May 2011 - mathfun@mailman.xmission.com

[math-fun] Slanted coverage
by Bill Gosper 07 Jul '11

07 Jul '11

In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2], which doesn't even converge for h < 2 r. The area element for the general x,y,z parametric surface is just Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]] (I don't recall this from 18.03.) E.g., for the torus with radii r and R, In[116]:= % /. {x :> (Cos[#2]*(R + r*Sin[#1]) &), y :> (Sin[#2]*(R + r*Sin[#1]) &), z :> (r*Cos[#1] &)} Out[116]= Norm[Cross[({Cos[#2] (R + r Sin[#1]) & , Sin[#2] (R + r Sin[#1]) & , (1,0) r Cos[#1] & }) , ({Cos[#2] (R + r Sin[#1]) & , (0,1) Sin[#2] (R + r Sin[#1]) & , r Cos[#1] & }) ]] In[117]:= % /. f_[L_List] :> f /@ L /. Abs[w_] -> w Out[117]= Sqrt[Power[-(-r Sin[#1] & ) (Cos[#2] (R + r Sin[#1]) & ) + (0 & ) (r Cos[#1] Sin[#2] & ), 2] + Power[-(0 & ) (r Cos[#1] Cos[#2] & ) + (-r Sin[#1] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2] + Power[(r Cos[#1] Cos[#2] & ) (Cos[#2] (R + r Sin[#1]) & ) - (r Cos[#1] Sin[#2] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2]] In[119]:= % /. (f_ &) :> (f &)[t, u] Out[119]= 2 2 2 2 Sqrt[r Cos[u] Sin[t] (R + r Sin[t]) + 2 2 2 2 r Sin[t] (R + r Sin[t]) Sin[u] + 2 2 2 (r Cos[t] Cos[u] (R + r Sin[t]) + r Cos[t] (R + r Sin[t]) Sin[u] ) ] In[120]:= FullSimplify[%] Out[120]= 2 2 Sqrt[r (R + r Sin[t]) ] In[125]:= Assuming[R > r > 0, Integrate[%120, {t, 0, 2*Pi}, {u, 0, 2*Pi}]] Out[125]= 2 4 Pi r R Did Archimedes know this? --rwg

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Re: [math-fun] surfaces of revolution & differential equations
by Allan Wechsler 31 May '11

31 May '11

I think this is a forlorn hope in the case of minimization problems in general. Although this particular problem has a solution with circular symmetry, such problems in general won't. A minimization problem with a completely symmetrical formulation might have a solution which breaks the symmetry. On Tue, May 31, 2011 at 10:56 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: > I was hoping that there was some sort of generic way to interconvert 2D <-> > 3D problems like this, rather than having to solve each one individually. I > noticed from MathWorld & Wikipedia that there is a generic way to compute > surfaces & volumes of revolution. > > At 07:49 AM 5/31/2011, Allan Wechsler wrote: > >Oh, OK, that's much clearer, thank you. > > > >I think the problem is still not well-formed, though for a much more > subtle reason. In the 2-dimensional version of this problem, we are to > preserve the length of the chain. What are we to preserve in the > 3-dimensional case? Area? There is no corresponding physical model, since > no physical fabric can shunt curvature around in such a way as to preserve > area. And if we pre-guess the curvature at each point, we have essentially > pre-guessed the solution. > > > >I suppose the area formulation is well-defined: what surface of a given > area, bounded by a horizontal surface, has minimum gravitational potential? > We need to clarify that the surface has a constant density in mass per unit > area. > > > >If we are completely satisfied that the solution will be rotationally > symmetric (I am, but don't know how to prove it), then certainly we can come > up with an r-z formulation to replace the x-y-z one. It's a classic > Calculus of Variations problem (so is the full 3D version, for that matter). > In the r-z formulation we have to be careful to calculate the potential > correctly, since length-elements near the periphery will contribute more > area than ones near the center. If we have z = f(r), then I think the > potential element is r dr sqrt( 1 + (dz/dr)^2 ), but I don't remember enough > Calculus of Variations to know what to do next. > > > >On Tue, May 31, 2011 at 10:30 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >You are right; I didn't do a very good job of explaining. > > > >Here's an example: in 2D, a catenary models a hanging chain. What is the > equivalent 3D structure (presumably some sort of hanging soap bubble) ? Is > it a revolved catenary or something else? > > > >At 07:24 AM 5/31/2011, Allan Wechsler wrote: > >>Somehow this problem statement is confusing me. Is the solution of your > >>system a function f(x,y,z), and then the "surface of revolution" of which > >>you speak is the surface f(x,y,z)=0? If I have gotten this wrong, please > >>clarify with the correct form of the solution. > >> > >>Intuitively, it would certainly seem as if the answer to your question > >>should be "yes". Suppose the axis of rotational symmetry to be the z > axis. > >> Then new coordinates would be z and r = sqrt(x^2 + y^2). > >> > >>On Tue, May 31, 2011 at 9:24 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >> > >>> Suppose I have a system of partial differential equations in 3D with > >>> rotational symmetry about an axis. Is it trivial to convert this > system > >>> into an ordinary differential equation in 2D, such that the solution, > >>> considered as a surface of revolution, is the 3D solution? > >>> > >>> Ditto in the reverse direction -- from 2D to 3D. > >

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Re: [math-fun] surfaces of revolution & differential equations
by Allan Wechsler 31 May '11

31 May '11

It's often possible to prove that the solutions retain the symmetry of the problem; it may be that you could prove this for the class of problems that is interesting you, and automate the transform to 2D for this class only. On Tue, May 31, 2011 at 11:35 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: > Hmmmm... I guess that's both good news & bad news: bad news because it > involves more work, but good news because it may produce more interesting > results. > > At 08:29 AM 5/31/2011, Allan Wechsler wrote: > >The p-orbitals of electrons, and some of the vibration modes of a circular > drumhead, are examples that occur to me. I'm sure there are simpler > "textbook" examples, but I don't know them offhand. > > > >On Tue, May 31, 2011 at 11:07 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >Interesting! Can you think of any specific examples? > > > >At 08:05 AM 5/31/2011, Allan Wechsler wrote: > >>I think this is a forlorn hope in the case of minimization problems in > general. Although this particular problem has a solution with circular > symmetry, such problems in general won't. A minimization problem with a > completely symmetrical formulation might have a solution which breaks the > symmetry. > >> > >>On Tue, May 31, 2011 at 10:56 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >>I was hoping that there was some sort of generic way to interconvert 2D > <-> 3D problems like this, rather than having to solve each one > individually. I noticed from MathWorld & Wikipedia that there is a generic > way to compute surfaces & volumes of revolution. > >> > >>At 07:49 AM 5/31/2011, Allan Wechsler wrote: > >>>Oh, OK, that's much clearer, thank you. > >>> > >>>I think the problem is still not well-formed, though for a much more > subtle reason. In the 2-dimensional version of this problem, we are to > preserve the length of the chain. What are we to preserve in the > 3-dimensional case? Area? There is no corresponding physical model, since > no physical fabric can shunt curvature around in such a way as to preserve > area. And if we pre-guess the curvature at each point, we have essentially > pre-guessed the solution. > >>> > >>>I suppose the area formulation is well-defined: what surface of a given > area, bounded by a horizontal surface, has minimum gravitational potential? > We need to clarify that the surface has a constant density in mass per unit > area. > >>> > >>>If we are completely satisfied that the solution will be rotationally > symmetric (I am, but don't know how to prove it), then certainly we can come > up with an r-z formulation to replace the x-y-z one. It's a classic > Calculus of Variations problem (so is the full 3D version, for that matter). > In the r-z formulation we have to be careful to calculate the potential > correctly, since length-elements near the periphery will contribute more > area than ones near the center. If we have z = f(r), then I think the > potential element is r dr sqrt( 1 + (dz/dr)^2 ), but I don't remember enough > Calculus of Variations to know what to do next. > >>> > >>>On Tue, May 31, 2011 at 10:30 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >>>You are right; I didn't do a very good job of explaining. > >>> > >>>Here's an example: in 2D, a catenary models a hanging chain. What is > the equivalent 3D structure (presumably some sort of hanging soap bubble) ? > Is it a revolved catenary or something else? > >>> > >>>At 07:24 AM 5/31/2011, Allan Wechsler wrote: > >>>>Somehow this problem statement is confusing me. Is the solution of > your > >>>>system a function f(x,y,z), and then the "surface of revolution" of > which > >>>>you speak is the surface f(x,y,z)=0? If I have gotten this wrong, > please > >>>>clarify with the correct form of the solution. > >>>> > >>>>Intuitively, it would certainly seem as if the answer to your question > >>>>should be "yes". Suppose the axis of rotational symmetry to be the z > axis. > >>>> Then new coordinates would be z and r = sqrt(x^2 + y^2). > >>>> > >>>>On Tue, May 31, 2011 at 9:24 AM, Henry Baker <hbaker1(a)pipeline.com> > wrote: > >>>> > >>>>> Suppose I have a system of partial differential equations in 3D with > >>>>> rotational symmetry about an axis. Is it trivial to convert this > system > >>>>> into an ordinary differential equation in 2D, such that the solution, > >>>>> considered as a surface of revolution, is the 3D solution? > >>>>> > >>>>> Ditto in the reverse direction -- from 2D to 3D. > >

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[math-fun] surfaces of revolution & differential equations
by Henry Baker 31 May '11

31 May '11

Suppose I have a system of partial differential equations in 3D with rotational symmetry about an axis. Is it trivial to convert this system into an ordinary differential equation in 2D, such that the solution, considered as a surface of revolution, is the 3D solution? Ditto in the reverse direction -- from 2D to 3D.

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[math-fun] Shape of optimum passive (non-tracking) solar arrays
by Henry Baker 30 May '11

30 May '11

Prof. Jeffrey Grossman of MIT is investigating 3D shapes for solar cells that optimize the performance of _non-tracking_ photovoltaic solar arrays. He is assuming that the cost of the PV material is no longer the limiting factor. He is using so-called "genetic algorithms" to optimize initially random patterns: http://dmse.scripts.mit.edu/news/?p=1753 His recent talk in Palo Alto (available somewhere on the web) leads me to pose the following question: Consider a 2D case. The sun rotates around a horizontal plane(actually line). We wish to come up with a 2D solar array (made up of plane(actually line) elements that maximizes the power generated _averaged over the whole day_. In our simplified 2D case, the sun only shines for 12 hours out of the 24 hour day, and during the daylight, the sun moves at a constant rate in a semicircle above the plane(actually line). We assume that the performance of small flat plane(line) element is proportional to length*cos(theta), where theta is the sun's angle with the element. We also assume that every element is large relative to the wavelength of the light, so diffractive effects can be ignored. We hold constant the area(actually length) of the _footprint_ (vertical projection) of the array, but allow any shape for the array itself. What would an optimal shape look like? (We also assume that any part in a shadow produces zero power.) Ditto, but now we also allow a combination of PV elements and _mirror_ elements -- i.e., an element can be either a PV element or a mirror element, but not both. However, an element could be a PV element on one side and a mirror element on the other. We also assume that the area outside the footprint of the array is non-reflective, so that any sun energy hitting outside won't bounce into the array. (A flat passive mirror outside the footprint could be part c of this question.) I presume that the optimum shape would be some kind of fractal, but I haven't been able to figure out how to visualize the problem yet. It would appear that the answer to part c might be a simple circle at some distance above the plane/line, because the mirror effect of the plane/line would make the solar array think that there were _two_ suns at the same time -- at angles +-phi above/below the horizon.

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[math-fun] The 11-cell
by Dan Asimov 27 May '11

27 May '11

I don't recall if we ever discussed the 11-cell, one of only two amazing 4-dimensional "abstract regular polytopes" (ARP) built from projective planes. See < http://en.wikipedia.org/wiki/11-cell > and < http://en.wikipedia.org/wiki/Abstract_polytope > and < http://en.wikipedia.org/wiki/Abstract_polytope#Regular_abstract_polytopes > for the precise definitions of ARP and 11-cell. But roughly, an ARP is a combinatorial generalization of a usual regular polytope that may seem too abstract at first but can grow on you. Anyhow, the 11-cell was discovered (apparently) independently by both Branko Grunbaum in 1977 and H.S.M. Coxeter in 1984. It consists of 11 projective planes P^2, represented as hemi-icosahedra -- the quotient of a regular icosahedron by its antipodal map -- so each P^2 is 10 equilateral triangles. (In fact, as a combinatorial object, metric properties are irrelevant to the 11-cell, but I like to make everything as regular as possible.) Each triangular face of each P^2 is identified with another such face of another P^2, so the 11-cell has 55 triangular faces in all. When this is done right, the 11-cell has a combinatorial automorphism group that is transitive on flags. A flag is a quadruple (F, T, E, V) where F is any of the 11 P^2's (so-called "3-dimensional faces") of the 11-cell, T is any triangle of F, E is any edge of T, V is any vertex of E. So there are 11x10x3x2 = 660 automorphisms in all. Also, this bizarrely symmetrical object is self-dual. I mean, who would've thought there could be such a thing with *11* faces. (The only other such thing has 55 faces -- it's 55 hemi-dodecahedra.) I'd like to know more about the 11-cell as a topological object: What is its universal covering space, its homology and its homotopy? --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] topology question
by Dan Asimov 25 May '11

25 May '11

In CP^2, any homogeneous polynomial in the coordinates, like P(X,Y,Z) = X^3 + Y^3 + Z^3, defines a compact Riemann surface (or what algebraic geometers like to call a "projective curve"). (P isn't well-defined on CP^2, but the locus P^(-1)(0) is.) If d := deg(P), then the genus g of this surface is given by the standard formula g = (d-1)(d-2)/2. (Assuming a technical condition that's satisfied for this P(X,Y,Z).) So, (3-1)*(3-2)/2 = 1 and hence P^(-1)(0) is a torus. Projective curves defined by X^n + Y^n + Z^n = 0 for some n >= 1 are called Fermat curves, after FLT. --Dan << On May 25, 2011, at 4:17 PM, "Cordwell, William R" <wrcordw(a)sandia.gov> wrote: > So, starting with the set CxCxC minus the origin, define the equivalence class of a point x to be all nonzero (complex) multiples of x, and "mod out" by that equivalence relation. This forms the complex projective space CP^2. I'm not sure how the topology is inherited or defined (topologists, please jump in)--maybe it's the largest one that keeps the mapping continuous? > > For this problem, one also wants initially to restrict to the subset of the original points in C^3 where x^3 + y^3 + z^3 = 0. The claim is that the final subset of CP^2 is homeomorphic to the torus. >> Sometimes the brain has a mind of its own.

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[math-fun] Electrons are fantastically round, say British scientists
by Henry Baker 25 May '11

25 May '11

FYI -- http://www.guardian.co.uk/science/2011/may/25/electrons-round-cosmos/print Electrons are fantastically round, say British scientists Most precise measurements of the roundness of the electron to date provide a window 'on the high-energy soul of the cosmos' Ian Sample, science correspondent guardian.co.uk, Wednesday 25 May 2011 18.06 BST After three months of experiments in a basement laboratory in London, scientists can confirm Â with more confidence than ever Â that the electron is very, very round. In the most exquisite measurements yet, researchers declared the particle to be a perfect sphere to within one billionth of a billionth of a billionth of a centimetre. Were the electron scaled up to the size of the solar system, any deviation from its roundness would be smaller than the width of a human hair, the team said. Abstruse as the experiment might seem, the work has profound implications for scientists wrestling with the mysteries of the cosmos. Even the slightest elongation of the electron can reveal what unknown particles might exist in nature, and even explain why matter won out over antimatter in the universe we observe. The findings, published in the journal Nature, already rule out some kinds of particles that theories suggested could pop into existence at the Large Hadron Collider at Cern, the European particle physics laboratory near Geneva. "It's been hard work. We've been working on this for a long time and we've had a lot of ups and downs," said Jony Hudson, a physicist at Imperial College London. "We have measured the shape really precisely. The deviations we were looking for are much smaller than the size of the electron. It is very, very round." The concept of shape might seem obscure when it comes to a subatomic particle, but the rules are the same as for everyday objects. Pick up a pen, for example, and you feel its shape because electrons in the pen push back against the electrons in your hand. And so it is with the electron itself. The particle is negatively charged, and the more evenly distributed the charge is around the centre of the particle, the more spherical it appears to be. Scientists pursue ever more accurate measurements of the electron's roundness because any sign of it being mishapen could herald a major discovery. One leading idea known as supersymmetry, which says that every kind of particle we know has a heavy twin, requires the electron to have a slightly distorted shape. "What's interesting is that the electron is so round it is becoming difficult for theories like supersymmetry to explain it," said Hudson, whose finding already rules out the existence of some supersymmetric particles. Evidence that the electron is mishapen on a minuscule scale might also explain why the universe we see is made of matter instead of antimatter. At the birth of the cosmos, both were made in equal measure, but some subtle difference between the two caused antimatter to disappear. If the electron is elongated, it will behave differently to its antimatter counterpart, the positron. For example, each would wobble differently in an electric field. "There must be a difference in the behaviour of matter and antimatter that we've not observed, and amazingly, the shape of the electron might just be enough to explain how the matter-antimatter imbalance built up over billions of years," Hudson said. His team studied the roundness of electrons by measuring how much, or how little, the particles wobbled in an electric field. The rounder the electron, the less wobble it will display. In the experiment, electrons were anchored to a molecule called ytterbium fluoride and examined with a laser beam. Each measurement took only one thousandth of a second. Running their experiment non-stop for more than three months, Hudson's team took 25 million measurements of electrons and averaged them out. They found no sign of the electron wobbling in the field, meaning it is more spherical than any previous experiment had shown. "To the best of our knowledge, with the experimental precision we have, the electron appears to be round," Hudson said. In an accompanying article, Aaron Leanhardt at the University of Michigan, Ann Arbor, said the work provided a window "on the high-energy soul of the cosmos". "This work has important ramifications for the types of particles that can be discovered at high-energy accelerators, and may eventually help to explain the composition of the observable universe," Leanhardt wrote. With improvements to their equipment, Hudson's team hopes to make even more precise measurements of the electron in the coming years. "If we could make it 10 times better we could pretty conclusively rule in or out supersymmetry, and that would be a huge result for us," he said. guardian.co.uk Â© Guardian News and Media Limited 2011

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[math-fun] topology question
by Cordwell, William R 25 May '11

25 May '11

Starting with C^3 \ (0,0,0), construct the complex projective space CP^2 with the constraint x^3 + y^3 + z^3. Is this homeomorphic to the torus? ------------------------------------- William R. Cordwell Sandia National Laboratories 505-844-0660 cordwell(a)sandia.gov

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[math-fun] Tiling puzzle
by Adam P. Goucher 25 May '11

25 May '11

I have a nice, fun tiling puzzle: 1. Prove that there is no way to fill a 2*2*2 cube with four 2*1*1 dominoes, such that there is no 2*2*1 subset containing two dominoes. 2. Fill a 2*2*2*2 hypercube with eight 2*1*1*1 dominoes, such that there is no 2*2*1*1 subset containing two dominoes. 3. Prove that your solution is unique, up to automorphism of the hypercube. (This is inspired by the fact that in two dimensions, every possible domino tiling of a genus-0 (hole-less) polyomino can be reached from any other by rotating 2*2 blocks of two dominoes. This shows that it cannot be generalised to four dimensions.) Sincerely, Adam P. Goucher

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