PURE Math Residents Program Gröbner Bases and Applications Week 3 Lectures

Transcription

1 PURE Math Residents Program Gröbner Bases and Applications Week 3 Lectures John B. Little Department of Mathematics and Computer Science College of the Holy Cross June 2012

2 Overview of this week The research project topics you will be working on will apply the algebra and geometry we have seen (e.g. Gröbner bases, varieties, etc.) to some interesting questions in celestial mechanics (and possibly related question in fluid mechanics

3 Overview of this week The research project topics you will be working on will apply the algebra and geometry we have seen (e.g. Gröbner bases, varieties, etc.) to some interesting questions in celestial mechanics (and possibly related question in fluid mechanics In this week s activities, we ll develop some background and look at basic cases to provide context

4 Overview of this week The research project topics you will be working on will apply the algebra and geometry we have seen (e.g. Gröbner bases, varieties, etc.) to some interesting questions in celestial mechanics (and possibly related question in fluid mechanics In this week s activities, we ll develop some background and look at basic cases to provide context Fortunately, not a lot of physics is required, but if you have seen some, that will help

5 Newtonian gravitation Everyone should know about Newton s gravitational law

6 Newtonian gravitation Everyone should know about Newton s gravitational law Says (in basic form) that magnitude of gravitational force exerted by one mass m 1 on another m 2 is proportional to the product m 1 m 2, and inversely proportional to the square of the distance:

7 Newtonian gravitation Everyone should know about Newton s gravitational law Says (in basic form) that magnitude of gravitational force exerted by one mass m 1 on another m 2 is proportional to the product m 1 m 2, and inversely proportional to the square of the distance: F = Gm 1m 2 r 2

8 Newtonian gravitation Everyone should know about Newton s gravitational law Says (in basic form) that magnitude of gravitational force exerted by one mass m 1 on another m 2 is proportional to the product m 1 m 2, and inversely proportional to the square of the distance: F = Gm 1m 2 r 2 Suppose the masses are located at q 1, q 2 R 3.

9 Newtonian gravitation Everyone should know about Newton s gravitational law Says (in basic form) that magnitude of gravitational force exerted by one mass m 1 on another m 2 is proportional to the product m 1 m 2, and inversely proportional to the square of the distance: F = Gm 1m 2 r 2 Suppose the masses are located at q 1, q 2 R 3. The gravitational force exerted by m 1 on m 2 is a vector directed from q 2 back to q 1 : F = Gm 1m 2 (q 2 q 1 ) r 3 where r = q 2 q 1 is the distance (note cancelation in magnitude formula!)

10 Some history Building on astronomical observations by Tycho Brahe ( ), Johannes Kepler ( ) developed three empirical laws of planetary motion in a simplified planet, star (much larger mass) system:

11 Some history Building on astronomical observations by Tycho Brahe ( ), Johannes Kepler ( ) developed three empirical laws of planetary motion in a simplified planet, star (much larger mass) system: Kepler 1: The planet follows a planar, elliptical orbit with the star at one focus

12 Some history Building on astronomical observations by Tycho Brahe ( ), Johannes Kepler ( ) developed three empirical laws of planetary motion in a simplified planet, star (much larger mass) system: Kepler 1: The planet follows a planar, elliptical orbit with the star at one focus Kepler 2: The radius vector from star to planet sweeps out equal areas in equal times as the planet orbits

13 Some history Building on astronomical observations by Tycho Brahe ( ), Johannes Kepler ( ) developed three empirical laws of planetary motion in a simplified planet, star (much larger mass) system: Kepler 1: The planet follows a planar, elliptical orbit with the star at one focus Kepler 2: The radius vector from star to planet sweeps out equal areas in equal times as the planet orbits Kepler 3: The square of the period of orbit is proportional to the cube of the semimajor axis of the ellipse: T 2 = ka 3 ; see Wikipedia entry for Kepler s laws of planetary motion

14 Some history Building on astronomical observations by Tycho Brahe ( ), Johannes Kepler ( ) developed three empirical laws of planetary motion in a simplified planet, star (much larger mass) system: Kepler 1: The planet follows a planar, elliptical orbit with the star at one focus Kepler 2: The radius vector from star to planet sweeps out equal areas in equal times as the planet orbits Kepler 3: The square of the period of orbit is proportional to the cube of the semimajor axis of the ellipse: T 2 = ka 3 ; see Wikipedia entry for Kepler s laws of planetary motion Isaac Newton ( ) basically developed his form of differential calculus to show that Kepler s laws follow from the inverse square law of gravitational attraction (and his own law F = ma)

15 The Newtonian n-body problem Basic problem in celestial mechanics is to understand what happens when we have any number n of masses m i, each attracting all the others according to Newtonian law. Say the masses are located at q i (functions of t).

16 The Newtonian n-body problem Basic problem in celestial mechanics is to understand what happens when we have any number n of masses m i, each attracting all the others according to Newtonian law. Say the masses are located at q i (functions of t). Then F = ma for each mass leads to a system of 2nd order differential equations like this m i d 2 q i dt 2 = j i Gm i m j (q j q i ) r 3 ij

17 The Newtonian n-body problem Basic problem in celestial mechanics is to understand what happens when we have any number n of masses m i, each attracting all the others according to Newtonian law. Say the masses are located at q i (functions of t). Then F = ma for each mass leads to a system of 2nd order differential equations like this m i d 2 q i dt 2 = j i Gm i m j (q j q i ) r 3 ij It s very difficult to understand all properties of the solutions(!)

18 The Newtonian n-body problem, continued For instance, even in the full 3-body problem, it is known that some solutions exhibit chaotic behavior

19 The Newtonian n-body problem, continued For instance, even in the full 3-body problem, it is known that some solutions exhibit chaotic behavior Can be random-looking even though completely deterministic

20 The Newtonian n-body problem, continued For instance, even in the full 3-body problem, it is known that some solutions exhibit chaotic behavior Can be random-looking even though completely deterministic Have sensitive dependence on initial conditions

21 The Newtonian n-body problem, continued For instance, even in the full 3-body problem, it is known that some solutions exhibit chaotic behavior Can be random-looking even though completely deterministic Have sensitive dependence on initial conditions Virtually impossible to make long-term predictions in those cases since we never know the initial conditions to arbitary precision in real-world measurements(!)

22 Central Configurations We will focus mostly on a special class of solutions of the n-body problem called central configurations

23 Central Configurations We will focus mostly on a special class of solutions of the n-body problem called central configurations One way to define them: A central configuration (with center of mass at q) is a configuration such that the acceleration vector of each body is proportional to the position vector from center of mass, all with the same (negative) proportionality constant ω 2

24 Central Configurations We will focus mostly on a special class of solutions of the n-body problem called central configurations One way to define them: A central configuration (with center of mass at q) is a configuration such that the acceleration vector of each body is proportional to the position vector from center of mass, all with the same (negative) proportionality constant ω 2 In other words, for all i, (setting G = 1) m j (q j q i ) + ω 2 (q i q) = 0 j i r 3 ij

25 Comments, and first properties Note: the central configuration equations are algebraic equations on the coordinates of the position vector (not differential equations)

26 Comments, and first properties Note: the central configuration equations are algebraic equations on the coordinates of the position vector (not differential equations) The q i, q j are now thought of as constant vectors, not functions of t

27 Comments, and first properties Note: the central configuration equations are algebraic equations on the coordinates of the position vector (not differential equations) The q i, q j are now thought of as constant vectors, not functions of t If n masses are released from rest at positions satisfying the central configuration equations, they will collapse homothetically to a complete collision in finite time

28 Comments, and first properties Note: the central configuration equations are algebraic equations on the coordinates of the position vector (not differential equations) The q i, q j are now thought of as constant vectors, not functions of t If n masses are released from rest at positions satisfying the central configuration equations, they will collapse homothetically to a complete collision in finite time This means that at all times the configuration will be the same as the original, up to scaling

29 Comments, and first properties Note: the central configuration equations are algebraic equations on the coordinates of the position vector (not differential equations) The q i, q j are now thought of as constant vectors, not functions of t If n masses are released from rest at positions satisfying the central configuration equations, they will collapse homothetically to a complete collision in finite time This means that at all times the configuration will be the same as the original, up to scaling Planar central configurations are also relative equilibria given the correct initial velocities, they can produce solutions of the n-body equations that undergo rigid rotation about the center of mass

30 An alternative definition Let U = 1 j<k n Gm j m k r jk be the gravitational potential energy, and

31 An alternative definition Let U = 1 j<k n Gm j m k r jk be the gravitational potential energy, and Let I = 1 2 be the moment of inertia n m j q j 2 j=1

32 An alternative definition Let U = 1 j<k n Gm j m k r jk be the gravitational potential energy, and Let I = 1 2 be the moment of inertia n m j q j 2 j=1 Then the bodies form a central configuration if U + ω 2 I = 0

33 An alternative definition Let U = 1 j<k n Gm j m k r jk be the gravitational potential energy, and Let I = 1 2 be the moment of inertia n m j q j 2 j=1 Then the bodies form a central configuration if U + ω 2 I = 0 Lagrange multiplier method c.c. s are critical points of U subject to I = const

34 Practical applications Central configurations in the 3-body problem

35 Practical applications Central configurations in the 3-body problem At each given time a planet-sun (e.g. Earth-Sun) system has 5 Lagrange points which are the positions where a third mass could be placed to form a central configuration: (image credit: NASA Wilkinson Microwave Aniosotropy Probe project)

36 Meaning of Lagrange points A satellite placed at rest at one of these Lagrange points will stay in a rotating orbit about the center of mass of the planet-sun system (relative equilibrium property) due to gravitational attraction exerted on it by the planet and sun

37 Meaning of Lagrange points A satellite placed at rest at one of these Lagrange points will stay in a rotating orbit about the center of mass of the planet-sun system (relative equilibrium property) due to gravitational attraction exerted on it by the planet and sun Same phenomenon has been observed for many natural objects, too

38 Meaning of Lagrange points A satellite placed at rest at one of these Lagrange points will stay in a rotating orbit about the center of mass of the planet-sun system (relative equilibrium property) due to gravitational attraction exerted on it by the planet and sun Same phenomenon has been observed for many natural objects, too For instance each cloud of Trojan asteroids has center of mass approx. at a Lagrange point of Sun and Jupiter

39 Meaning of Lagrange points A satellite placed at rest at one of these Lagrange points will stay in a rotating orbit about the center of mass of the planet-sun system (relative equilibrium property) due to gravitational attraction exerted on it by the planet and sun Same phenomenon has been observed for many natural objects, too For instance each cloud of Trojan asteroids has center of mass approx. at a Lagrange point of Sun and Jupiter You will study the Lagrange points in this week s lab assignment (via consideration of central configurations for the 3-body problem in general)

40 Meaning of Lagrange points A satellite placed at rest at one of these Lagrange points will stay in a rotating orbit about the center of mass of the planet-sun system (relative equilibrium property) due to gravitational attraction exerted on it by the planet and sun Same phenomenon has been observed for many natural objects, too For instance each cloud of Trojan asteroids has center of mass approx. at a Lagrange point of Sun and Jupiter You will study the Lagrange points in this week s lab assignment (via consideration of central configurations for the 3-body problem in general) More generally, rings of Saturn are a central configuration with a large number of smaller masses orbiting the large mass of Saturn

41 Saturn and its rings

42 The big problem Major question about central configurations is

43 The big problem Major question about central configurations is Given n masses m 1,..., m n, at how many different locations can these be placed to get central configurations? In particular, is the set of possible configurations finite? On Smale s 21st century problem list.

44 The big problem Major question about central configurations is Given n masses m 1,..., m n, at how many different locations can these be placed to get central configurations? In particular, is the set of possible configurations finite? On Smale s 21st century problem list. To make this meaningful, need some further explanations or definitions, since we can translate, rotate, and scale central configurations and the results are again central configurations

45 The big problem Major question about central configurations is Given n masses m 1,..., m n, at how many different locations can these be placed to get central configurations? In particular, is the set of possible configurations finite? On Smale s 21st century problem list. To make this meaningful, need some further explanations or definitions, since we can translate, rotate, and scale central configurations and the results are again central configurations Convention: We will consider two central configurations to be equivalent if there exists a mapping of R k taking one into the other that can be obtained as a composition of a rigid motion (translation, rotation) and a scaling

46 Status of the big problem Answer is known to be yes for n = 3, 4

47 Status of the big problem Answer is known to be yes for n = 3, 4 Some fairly strong results for n = 5

48 Status of the big problem Answer is known to be yes for n = 3, 4 Some fairly strong results for n = 5 Only fairly limited special cases known in general

49 Status of the big problem Answer is known to be yes for n = 3, 4 Some fairly strong results for n = 5 Only fairly limited special cases known in general Extremely subtle problem in general(!)

50 Status of the big problem Answer is known to be yes for n = 3, 4 Some fairly strong results for n = 5 Only fairly limited special cases known in general Extremely subtle problem in general(!) For instance, it s known that there are collections of n = 5 masses, including one negative value, for which there is a whole curve of central configurations (not just a finite number, even taking rigid motions and scaling into account)

51 Mutual distances as coordinates Because of the way we want to consider equivalence, it makes sense to try to set up the problem so that:

52 Mutual distances as coordinates Because of the way we want to consider equivalence, it makes sense to try to set up the problem so that: A fixed distance scale is used, and

53 Mutual distances as coordinates Because of the way we want to consider equivalence, it makes sense to try to set up the problem so that: A fixed distance scale is used, and The mutual distances r ij become the coordinates for describing the configuration

54 Mutual distances as coordinates Because of the way we want to consider equivalence, it makes sense to try to set up the problem so that: A fixed distance scale is used, and The mutual distances r ij become the coordinates for describing the configuration This leads to some subtleties that we will need to understand, though

55 Mutual distances as coordinates Because of the way we want to consider equivalence, it makes sense to try to set up the problem so that: A fixed distance scale is used, and The mutual distances r ij become the coordinates for describing the configuration This leads to some subtleties that we will need to understand, though For instance, what are all triples (r 12, r 13, r 23 ) that can represent the mutual distances of three points in R 2?

56 Some obvious restrictions All r ij R (they re distances!) and any nonreal solutions of the c.c. equations will not be meaningful for the physical problem

57 Some obvious restrictions All r ij R (they re distances!) and any nonreal solutions of the c.c. equations will not be meaningful for the physical problem All r ij 0 and = 0 only in a partial or total collision case

58 Some obvious restrictions All r ij R (they re distances!) and any nonreal solutions of the c.c. equations will not be meaningful for the physical problem All r ij 0 and = 0 only in a partial or total collision case By triangle inequalities, we also have r 12 + r 23 r 13 0, r 12 + r 13 r 23 0, and r 13 + r 23 r 12 0 with equality in one if and only if the three points are collinear

59 Some obvious restrictions All r ij R (they re distances!) and any nonreal solutions of the c.c. equations will not be meaningful for the physical problem All r ij 0 and = 0 only in a partial or total collision case By triangle inequalities, we also have r 12 + r 23 r 13 0, r 12 + r 13 r 23 0, and r 13 + r 23 r 12 0 with equality in one if and only if the three points are collinear Which one gives zero depends on which point is between the other two along the line

60 Some obvious restrictions All r ij R (they re distances!) and any nonreal solutions of the c.c. equations will not be meaningful for the physical problem All r ij 0 and = 0 only in a partial or total collision case By triangle inequalities, we also have r 12 + r 23 r 13 0, r 12 + r 13 r 23 0, and r 13 + r 23 r 12 0 with equality in one if and only if the three points are collinear Which one gives zero depends on which point is between the other two along the line The set of all points in R 3 satisfying these inequalities is a triangular cone C with vertex at the origin

61 The configuration space of triangles We claim that every point in C represents a triangle (including degenerate cases in which the three points are collinear, and two of them coincide)

62 The configuration space of triangles We claim that every point in C represents a triangle (including degenerate cases in which the three points are collinear, and two of them coincide) For instance, consider the cases where r 13 r 12.

63 The configuration space of triangles We claim that every point in C represents a triangle (including degenerate cases in which the three points are collinear, and two of them coincide) For instance, consider the cases where r 13 r 12. Then we can place point 1 at the origin, point two at (r 12, 0)

64 The configuration space of triangles We claim that every point in C represents a triangle (including degenerate cases in which the three points are collinear, and two of them coincide) For instance, consider the cases where r 13 r 12. Then we can place point 1 at the origin, point two at (r 12, 0) Then point 3 can be any point on the circle x 2 + y 2 = r 2 13

65 The configuration space of triangles We claim that every point in C represents a triangle (including degenerate cases in which the three points are collinear, and two of them coincide) For instance, consider the cases where r 13 r 12. Then we can place point 1 at the origin, point two at (r 12, 0) Then point 3 can be any point on the circle x 2 + y 2 = r 2 13 Can see all r 23 with r 13 r 12 r 23 r 13 + r 12 are attained (as in inequalities describing C)

66 Dimensions Note that n points in R k span at most an n 1-dimensional space.

67 Dimensions Note that n points in R k span at most an n 1-dimensional space. For instance, three points always lie on a plane, and may be collinear. If they are not collinear, then they form the vertices of a triangle with nonzero area.

68 Dimensions Note that n points in R k span at most an n 1-dimensional space. For instance, three points always lie on a plane, and may be collinear. If they are not collinear, then they form the vertices of a triangle with nonzero area. Four points always lie in a 3-dimensional space and may be coplanar or collinear, etc. If they are not coplanar, then they form the vertices of a tetrahedron with nonzero volume.

69 Cayley-Menger determinants In working with mutual distance coordinates there are certain expressions that come up repeatedly and have significant geometric meaning.

70 Cayley-Menger determinants In working with mutual distance coordinates there are certain expressions that come up repeatedly and have significant geometric meaning. For instance, consider configurations of n = 3 points. Then it can be shown (by you in today s discussion, for instance :) that CM = det 1 0 r12 2 r r r r13 2 r is a constant multiple of the square of the area of the corresponding triangle if (r 12, r 13, r 23 ) C, and is zero if the points are collinear. So det CM 0.

71 Cayley-Menger determinants, continued In general, for a configuration of n points, the CM determinant is the (n + 1) (n + 1) determinant r12 2 r1,n 1 2 r 2 1n 1 r r2,n 1 2 r 2 2n CM = det r1,n 1 2 r2,n rn 1,n 2 1 r1n 2 r2n 2 rn 1,n 2 0 (Note: the matrix is symmetric.)

72 Configuration spaces in general The mutual distance description for configurations of n 4 points is more subtle

73 Configuration spaces in general The mutual distance description for configurations of n 4 points is more subtle For instance with n = 4, we have ( 4 2) = 6 mutual distances

74 Configuration spaces in general The mutual distance description for configurations of n 4 points is more subtle For instance with n = 4, we have ( 4 2) = 6 mutual distances Can ask which (r 12, r 13, r 14, r 23, r 24, r 34 ) R 6 + can be realized as mutual distances for configurations of n = 4 points.

75 Configuration spaces in general The mutual distance description for configurations of n 4 points is more subtle For instance with n = 4, we have ( 4 2) = 6 mutual distances Can ask which (r 12, r 13, r 14, r 23, r 24, r 34 ) R 6 + can be realized as mutual distances for configurations of n = 4 points. Must have r ij r ik + r kj for all distinct triples {i, j, k} {1, 2, 3, 4}, by triangle inequality

76 Configuration spaces in general The mutual distance description for configurations of n 4 points is more subtle For instance with n = 4, we have ( 4 2) = 6 mutual distances Can ask which (r 12, r 13, r 14, r 23, r 24, r 34 ) R 6 + can be realized as mutual distances for configurations of n = 4 points. Must have r ij r ik + r kj for all distinct triples {i, j, k} {1, 2, 3, 4}, by triangle inequality Must also have det CM 0 (this is automatic in n = 3 case, but not here)

77 Configuration spaces, continued Example: Let (r 12, r 13, r 14, r 23, r 24, r 34 ) = (1, t, 1, 1, 2t ), 1

78 Configuration spaces, continued Example: Let (r 12, r 13, r 14, r 23, r 24, r 34 ) = (1, t, 1, 1, 2t ), 1 There is an interval of t values for which all the triangle inequalities are satisfied, namely 1 < t < 2, but

79 Configuration spaces, continued Example: Let (r 12, r 13, r 14, r 23, r 24, r 34 ) = (1, t, 1, 1, 2t ), 1 There is an interval of t values for which all the triangle inequalities are satisfied, namely 1 < t < 2, but det CM = 8(t2 2) < 0 unless t = 2, which gives the unit t 2 square.

80 Configuration spaces, continued Example: Let (r 12, r 13, r 14, r 23, r 24, r 34 ) = (1, t, 1, 1, 2t ), 1 There is an interval of t values for which all the triangle inequalities are satisfied, namely 1 < t < 2, but det CM = 8(t2 2) < 0 unless t = 2, which gives the unit t 2 square. In general a collection of distances r ij R + can be realized by a configuration if and only if all the triangle inequalities hold, and det CM 0.

81 The Albouy-Chenciner equations In mutual distance coordinates, the equations for central configurations can be written in the following form (asymmetric Albouy-Chenciner):

82 The Albouy-Chenciner equations In mutual distance coordinates, the equations for central configurations can be written in the following form (asymmetric Albouy-Chenciner): For each pair 1 i, j n: G ij = n k=1 m k S ki (r 2 jk r 2 ik r 2 ij ) = 0 where S ki = { r 3 ki 1 if k i 0 if k = i

83 The Albouy-Chenciner equations In mutual distance coordinates, the equations for central configurations can be written in the following form (asymmetric Albouy-Chenciner): For each pair 1 i, j n: G ij = n k=1 m k S ki (r 2 jk r 2 ik r 2 ij ) = 0 where S ki = { r 3 ki 1 if k i 0 if k = i Note: r lm = r ml for all l, m. S ki = S ik.) One nontrivial equation for each pair 1 i, j n with i j (everything cancels out if i = j).

84 A comment about AC equations The form of the AC equations we are using here comes by setting ω 2 = 1 in the definition

85 A comment about AC equations The form of the AC equations we are using here comes by setting ω 2 = 1 in the definition This has the effect of imposing a particular distance scale so that

86 A comment about AC equations The form of the AC equations we are using here comes by setting ω 2 = 1 in the definition This has the effect of imposing a particular distance scale so that We never get two central configurations that are homothetic.

87 A comment about AC equations The form of the AC equations we are using here comes by setting ω 2 = 1 in the definition This has the effect of imposing a particular distance scale so that We never get two central configurations that are homothetic. In particular, for instance if three masses m 1, m 2, m 3 at vertices of a triangle form a central configuration, then there are no similar triangles that will also give solutions of the AC equations.

88 A comment about AC equations The form of the AC equations we are using here comes by setting ω 2 = 1 in the definition This has the effect of imposing a particular distance scale so that We never get two central configurations that are homothetic. In particular, for instance if three masses m 1, m 2, m 3 at vertices of a triangle form a central configuration, then there are no similar triangles that will also give solutions of the AC equations. OK for our purposes!

89 What the AC equations look like Let s see what the AC equations look like for 4 masses

90 What the AC equations look like Let s see what the AC equations look like for 4 masses The AC equation for i, j = 1, 2 is 2m 2 S 12 r m 3S 13 (r 2 23 r 2 13 r 2 12 )+m 4S 14 (r 2 24 r 2 14 r 2 12 ) = 0 where S jk = 1 r 3 jk 1 as before

91 What the AC equations look like Let s see what the AC equations look like for 4 masses The AC equation for i, j = 1, 2 is 2m 2 S 12 r m 3S 13 (r 2 23 r 2 13 r 2 12 )+m 4S 14 (r 2 24 r 2 14 r 2 12 ) = 0 where S jk = 1 1 as before rjk 3 Note: can clear denominators to get a polynomial

92 An example Say we want to study collinear cc s of n = 4 bodies. Use original form of cc equations (not AC).

93 An example Say we want to study collinear cc s of n = 4 bodies. Use original form of cc equations (not AC). m 1 m 2 m 3 m 4 a b c

94 An example Say we want to study collinear cc s of n = 4 bodies. Use original form of cc equations (not AC). m 1 m 2 m 3 m 4 a b c Fix the first mass at q 1 = (0, 0), then place the others at q 2 = (a, 0), q 3 = (a + b, 0), q 4 = (a + b + c, 0) (with a, b, c > 0).

95 An example Say we want to study collinear cc s of n = 4 bodies. Use original form of cc equations (not AC). m 1 m 2 m 3 m 4 a b c Fix the first mass at q 1 = (0, 0), then place the others at q 2 = (a, 0), q 3 = (a + b, 0), q 4 = (a + b + c, 0) (with a, b, c > 0). This gives r 12 = a, r 13 = a + b, r 14 = a + b + c, r 23 = b, r 24 = b + c, and r 34 = c.

96 An example Say we want to study collinear cc s of n = 4 bodies. Use original form of cc equations (not AC). m 1 m 2 m 3 m 4 a b c Fix the first mass at q 1 = (0, 0), then place the others at q 2 = (a, 0), q 3 = (a + b, 0), q 4 = (a + b + c, 0) (with a, b, c > 0). This gives r 12 = a, r 13 = a + b, r 14 = a + b + c, r 23 = b, r 24 = b + c, and r 34 = c. The center of mass is at q = i m iq i i m i

97 Example, continued For instance, if the masses are all equal (so we can set m i = 1, all i), and we make ω 2 = 1 (this fixes the distance scale) the central configuration equations become:

98 Example, continued For instance, if the masses are all equal (so we can set m i = 1, all i), and we make ω 2 = 1 (this fixes the distance scale) the central configuration equations become: 0 = a 2 + (a + b) 2 + (a + b + c) 2 3/4 a 1/2 b 1/4 c 0 = a 2 + b 2 + (b + c) 2 +1/4 a 1/2 b 1/4 c 0 = (a + b) 2 b 2 + c 2 +1/4 a + 1/2 b 1/4 c 0 = (a + b + c) 2 (b + c) 2 c 2 +1/4 a + 1/2 b + 3/4 c

99 Example, continued Clear denominators to get polynomial equations

100 Example, continued Clear denominators to get polynomial equations When we do that, get factors of a + b, b + c, a + b + c in some polynomials; since we want a, b, c > 0, those factors cannot be zero either

101 Example, continued Clear denominators to get polynomial equations When we do that, get factors of a + b, b + c, a + b + c in some polynomials; since we want a, b, c > 0, those factors cannot be zero either ( Trick ) to find only solutions with a, b, c, a + b, b + c, a + b + c 0 we can add a variable t and a new equation 1 tabc(a + b)(b + c)(a + b + c) = 0

102 Example, continued Clear denominators to get polynomial equations When we do that, get factors of a + b, b + c, a + b + c in some polynomials; since we want a, b, c > 0, those factors cannot be zero either ( Trick ) to find only solutions with a, b, c, a + b, b + c, a + b + c 0 we can add a variable t and a new equation 1 tabc(a + b)(b + c)(a + b + c) = 0 Compute a Gröbner basis for the ideal generated by these with respect to a monomial order set up to eliminate t, break ties with grevlex. [Demo with Sage]

103 How many solutions? Note that the leading terms of the Gröbner basis polynomials contain powers of each of the variables: t, a 5, b 8, c 10

104 How many solutions? Note that the leading terms of the Gröbner basis polynomials contain powers of each of the variables: t, a 5, b 8, c 10 This implies that the set of all complex solutions of the system is finite (hence the same is true for the real solutions!)

105 How many solutions? Note that the leading terms of the Gröbner basis polynomials contain powers of each of the variables: t, a 5, b 8, c 10 This implies that the set of all complex solutions of the system is finite (hence the same is true for the real solutions!) Theorem 1 (Th. 6 in Chapter 5, 3 of IVA) Let I be an ideal in C[x 1,..., x n ] and > a monomial order. Then TFAE: i. V (I) is a finite set ii. Let G be a GB of I wrt >. For each i, there is g i G with LT (g i ) = x m i i for some m i

106 Sketch of proof of this theorem Proof. ii i: Think about the pictures we drew in the proof of Dickson s Lemma. If ii is true, there are only finitely many monomials that are not in LT (I). Hence for each i, the remainders of the powers xi n, n 0 on division by G must be linearly dependent. This implies that there is some h i (x i ) I for each i. It follows that V (I) is finite.

107 Sketch of proof of this theorem Proof. ii i: Think about the pictures we drew in the proof of Dickson s Lemma. If ii is true, there are only finitely many monomials that are not in LT (I). Hence for each i, the remainders of the powers xi n, n 0 on division by G must be linearly dependent. This implies that there is some h i (x i ) I for each i. It follows that V (I) is finite. i ii: If V is a finite set, then for each i there is a univariate polynomial h i (x i ) I(V (I)). By a big theorem from Chapter 4 of IVA (the Nullstellensatz ), h i I so some power hi l I. The leading term of hi l is a power x lk i i. By the definition of a GB, there must be some g i G whose leading term divides x lk i i, so LT (g i ) is a power of x i.

108 Back to collinear 4-body cc s In our case, the ideal contains a univariate polynomial of degree 105 in a.

109 Back to collinear 4-body cc s In our case, the ideal contains a univariate polynomial of degree 105 in a. This factors; the smaller factor has this form (all terms have exponent divisible by 3!) 16a a a a a

110 Back to collinear 4-body cc s In our case, the ideal contains a univariate polynomial of degree 105 in a. This factors; the smaller factor has this form (all terms have exponent divisible by 3!) 16a a a a a The method of Sturm sequences (see this week s lab) can be used to show this has exactly one real root, which is positive.

111 Back to collinear 4-body cc s In our case, the ideal contains a univariate polynomial of degree 105 in a. This factors; the smaller factor has this form (all terms have exponent divisible by 3!) 16a a a a a The method of Sturm sequences (see this week s lab) can be used to show this has exactly one real root, which is positive. Note that it must have at least one real root since it has odd degree.

112 Back to collinear 4-body cc s In our case, the ideal contains a univariate polynomial of degree 105 in a. This factors; the smaller factor has this form (all terms have exponent divisible by 3!) 16a a a a a The method of Sturm sequences (see this week s lab) can be used to show this has exactly one real root, which is positive. Note that it must have at least one real root since it has odd degree. The other factor has even degree and no real roots.

113 Collinear 4-body cc s, continued The ideal also contains univariate polynomials in b, c

114 Collinear 4-body cc s, continued The ideal also contains univariate polynomials in b, c These are very similar to the one for a. (In fact, the polynomial for c is exactly the same as the polynomial for a; can you see why this might be true?)

115 Collinear 4-body cc s, continued The ideal also contains univariate polynomials in b, c These are very similar to the one for a. (In fact, the polynomial for c is exactly the same as the polynomial for a; can you see why this might be true?) It follows with a bit of computation that there is exactly one real solution of the cc equations in this case,

116 Collinear 4-body cc s, continued The ideal also contains univariate polynomials in b, c These are very similar to the one for a. (In fact, the polynomial for c is exactly the same as the polynomial for a; can you see why this might be true?) It follows with a bit of computation that there is exactly one real solution of the cc equations in this case, In fact it has been proved (by other methods!) that for all choices of m i > 0, there is exactly one collinear cc of the form we are studying (Moulton s theorem).