lectures iii and iv - connelly/packingsiii.iv.pdf · 9 random close packings “is random close...

1

Packings of circles and spheres

Lectures III and IVSession on Granular Matter

Institut Henri Poincaré R. Connelly

Cornell UniversityDepartment of mathematics

2

What model is good forpackings?

3


• There must be a container of some sort forthe packing elements.

4



• Is the packing jammed?

5



• Is the packing jammed?• If so, what criteria do you want for

jamming?

6

Some containers

Put the packing in a box. Let the box shrink in sizeor let the packing elements grow in size until youare stuck.

7

More possible containers

• A rectangular box• A circle• A torus

8

Jammed packings

When the packing is locally most dense, there is arigid cluster that is jammed, but there could be“rattlers”, non-rigid pieces.

The most dense packing of 7 congruent circles in asquare. (Note the "rattler" in the upper right corner.)(Due to Shaer)

9

Random close packings

“Is Random Close Packing of Spheres WellDefined?” (Torquato, Truskett and Debenedetti 2000)Experimentally, if you pour ball bearings into alarge container, shake, and calculate the density,you get about 0.636. But this depends on themethod you use to “densify”. Other methods give0.642, 0.649, 0.68 … This paper suggests “orderparameters”…

The following is joint work with A. Donev, F. Stillinger, andS. Torquato. See also my papers in:

http://www-iri.upc.es/people/ros/StructuralTopology/

10

Definitions

If P is a packing in a container C, call it locallymaximally dense if every other packing of thesame packing elements, sufficiently close to P inC has a packing density (=packing fraction) nolarger than the density of P in C.

A packing P is (collectively) jammed in C if anyother packing sufficiently close to P in C is thesame as P. (I also say P is rigid in C.)

11

Why jammed?

Theorem: If a packing P of circles (or sphericalballs in 3-space) is locally maximally dense in apolygonal container C, then there is subset P0 of Pthat is collectively jammed in C.

Theorem (Danzer 1960): A packing P of circles (orspherical balls in 3-space) is collectively jammedin a polygonal container C, if and only if theunderlying strut tensegrity framework isinfinitesimally rigid.

12

Remarks

• If the packing P has no part that is collectivelyjammed, then the packing can be moved slightlyand the container shrunk slightly to improve theoverall density.

• Since a necessary condition for being jammed isinfinitesimal rigidity, circle packings in polygonscan be tested for being jammed with reasonablyefficient linear programming (feasability)algorithms and the nearby improved packingcalculated as well.

13

ProofsLet the configuration p = (p1, p2, … pn) be the

centers of the packing circles, including the pinnedpoints on the boundary. If the associated strutframework G(p) is infinitesimally rigid, even for asubset of the packing centers, then there is nonearby configuration satisfying the packingconstraints for that subset that increases density.

If G(p) is not infinitesimally rigid, let p¢= (p1¢, p2¢,… pn¢) be the (non-zero) infinitesimal flex ofG(p). Then for each i define pi(t)= pi + t pi¢, 0 ≤ t .

16

Remarks• The motion provided by the linear programming

algorithm (LP) can be quite complicated, and itseems best to use it to clear up log jams, ratherthan ongoing “densification”.

• For example, the Lubachevsky-Stillingeralgorithm of slow densification (bump-and-slowdown) is more efficient in converging to a jammedstate than LP.

• The (LP) motion does not continue forever. Somenon-incident packing disks will get closer togetherand collide, sometimes quite soon.

17

StaticsWhen the packing is collectively jammed, it must be

statically rigid, which means it must be able toresolve arbitrary loads. But such tensegrityframeworks are never statically determinant, sinceby the Roth-Whiteley Theorem, there is always anon-zero proper self stress. On the other hand, aphysical system must resolve loads in some way…Later more about this. The final resolution of anexternal load is a function of the relative elasticityin the packing elements. Geometry, alone, cannot“resolve” this.

18

The fundamental conundrumDo numerical simulations make sense for large numbers of

packing elements? For example, for monodispersed (i.e.congruent) disks in the plane, the following configurationoften appears in some simulations, which is impossible.

19

Infinite packings

How can infinite packings be “jammed”?

20

Infinite packings

How can infinite packings be “jammed”?They can’t. We have to keep them from

flying apart. Some ideas…

21

Infinite packings


flying apart. Some ideas…1. Insist that each packing element be fixed

by its neighbors. (Locally jammed.)

22

Infinite packings


flying apart. Some ideas…1. Insist that each packing element be fixed

by its neighbors. (Locally jammed.)2. Pin all but a finite number. (Finite

stability=collectively jammed)

23

Infinite packings

How can infinite packings be “jammed”?They can’t. We have to keep them from flying apart.

Some ideas…1. Insist that each packing element be fixed by its

neighbors. (Locally jammed.)2. Pin all but a finite number. (Finite stability

=collectively jammed)3. Enforce periodicity. This amounts to putting the

packing on a torus. (Periodic stability)

24

Difficulties

The locally jammed property is very weak. K.Böröczky (1964) showed that there is apacking of congruent disks in the plane with0 density that is locally jammed. If youmove 8 disks at a time, the packing fallsapart.

25

Difficulties

The collectively jammed property is better,but as the number of packing elements thatare allowed to move gets large, the packingcan be very fragile. For example, thefollowing square lattice packing in a squareseems to be not collectively jammed ... (A.Bezdek, K. Bezdek, R. Connelly 1998)

26

A near unjamming

27

A near unjamming

28

A near unjamming

29

A near unjamming

30

A near unjamming

31

A near unjamming

32

A near unjamming

33

A near unjamming

34

The size of the square

35

Discrete rigidityTheorem: Suppose that G(p) is a finite (tensegrity)

framework, with pinned vertices, that is rigid inEd. Then there is an e > 0 such that when |pi-qi| <e for all i, and the same member constraints holdfor G(q), then p=q.

Even when the member constraints are relaxed byd > 0, say, then the configuration can jiggle about,but not too far.

The e for the square lattice packing in a squareseems quite small.

36

Calculations for the square latticepacking in a square

Let s be the length of the side of the boundingsquare during the deformation shown. Then whenthe first pair of circles are rotated by q, then s iscalculated as

s = 4n cos q + 2 sin q + 2,where the radius of the circles are 1, and there are

n+1 circles on a side, hence (n+1)2 circles in all.So the square expands less than 1/(2n), ands < 4n +2, the side length of the square, forq > 2/n.

37

The point

The e needed to insure rigidity converges to 0as n -> • for the square lattice packing. Inother words, you have to decrease thetolerances to 0 as you release more andmore of the packing elements and allowthem to move.

A suggestion to address this problem …

38

Uniform stabilityAn infinite packing of circles in the plane is uniformly stable

if there is an e > 0 such that the only finite rearrangementof the disks as a packing, where each center is displacedless than e, is the identity. (This is related to a different, butsimilar, definition of L. Fejes Toth.)

39

Examples1. The square lattice packing (or the cubic lattice

packing in higher dimensions) is NOT uniformlystable.

2. The triangular lattice packing in the plane ISuniformly stable, even with the removal of onepacking element. (I. Bárány and N.P. Dolbilin (1988)as well as A. Bezdek, K. Bezdek, R.C. (1998))

3. Most of the candidates for the most densepackings of congruent spheres in higherdimensions are uniformly stable. (A. Bezdek, K.Bezdek, R. Connelly 1998)

Another suggestion for this problem:

40

How do you handle infinitepackings?

• Only allow a finite number of packings tomove. . . . or

• Consider periodic packings with a fixedlattice defining the periodicity . . . or

• Consider periodic packings with a variablelattice as well as a variable configuration,but such that the density is locally maximal.

41

Periodic Packings

A packing is periodic with lattice L if forevery packing element X, l + X is apacking element for every l in L. Regardthis as a packing of a finite number ofpacking elements in the torus R2/L. So sucha packing can be regarded as stable if thereis no perturbation of the configurationexcept trivial perturbations.

42

The problem with fixing thelattice defining the torus

The packing (with only 2 disks) defined on a torus, whosefundamental region is the small yellow square, is jammed.

The packing (with 8 disks) defined on a torus, whosefundamental region is the larger square, is NOT jammed.

43

Counting in Coverings

Let k be the order of the covering, so there areexactly kn disks in the covering and ke contacts. Ifthere are the minimal number of contacts e = 2n-1in the original rigid configuration, there will be

ke = k(2n-1) = 2kn-k < 2kn-1 contacts in the cover, and so the cover will NOT be

rigid.

44

Strictly Jammed

So we define a packing on a (flat) torus to bestrictly jammed if there is no non-trivialinfinitesimal motion of the packing, as wellas the lattice defining the torus, subject tothe condition that the total area of the torusdoes not increase during the infinitesimalmotion.

45

Counting for Strictly JammedPackings

• Variables:– Coordinates: 2n - Lattice vectors: 2 + 2 = 4

• Constraints:– Packing contacts: e - Area constraint: 1

• Trivial motions:– Translatons: 2 - Rotations: 1

The final count is thene +1 ≥ 2n + 4 - 3 + 1 or e ≥ 2n +1.

So the packing covers have a chance to be strictly jammed,although it is not guaranteed. The number of constraints ina cover is more than is needed for stability.

46

The Crystallization Conjecture

If a periodic packing is strictly jammed, whatdoes it look like?

Conjecture: The only strictly jammedperiodic packing in the plane is thetriangular lattice packing, possibly withsome disks missing.

47

The Crystallization Conjecture

If a periodic packing is strictly jammed, whatdoes it look like?

Conjecture: The only strictly jammedperiodic packing in the plane is thetriangular lattice packing, possibly withsome disks missing.

Sadly this is false.

48

Start with this tiling

49

The construction of the pentagonin the tiling

1

1

1

1

1

2a + b + 60 = 360

120120

b

aa

33

50

Split it apart and add some“bracing triangles” as well assome connecting rhombuses.

51

Split it apart and add some“bracing triangles” as well assome connecting rhombuses.

52

This what the packing looks like.

53

Another possible example of astrictly jammed packing

54

Low density strictly jammedpackings

Conjecture: The density of a strictly jammedperiodic packing is greater than (3/4)p/√12, thedensity of the Kagome packing.

55

“In fact, in a flat two-dimensional space it isbelieved that only the triangular latticeconfiguration … is stable in the alter case.” (Einar L.Hinrichsen, Jens Feder, and Torstein Jossnag, Randompacking of disks in two-dimensions, Phys. Rev. A, Vol 41,No. 8, 15 April 1990, p.4200.)

lectures iii and iv - connelly/packingsiii.iv.pdf · 9 random close packings “is random close...

Documents