Register to reply

KAM curves

by kushal
Tags: curves
Share this thread:
Jul21-08, 05:00 AM
P: n/a

I have a question regarding KAM curves.

KAM theory says that if we add a very small non-integrable Hamiltonian
perturbation to an integrable Hamiltonian system, then most of the
invariant tori are preserved if the perturbation is very small and the
measure of the destroyed invariant tori scales with the magnitude of
the perturbation.

Now, are the KAM curves for perturbed Hamiltonian systems "exact
invariants" or "invariants upto all orders". Since the perturbation is
non-integrable, I would believe that the KAM curves would only be
"invariants upto all orders" and not really be "exact invariants". Is
this right?

If the above is right, then, in addition to Arnold Diffusion, should
not there be an exponentially slow diffusion across the KAM curves
also since they are not exact invariants? And if the above statement
is wrong, then how does a non-integrable perturbation happen to
conserve the integrability of the Hamiltonian in certain regions in


Phys.Org News Partner Physics news on
Physicists unlock nature of high-temperature superconductivity
Serial time-encoded amplified microscopy for ultrafast imaging based on multi-wavelength laser
Measuring the smallest magnets: Physicists measured magnetic interactions between single electrons

Register to reply

Related Discussions
Polynomial curve problem Calculus & Beyond Homework 5
Sketch the curves y =|x| and y = 2 - x^2 on the graphs Precalculus Mathematics Homework 5
Gradient of the curve help Precalculus Mathematics Homework 5
free-body diagram to identify the forces acting on the car Introductory Physics Homework 7
Vector function for the curve of intersection of the paraboloid Introductory Physics Homework 2