Re: Special Relativity in a Finite Universe - apparent paradox

Dear sal:

sal wrote:
On Thu, 11 Jan 2007 07:21:18 -0800, dlzc wrote:

Deat sal:

sal wrote:
On Thu, 11 Jan 2007 03:03:41 -0800, Eric Gisse wrote:
...
Whoa. That is not necessarily true.

Take a hypercube, equivalence opposite sides, and you get a space
which "wraps back" on itself but the "glue job" does not affect the
metric. Hence Riemann will still be zero everywhere and it won't be
curved.

In a multiply connected universe something else breaks down, but it's
not due to the curvature.

A simpler example which is really easy to picture is a cylinder.
It's a 2-d manifold, and it is "flat" in the sense that its metric is
the constant diag(1,1) (or (diag(-1,1) if you want to make it into a
spacetime with 1 spacial dimension).

The surface of a sphere has "intrinsic curvature"; the surface of a
cylinder does not.

I flat out did not believe this.

I'm too lazy to seriously study GR seriously and to the level I want
without help, but I am not lazy enough that I won't spend time working
on little things that interest me. Like how to use the math of
differential geometry to compute gradient/curl/laplacian/divergence
in a particular coordinate system.

While playing with that, it struck me - is it really true that a
cylindrical manifold has no intrinsic curvature? I checked.

There is only one nonzero connection coefficient: r,r,r [it doesn't
matter what index is which] and it is equal to r^3. Toss that through
the Riemann curvature tensor and it turns out it is equal to zero!

That _surprised_ me. A lot. I actually expected there to be some kind
of curvature, though I was not sure exactly what it would be.

Here's the quick and dirty intuitive deal, for deciding if a 2-d
manifold is curved:

Can you take an ordinary flat *** of paper and bend it into the shape
of the manifold, _without_ tearing or stretching it?

No, you cannot.

I was talking about _real_ paper.

So was I, and I was not trying to be a smartass.

Can't you roll a *** of paper into a cylinder?
I can.
So can the people who ship posters in mailing tubes.
Can you take a flat *** of paper and cover a sphere
with it, without crumpling or tearing it? I can't.
Neither can the manufacturers of globes; they need to
cut the paper into strips and warp each one individually.
Take a look at a (cardboard) globe some time; you'll see
what I mean. (The newer ones are all plastic and are
molded in one piece, of course.)

Perhaps you misunderstood what I meant by "stretching".
I meant, without stretching it in a gross way, by, say,
steaming the paper and then pulling slowly while it's hot
and wet, or by cheating and using a balloon or ***
of saran wrap instead of paper. In other words, no first-
order stretching allowed.

This is supposed to provide an intuitive, hands-on
approach to seeing the difference between "intrinsic
curvature" and "extrinsic curvature". If you want to fight
it and object on the grounds that paper isn't really
a 2-d manifold that's your prerogative but it kind of misses
the point.

I just wanted to point out that there is some stretching involved in
deforming *real* paper to act as you can do without this "thickness
effect" to a mathematical ***. You invoked an "ordinary flat ***
of paper". There is a bit of approximation here... namely "deformable
solids".

All real papers are non-zero thickness, and any such
manipulation will compress the inner fibers and stretch
the outer. If you will note, the paper will retain some of
the curled shape after you let go. In the absence of
gravity or static attraction, even the thinnest rice paper
and the largest possible *** will retain curvature.
Neglecting these "details"...

You can for a cylinder, so it's not curved.

For the OP's gedanken of two parallel trains

This has little to do with the OP, and nothing to do with
his trains. It has to do with a question which was raised
as to whether a multiply connected space can have zero
intrinsic curvature. It can.

Zero intrinsic curvature implies that travelling around it yields net
zero acceleration, no?

oriented with their axis of symmetry parallel to the
self-same axis of the "cylindrical space", where does
one not see acceleration of either of the trains?
Anywhere short of z = oo *on the ****, and one
point on either train will move towards you and then
away from you on a full rotation.

For a sphere, you can't. So, it's curved.

Maybe I'm not seeing something about the analogies
of "cylindrical space".

I'd agree with that.

I think I missed the observer standing "next to" the train. At no time
does any car get non-uniformly further away or closer to this observer
over a full cycle. With all distances measured on the ***, and "next
to" meaning actually having the train coincident with him.

So that leaves two or three classes of observers that see zero
acceleration (z = 0, +/-oo). Everyone else gets non-zero acceleration.
Can you explain that so a mechanical engineer can understand, or is it
just "tough luck" for me?

David A. Smith

.

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