Simple question about SR paradox

Tom Roberts says...

Ann O'Nymous wrote:
I once read somewhere that if the Big Bang is true, it is possible for
light to travel completely across the expanding universe and appear from
the other side.

This is not true in the FRW manifolds that are the basis of big bang cosmology.
There are three classes of FRW manifolds, classified by their spatial curvature
at constant cosmological time. Those with flat or hyperbolic spatial curvature
have infinite spatial extent at all times and no "crossing" is possible. Those
with 3-sphere spatial surfaces have a closed space, but they all end in a "big
crunch" that is the time reversal of the big bang; a light ray can barely
circumnavigate space between the big bang and the big crunch.

I thought I remembered seeing once that there is a torus-like solution
that has a finite but expanding universe. In this universe, you actually
can circumnavigate the universe. I don't remember any details, though.

--
Daryl McCullough
Ithaca, NY

Steve Willner wrote:
I don't think SR makes any assumption about global topology, as you
say next.

Yes, it does. SR implicitly assumes the topology of the manifold is R^4 -- this
is inherent in the Newtonian meaning of "inertial frame" that SR inherited. SR
also assumes the manifold is flat (i.e. the Riemann curvature tensor is
identically zero everywhere and everywhen) -- that is required for it to be
possible to extend inertial coordinates throughout the manifold.

GR does not inherit inertial frames; in general it has none, it
has only locally-inertial frames that are only approximately
inertial over a limited region of the manifold.


Daryl wrote:
However,
there is no problem with generalizing SR to other topologies. You can
imagine a "cylindrical" universe in which traveling far enough in one
direction returns you back to where you started.

I don't see why it would be cylindrical and not (hyper-)spherical.

Because only a cylindrical space can be flat; the spherical and hyperspherical
topologies, S^2 and S^3, are inconsistent with being flat.


In such a universe, there is a preferred frame,

If there's a preferred frame, SR is invalid.

Yes, in the usual SR with topology R^4. But he has "bent" the theory by changing
the topology, and in doing so introduced a preferred frame. This is not SR, but
it is rather close.


But I don't see why
there should be a preferred frame, any more than the CMB frame is
"preferred" in our Universe.

In the version of SR with topology Rx(R^2xS) there is a TOPOLOGICALLY PREFERRED
class of inertial frames -- ones that do not go around the circle (topology S).


I think the answer is much simpler: each twin sees the other
redshifted as they separate and blueshifted as they return to each
other. The redshifts and blueshifts cancel, so absent further
complications, they will be the same age.

Actual computations in such a modified SR show otherwise. Daryl is correct in
that the twin that "goes around" returns younger to her sibling.


Tom Roberts

Steve Willner says...

(Daryl McCullough) writes:

However,
there is no problem with generalizing SR to other topologies. You can
imagine a "cylindrical" universe in which traveling far enough in one
direction returns you back to where you started.

I don't see why it would be cylindrical and not (hyper-)spherical.

Because a cylinder (or a torus) is flat, while a sphere (or hypersphere)
is not.

In such a universe, there is a preferred frame,

If there's a preferred frame, SR is invalid.

That's why I called it a "generalization" of SR. In such
a cylindrical spacetime, there is a preferred frame, but
it's not *locally* observable.

It's the same sort of situation as an ant crawling on a huge piece
of paper that is wrapped around into a tube. There is no experiment
the ant can do locally that can tell the difference between life
on a tube and life on a flat piece of paper. But on a flat piece
of paper, all directions are equivalent, while on a tube, there
is a big difference between traveling parallel to the axis of the
tube and traveling *around* the tube. The first kind of travel
will never get you back to where you started, while the second
kind will.

But I don't see why there should be a preferred frame, any more
than the CMB frame is "preferred" in our Universe.

I'm just saying that if you try to do a "twin paradox" in a
cylindrical spacetime, you can tell which twin traveled "around"
the universe and which one stayed put, because the one that
traveled will be younger when they get back together.

I think it's a different situation from the CMB, in that it
is a property of space that singles out one rest frame, rather
than a property of matter/radiation.

I think the answer is much simpler: each twin sees the other
redshifted as they separate and blueshifted as they return to each
other. The redshifts and blueshifts cancel, so absent further
complications, they will be the same age.

That's a nice simple way to reason about it, but it gives
you the wrong answer.

--
Daryl McCullough
Ithaca, NY

On May 25, 3:11 pm, wrote:
(Daryl McCullough) writes:

Well, you're right, that SR usually assumes a "simply connected" universe
in which it is impossible for two travelers to separate and get back
together again without one or the other (or both) accelerating.

I don't think SR makes any assumption about global topology, as you
say next.

There is no mysticism in this acceleration. If claimed so, show Yours
Truly the math. shrug

However,
there is no problem with generalizing SR to other topologies. You can
imagine a "cylindrical" universe in which traveling far enough in one
direction returns you back to where you started.

I don't see why it would be cylindrical and not (hyper-)spherical.

All space is observed to be flat by anyone on his own. Curved space
only applies to observing someone else’s space. Curvature of space is
all relative. It is indeed very lonely to be the only person to have
understood this curved space business after Riemann. shrug

In such a universe, there is a preferred frame,

If there's a preferred frame, SR is invalid.

Yes. applause

But I don't see why
there should be a preferred frame, any more than the CMB frame is
"preferred" in our Universe.

All solutions (transforms) that satisfy the null results of the MMX
must be reference back to the absolute frame of reference. Thus, the
MMX actually proved the existence of the Aether unlike the myth spun
after Poincare. shrug

This so-called preferred frame is found through the Doppler shift in
CMBR as you have pointed out. That would be the second time the self-
styled physicists have turned their backs on the Aether. Well, Peter
did not recognized Christ three times. shrug

I think the answer is much simpler: each twin sees the other
redshifted as they separate and blueshifted as they return to each
other. The redshifts and blueshifts cancel, so absent further
complications, they will be the same age.

Totally wrong. Doppler effect is not time dilation. Effect from
Doppler shift is not accumulative in anomaly, but time dilation is
accumulative. Please try to understand the Lorentz transform first.
shrug

http://groups.google.com/group/sci.p...1209448d?hl=en

Of course the twins are and remain in relative motion as they pass
each other and compare clocks both times. If either accelerates, we
have to take that into account, and then their clocks may not match.

Under the Lorentz transform, each twin will observe the other one
slowing in aging regardless who travels or not. The myth about the
turn-around is totally bull****. Bull**** is not truth. If claimed
so, show Him (Yours Truly) the math. shrug

The study of physics seem to follow the Orwellian school of thought
where

** FAITH IS THEORY
** LYING IS TEACHING
** NITWIT IS GENIUS
** OCCULT IS SCIENCE
** PARADOX IS KOSHER
** FUDGING IS DERIVATION
** BULL**** IS TRUTH
** BELIEVING IS LEARNING
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** CONJECTURE IS REALITY
** PLAGIARISM IS CREATIVITY
** MATHEMAGICS IS MATHEMATICS

And it is time for the Relativity play.

http://groups.google.com/group/sci.m...a0f3c305008773
* * * * *

In article ,
"Androcles" writes:
"If there's a preferred frame, SR is invalid." -- Willner.

Admittedly I'm breaking my own rule here, but it's worth noting that
the Universe we live in doesn't seem to have a preferred frame.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

"Steve Willner" wrote in message
...
| In article ,
| "Androcles" writes:
| "If there's a preferred frame, SR is invalid." -- Willner.
|
| Admittedly I'm breaking my own rule here, but it's worth noting that
| the Universe we live in doesn't seem to have a preferred frame.
|
SR doesn't seem to be mathematics either, probably because it isn't.

And before you say "experimental evidence", here it is:
http://www.androcles01.pwp.blueyonde...uons/Muons.htm

I'd be cautious about breaking my own rule if I were you, but I'm not you,
so I won't break mine.

--
"Let there be given a stationary rigid rod; and let its length be L as
measured by a measuring-rod which is also stationary. We now imagine
the axis of the rod lying along the axis of x of the stationary system of
co-ordinates, and that a uniform motion of parallel translation with
velocity v along the axis of x in the direction of increasing x is then
imparted to the rod. We now inquire as to the length of the moving rod" --
Einstein
"The length to be discovered by the operation (b) we will call ``the length
of the (moving) rod in the stationary system.''"-- Einstein

"This we shall determine on the basis of our two principles, and we shall
find that it differs from L." -- Einstein.

AND THE ANSWER IS...

"xi = (x-vt)/sqrt(1 - v^2/c^2)" -- Einstein.

Yep, xi differs from L, Greek letters differ from Roman letters.

In agreement with experience we further assume the deranged babbling
incompetent cretin couldn't answer his own inquiry, he was too stupid
to realise xi is greater than L when he wrote 'for v=c all moving
objects--viewed from the "stationary'' system--shrivel up into plane
figures', whereas his own equation shows they stretch to infinity...
sqrt(1-c^2/c^2) = 0.


"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v" - Einstein
"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity" - Einstein.
"In agreement with experience we further assume the quantity
2AB/(t'A -tA) = c to be a universal constant--the velocity of light in
empty space." -- Einstein
He was right. The distance from A to A divided by the time it takes
to get there is undefined. Anyone that divides by zero is a lunatic.

SWI think the answer is much simpler: each twin sees the other
SWredshifted as they separate and blueshifted as they return

In article ,
(Daryl McCullough) writes:
That's a nice simple way to reason about it, but it gives
you the wrong answer.

I'd come to that conclusion myself after another day's thought, and I
appreciate your and Tom's detailed answers.

I'm still trying to see what happens in a hyper-spherical space,
where SR doesn't apply except locally. There must be some sort of
preferred frame, but I'm having trouble figuring out why there should
be one or how one would do an experiment to determine it. You could
measure space curvature by measuring triangles (large ones!), but how
would you establish a "universal standard of rest" for velocities?
The experiment would have to be non-local, of course, but I wouldn't
think you'd have to "go all the way around."

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

On 5/25/11 5/25/11 - 8:40 PM, Daryl McCullough wrote:
Tom Roberts says...
Ann O'Nymous wrote:
I once read somewhere that if the Big Bang is true, it is possible for
light to travel completely across the expanding universe and appear from
the other side.

This is not true in the FRW manifolds that are the basis of big bang cosmology.
There are three classes of FRW manifolds, classified by their spatial curvature
at constant cosmological time. Those with flat or hyperbolic spatial curvature
have infinite spatial extent at all times and no "crossing" is possible. Those
with 3-sphere spatial surfaces have a closed space, but they all end in a "big
crunch" that is the time reversal of the big bang; a light ray can barely
circumnavigate space between the big bang and the big crunch.

I thought I remembered seeing once that there is a torus-like solution
that has a finite but expanding universe. In this universe, you actually
can circumnavigate the universe. I don't remember any details, though.

I know nothing about this.


Tom Roberts

In article , Tom Roberts says...

On 5/25/11 5/25/11 - 8:40 PM, Daryl McCullough wrote:
Tom Roberts says...
Ann O'Nymous wrote:
I once read somewhere that if the Big Bang is true, it is possible for
light to travel completely across the expanding universe and appear from
the other side.

This is not true in the FRW manifolds that are the basis of big bang cosmology.
There are three classes of FRW manifolds, classified by their spatial curvature
at constant cosmological time. Those with flat or hyperbolic spatial curvature
have infinite spatial extent at all times and no "crossing" is possible. Those
with 3-sphere spatial surfaces have a closed space, but they all end in a "big
crunch" that is the time reversal of the big bang; a light ray can barely
circumnavigate space between the big bang and the big crunch.

I thought I remembered seeing once that there is a torus-like solution
that has a finite but expanding universe. In this universe, you actually
can circumnavigate the universe. I don't remember any details, though.

I know nothing about this.

There is a discussion of it in an old Scientific American
http://cosmos.phy.tufts.edu/~zirbel/...paceFinite.pdf

--
Daryl McCullough
Ithaca, NY