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A new solution to the three body problem - and more

by Bill Casselman

 

 

3. The triangle construction

In order to explain what Chenciner and Montgomery have (and haven't) proven, I need to include a short digression about the topological nature of the system.

Richard Moeckel was apparently the first to suggestthata relatively simple but effective way to understand what is going on topologicallyis to track the triangle whose vertices are located atthe centers of the three bodies.With any legitimate choreography orbit, the bodiesfollowing the path will avoid collisions, which means thatat any moment all three of the sides of this triangle will have non-zero length.It turns out to be useful to focus attention onthe shapes of the triangles,that is to say, to consider similar triangles as being equivalent.Furthermore,it is useful to keep track of labels for the vertices.There is a very pretty way to parametrize such configurations.

We are interested in parametrizinglabeled configurations of three points ABCin the Euclidean plane, which we identify with the plane ofcomplex numbers,such that at least one of the sides has non-zerolength. Such a configuration will be called here a labeledtriangle.Two such configurationsABC and A*B*C* will be called equivalentif the maptaking A to A* etc. is a similarity(so the labeling is an important part of the structure).Fix one particular labeled triangle abc tobe the equilateral triangle in the complex planewith vertices at the cube roots of unity,lettering counter-clockwise starting with 1.

Now some complex analyis.If ABC is a triangle with all vertices distinct,there exists a unique Möbius transformation

 

taking ABC toabc. Then the map associating to ABC the image dunder T of the point at infinity extends to alllabeled triangles, and establishes a bijection between the space of equivalence classesof non-singular triangles and the points of the Riemann sphere (the complexplane together with the point at infinity).The easiest way to find that complex number d is to solve thisequation, which expresses that the cross-ratio of four pointsis invariant under Möbius transformations.

 

We get

 

The equation shows that apositively oriented equilateral triangle corresponds to the point at infinity; this is because for such a trianglex=1; similarly a triangle with A=B maps to the point c, a triangle with A=C to the point b, a triangle with B=C to the point a. This is the important point for us:the triangles formed by a collisionless choreographywill always lie in the complement of a, b and c. Furthermorepositively oriented triangles map to the outside of the unit circle,where |d| > 1; degenerate triangles, with all vertices on a line,map to the unit circle; and negatively oriented ones to the inside, with a negatively oriented equilateral triangle goingto 0.

If we are given any periodic collisionless system of three bodies, as they move the point corresponding to the triangle they form will trace out a path in the complement, on the Riemannsphere, of a, b, and c. The homotopy class of this path isa strong topological invariant of the system. In a Lagrangian system, the three bodies always forman equilateral triangle, although it may grow, shrink, or rotate.Thus Lagrangiansystems map to 0 or to infinity, the two poles of the Riemann sphere.The Eulerian systems map to points on the unit circle (its equator). One getsa feel for the complexity of a system byconstructing its path traced out on the Riemann sphereby the triangle the bodies bound. Here is what happens for the figure eight:

 

 

The principal result of Chenciner and Montgomery is thatin the homotopy class of the track of the pathexhibited above there does exist the track of a three-body system.


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