Re: shell theorem

On Mon, 17 Nov 2008 09:26:31 -0800, Uncle Al <UncleAl0@xxxxxxxxxxxxx>
wrote:

eric gisse wrote:

On Thu, 13 Nov 2008 12:39:12 -0800, Uncle Al <UncleAl0@xxxxxxxxxxxxx>
wrote:
[...]


Agreed. If you'd like to play, e-mail me the physics and the flow
chart. We'll code it, run it, and look at the outputs. If you'd like
the point coordinates I can e-mail you. G = 1, all masses = 1.
Cartesian coordinates.

Sketch of how I would do it, tell me how this compares.

Since the atoms are free [no lattice holding them together], their
relations to one another are entirely defined through their
gravitational effects.

That is the Gedankenexperiment. Do left-left and left-right systems
of otherwise identical, energy-minimized mass distributions diverge?

What's your definition of "diverge" ?

An arbitrary arrangement of objects will gravitationally arrange
themselves differently than a slightly different arrangement of the
same number of objects. Initial conditions matter greatly - think
Cauchy problem.


Define a chiral vacuum with a big hollow CHI=1 fullerene bubble. The
biggest we have is C980. Probe it with the smallest rigd CHI=1 lump.
That would be [5.5]chiralane less its hydrogens. Given Kuratowkski's
theorem, no Schlegel diagram is possible! This is cute (and
irrelevant to the physics) if you are a chemist.

All I get out of that is that you can't project the solids down upon
another dimension without losing information. Not sure how that
relates to chirality.


Assuming equal massed constituents.

G=1, m=1. Only the difference matters.

The acceleration on the i'th atom from the other N - 1 atoms is after
summing from j = 1...N-1 while excluding i = j :

m a_i = -m G / | r_i - r_j | ^2

The force will be directed along the line of sight between the two
atoms.

That is a handshake. Given N atoms total for the system a whole,
N(N-1)/2 unique handshakes.

N^2 choices overall, N(N-1) since shaking your own hand is not
allowed, and N(N-1) / 2 since a shake b is the same as b shake a.


The only hinkey part is picking the coordinates, since there is no
especially obvious choice. Pick whatever is most convinient - I'd
suggest the common center of mass for the group since it will not
move.

Well now... two cases! Centers of mass gravitation (easy). Each
handshake projected upon the centers of mass line and summed for net
magnitude and dirction. Easy but laborious. Can be done with
direction cosines or vector dot products (end up as doing the same
thing via algebra or matrices).

I have heard nothing but horror stories about numerical linear
algebra. But at least that way you can frame the problem in a way that
amends itself to finding pathologies.

In my example, the group's center of mass does not move. So the
reference point defined by the center of mass is stationary - no
center of mass dynamics.

In your example, however, there is an attraction between the two
centers of mass. The equations of motion are altered greatly by using
center of mass coordinates, and I can't say offhand if it would
simplify things for you.

But if you are considering the two centers of mass, you have to take
into account the fact that the centers of mass will be attracted to
eachother.


Everybody knows the two answers must be identical. True for billiard
balls and their constituent atoms. Curiously NOT true on a small
scale. A homogenteity approximation fails. Summation in the small is
not integration.

Everyone knows it because it has the virtue of being true.

Rederiving Newtonian gravitation via Cartan by giving it the
Riemannian geometry treatment explicitly formulates something already
known to be true: Newtonian gravitation obeys Galilean invariance and
[well, rephrasing of the same thing] preserves the Euclidean metric.

Specifically, 3-distances are invariant.

Arbitrary CONSTANT coordinate displacements have no effect on the
acceleration term in Newton, as the time derivatives go away.
Furthermore, they have no effect on distance as | r_i - r_j |^2 is an
invariant. The dynamics don't care about different reference points.

No approximations. No integrations. This is exact - on all scales.

Getting a different answer is diagnostic of a bug in the code. No way
of knowing which implementation is wrong.


Exchange a_i for whatever computational implementation of the second
derivative you like. Don't use Euler, it diverges too easily. 4'th
order Runge-Kutta is one of the better choices.

Crunch it explicitly, both ways, by any way,

http://www.mazepath.com/uncleal/tettet2.png
stereogram. Note the mirror symmetries of the system.

G=1, m=1,

4
1.146388 0.4067684 0.3201515
0.04216192 1.348825 0.1133035
1.043776 1.151007 -0.9389371
1.407833 1.849292 0.2978272
4
4.817326 0.4826342 0.06582354
4.714714 1.226873 -1.193265
4.453269 -0.2156502 -1.170941
5.81894 0.2848162 -0.9864171

What am I looking at?


Calculate the centers of mass gravitation between the two masses,
(4)(4)/r^2. Calculate all (8)(8-1)/2 = 28 unique handshakes and sum
their forces parallel to the centers of mass line, respecting
direction.

You can write it all out by hand, but you can't solve it by hand. All
N-body problems [N >= 3] are kinda...proven analyitically unsolvable.


March through each of the a_i and obtain the positions of the i'th
atom. Store. Move to the next time step. Lather, rinse, repeat.

Your flow chart will be quite circular for many time steps.

Loop within a loop. Tried it three ways. Answers are consistent to
five decimal places - and different for averaged vs. discrete cases by
under a percent.

Oh, so the exact cases match while diverging with the average cases?
Kewl - good code then. No surprises on the divergernce front.


Kewl. Conceptualization conflicts are resolved. We'll do it the
exact way.

So is the next step a large amount of atoms?
.

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