A Tour Around Central Configurations

A tour around Central Conﬁgurations

Josep M. Cors

Departament de Matem`atiques Universitat Polit`ecnicade Catalunya Campus Manresa

GSDUAB Online Seminar December 14, 2020

J.M. Cors Central Conﬁgurations A tour with...

Martha Alvarez Esther Barrabés Montserrat Corbera JoséLino Cornelio Antonio Fernandes Dick Hall Jaume LLibre Rick Moeckel MercèOllé Ernesto Perez-Chavela Gareth Roberts Claudio Vidal

J.M. Cors Central Conﬁgurations Central conﬁgurations

A central conﬁguration is a special arrangement of point masses interacting by Newton’s law of gravitation with the following property: the gravitational acceleration vector produced on each mass by all the others should point toward the center of mass and be proportional to the distance to the center of mass.

Figure: A central conﬁguration of ﬁve equal masses with gravitational acceleration vectors. Figure by Rick Moeckel (2014), Scholarpedia, 9(4):10667.

J.M. Cors Central Conﬁgurations The role of Central Conﬁgurations

Central conﬁgurations (or CC’s) play an important role in the study of the Newtonian n-body problem.

they lead to the only explicit solutions of the equations of motion, they govern the behavior of solutions near collisions, they inﬂuence the topology of the integral manifolds.

J.M. Cors Central Conﬁgurations Explicit solutions of the n-body problem

Released from rest, a central conﬁguration maintains the same shape as it heads toward total collision (homothetic motion). lagrangehomothetic

Given the correct initial velocities, a central conﬁguration will rigidly rotate about its center of mass. Such a solution is called a relative equilibrium eulerRE

Any Kepler orbit (elliptic, hyperbolic, ejection-collision) can be attached to a central conﬁguration to obtain a solution to the full n-body problem. Above is an example of an asymmetric 8-body c.c. with elliptic homographic motion (eccentricity 0.8). homographic

Simulations by Rick Moeckel (2014), Scholarpedia, 9(4):10667.

J.M. Cors Central Conﬁgurations Approaches to Studying CC’s

Existence: Fix n. Find all possible c.c.’s and investigate how they depend on the masses. Existence for special cases: For a particular choice of masses (or set of masses), what are the c.c.’s and are there any interesting bifurcations? An Inverse Problem: Given a ﬁxed set of positions, what (if any) are the possible masses that make the conﬁguration central? (no restriction on the center of mass)

J.M. Cors Central Conﬁgurations Mass Mapping

k +n M : D ⊂ R 7→ R (suitably normalized)

Notice: central conﬁgurations are not isolated. It is standard practice to ﬁx a scaling and center of mass c, and then identify solutions that are equivalent under a rotation.

J.M. Cors Central Conﬁgurations Four-body problem: Quadrilaterals

Legend Quadrilaterals A, B, C, D are angles a, b, c, d are side lengths B All sides in one plane Not all sides in one plane b Skew a C Planar c A D No crossed sides d Two crossed Note that all properties are sides Simple inherited along the arrows

All angles < 180° Complex One angle > 180° Arrow Antiparallelogram (Self-intersecting) Concave (aka Dart/Chevron) Convex

Two pairs of Two pairs of opposite equal adjacent equal sides sides

Opposite angles sum to 180° Opposite side One pair Equal One pair lengths have Perpendicular of parallel diagonals of opposite equal sums diagonals sides equal sides Tangential Watt Cyclic Ex-tangential Orthodiagonal A + C = B + D = 180° (Inscriptible) a + c = b + d 2 2 2 2 Trapezium Equidiagonal a + c = b + d quadrilateral a + c = b + d A + B = C + D = 180°

One pair of Opposite side lengths parallel ides Non-equal Two pairs of equal have equal sums opposite adjacent sides sides parallel Two One pair of Non-parallel Two pairs J.M. CorsA second adjacentCentral Conﬁgurations parallel sides Opposite of equal sides equal (if circle r=∞) pair of parallel right sides Opposite adjacent sides angles equal sum angles sides equal sum Right-angled Isosceles Kite Tangential Parallelogram Bicentric (‘Rhomboid’ if adjacent trapezium trapezium sides are unequal) trapezium

Two Four sides Four Four sides Four right Four Four right One pair of opposite equal sides equal angles right angles opposite right equal angles equal sides angles Opposite side Third lengths have equal sums side equal Four right angles Right kite Rhombus Quadric Rectangle 3-sides-equal Isosceles (‘Oblong’ if adjacent trapezium tangential sides are unequal)

Four right angles Four Four equal equal sides Four right angles sides Four equal sides

Square Equilic

60°

All angles < 180° Complex One angle > 180° Arrow Antiparallelogram (Self-intersecting) Concave (aka Dart/Chevron) Convex

Two pairs of Two pairs of opposite equal adjacent equal sides sides

One pair of Four-body problem: QuadrilateralsOpposite side lengths parallel ides Non-equal Two pairs of equal have equal sums opposite adjacent sides sides parallel Two One pair of Non-parallel Two pairs A second adjacent parallel sides Opposite of equal sides equal (if circle r=∞) pair of parallel right sides Opposite adjacent sides angles equal sum angles sides equal sum Right-angled Isosceles Kite Tangential Parallelogram Bicentric (‘Rhomboid’ if adjacent trapezium trapezium sides are unequal) trapezium

Four right angles Four Four equal equal sides Four right angles sides Four equal sides

Square Equilic

60°

J.M. Cors Central Conﬁgurations Classifying Convex Quadrilaterals CC

Goal:

Classify the full set of convex central conﬁgurations in the Newtonian planar four-body problem

Answer

The set of four-body convex central conﬁgurations with positive masses is three-dimensional, a graph over a domain D that is the +3 union of elementary regions in R

J.M. Cors Central Conﬁgurations Four-body central conﬁgurations

J.M. Cors Central Conﬁgurations Convex four-body central conﬁgurations

J.M. Cors Central Conﬁgurations The set of convex central conﬁguartions

Recall Dziobek’s equation:

−3 0 −3 0 −3 0 −3 0 −3 0 −3 0 (r12 − λ )(r34 − λ ) = (r13 − λ )(r24 − λ ) = (r14 − λ )(r23 − λ )

Eliminating λ0 from the above, deﬁne F to be the function

3 3 3 3 3 3 3 3 3 3 3 3 F (a, b, c, θ) = (r24−r14)(r13−r12)(r23−r34)−(r12−r14)(r24−r34)(r13−r23)

Up to an isometry, relabeling, and rescaling, the set of all four-body convex central conﬁgurations with positive masses is given by

n +3 o E = s = (a, b, c, θ) ∈ R × (0, π): F (s) = 0, and r13, r24 > r12 ≥ r14, r23 ≥ r34

J.M. Cors Central Conﬁgurations Good coordinates

Three radial variables a, b, c > 0 and an angular variable θ ∈ (0, π)

J.M. Cors Central Conﬁgurations Deﬁning the domain D

Let D = D1 ∪ D2 denote the three-dimensional region, where

Main result D is the domain of the function θ = f (a, b, c) and the projection of E into abc-space.

J.M. Cors Central Conﬁgurations Plot of the boundary of D

J.M. Cors Central Conﬁgurations Conﬁgurations on the boundary D

these points correspond to conﬁgurations where two or more of mutual distances inequalities

r13, r24 > r12 ≥ r14, r23 ≥ r34 become equalities. Moreover, only points for which is true lie on the boundary of D.

J.M. Cors Central Conﬁgurations Back to special cases of central conﬁgurations

J.M. Cors Central Conﬁgurations What’s next

+3 Study the mass map from D into R and show that is injective

Given a particular ordering of the bodies, this would prove that there is a unique convex central conﬁgurations for any choice of four positive masses.

J.M. Cors Central Conﬁgurations The 1 + n–body problem

Of course this is the planetary case, which is very studied, but the investigation of the relative equilibria starts with an unpublished paper of G.R. Hall from 1988.

Hall’s Deﬁnition A relative equilibria of the planar 1 + n–body problem is a n–body conﬁguration 2 q¯ = (¯q0, q¯1, ..., q¯n)q ¯i ∈ R which is a limit of relative equilibria qε for the n–body problem with masses m0 = 1, mi = εµi , i = 1, ..., n for some sequence ε → 0.

The parameters µi determine the mass ratios of the small bodies.

J.M. Cors Central Conﬁgurations Approach to the limit

As ε → 0 Hall showed that Proposition All the relative equilibria of the planar 1 + n–body problem lie on a circle centered at q0 = 0.

Convergence with clustering (collinear) Convergence with clustering

J.M. Cors Central Conﬁgurations Approach to distinct limit positions

We are interested in the case where the limit positions of the small bodies are distinct. The positions can be described by n angles

0 ≤ θ1 < θ2 < ··· < θn < 2π

Convergence without clustering (convex) Convergence without clustering (convave)

J.M. Cors Central Conﬁgurations Antisymmetric matrix

For given angles θ = (θ1, . . . , θn), equations of relative equilibria can be viewed as an n × n system of linear equations

A(θ)µ = 0,

for the mass vector µ = (µ1, . . . , µn). Note that the matrix A(θ) is antisymmetric. It follows that if n is odd, there is always at least one nonzero mass vector µ satisfying the equation. It is interesting that for even n the situation is diﬀerent. A given vector of angles will not give a relative equilibrium for any choice of masses unless the determinant of A(θ) vanishes. If it does vanish, there will be at least a two-dimensional set of mass vectors.

J.M. Cors Central Conﬁgurations n small equal masses

Numerical results When n = 2, 3,. . . ,9, 10,. . . , respectively, Salo and Yoder (1988) found 3, 3, 3, 3, 5, 3, 1, 1,. . . distinct central conﬁgurations.

J.M. Cors Central Conﬁgurations n small equal masses

Two conjectures

Conjecture For n ≥ 9 there is only one central conﬁguration, the trivial one.

Conjecture All central conﬁgurations of the planar 1 + n–body problem are symmetric with respect to a straight line containing the large mass.

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

t1 = tan θ1/4, t2 = tan θ2/4, t3 = tan θ3/4. The assumptions on the angles imply that the new variables satisfy 0 ≤ t1 < t2 < t3 < ∞.

 0 c −b 1 A(t , t , t ) = −c 0 a 1 2 3 d   b −a 0

A(t2, t3)µ = 0

Lemma

The kernel of the matrix A(t2, t3) is spanned by the vector

µ = (a, b, c),

2 for all (t2, t3) ∈ R \ (0, 0).

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

Visualizing the mass map

With the help of Mathematica we get explicit formulas for its components (a, b, c). For example

2 2 2 2 2 2 3 4 b=t2 (t3−q)(t3−r)(t3−1)(t3+1)(t3−t2) (1+t2 ) (1+t2t3) (1+4t3+18t3 +4t3 +t3 )

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

Visualizing the mass map

Μ3 5

Μ2 Μ1

0 1 2 3 4 5

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

The singular curve and the bifurcation curve

4.0

3.5 Μ3

3.0

2.5

2.0

1.5 Μ1 Μ2

1.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2

J.M. Cors Central Conﬁgurations Plot of the boundary of D

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

Main results: Convex conﬁgurations

For every choice of µi > 0 there are exactly 6 convex relative equilibria of the 1 + 3–body problem, one for each cyclic order of the four bodies around the convex quadrilateral.

For every choice of µi > 0 and every suﬃciently small ε > 0 there are exactly 6 convex relative equilibria of the 4–body problem with masses m0 = 1, m1 = εµ1, m2 = εµ2, m3 = εµ3. The circular periodic orbits solutions associated with these convex relative equilibria of the 4–body problem are linearly stable, for ε > 0 suﬃciently small.

J.M. Cors Central Conﬁgurations n = 3 small unequal masses

Main results: Bifurcations and counts of relative equilibria Up to rotation, mass vector µ “inside” the bifurcation curve admit 14 relative equilibria of the 1 + 3–body problem (with distinct position for the small masses) while those “outside” admit 10 such relative equilibria.

Μ3

Μ1 Μ2

J.M. Cors Central Conﬁgurations Planar Central Conﬁguration of κn Bodies (Crowns)

Size of the n-gon: aj , j = 1, . . . , κ Argument of the ﬁrst particle in a group: $j , j = 1, . . . , κ First group: m1 = 1, a1 = 1 and $1 = 0

i$j i2π(k−1)/n qj1 = aj e , qjk = qj1e , k = 2,..., n, j = 1, . . . , κ.

J.M. Cors Central Conﬁgurations Equations

Using the symmetry of the problem, the CC equation reduces to ∂U 2 <(Eqj ) = 0 (Eqj ): + w mj qj1 = 0 −→ ∂qj1 =(Eqj ) = 0 for all j = 1, . . . , κ.

=(Eqj ) = 0 if and only if:

κ n X X sin($l − $j + 2πk/n) m a = 0 l l (a2 + a2 − 2a a cos($ − $ + 2πk/n))3/2 l=1 k=1 l j l j l j l6=j | {z } q 0 if $k =0,π/n, ∀k

J.M. Cors Central Conﬁgurations Nested and Twisted

Nested rings Twisted rings

Vertices align Rotation of π/n

J.M. Cors Central Conﬁgurations Two twisted rings: Mass mapping

2 a (Sna − C21(a)) m = H(a) = 3 , a > 0. Sn − a C12(a) where n−1 1 X 1 S = , C (a) = −φ0(a), C (a) = φ(a)+aφ0(a) n 4 sin(kπ/n) 21 12 k=1 and n X 1 φ(a) = 2 (2k−1)π 1/2 k=1 (1 + a − 2a cos n ) H(1/a) = 1/H(a). If (m, a) corresponds to a CC −→ (1/m, 1/a) is also a CC The CC is convex if π 1 cos ≤ a ≤ π n cos n J.M. Cors Central Conﬁgurations Results for n = 3

The set of admissible values of a such that there exist a CC is

(0, z1) ∪ (1/z2, z2) ∪ (1/z1, ∞),

where z1 < 1 < z2 are the only two positive zeros of m = H(a) = 0, and z1z2 < 1

convex

J.M. Cors Central Conﬁgurations Results for n = 3

Given a value of the mass ratio m ≥ 1, the number of diﬀerent CC are three for any m ∈ (1, m∗) ∪ (m∗∗, ∞) one for any m ∈ (m∗, m∗∗) two for m = 1, m∗, m∗∗. m∗ ' 1.000768, m∗∗ ' 35.700177

J.M. Cors Central Conﬁgurations Convex CC for the 6-body problem

Moeckel: numerical exploration of CC for the N-body problem with equal masses (N = 4,..., 8)

a = 1 a ' 1.077171777

For any m ∈ [1, m∗) there exist two convex CC of the 6-body problem

J.M. Cors Central Conﬁgurations Results for n ≥ 4

Conjecture

For n ≥ 4 the equation H(a) = 0 has only two solutions z1, z2 in (0, ∞) with z1 < 1 < z2 and and z1z2 > 1.

Two nested rings with n ≥ 4 bodies in each ring, and for any positive value of the mass ratio m there exists at least three diﬀerent central conﬁgurations.

J.M. Cors Central Conﬁgurations Convex CC for n = 4

(0.69738051, 0.707106781) (0.707106781, 1.414213563) (1.414213563, 1.433937407) convex

not all the conﬁgurations a ∈ (z1, 1/z1) are convex. convex conﬁgurations correspond to m ∈ (0.062278912, 16.05679941) there are not convex CC of twisted rings for certain values of m

J.M. Cors Central Conﬁgurations n Central admissible interval A2C (n) cos(π/n) 3 (0.617364128382, 1.619789608802) 0.5 4 (0.697380509876, 1.433937406966) 0.707106781187 5 (0.822182869908, 1.216274428233) 0.809016994375 6 (0.884321138125, 1.130810920250) 0.866025403784 7 (0.918990363772, 1.088150691695) 0.900968867902 8 (0.940138179122, 1.063673428234) 0.923879532511 9 (0.953949939513, 1.048273036749) 0.939692620786 10 (0.963459881269, 1.037925936971) 0.951056516295 ...... 100 (0.999674025507, 1.00032608079) 0.999506560366 500 (0.999986989988, 1.00001301018) 0.999980339576 1000 (0.999996754292, 1.00000324572) 0.999995075057 5000 (0.999999869916, 1.00000013008) 0.999999802608 Table: The central admissible interval and cos(π/n) for diﬀerent the values of n.

J.M. Cors Central Conﬁgurations Stacked central conﬁgurations

Definition A Stacked central configuration is a central configuration such that a proper subset of the n bodies also form a central configuration.

J.M. Cors Central Conﬁgurations Five body problem: Lagrange plus two

Two Questions:

Are there convex central configurations of the five body problem formed by piling up two bodies to the Lagrange central configuration? Are there non-symmetric central configurations of the five body problem formed by piling up two bodies to the Lagrange central configuration?

J.M. Cors Central Conﬁgurations Five body problem: Lagrange plus two

Answers:

Theorem

Let r = (r1, r2,..., r5) be a planar convex central conﬁguration of the ﬁve-body problem. Then there are no three vertices of the pentagon forming an equilateral triangle.

J.M. Cors Central Conﬁgurations Five body problem: Euler plus two

J.M. Cors Central Conﬁgurations Five body problem: Lagrange plus Euler in one

J.M. Cors Central Conﬁgurations Counting stacked central conﬁgurations

Question Given a central configuration of the n–body problem, how many different subsets of bodies, A, do exist with cardinality |A| = k, k = 1, ..., n − 3, such that the n − k bodies form a central configuration of the (n − k)–body problem?

Question Using the notation, (n, k)–stacked central configuration, where n is the number of bodies of the original central configuration and k = 1, ..., n? is the number of the removed bodies, previous question can be stated as following: What is the number of (n, k)–stacked central configurations, for all k = 1, ..., n − 3?

J.M. Cors Central Conﬁgurations Counting stacked: Five body problem

Theorem In the non–collinear planar ﬁve–body problem, the number of (5, k)–stacked central conﬁguration, for all k = 1, 2 up to similarity is at most two.

J.M. Cors Central Conﬁgurations Counting stacked: Six body problem

J.M. Cors Central Conﬁgurations Counting stacked

Glubs stucked with stacked

keep traveling

J.M. Cors Central Conﬁgurations