Thank you. I would not be surprised if Hawking radiation would have a limit cycle. One probably can measure this.

No. Inertial manifolds are relevant for dissipative systems, Newtonian gravitational systems aren't dissipative, they conserve energy: They're invariant under time translations.

Stam Nicolis

In my dynamical system, an inertial manifold has free motion, which is rotation and translational motion, therefore the system is not dissipative but conservative.

Preston Guynn

From the point of view of a local observer, there is only locally Euclidean geometry, and from the point of view of a global observer there is global Euclidean geometry.

Igor Bayak If the system isn't dissipative, the notion of an inertial manifold doesn't make sense. Hilbert's 16th problem isn't of relevance for the Newtonian gravitational potential, either, because it pertains to the number of limit cycles of polynomial vector fields in two dimensions.

Stam Nicolis

By inertial manifolds I mean a set of limit cycles. As for Hilbert's 16th problem, I am just asking if it is possible to match some polynomial vector field to this (gravity-like) vector field, which generates n cycles on the plane.

Inertial manifolds habe to do with dissipative systems, once more. And Newton's equations for the gravitational potential aren't dissipstive. Nor do they define polynomial vector fields in two dimensions, which is the f9cus of Hilbert's 16th problem.

Stam Nicolis

I am referring to the geometric definition of an inertial manifold as a manifold on which limit cycles fall. As for the dissipativity of an inertial manifold, add a vacuum vector field to the dynamical system and it will become conservative. You mention Newton's equations in vain. Still, look at the text on the link.

Okay, let me ask you straight out - is this a polynomial dynamical systems?

Σn ∂ρi ∕ ∂φi = Σn 1 ∕ (ρi-1)

where

∂φi = (y-yi)∂x-(x-xi)∂y

∂ρi = (x-xi)∂x+(y-yi)∂y

ρi = ((x-xi)2+(y-yi)2)½