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Ludwik wrote:
On Dec 30, 2007, at 1:25 PM, John Mallinckrodt wrote:
Imprecise language often results in a lot of misunderstanding
and "speaking past each other." . . . By the way, I am willing
to absolutely guarantee that you will not find a system like this
anywhere in this universe. First, it would require a perfect
realization of the initial conditions. Any single deviation, no
matter how minor from those conditions would entirely
obliterate any semblance of orbital perfection. Furthermore,
even if you could realize those perfect initial conditions (and
you couldn't), external perturbations would still entirely
obliterate the orbital perfection.
The issue is not orbital perfection; it is persistence of the average
state of motion; one circle after another. Tiny osillations (random or
periodic) of distances between the stars would not matter to me.
I guess I still haven't said it strongly enough, so let me try again:
Any random perturbation whatsoever--no matter how minor--will lead
inevitably to a fantastically different state of motion that bears
essentially no resemblance at all to the nice circles that you get in
the absence of the perturbation. You will not get "tiny
oscillations" about those circles or anything that could possibly be
mistaken for some kind of average persistence of the unperturbed
motion. This is because the orbits you have constructed are
stupendously unstable. An IP simulation will easily demonstrate this
fact in a minute or two and it will not be a computational artifact.
A little physical intuition should be all it takes to see the following:
After the complete (and, realistically, entirely unavoidable) breakup
of the initial orbit, the only possible definitely persistent final
state would occur if one of the particles were ejected from the
system (which is certainly possible energetically and probably pretty
likely) leaving the other two locked forever in a close and drifting
Keplerian orbit.
Alternatively, you may see a quasi-stable situation develop in which
one of the objects goes into a relatively large orbit about the other
two as they, in turn, orbit each other relatively closely. Call it
an "orbit about the orbits" state.
The mutual perturbations, however, will lead these orbits to evolve
over time and, possibly, break up again leading to either an ejection
or another, different, quasi-stable "orbit about the orbits."
It may be that an ejection is unavoidable in the long term, but my
intuition fails me here.
I do not think that tiny oscillations will obliterate the system
(three
identical stars at the rotating diameter).
I don't understand this description of the system at all, but there
will be no "tiny oscillations."
The system is gravitationally bound.
No it isn't. It has negative total energy, but it is also a three
particle system.
The binding energy is 1.25*G*M^2 / R. That is
equal to work an outside agent would have to do to completely separate
three mutually-attracting particles. Is it OK to think the to
“obliterate” means to “put particles infinitely far away from each
other?”
That's not what I meant, but as I mentioned above it is possible,
perhaps even unavoidable, that in the long run one (and only one) of
the particles will end up infinitely far from the other two.
In the case of three stars of solar mass (m=2e30 kg) orbiting a circle
whose diameter is 200 AU (R=100 AU = 1.5e13 m), the binding energy
would be about 2e37 joules. Where would the energy needed to
obliterate
the system come from?
No additional energy is required.
I suspect that my interpretation of the term
“obliterate” is wrong. We are probably again speaking past each other.
Do you agree, John, that my hypothetical system will not disintegrate
spontaneously into three no-longer-interacting particles?
That we can agree on! This system has negative total energy and,
thus, requires at least two of the particles to be interacting
attractively.
That what I had in mind in using the term "stable."
If you want "stable" to mean simply that there is no way for all
three particles to spontaneously end up infinitely far away from each
other, then the system is indeed "stable," but that seems to me to be
a very impoverished meaning for the word "stable," especially since
this system isn't even "bound."