science of spherical arrangements l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Science of Spherical Arrangements PowerPoint Presentation
Download Presentation
Science of Spherical Arrangements

Loading in 2 Seconds...

play fullscreen
1 / 20

Science of Spherical Arrangements - PowerPoint PPT Presentation


  • 272 Views
  • Uploaded on

Science of Spherical Arrangements Well-distributed points on the sphere Motivation from Chemistry, Biology, Physics Survey of results in the literature New results for logarithmic points Peter Dragnev Mathematical Sciences, IPFW Well-distributed points on the sphere

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Science of Spherical Arrangements' - Sharon_Dale


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
science of spherical arrangements

Science of Spherical Arrangements

  • Well-distributed points on the sphere
  • Motivation from Chemistry, Biology, Physics
  • Survey of results in the literature
  • New results for logarithmic points

Peter Dragnev

Mathematical Sciences, IPFW

slide2

Well-distributed points on the sphere

Let N ={X1 , … , XN} Sd-1 - the unit sphere in

d-dimensional space Rd : { X = (x1 , x2 , … , xd ) }.

Sd-1 = { X : x12 + x22 + … + xd2=1 }

How do we distribute “well” the points?

  • For d=2, this problem is simple.The solution is up to rotations the roots of unity.

Reason - direction and order

slide3

For d3 - no direction or order exists. Other methods and criteria are needed.

To well-distribute means to minimize some energy.

We distinguish

- Best packing points;

- Fekete points;

- Logarithmic points, etc.

slide4

Minimum Energy Problem on the Sphere

GivenanN-point configuration N ={X1 , … , XN}on the sphere S2 we define its generalized energy as

E(N )=Σij |Xi - Xj|.

  • MaximizeE(N ) when >0;
  • Minimize it when <0;
  • When =0minimize the logarithmic energy
  • E0( N )=Σij log(1/|Xi - Xj|)
  • or maximize the product P( N )=exp(-E0 ( N )).

Denote the extremal energy with E(N,d) .

slide5

 = 1Except for small N a long standing open problem in discrete geometry (L. Fejes Toth - 1956)

  -Tammes problem; maximize the minimum distance between any pair of points. Known for N= 1-12 and 24.(Best packing points)

 = -1Thompson problem; recent discovery of fullerenes attracted the attention of researchers in chemistry, physics, crystallography. Answer known for N=1-4, 6, 12. (Fekete Points)

 = 0 The problem was posed by L. L. Whyte in 1952. Until recently the answer was known only for N=1-4. (Logarithmic Points)

slide6

C60

C70

Motivation from Chemistry

  • Large carbon molecules discovered in 1985 by Richard Smalley et. el.
  • Fullerenes (Buckyballs)
slide7

Nanotechnology

Nanowire -- “a giant single fullerene molecule”, “a truly metallic electrical conductor only a few nanometers in diameter, but hundreds of microns (and ultimately meters) in length”, “expected to have an electrical conductivity similar to copper's, a thermal conductivity about as high as diamond, and a tensile strength about 100 times higher than steel” (R. Smalley).

slide8

Motivation from Biology

  • Problem of Tammes: (  -)
  • Questions, raised by the Dutch botanist Tammes in 1930 in connection with the distribution of pores on pollen grains.
  • What is the largest diameter of n equal circles that can be packed on the surface of a unit sphere without overlap?
  • How to arrange the circles to achieve this maximum, and when is the arrangement essentially unique?
  • This is the same as to ask to maximize the minimum distance between the points in the arrangement.
slide9

A “disco ball” in space

Starshine 3 satellite was launched in 2001 to study Earth’s upper atmosphere. The satellite was covered by 1500 small mirrors, which reflected the sun light during its free fall, allowing a large group of students nationwide to track the satellite.

Image credit: Michael A.Savell and Gayle R. Fullerton.

slide10

Motivation from Physics

Electrons in Equilibrium

  • N electrons orbit the nucleus
  • Electrons repel
  • Equilibrium will occur at minimum energy

Problem:

If Coulomb’s Law is assumed, then we minimize the sum of the reciprocals of the mutual distances, i. e.

Σij|Xi - Xj|-1

over all possible configurations of N points X1 , … , XN on the unit sphere. This corresponds to the case  = -1.

slide11

32 Electrons

122 Electrons

In Equilibrium

slide12

Distribution of Dirichlet Cells (School Districts)

Dj:={Xє S2: |X-Xj|=mink |X-Xk|} j=1,…,N

The D-cells of 32 electrons at equilibrium are the tiles of the Soccer Ball. Soccer Ball designs occur in Nature frequently. The vertices of the Soccer Ball form C60.

There has to be exactly 12 pentagons in a soccer ball design.

Q. How are D-cells distributed for large N?

survey of results for small n
Tammes problem (best packing) = -

The solution is known for N=1,2, …, 12, 24.

N=4 - regular tetrahedron

N=5,6 - south, north and the rest on the equator

N=8 - skewed cube

N=12 - regular icosahedron (12 pentagons)

Survey of results for small N

Thompson problem  = -1 (known for N=4,6,12)

N=4 - regular tetrahedron

N= 6 - regular octahedron

N=12 - regular icosahedron

slide15

Whyte’s problem  = 0 (Logarithmic points)

N=4 - regular tetrahedron - {1,3}

N=12 - regular icosahedron - {1,5,5,1} [A] ‘96

N=6 - regular octahedron - {1,4,1} [KY] ‘97

N=5 - D3h {1,3,1} - [D+Legg+Townsend] ‘01

Remark: The method of proof is different from the other two results - a mixture of analytical and geometrical methods.

Definition:A collection of points which minimizes the logarithmic energy E0( N )=Σij log(1/|Xi - Xj|)is calledoptimal configuration. The points are referred to as logarithmic points.

slide16

Theorem 2 (Dragnev ‘02)

Remark:Rakhmanov, Saff, Zhou were first to show the separation condition with constant 3/5. Dubickas obtained a constant 7/4. It is not known whether exists.

Logarithmic points - new results

Let N ={X1 , … , XN } be an optimal configurationon the sphere S2. Define dN :=minij |Xi - Xj|.

Theorem 1(Dragnev, Legg, Townsend - ‘01)

{1,3,1} is the only optimal 5-point configuration.

slide17

(i)

O is center of mass for {X1, X2,… , XN }.

(ii)

Proposition (Properties of optimal configurations)

Let N={X1, X2 … , XN } be an optimal configuration. Then

(iii)

d+2 configuration on Sd-1

Let N={X1, X2 … , XN } Sd-1be optimal, i.e.

E0 ( N )= E0(N,d). Derivative conditions on the energy functional (when N-1 points are fixed) yield:

Corollary:Theregular d-simplex is the only optimal d+1-configuration on Sd-1 for any  0.

slide18

Definitions: Configuration is called

  • critical, if it satisfies (i) (recall (i) implies (ii) and (iii)).
  • degenerate, if it does not span Rd.
  • A vertex Xi is mirror related toXj (Xi ~ Xj) if Xi Xk= Xj Xkfor all k i,j. Then N\ {Xi , Xj } lie in the orthogonal bisector hyperspace of XiXj.

Note: Mirror relation is equivalence relation.

Theorem 3 (Dragnev ‘02)For fixed N and 0 the extremal energy E(N,d) is strictly decreasing for d<N, and for dN, E(N,d) = E(N,N-1).

CorollaryOptimal configurations are non-degenerate.

slide19

Theorem 4 (Dragnev ‘02)If N=d+2, then any critical configurations satisfies at least one of the following:

(a) it is degenerate;

(b)  a vertex with all edges stemming out equal;

(c) every vertex has a mirror related partner.

Example:Let N=5, d=3. By Theorem 3, optimal configurations will satisfy at least one of the following:

(a) degenerate  {5}

(b) EA=EB=EC=ED  {1,4}

(c) Every vertex has a mirror related partner. In this case we arrive at A~B~C and D~E  {1,3,1}

Comparing {5}, {1,4}, and {1,3,1} proves Theorem 2.

slide20

Examples

    • N=4, d=2; diagonals of a square, form orthogonal simplexes.
    • N=5, d=3; equilateral triangle on Equator and  diameter.
    • N=6, d=4; (cos k/3, sin k/3,0,0); (0,0,cos k/3, sin k/3), k=0,1,2;

In progress: If N=d+2 the optimal configurations is unique up to rotations and consists of two mutually orthogonal regular [d/2]- and [(d+1)/2]-simplexes. So,