January 2013 - mathfun@mailman.xmission.com

[math-fun] why string theory can have high-spin particles without inconsistency (allegedly)
by Warren Smith 17 Jan '13

17 Jan '13

There are various arguments, some dating back to the 1960s, which show that quantum field theories cannot allow any fundamental particle to exist with spin>2. But "string theory" has arbitrarily high spin "particles." Contradiction? I had asked several physicists about this over many years & none provided a good answer. However, allegedly the answer is this. The inconsistency arguments assume (sometimes implicitly) that the number of particle/field types in the quantum field theory is finite. String theory escapes by having infinite set. Essentially, each time there would be an inconsistency, it is canceled out by putting in an interaction with even-higher spin particles which cancel out the inconsistency. Due to infiniteness, the chain of such fudge-corrections need never end. Still, I think this represents a very important/frightening way in which string theory is fundamentally different from quantum field theories, deserving of note. This all is not mentioned in most (all?) of the popular books "explaining" string theory to the unwashed.

1 0

[math-fun] Best explanation of the Higgs particle
by Warren Smith 17 Jan '13

17 Jan '13

> Here's the download link for the full-motion video: http://online.itp.ucsb.edu/download/bblunch/arkanihamed2/snd/Arkanihamed2_B… --I listened to this Nima Arkani-Hamed lecture, my computer was able to handle it although took an hour to download. One thing N.A-H mentioned as a key building block was that it is impossible for a particle to have spin>2. However, in "string theory" just spin a string harder. You can get whatever spin you want, no matter how large. No? So: How are these two claims to be reconciled? [I have asked this question of physicists before, by the way. So far, I have never got any answer back I considered reasonable.]

4 3

[math-fun] Low Rollers: Bodies with Minimal Constant Width and Tetrahedral Symmetry
by Fred lunnon 15 Jan '13

15 Jan '13

Bodies with Minimal Constant Width and Tetrahedral Symmetry ___________________________________________________________ Fred Lunnon, Jan 2013 Ernst Meissner was an eminent applied mathematician in his day (1912). On first encountering this topic --- some 30 years ago I suppose, perhaps excusing numerous recent lapses of memory --- I wondered why a good engineer would be satisfied with something as ugly as the asymmetrical solutions now given his name. Failing to to appreciate the part played by minimal surface area and volume in his investigations --- discussed by Kawohl & Weber --- and preferring full tetrahedral symmetry, in I had waded; just as Patrick Roberts has, and no doubt a few more folk besides. Failing somehow to persuade my construction to hang together, I eventually was obliged to put it aside. This time round, Roberts has proved doughtier; sadly however, despite provision of a plethora of painstakingly prepared pictures, his report manages to follow such an circuitous logical path, that just having a another shot myself looked easier than attempting to check his. The process proves so gratifyingly straightforward that I can no longer imagine what my earlier difficulty could have been --- which must after all be preferable to the alternative! Stage (O): inflate the 4 faces of the unit-edge tetrahedron out to congruent caps of unit-radius spheres, meeting the neighbouring vertices, but with great circle boundaries to be specified later. The gaps over the 6 edges are to be filled by congruent lunes of cyclides: envelopes of pencils of spheres in two orthogonal ways, having circles of principal curvature, and 9 degrees of freedom. Stage (A): nail down the cyclide. Take a section through a plane of symmetry of the tetrahedron, and inflate the isosceles triangle out to constant width. Attach arcs of radius 1 over two sides, centred on triangle corners; and arcs of radius b over the base, c over the opposite corner, both with centre P at distance d along the altitude and outside the corner. [The disarmingly detailed diagram on page 2 of Roberts assumes without justification that d = 0.] Inspection of the lines joining P to the mid-point and corners of the base, together with Pythagoras and constant unit width, establishes b + c = 1 ; 2(c + d) + rt2 / 2 = 1 ; b^2 = 1/4 + (d + rt2 / 2)^2 ; which solve easily yielding c = d = (2 - rt2) / 8 ; b = 1 - d = (6 + rt2) / 8 . In particular, the completed small circle actually passes through the nearby triangle corner. It is a little harder to justify an assumption that the cyclide has pinch points at tetrahedron vertices, but we're going to assume that anyway. Now the cyclide is determined, and furthermore the tetrahedron edge lies on its surface --- a bonus feature suggesting that we're on the right track. Stage (B): take x-axis along the base, z-axis along the altitude, y-axis along the edge through the opposite corner. Using Pythagoras, the pencil sphere with centre (x, 0, z) and radius f(x) satisfies f(x) = z ; x^2 + (z + rt2 / 2 + d)^2 = (b - z)^2 ; whence f(x) = (1 - 4 x^2) d . Stage (C): consider a general diameter joining points on cyclides at opposite tetrahedral edges. This passes through the centres (x, 0, f(x)), (0, y, -f(y) - rt2 / 2) of pencil spheres tangent to the envelope at those points. By Pythagoras, the distance between their centres equals the square root of x^2 + y^2 + (f(x) + f(y) + rt2 / 2)^2 = (1 - f(x) - f(y))^2 , using f(x) definition, with (f(x) + f(y))^2 conveniently cancelling. Hence the width along that diameter --- the distance between cyclide surface points on the spheres --- equals constant 1 . Stage (D): inspect the boundary between cyclide and unit radius spherical cap over a tetrahedral face. This is a longtitudinal circle on the cyclide at x = +/- 1/2 with radius 1 - f(1/2) = 1 , coincident with a great circle of the sphere passing through the tetrahedral vertices. Furthermore cyclide and sphere meet tangentially, since both are perpendicular to radii from the vertex; so the composite surface remains differentiable, away from vertices. Bingo! Now what about the Minkowski mean of a pair of non-congruent Meissner bodies, mentioned by Kawohl & Weber? Elementary properties of Minkowski sums suggest --- now WFL has at last got his head around some, courtesy of strenuous combined efforts by DA and WDS --- that this also has minimal constant width, and composes spheres and cyclides as before. Any local alteration in curvature of such a surface would entail some counter-balancing alteration at its opposite side, in order to maintain the width at the expense of destroying the symmetry. Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist? The singularities at the vertices may be avoided via expansion to a parallel surface at the cost of sacrificing minimal width, the larger surface composed of segments from 8 spheres and 6 cycloids. Though differentiable, this solution is less elegant than the analytical surface proposed by Fillmore. It would be nice to see analytical parametric forms or implicit equations for the latter, provided some heroic geek is prepared to decipher the spherical harmonic technicalities involved! [The Asimov discussion thread critique of the Fillmore n-space simplex corollary is inapplicable here, since we already are in possession of a continuous body for submission to Fillmore's finite analytic smoothing procedure.] References contributed by various math-fun sources: General page: en.wikipedia.org/wiki/Reuleaux_tetrahedron Discussion thread: http://mathoverflow.net/questions/54252/are-there-smooth-bodies-of-constant… Newslist thread: [math-fun] rounded tetrahedron Demo video: http://www.youtube.com/watch?v=jYf3nOYM_mQ Patrick Roberts "Proof of Constant Width of Spheroform with Tetrahedral Symmetry" Corvallis Oregon, (August 2012) http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf Bernd Kawohl & Christof Weber "Meissner's Mysterious Bodies" Mathematical Intelligencer vol 33 (2011) 94--101 http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf Jay P. Fillmore "Symmetries of surfaces of constant width" J. Differential Geom. vol 3 (1969) 103--110. intlpress.com/JDG/archive/1969/3-1&2-103.pdf Thomas Lachand-Robert & Edouard Oudet "Bodies of constant width in arbitrary dimension" Mathematische Nachrichten vol. 7 (2005) 740--750 http://www.lama.univ-savoie.fr/~lachand/pdfs/spheroforms.pdf ___________________________________________________________

2 4

Re: [math-fun] Best explanation of the Higgs particle
by Henry Baker 13 Jan '13

13 Jan '13

Thanks, Robert. I must have had a transmission error while downloading the full-motion video. Yes, the full-motion video (".mov") plays fine with VLC. The version that _doesn't_ play video with VLC is the ".m4v" (if you click on "Download the Podcast") version, which is only 39 MBytes, and MediaInfo reports that this file has AVC-encoded video at a frame rate of .1 fps = 6 frames per minute. Many of the UCSB KITP podcast videos have this 6fpm format, which is fine for slides and (most) blackboard work. The VLC people don't seem to be interested in fixing this problem, which is a pity, because there are thousands/hundreds of thousands of "enhanced" podcasts available on the internet with very low or variable frame rate video. Windows Quicktime is the only player I've found capable of playing these types of videos correctly. At 12:31 PM 1/13/2013, Robert Munafo wrote: >I downloaded the full-motion video from your link, and I was able to >pay it all the way through in VLC, and without any problems near the >end: > >On 1/13/13, hbaker1 <hbaker1(a)pipeline.com> wrote: >> [...] >> Here's the download link for the full-motion video: >> http://online.itp.ucsb.edu/download/bblunch/arkanihamed2/snd/Arkanihamed2_B… > >> I don't know if this file is corrupted or whether I had an error while >> downloading, but my downloaded video failed after about 60 minutes into this >> 80-minute video. > >My version of VLC is version 1.1.10. (Intel 64-bit) for Mac, compiled >by fpk with llvm-gcc 4.2.1 based on Apple build 5658, LLVM build >2335.6, and I'm running MacOS X 10.6.8 "Snow Leopard". > >I was also able to use FFMPEG to recompress (transcode) the video into >an M4V/MP4, which also plays in VLC with no troubles. This >recompressed version is 752 MB (a little smaller than the original) >and if you want I can try to upload it to Rapidshare or something. > >> Note that the VideoLAN "VLC" player will not play this video correctly; you >> have to use Apple Quicktime player (either Mac or Windows). Even iTunes >> won't play it correctly. >> >> VLC will play the sound, but I found the pictures necessary to understand >> what was going on. > >-- > Robert Munafo -- mrob.com > Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - >mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

2 1

[math-fun] constant-width convex objects
by Warren Smith 13 Jan '13

13 Jan '13

1. The Minkowski sum of two constant-width convex objects, is another. Scaling and/or rotating one, yields another. 2. Consequently, once you find a few, then there should automatically be an infinite set of them. Also, one can perform a Minkowski sum of one object, plus itself rotated by an amount which is distributed via a "normal" distribution with very small "typical angular width" (infinite Minkowski sum) and in this way obtain a constant-width object which is Cinfinity-smooth, but is arbitrarily near in shape to any starting const-width object. 3. Video: http://www.youtube.com/watch?v=jYf3nOYM_mQ 4. One of the cites that has been mentioned: Bernd Kawohl, Christof Weber: Meissner’s Mysterious Bodies, The Mathematical Intelligencer 33,3 (September 2011) 94-101 http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf cites many other papers and mentions some remarkable theorems.

3 3

[math-fun] Best explanation of the Higgs particle
by hbaker1 13 Jan '13

13 Jan '13

The coverage of the Higgs in the press has been abysmal. This lecture from UCSB will actually give you some idea of what the Higgs is all about. The Inevitability of the Higgs Nima Arkani-Hamed, IAS & KITP Here's the web page for the UCSB KITP series: http://online.itp.ucsb.edu/online/bblunch/ Here's the web page for this particular talk: http://online.itp.ucsb.edu/online/bblunch/arkanihamed2/ Here's the download link for the full-motion video: http://online.itp.ucsb.edu/download/bblunch/arkanihamed2/snd/Arkanihamed2_B… I don't know if this file is corrupted or whether I had an error while downloading, but my downloaded video failed after about 60 minutes into this 80-minute video. Here's the download link for the 3 frames-per-minute video: http://online.itp.ucsb.edu/download/bblunch/arkanihamed2/snd/Arkanihamed2_B… Note that the VideoLAN "VLC" player will not play this video correctly; you have to use Apple Quicktime player (either Mac or Windows). Even iTunes won't play it correctly. VLC will play the sound, but I found the pictures necessary to understand what was going on.

1 0

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

13 Jan '13

MLB> And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? ------ I nearly slid through a stopsign recently. It was on a sharp downgrade, preceded by gravel washed off a hillside. Had the gravel been perfectly spherical, I wouldn't even have slowed down. Except spherical gravel would have rolled away. What about Reuleaux tetrahedra (http://www.youtube.com/watch?v=jK7xPo1YXzY 3:18)? Neil 3D printed a lovely one, ~1.5", and then split for CES w/o photographing it. It's eerie to roll a book over. Why aren't they sold at science museum gift shops? But would one stay put on a hill? Neil claims, with sufficient friction, an amazing 60°! They're much less round than you might think. This has 007esque possibilities, not just for foot chases and booby-traps. Imagine pursuing Bond on a winding road where he releases a few pounds of Reuleaux gravel. You say http://www.zombiesurvivalwiki.com/thread/3341691/Liquid+Banana+Peel+-+Make+… is cheaper? OK, but it leaves a residue. For the perfect crime, use dry ice Reuleaux gravel. --rwg http://gosper.org/reuleauxtet.png

7 23

[math-fun] three-body (and begin for 4-body) problem with stationary shape; "opposite" jujitsu
by Warren Smith 12 Jan '13

12 Jan '13

Stockton also in another of his pages claims without proof that the 3-body eq.triangle stationary-shape solution is stable if the masses A,B,C obey 27*(A*B+B*C+A*C) < (A+B+C)^3. This was shown by Gascheau in 1843. Stockton cites this paper Bruno Sicardy: Stability of the triangular Lagrange points beyond Gascheau’s value, http://www.lesia.obspm.fr/perso/bruno-sicardy/biblio/biblio/Sicardy_Gaschea… which shows the stability problem is really quite deep and nasty! -- and J M A Danby: The Stability of the Triangular Lagrangian Points in the General Problem of Three Bodies, The Astronomical Journal 69.4 (1964) 294-296.

1 0

[math-fun] three-body (and begin for 4-body) problem with stationary shape; "opposite" jujitsu
by Warren Smith 12 Jan '13

12 Jan '13

http://www.merlyn.demon.co.uk/gravity6.htm is a somewhat strange web page by "Dr J.R. Stockton," also with contributions by Michael Behrend, plus looking into original works by Euler & Lagrange, which outlines a simple proof that with 3 point bodies of arbitrary positive masses A,B,C the only solutions of the 3 body problem in which the configuration stays unaltered in shape, is where the bodies are (i) collinear or (ii) vertices of equilateral triangle. Note, the masses not required to be equal, but you still get equilateral triangle (it rotates about its center of mass) and the force not required to be inverse-square (any of a wide class of functions of distance will do), and his page has scroll-windows within scroll-windows making it a bit strange to read. He does not include a stability analysis. It seems to me the collinear solutions are always unstable even to perturbations staying on the line. The triangle solutions might sometimes be stable. It's not obvious how to prove stability. Proving instability can sometimes be easy, since you just present some instability mechanism; but stability seems much harder, since need to disprove every possible instability. Stockton also has a few clearly incomplete mutterings about stationary-shape 4-body problem. For example, given 4 bodies of masses A,A,B,B you can get a rotating rhombus stationary-shape 4-body solution. He does not mention, but for any 4 masses A,B,C,D you can put A,B,C in the equilateral triangle solution and D at the center of mass. Another he does not mention: given 4 bodies A,B,C,C, you can place C,C at the 2 far image-corners of a bilaterally-symmetric quadrilateral and A,B at the other two corners and rotate thing about its center of mass, and this will work if rotation-speed and quad-shape are chosen right. For this one you can also place a fifth body D at the center of mass. --- EQUIVALENT PROBLEM: It actually has advantages to think about this whole problem in the "opposite" way. That is, instead of the centrifugal repulsive force, consider an attractive "spring" potential of form r*r*k for r=distance to rotation-center and k=positive constant. And instead of the gravitational inter-body attractions from positive mass-points, consider Coulomb repulsions (of positive charge-points). If you just attach your charges to a center point with a spring (or place in a shallow parabolic dish) and let them settle, they clearly must+will find a position of local-min energy. This proves EXISTENCE THEOREM: for any N>0, and any N positive real masses M1,M2,M3,...,MN, there exists at least one configuration in the plane (i.e. locations for each mass) such that that configuration is a solution of the Newton gravity+motion laws which just rotates bodily, preserving its shape & size. This is very cute piece of jujitsu, eh? In the original problem formulation, this is not obvious, but I simply negate the potential function, and voila, existence becomes obvious. The bad news is: every solution generated in this "negated" way, since a min of the negated-potential, automatically is a max of the original pseudo-potential, hence automatically is unstable in the original context? There are two reasons I am putting "?": 1. conservation of angular momentum and center of mass perhaps will enable stability even though without those conservation laws we would have instability. 2. A max of a potential is clearly unstable, but for a PSEUDO-potential, perhaps you can still sometimes survive. This "2" represents a flaw in my original "Theorem" that every N-body stationary-shape solution if N>6 is automatically unstable -- that proof has now been murdered by several such flaws (groan!) but might still be correct! (And stable solutions in the original Newton context are necessarily unstable in the Coulombized opposite version? But instability in the Coulomb setting might not imply stability in the Newton setting!) So this suggests, but in view of caveat 1 & 2 does not prove, that: The equilateral triangle 3-body solutions always are unstable.

1 0

Re: [math-fun] As long as we're talking astronomy
by Bill Gosper 12 Jan '13

12 Jan '13

See last link for comparison views. Fred> RWG's incomprehension is way ahead of mine --- I cannot make visual sense of this abstractly impressive graphic at all! WFL rwg> Confession: I had Neil explain it to me first. But now I really get it. It's majestic! The northern hemisphere, lit by the bright upper surface of the rings, soars above them. The central dark smudge is due to the planet's shadow on the rings. The southern hemisphere is faintly lit by the light we see filtering through the underside of the rings. MRob> Part of that may be the fact that they look different from in front and from behind. rwg> But we're not seeing any front. MRob> To first approximation, the dark shadow pattern agrees with the bright Sunward-facing reflection pattern (what we usually see) but not necessarily with the backscatter pattern (which involves diffraction and other messy stuff). rwg> No, all we need is scatter and density. A backlit ring can be faint if too tenuous, or too dense to transmit. There's a density of maximum brightness. The apparent density varies with viewing angle. Note how the rings brighten with distance. Backscatter varies even more strongly. Look how the relatively tenuous inner rings light up the area just above the equator. MRob> I think it's the E ring, but maybe it's the G ring. See: http://en.wikipedia.org/wiki/File:Saturn_outer_rings_labeled.jpg Ah, I'd somehow forgotten about those distant, fuzzy ones <http://gosper.org/backlit.png>. --rwg Consider neutrinium: a gravitationally closed ball of neutrinos.

1 1