# Arnold´s Math Methods of Classical Mechs - a question on Newtonian Mechanics

There is the following question/answer on Arnold´s book Mathematical Methods of Classical page 10.

Arnold's Question A mechanical system consists of two points. At the initial moment their velocities (in some inertial coordinate system) are equal to zero. Show that the points will stay on the line which connected them in the initial moment.

Proof A solution to this problem would be: Any rotation of the system around the line connecting the initial configuration is a Galilean transformation and so sends a solution to the differential equation of motion to another solution. Since these rotations fix the initial conditions, by the uniqueness of solutions of differential equations, it is easy to see that the motion of the points must be constrained to the afore mentioned line. qed.

My qualms with this solution is that it assumes the equation of motion is nice enough and hence has unique solutions. My question is then:

Question Is there a force field with a solution to the equation of motion that contradicts the above exercise question of Arnold? This force field will have to be "pathological" enough so that the solutions to the equations of motion are not unique - is there a known physical configuration (i.e. an existing real physical configuration) with this property?

I came up with the given solution after trying, and not suceeding, to solve this problem via the creation of some conserved quantity (inner product) that would restrict the motion to the given line. Perhaps such an argument would rule out the above pathologies, and hence would be a stronger/better argument.

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You can give a proof using only conservation of center of mass and angular momentum. The conservation of center of mass gives a one-degree of freedom motion in terms of the relative separation, which can be taken along the z axis. Then the vanishing of the x and y angular momentum tells you that the transverse velocity must be zero at all times before collision and after, which establishes the result. The motion is collinear away from collisions, just by conservation laws.

But there is a simple counterexample to the theorem consisting of exact point collisions--- you can have the particles nonuniquely veer off the line of separation at the instant they are on top of each other. To see that this is allowed, consider the limiting motion in a central force law for infinitesimal noncollinearities, where the central force is repulsive at infinitesimally short distances and attracting at large ones. you can get an undetermined sharp turn at collision, which becomes a kink in the limit. The kink angle is determined by the exact infinitesimal displacements away from the z axis, and the exact infinitesimal repulsive force structure, which you can specify arbitrarily. Then you can take the point limit, keeping the kink-angle fixed, and you get collisions which go at a different angle.

There is a physical example of this phenomenon in monopole soliton collisions in field theory, which are Hamiltonian motions of point particles for slow velocities, but which go off at 90 degrees after a collinear collision. This scattering is described by the geometry of Atiyah Hitchin space. This shows that there are cases where the kink-angle is not arbitrary in the pointlike limit, but is a well defined property of the theory.

### A nonuniqueness example

There are no examples for a rotationally invariant potential, no matter how horrible, because the force law is then a function of $x^2 + y^2$, and you can prove conservation of angular momentum for solutions in the weakest sense you can think of, because it follows algebraically from the equations of motion.

But if you allow a 2-d potential which is of the seperable form $V(x) + V(y)$, you can also prove the collinear theorem for starting using reflection symmetry in y. But for $V(y)=-\sqrt{y}$, there are two solutions, $y=0$ and $y= (t +C)^{4/3}$, where the linear constant multiplying t has been made 1 by a judicious choice of mass.

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thank you for your input! – M. Otts Oct 30 '11 at 0:49