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Re: Partitioning 4 space with ultraskew lines, and the three body
problem.
Posted:
Nov 2, 2004 10:31 PM
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Richard L. Peterson wrote:
> Suppose we are in 4 space. Two lines are "skew" if they do not lie in
> the same plane(2 space). Skewness is a binary relation on lines. Now
> two skew lines will lie in the same 3 space. But it is possible for
> three lines to not lie in the same 3 space. This is a ternary relation
> on lines. I don't know if there's a name for this relation, so let's
> call it "ultraskewness" until we find out.
>
> 1) Can one foliate, or at least partition, 4 space with lines that
> are tripletwise ultraskew? I mean EACH 3 line subset of the partition
> must not lie in the same 3 space.
>
> 2) In the three-body problem, one could approximate the trajectory
> of each of the 3 masses as a straight lines until they became close
> enough to each other for their gravitation to have an appreciable
> effect. I think much of the work on this problem assumes all three
> trajectories lie in the same plane.(The restricted 3-body problem.)
> But some work has been where the trajectories are not coplanar.
> Wouldn't it be fun to explore the three-body problem when the
> trajectories are not cospatial?
>
> Richard Peterson, CSU Sacramento
>
I don't know about the three-body problem, or about foliations, but
I think one can partition 4-space into "ultraskew" lines without much
trouble. The proof is via a transfinite induction of c (the cardinality
of the continuum) many steps; at step k one considers the k-th point p
in a fixed enumeration of 4-space, and one has already constructed
a collection L of |k|-many (fewer than c) lines. If p is in the union
of L there is nothing to do. Otherwise one need only find a unit
tangent vector u at p so that the line through p in direction u is
(a) disjoint from each line in L and (b) ultraskew to every pair of
lines in L. Since p lies on no line in L (a) is satisfied so long as u
does not lie in any plane containing both p and a line in L. Also,
(b) is satisfied so long as u does not lie in any translate containing
p of an (affine) 3-space generated by a pair of lines in L.
So we need a point on the unit 3-sphere in R^4 not lying in any of
a collection of fewer than c many subspaces of R^4 of dimension
at most 3. Without loss of generality we may assume all the
subspaces have dimension 3, so each has a perp that meets the 3-sphere
in at most 2 points. As we have fewer than c subspaces, there is
a point w on the 3-sphere not in the perp of any of them. Then the
perp P of the line spanned by w is a 3-dimensional space meeting each
of the spaces we want to avoid in a subspace of dimension at most 2.
Thus it suffices to find a point on the intersection of the unit
3-sphere with P (i.e., a unit 2-sphere) not in any of a collection
of fewer than c subspaces of P of dimension at most 2. By a
similar argument we can drop the dimension once again, and then
we need to find a point on the unit circle avoiding fewer than c
lines through the origin. As the circle has c points, and each
line meets it in two point, that is easy.
Well, it's late and I'm hurrying, but I think this argument holds
water.
Bob Beaudoin
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