On Dec 29, 2007, at 2:08 PM, John Denker wrote:
> . . . I suspect stability and instability are the wrong concepts,
> or at least the wrong terminology. You might be much better
> off asking about chaotic versus non-chaotic. . . .
The fact that a moving system can possibly be unstable (or chaotic)
does not mean that it actually is unstable or chaotic. How to
distinguish a set of initial conditions that produces a stable moving
system from a set of conditions that produces an unstable system? Keep
in mind that I am asking this questtion in the context of a specific
problem. Three identical stars are initially at rest on the diameter of
a circle (2*R) one at the center and others at distances R from it. At
time zero the stars located at the opposite ends of the diameter are
given identical speeds v along clockwise directions. The central star
remains at rest. The speeds v are chosen to satisfy the F=m*v^2/R
condition, where F is the net force on each of the outer stars. It is
easy to show that F=1.25*G*m^2/R^2, where m is the mass of each star.
3) What is wrong with saying that an orbiting system, described above,
is stable when it is energetically bound? Using a common convetion --
potential energy is zero when interacting particles are infinitely far
from each other -- one can say, that stability or instability is
determined by the sign of the total mechanical energy (kinetic and
potential), E=K+U. A state of a system is statble when E is negative
(system is bound), otherwise a stable circular orbit will not be
produced. For the above problem U=-1.5*G*m^2/R, K=1.25**G*m^2/R and
E=-0.25*G*m^2/R. On that basis (negative E) I am saying the created
moving system (two stars on the same circular orbit) will be stable.
The stars will continue to revolve along the same circle forever.
4) Why am I insisting on limiting the discussion to a specific
three-body system? Because the circular orbit system is mathematically
simple. If my proposed definition of stability is rejected for this
simple case then it is not acceptable "in general." But the opposite
is not true. Let us not discuss elliptical orbits, etc. at this time.
5) So what is wrong with saying that the above dynamic system will be
stable? Yes, I am not inventing anything new, the idea of bound energy
states of atoms and molecules was formulated before I was born. And the
concept of escape velocity can be found in many introductory physics
textbooks.
6) Will the described set of three stars be stable (persistent orbiting
along the same trajectory)? I think it will? Have similar stable
systems been observed by astronomers ? I do not know. That would be the
most convincing argument that a similar moving system can be stable.
On Dec 29, 2007, at 2:08 PM, John Denker wrote:
> . . . I suspect stability and instability are the wrong concepts,
> or at least the wrong terminology. You might be much better
> off asking about chaotic versus non-chaotic. . . .
The fact that a moving system can possibly be unstable (or chaotic)
does not mean that it actually is unstable or chaotic. How to
distinguish a set of initial conditions that produces a stable moving
system from a set of conditions that produces an unstable system? Keep
in mind that I am asking this questtion in the context of a specific
problem. Three identical stars are initially at rest on the diameter of
a circle (2*R) one at the center and others at distances R from it. At
time zero the stars located at the opposite ends of the diameter are
given identical speeds v along clockwise directions. The central star
remains at rest. The speeds v are chosen to satisfy the F=m*v^2/R
condition, where F is the net force on each of the outer stars. It is
easy to show that F=1.25*G*m^2/R^2, where m is the mass of each star.
3) What is wrong with saying that an orbiting system, described above,
is stable when it is energetically bound? Using a common convetion --
potential energy is zero when interacting particles are infinitely far
from each other -- one can say, that stability or instability is
determined by the sign of the total mechanical energy (kinetic and
potential), E=K+U. A state of a system is statble when E is negative
(system is bound), otherwise a stable circular orbit will not be
produced. For the above problem U=-1.5*G*m^2/R, K=1.25**G*m^2/R and
E=-0.25*G*m^2/R. On that basis (negative E) I am saying the created
moving system (two stars on the same circular orbit) will be stable.
The stars will continue to revolve along the same circle forever.
4) Why am I insisting on limiting the discussion to a specific
three-body system? Because the circular orbit system is mathematically
simple. If my proposed definition of stability is rejected for this
simple case then it is not acceptable "in general." But the opposite
is not true. Let us not discuss elliptical orbits, etc. at this time.
5) So what is wrong with saying that the above dynamic system will be
stable? Yes, I am not inventing anything new, the idea of bound energy
states of atoms and molecules was formulated before I was born. And the
concept of escape velocity can be found in many introductory physics
textbooks.
6) Will the described set of three stars be stable (persistent orbiting
along the same trajectory)? I think it will? Have similar stable
systems been observed by astronomers ? I do not know. That would be the
most convincing argument that a multi-star system can be stable.
_______________________________________________________
Ludwik Kowalski, a retired physicist
5 Horizon Road, apt. 2702, Fort Lee, NJ, 07024, USA
Also an amateur journalist at http://csam.montclair.edu/~kowalski/cf/