From: buhr@umanitoba.ca (Kevin Andrew Buhr)
Subject: Re: Gravity & Rubber Sheet Analogy Problem
Organization: University of Manitoba, Canada
	<1iv7cvINN7l1@moe.ksu.ksu.edu>
Distribution: usa
Date: Wed, 13 Jan 1993 04:07:17 GMT

In article <1iv7cvINN7l1@moe.ksu.ksu.edu> diverson@phys.ksu.edu (David
Iverson) writes:
|
|  O.K. I'll put a new twist on this, not an answer, more like a continuation
|  of the question.
|
|  Take two balls (you choose the size) and sandwich them between two rubber
|  sheets.  Do this in the SpaceShuttle to avoid the nasty business of an
|  external gravity.  I imagine that the balls would be pushed towards each
|  other as the elastic sheets attempt to minimize their energy.  Encompassing
|  both balls in one deformation would take less energy than placing each one
|  in it's own.

I don't think you need two rubber sheets.  If you place two large
masses reasonably close to one another on a single rubber sheet, they
*WILL* fall towards each other.  No special extra force is needed.

Consider the following cross-section of two masses on the rubber
sheet:

	        
     b	        		B                            b
   -----...    	 MMMM      __........__      MMMM      ...------
	    \    1111    /      	\    2222    /
	 A--> \  MMMM  / <--a  	     a--> \  MMMM  / <--A
	        ~~~~~~			    ~~~~~~

Let's assume that the rubber sheet is very taut equally at all points
and in all directions, at least without any masses sitting on it; that
it deforms "efficiently" without energy loss; and that it is
frictionless as masses move across it.

Then the rubber is deformed more between the two masses than in the
opposite direction.  Why?  Well, rougly speaking, the deformation
caused by the two masses is a decreasing function of the distance from
the mass center.  The deformation at any point (including the point
between the two masses) is the "sum" of all deformations caused by all
masses.  The point marked "B" is relatively close to both masses,
whereas the points marked "b" are close to one mass and much further
away from the other.  Hence, the deformation is greater at "B" than at
"b" as these deformations are added.

Arbitrarly close to either one of the masses, we get the same trend in
the deformation: the rubber closer to the "other" mass is deformed
more than the rubber on the opposite site.  Let me repeat the diagram
again:

     b	        		B                            b
   -----...    	 MMMM      __........__      MMMM      ...------
	    \    1111    /      	\    2222    /
	 A--> \  MMMM  / <--a  	     a--> \  MMMM  / <--A
	        ~~~~~~			    ~~~~~~

Suppose the masses are at rest.  The deformation immediately to the
left of mass 1 is relatively slightly smaller compared to the
deformation immediately to its right.  As a result, the rubber
produces a slightly stronger force "A" to the left of mass 1 than the
force "a" applied to its right.  Why?  Well, this is *real* force of
gravity at work (i.e.  the pull on the masses from the large object,
probably the Earth, situated below the rubber sheet).  Essentially,
the mass moves to the right because, as it does, the rubber beneath
the first mass is deformed--by the resulting increased proximity to
the other mass--to a larger degree than it is in the initial state,
trading gravitational potential energy for kinetic energy.  For the
same reason, the mass 2 is pushed toward mass 1.

If you put two masses on a rubber sheet and they *don't* move
together, it's because your rubber sheet is too frictionful or (more
likely) because your rubber sheet deforms inefficiently: it requires a
constant supply of energy to deform in a given direction.

Well...  That's my opinion on the matter.  And, since I don't have the
rubber sheet handy (and my waterbad has an incredibly inefficient
deformation) it is far too Aristotelian for this group...

Kevin <buhr@ccu.UManitoba.CA>
Newsgroups: sci.physics