 Are central forces the only ones giving planar orbit for any initial condition?

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It is textbook knowledge that a massive point subject only to a central force has a planar trajectory for any initial condition — here I consider a force field with a constant direction as a central force field with the centre of attraction at infinity. But what about the converse statement?

Precisely, using the usual convention that bold symbols are vectors, denoting by $\def\vector#1{\mathbf{#1}} \vector{x}$ the position of the point and by $m$ its mass, if all solutions of $m\ddot{\vector{x}} = \mathbf{F}(\vector{x})$ are planar, does it imply that $\vector{F}(\vector{x})=\lambda\frac{\vector{x} - \vector{o}}{\|\vector{x}-\vector{o}\|}?$ Where $\vector{o}$ would be the centre of attraction, a constant point, possibly moved to infinity (in which case, $\vector{F}=\lambda \vector{n}$ for some constant unit vector $\vector{n}$), and $\lambda$ a scalar, only function of the position of the massive point (i.e. not of its speed).

Note: I did ask the same question on physics.stackexchange.

asked May 5, 2017

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Consider a set of Cartesian coordinates $x,y,z$, and a force field which points only in the direction of $z$: $\vec{F} = (0,0,F_z)$. Now consider a particle with a general velocity $\vec{v}$, we can always rotate the coordinate system around the $z$ axis so that the $x$ component of velocity is eliminated and we have $\vec{v} = (0,v_y,v_z)$. Now the equations of motion read

$$\ddot{z} = \frac{F_z}{m},\; \ddot{x} = \ddot{y} = 0$$

That is, $v_y = const.$ and $v_x = const$. Obviously, this particle will then stay in the $y-z$ plane and never move in the $x$ direction. This means that the motion is planar and this is a counterexample to your question.

answered Sep 22, 2017 by (1,635 points)

Nice! Haha sometimes my first instinct is to think too abstractly. :P

This seems to conflict with the conclusion of the paper

http://iopscience.iop.org/article/10.1088/0143-0807/7/3/001/meta

which claims that only central force fields work. I haven't read the paper in detail, but found that the second, detailed proof given there produces at the end the central force field $F=\nabla V=\frac{x-c}{|x-c|}$, while the first, elementary argument (on p.157) seems to allow for a center at infinity, which (after rotating this center at infinity to be along the $z$ axis) seems to correspond to your counterexample, which then would be a limiting case of the central force field. But I haven't been able to make this explicit.

So I wonder what the real truth is....

@ArnoldNeumaier Nice, the paper does mention the force field pointing in one constant direction as a special subcase under equation (10).

I had completely forgotten I had asked the question on this forum too. Yes, I have found Urbankte's paper a long time ago! It is amusing how simple is the geometric demonstration whereas the one based on calculus is quite lengthy.

@LucJBourhis: Strictly speaking, the conclusion of the paper is that there are precisely two classes of models - the central fields and the fields pointing in a constant direction, the latter being a limiting case of the former.

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You can think of a system which always lies in a plane as having a conserved quantity which is to specify that plane. This is implied by conservation of angular momentum (which requires a center of force around which the potential has rotation symmetry) but I believe conservation of angular momentum is too strong a requirement, since any torque in the direction of the angular momentum will preserve the plane of motion. To determine the symmetry implied by this conservation law, one needs to write the conserved quantity as a function on the phase space of the particle and consider the canonical transformation generated by it (a la Noether's theorem).

answered Sep 22, 2017 by (1,895 points)

Indeed, only central force fields (and the fields pointing in a constant direction, a limiting case) work:

http://iopscience.iop.org/article/10.1088/0143-0807/7/3/001/meta

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