually a win for Left if m/n converges to any value > 2 and is m,n-eventually a win for RIght if m/n converges to any value < 2. Cincotti's theory of multi-player Hackenbush might be pertinent here (since double-move 2-player Hackenbush has similarities to 4-player Hackenbush), but I've never understood Cincotti's theory, and to the extent that I've been able to glean what he's up to, it seems to me that he skirts issues like this. Jim Propp

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Re: [math-fun] The highest mountain in the solar system.
by Henry Baker 20 Jul '15

20 Jul '15

Similar question, inspired by ancient Egyptians & Romans: The definition of a planet(oid) seems to be a body large enough to force itself into a nearly spherical shape. Suppose you wanted to hack this defn. What weird shape could you build out of standard materials (bricks, cement, steel, carbon fiber?) that would have the same mass as a real planet but retain its weird shape. Alternatively, assuming the exact same recipe of elements as found in the entire Earth, how big of an object could be built that retained its shape under a) self gravity; and b) rotational stresses; and b) the various tidal forces due to the Moon, planets, Sun. Fractal structures aren't so farr-fetched: http://www.london-institute.org/people/farr/fractals.shtml At 10:23 AM 7/20/2015, Bill Gosper wrote: >Maybe one of the deviations from sphericity of >https://en.wikipedia.org/wiki/21_Lutetia ? How would you rule it out? >Related: An earth-size cube of "solid" rock would immediately collapse to >a sphere with eight mountains. About >how tall? Maybe mountains should be ranked by gravitational potential >energy. --rwg

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[math-fun] "optimum" rigid-rotating masses (ala Saari spiderwebs)
by Warren D Smith 20 Jul '15

20 Jul '15

Earlier, I had wondered what it might mean for a Saari spiderweb to be "optimal." I now have a suggestion. 1. Consider a set of N point masses in the xy plane which under Newton's laws of gravity & motion act like a rigid rotator rotating about the center of mass with axis in the z direction. If we write the potential & kinetic energies PE = -G * SUM(0<i<j<=N) m_i * m_j / |x_i - x_j| KE = (w^2 / 2) * SUM(0<i<=N) m_i * |x_i|^2 where G=Newton constant, w=constant angular velocity, m_i = positive mass of ith planet, x_i = position of ith planet in rotating reference frame (which therefore is "stationary") which is a 2-vector... Then one way to find suitable coordinates is MINIMIZE KE-PE by choice of x_1, x_2, ..., x_N. Note, this minimizes the "action." https://en.wikipedia.org/wiki/Principle_of_least_action Also note, KE is a concave-U quadratic, while -PE has spikes to +infinity but is bounded below by 0. That makes it completely obvious that a minimum of KE-PE always exists for ANY number N of planets and ANY set of positive masses. That makes it clear this is a wider class of solutions than Saari spiderwebs because those always involve exactly 2*N*K masses coming in K sets of 2*N equal masses, plus one optional central mass. Any such action-min always is an exact rigid rotator solution of Newton laws because grad(PE) = grad(KE) necessarily happens at any such min, and this causes negated gravitational forces = centrifugal forces to precisely balance so that each body feels zero net force in the rotating frame. The global min is presumably "more optimum" than nonglobal mins and also more optimal than non-min flat spots -- in the sense that it presumably corresponds to less-unstable motion. 2. Now, about stability. If an infinitesimal perturbation of the bodies exists which would decrease the action, then the rigid rotator motion is unstable. But such a perturbation is always available: perturb the x_i almost arbitrarily in the third (z) dimension. This leaves KE unaltered (also leaving the momentum and angular momentum and kinetic energy of each mass unaltered) but increases PE, thus decreasing the action KE-PE per unit time. Also true if we restrict attention to the codimension-1 subclass of these perturbations which leave the center of mass unaltered. Now you might complain that this perturbation alters the total energy KE+PE, but we could if desired rescale all the coordinates by a factor s infinitesimally close to 1 [since KE is proportional to s^2 while PE is proportional to -1/s] to restore the old value of KE+PE while still decreasing KE-PE... we also can simultaneously adjust w to keep the angular momentum unaltered... So in view of these possibilities, it would seem that the rigid-rotator motion for ALL of these configurations is ALWAYS unstable if N>=4. When N=1 or 2 or perhaps 3 you might be able (and presumably are able when N=1 and 2) to argue we have stability because the s and w counter-adjustments sort of reduce the dimension of the space of the perturbations I described, to zero. But you can't hide from instability in that way if N is large enough, and N>=4 seems large enough. Wonderful result, except there is something wrong with that analysis by me: Moeckel showed (correcting J.C.Maxwell 1885) that N planets of equal small-enough masses co-orbiting a unit-mass sun as a rigidly rotating regular N-gon, are stable if N>=7. Richard MOECKEL: Linear stability analysis of some symmetrical classes of relative equilibria, in Hamiltonian Dynamical Systems: History, Theory and Applications, IMA vol.63, Springer Verlag 1995. http://link.springer.com/chapter/10.1007/978-1-4613-8448-9_20 To get stability, the sun mass needs to grow at least like 0.435*N^3 for unit mass planets, and N>6 is necessary: Gareth E. Roberts: Linear stability of the n+1-gon relative equilibrium, Hamiltonian Syst. Celest. Mech., World Scient. Monogr. Ser. Math. 6 (2002) 303-330. http://mathcs.holycross.edu/~groberts/Papers/HAMSYS-98.pdf Richard Moeckel: Linear stability of relative equilibria with a dominant mass Journal of Dynamics and Differential Equations 6,1 (January 1994) 37-51 gives conditions under which the rigid-rotators consisting of N+1 masses in a limit when there is 1 mass much larger than all the others, will be stable. I think my problem was that "being able to decrease the action" is NOT the same thing as "instability" even though some sources act as though it is the same thing. Really, you need to prove all eigenvalues of the time-evolution operator have non-positive real part, to show (a commonly accepted kind of) "stability." GE Roberts: Spectral instability of relative equilibria in the planar n-body problem. Nonlinearity 12 (1999) 757-769. http://mathcs.holycross.edu/~groberts/Papers/EqualMass.pdf showed that if N>24305 then any rigid rotator Newton solution made of N EQUAL-mass stars is spectrally unstable. (And he conjectures N>2 suffices.) That seems to destroy any hope these configurations could be reasonable models of any kind of galaxy. This analysis is purely in the plane, no 3D perturbations are considered. If I understand Roberts EQ8 correctly, then for a rigid-rotator N-body Newton solution, he constructs the 4Nx4N matrix [ wK Minv ] [ S wK ] where each block is 2Nx2N, K=BlockDiag(J,J,...J) and J is the 2x2 matrix [ 0 1] [-1 0] and M is the diagonal matrix with entries m1,m1, m2,m2, m3, m3, ..., mN,mN and Minv is its inverse, and finally S is the symmetric Hessian matrix of the potential energy function of x1,y1,x2,y2,x3,y3,...,xN,yN. He then claims we have spectral stability in 2D if and only if the eigenvalues of this 4Nx4N matrix, aside from the 8 eigenvalues 0,0,,w*i,w*i,w*i,-w*i,-w*i,-w*i all have zero real part. Spectral stability in 2D is a necessary condition for stronger forms of stability. 3. Now, about the 3-body case N=3. Euler realized that collinear solutions exist L. EULER: De moto rectilineo trium corporum se mutuo attahentium, Novi Comm. Acad. Sci. Imp. Petrop., 11 (1767) 144- (and the body-locations arise from solving a quintic) but it has been shown (Andoyer 1906, Meyer 1933, Conley) those solutions are always unstable. Lagrange proved that, for 3 generic masses, the only non-collinear solutions involve the 3 masses lying at the vertices of an EQUILATERAL triangle. It is a quite remarkable theorem of geometry that no matter what the 3 masses are, an equilateral triangle always yields a force balance. (For three non-generic masses such as m1=m2, further solutions can exist, such as certain isoceles triangles.) Gascheau then proved Lagrange's solution is STABLE if 0 < (m1*m2+m2*m3+m3*m1)/(m1+m2+m3)^2 < 1/27. For example, if one of the masses is large and the other two small, we get stability. J. L. LAGRANGE: Essai sur le probleme de trois corps, Oeuvres, 6 (1772) 229-. G. GASCHEAU: Examen d’une classe d’equations differential et applications a un cas particulier du probleme des trois corps, Comp. Rendus. Acad. Sci, 16 (1843) 393-. (E.J.Routh 1875 also proved stability and solved the question for general power-law forces, not just inverse-square law.) 4. It turns out I was not the first to suggest considering "post Newtonian" relativistic corrections. Kei Yamada, Hideki Asada: Triangular solution to general relativistic three-body problem for general masses Phys.Rev.D 86 (2012) 124029 http://arxiv.org/abs/1212.0754 apparently Yamada can show that perturbations of the Lagrange equilateral triangle solution to Newton laws, exist which are exact solutions of the first-order post-Newtonian more-relativistic laws of motion; and that these are stable if a perturbed version of the Gascheau condition is satisfied. 5. It occurs to me Saari's spiderwebs can be generalized to allow the concentric N-gons to be dovetailed if desired; and to allow N to be odd if desired.

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Re: [math-fun] LOOP: it's pool, but backwards.
by Bill Gosper 20 Jul '15

20 Jul '15

On 2015-07-17 16:43, Allan Wechsler wrote: > Ooh. There is a continuum of cones that have a given ellipse as > cross-section. This seems a fine time to reprofess that the apices of these cones all lie on the same hyperbola, and the apices of cones through that hyperbola all lie on your original ellipse. gosper.org/conethm.png Each curve contains the other's foci. --rwg > In the limit, with the apex of the cone at infinity, the > cone becomes a cylinder. Intuitively, the Dandelin construction must still > work with a cylinder. And proving that the cross-section of a cylinder is a > stretched circle seems like it should be easy. I bet there is a good proof > hiding here. > > On Fri, Jul 17, 2015 at 5:04 PM, Andy Latto <andy.latto(a)pobox.com> wrote: > >> That's a great geometrical proof that "Set of points with a constant >> distant sum from two foci" is the same as "intersection of a circular >> cone and a plane". But what's the geometrical argument that either of >> those is the same as a stretched circle? >> >> Andy >> >> >> On Fri, Jul 17, 2015 at 3:05 PM, Adam P. Goucher <apgoucher(a)gmx.com> >> wrote: >> >> Is there a proof this short and simple that the ellipse defined by >> >> these definitions is related to a circle by an affine transformation? >> > >> > Yes: https://en.wikipedia.org/wiki/Dandelin_spheres >> >

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[math-fun] The highest mountain in the solar system.
by Bill Gosper 20 Jul '15

20 Jul '15

Maybe one of the deviations from sphericity of https://en.wikipedia.org/wiki/21_Lutetia ? How would you rule it out? Related: An earth-size cube of "solid" rock would immediately collapse to a sphere with eight mountains. About how tall? Maybe mountains should be ranked by gravitational potential energy. --rwg

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[math-fun] Fwd: Gerson Robison's son
by James Propp 20 Jul '15

20 Jul '15

Were any of you ever in touch with Gerson Robison? Jim Propp ---------- Forwarded message ---------- From: *Diego Dominici* <dominicd(a)newpaltz.edu> Date: Monday, July 20, 2015 Subject: Gerson Robison's son To: James Propp <jamespropp(a)gmail.com> Cc: dominicd(a)newpaltz.edu Dear Jim Propp, I've asked around, but haven't received any positive answers yet. I will let you know if get any information. In the meantime, I found this piece of news about him: http://www.nytimes.com/1964/06/07/clerk-sentenced-for-attack-on-a-new-paltz… Best wishes, Diego Dominici Chair, Department of Mathematics SUNY New Paltz http://www2.newpaltz.edu/~dominicd/ On 7/18/2015 11:39 PM, James Propp wrote: > Dear Diego Dominici, > > Do you know if anyone in your department is in touch with the family of > Gerson Robison, who I believe taught at SUNY New Paltz several decades ago? > > I ask because Prof. Robison's son played a part in the chain of events > that led to the invention of the square-wheeled tricycle currently on > display at the Museum of Mathematics in New York City (by way of other > incarnations of the idea at the San Francisco Exploratorium and Macalester > College). > > If the junior Robison is still alive, and lives in the New York area, he > might enjoy a visit to the exhibit his messiness inspired, so I wanted to > contact him. > > Thanks, > > Jim Propp > > (mathematician at UMass Lowell, math-blogger, and member of the Museum of > Mathematics Advisory Council) > >

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[math-fun] Dark Matter 'Bridge' Between Milky Way & Virgo Cluster
by Henry Baker 20 Jul '15

20 Jul '15

FYI -- Curiouser & curioser... http://motherboard.vice.com/read/theres-a-dark-matter-bridge-from-the-milky… There's a Dark Matter 'Bridge' Between the Milky Way and the Virgo Cluster Written by Michael Byrne, Editor July 19, 2015 // 04:03 PM EST In the process of charting the movements of nearby galaxies, researchers at the Leibniz Institute for Astrophysics have reached a curious conclusion: There's a narrow band of dark matter extending from our Local Group of galaxies, which includes the Milky Way, to the Virgo Cluster some 65 million light-years away. On either side of it are just bubbles of nothingness. The revelation, which is described in the current Monthly Notices of the Royal Astronomical Society, could help explain a lingering mystery relating to the unusual distributions of dwarf galaxies often found scattered just beyond the boundaries of galaxies like the Milky Way. Rather than the expected random distribution, like bees buzzing around a hive, the dwarves seemed to be confined to a flat plane or disc of space. In the case of the Andromeda galaxy, that plane is nearly a million light years across but only 30,000 light years thick. A stellar pancake. This is unexpected. The arrangement surely isn't random, but the dwarf galaxies should be orbiting independently of each other. Simulations have reached this same conclusion: bees buzzing. So, there is an unknown at work. Current theories don't account very well for an anisotropic (directionally dependent) distribution of dwarf galaxies. "Although not identically defined, the evidence for the ubiquitous anisotropic distribution of satellite galaxies has by now a strong observational foundation," the current paper notes. "The abundance of evidence pointing towards the anisotropic distribution of satellites around large central host galaxies calls for a comprehensive and robust explanation, within the framework of the cold dark matter with dark energy (?CDM) model." And this is where we find our intergalactic dark matter bridge. "This is the first time we have had observational verification that large filamentary super highways are channeling dwarf galaxies across the cosmos along magnificent bridges of dark matter." More generally, what's needed to explain the satellite disc distributions is some variety of asymmetry. If the gravitational forces surrounding the Milky Way are all about the same, on average, we'd expect the aforementioned bees. But if something were out there exerting some additional force, it's easier to imagine such an extreme flattening. This is where a dark matter bridge starts to make sense: an additional outside force to begin the process of structure formation, a seed for anisotropy. Once evenly distributed galaxies begin to align around a single axis. It's not all that weird really. The distribution of stuff across the universe is on very large scales evenly spread around, but on smaller scales is more of a collection of galaxy clusters surrounded by deep voids. These voids are bridged by thin filaments of material, which are the dark matter bridges in question. In the paper's words: "The analysis reveals that the Local Group and Centaurus A reside in a filament stretched by the Virgo cluster and compressed by the expansion of the Local Void. Four out of five thin planes of satellite galaxies are indeed closely aligned with the axis of compression induced by the Local Void." In a statement, Noam Libeskind, an astrophysicist and the study's lead author, puts it a bit more poetically. "This is the first time we have had observational verification that large filamentary super highways are channeling dwarf galaxies across the cosmos along magnificent bridges of dark matter," he says. The filaments offer dwarf galaxies an "off ramp" of sorts along which they can be beamed towards the Milky Way and other galaxies sporting similar satellite accretion discs. "The fact that this galactic bridge can affect the dwarf galaxies around us is impressive, given the difference in scale between the two: the planes of dwarfs are around 1 percent of the size of the galactic bridge to Virgo." The current study can also be read in free, open-access form on the arXiv preprint server. http://arxiv.org/pdf/1503.05915v2.pdf Topics: astrophysics, space, Andromeda, dark matter, local group, virgo, Leibniz Institute, interstellar medium, discoveries, dwarf galaxy

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