Re: Constraints on spatial flatness.

[Chalky]

On Dec 30, 11:20Â*pm, carlip-nos...@physics.ucdavis.edu wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [...]
>
> > I was merely enquiring whether that spatial curvature (or lack of it)
> > also applies close to, for example, the surface of the Sun, or the
> > event horizon of a black hole.
>
> The spatial curvature people refer to when they say "the universe is
> nearly spatially flat" is the average curvature, averaged on the same
> scale that one can say "the universe is nearly sptially homogeneous." Â*

I have given this point some further thought since submitting my first
response, and your statement appears to imply that there are smaller
regions of positively curved space, and regions of negatively curved
space which happen to average out to close to zero, on the large
scale.

If that was the intended implication, I would like to ask.....Where
are they?

Lets start at home with the solar system, since we all know where we
stand here. I can easily understand spacetime curvature caused by the
Sun, in the sense that, after a year of time, the Earth returns to
pretty much where it started, in space (assuming Heliocentric
coordinates). I can also understand that spacetime curvature is
stronger closer to the Sun, since Mercury returns to the same point in
space more quickly. However, this does not alter the fact that space,
per se, still looks remarkably flat to me, from where I am standing,
and I doubt that this would change substantially if I moved to
Mercury. Hence my question. Where are these curved regions of space
supposed to be located?

[Tom Roberts]

Chalky wrote:
> To what extent can we say that the effects of spatial curvature [as
> opposed to spacetime curvature] correspond to the effects of Newtonian
> gravity.....or would such a suggestion be completely bonkers?

It's in some sense backwards -- speaking rather loosely, Newtonian
gravity is due to "time curvature" (not really a good name for it, but
there isn't a good name for it).

That is, in the Newtonian approximation to GR, the line element for
"Newtonian coordinates" is to excellent approximation:

ds^2 = -(1 - 2\phi) dt^2 + dx^2 + dy^2 + dz^2
where \phi is the Newtonian gravitational potential, c=1

The only deviation from Minkowski spacetime is in the time coordinate --
using these coordinates the 3-space corresponding to a given value of t
is Euclidean (flat).


Tom Roberts

[Chalky]

On Jan 4, 2:56Â*pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> Chalky wrote:
> > To what extent can we say that the effects of spatial curvature [as
> > opposed to spacetime curvature] correspond to the effects of Newtonian
> > gravity.....or would such a suggestion be completely bonkers?
>
> It's in some sense backwards -- speaking rather loosely, Newtonian
> gravity is due to "time curvature" (not really a good name for it, but
> there isn't a good name for it).
>
> That is, in the Newtonian approximation to GR, the line element for
> "Newtonian coordinates" is to excellent approximation:
>
> Â* Â* Â* Â* ds^2 = -(1 - 2\phi) dt^2 + dx^2 + dy^2 + dz^2
> Â* Â* Â* Â* where \phi is the Newtonian gravitational potential, c=1
>
> The only deviation from Minkowski spacetime is in the time coordinate --
> using these coordinates the 3-space corresponding to a given value of t
> is Euclidean (flat).


That is extremely interesting. Can we conclude from this that, within
the context of GR, the observable consequences of gravity will
accurately obey the laws of Newtonian physics on the large scale if,
on that large scale, the universe is perfectly flat?

[carlip-nospam@physics.ucdavis.edu]

Chalky <chalkyspam@bleachboys.co.uk> wrote:

[...]
> However, the new side issue which intrigues me right now, is this:

> To what extent can we say that the effects of spatial curvature
> [as opposed to spacetime curvature] correspond to the effects of
> Newtonian gravity.....or would such a suggestion be completely
> bonkers?

As others have said, in almost this context, the question doesn't make
sense. It runs into the standard problem that lies behind most
relativity "paradoxes," the relativity of simultaneity.

To talk about "spatial curvature [as opposed to spacetime curvature],"
you have to specify what you mean by "space," or more precisely, "space
at a fixed time." In a theory with an absolute tiome, this makes sense.
In relativity, it doesn't: there are an infinite number of time
coordinates, that is, an infinite number of ways of synchronizing
clocks, and none of them is better than any other.

For the Schwarzschild metric, for example -- the metric that describes
the spacetime outside a static, spherical mass -- one can use standard
"Schwarzschild coordinates," the ones you are most likely to see in a
textbook. In these coordinates, there is a sense in which the curvature
is half "purely spatial" and half "spacetime." (To be technical, the
square of the spatial curvature tensor is, I believe, half of the square
of the fiull spacetime curvature tensor, where by "square " I mean
R_{abcd}R^{abcd}.) On the other hand, one can instead choose to use
Painleve-Gullstrand coordinates. In these coordinates, space at a fixed
time is flat.

Now, if you want to forget general relativity, it is also possible to
formulate Newtonian gravity -- with its absolute time -- in terms of
spacetime curvature. This was first done by Cartan, and is described in
detail in chapter 12 of Misner, Thorne, and Wheeler. In Cartan- Newton
gravity, "space at a fixed time" makes sense, since there is an absolute
time, and it turns out that the relevant geometry is spatially flat --
all the curvature is in spacetime [as opposed to space]." But this is
*not* general relativity, and is also not our universe.

Steve Carlip

[Chalky]

On Jan 6, 5:05 pm, carlip-nos...@physics.ucdavis.edu wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [...]
>
> > However, the new side issue which intrigues me right now, is this:
> > To what extent can we say that the effects of spatial curvature
> > [as opposed to spacetime curvature] correspond to the effects of
> > Newtonian gravity.....or would such a suggestion be completely
> > bonkers?
>
> As others have said, in almost this context, the question doesn't make
> sense. It runs into the standard problem that lies behind most
> relativity "paradoxes," the relativity of simultaneity.
>
> To talk about "spatial curvature [as opposed to spacetime curvature],"
> you have to specify what you mean by "space," or more precisely, "space
> at a fixed time." In a theory with an absolute tiome, this makes sense.
> In relativity, it doesn't: there are an infinite number of time
> coordinates, that is, an infinite number of ways of synchronizing
> clocks, and none of them is better than any other.

I don't see that this need be a problem in any system of coordinates
constructed around the observer at the (thus defined) spatial origin.
You then only need one clock, the observer's, to describe the dynamism
of the observable universe, relative to that observer. Space at a
fixed time difference from that clock is then space at a fixed light
travel time, relative to that observer.

I have just looked up "Painleve-Gullstrand coordinates" and this
appears to be one (limiting) example of such a coordinate system.

That would seem to confirm that space is, indeed, flat relative to all
such free falling observers.

Now, given Tom Roberts' point, can we thus conclude that, relative to
such observers, the obsevable dynamism will, consequently, obey
Newtonian laws to a high level of accuracy?

[cmaj10@yahoo.com]

On Jan 6, 9:05 am, carlip-nos...@physics.ucdavis.edu wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [...]
>
> > However, the new side issue which intrigues me right now, is this:
> > To what extent can we say that the effects of spatial curvature
> > [as opposed to spacetime curvature] correspond to the effects of
> > Newtonian gravity.....or would such a suggestion be completely
> > bonkers?
>
> As others have said, in almost this context, the question doesn't make
> sense. It runs into the standard problem that lies behind most
> relativity "paradoxes," the relativity of simultaneity.
>
> To talk about "spatial curvature [as opposed to spacetime curvature],"
> you have to specify what you mean by "space," or more precisely, "space
> at a fixed time." In a theory with an absolute tiome, this makes sense.
> In relativity, it doesn't: there are an infinite number of time
> coordinates, that is, an infinite number of ways of synchronizing
> clocks, and none of them is better than any other.
>
> For the Schwarzschild metric, for example -- the metric that describes
> the spacetime outside a static, spherical mass -- one can use standard
> "Schwarzschild coordinates," the ones you are most likely to see in a
> textbook. In these coordinates, there is a sense in which the curvature
> is half "purely spatial" and half "spacetime." (To be technical, the
> square of the spatial curvature tensor is, I believe, half of the square
> of the fiull spacetime curvature tensor, where by "square " I mean
> R_{abcd}R^{abcd}.) On the other hand, one can instead choose to use
> Painleve-Gullstrand coordinates. In these coordinates, space at a fixed
> time is flat.
>
> Now, if you want to forget general relativity, it is also possible to
> formulate Newtonian gravity -- with its absolute time -- in terms of
> spacetime curvature. This was first done by Cartan, and is described in
> detail in chapter 12 of Misner, Thorne, and Wheeler. In Cartan- Newton
> gravity, "space at a fixed time" makes sense, since there is an absolute
> time, and it turns out that the relevant geometry is spatially flat --
> all the curvature is in spacetime [as opposed to space]." But this is
> *not* general relativity, and is also not our universe.
>
> Steve Carlip

I have a few questions.

Assuming that the universe is truly spatially flat (in the common
astrophysical sense that corresponds to the empirical observations
discussed above in this thread), in what GR coordinate(s) would this
be true?

Chalky wrote "On this large scale we know that space is flat, or
nearly so. On the large scale, however, we also know that spacetime is
strongly curved" and nobody argued with that. So I guess this is a
correct statement (assuming the empirical evidence is 100% correct).
Now I wonder if it is too naive of me to conclude that---on the large
scale---only time is strongly curved. Also I understand that local
time curvature manifests itself as gravitational effects, but what is
its manifestation on the large scale? [I guess it must be observable
in the so-called big-bang, but I'm not quite sure I really get it.]

cmaj10

[Igor Khavkine]

On Jan 10, 5:22 pm, cma...@yahoo.com wrote:

> Assuming that the universe is truly spatially flat (in the common
> astrophysical sense that corresponds to the empirical observations
> discussed above in this thread), in what GR coordinate(s) would this
> be true?

The common astrophysical (or rather cosmological) sense refers to
spatial slices at constant cosmological time. This cosmological time
is measured from the Big Bang along the world lines of observers that
observe the universe to be isotropic, that is they are co-moving with
the background matter/energy density. Take the cosmological time as
one coordinate, and any set of coordinates on these spatial slices and
you've got yourself a coordinate system for the whole space-time.

> Chalky wrote "On this large scale we know that space is flat, or
> nearly so. On the large scale, however, we also know that spacetime is
> strongly curved" and nobody argued with that. So I guess this is a
> correct statement (assuming the empirical evidence is 100% correct).
> Now I wonder if it is too naive of me to conclude that---on the large
> scale---only time is strongly curved. Also I understand that local
> time curvature manifests itself as gravitational effects, but what is
> its manifestation on the large scale? [I guess it must be observable
> in the so-called big-bang, but I'm not quite sure I really get it.]

Unless you really know what you are doing, it's best not to invoke
"space curvature" and "time curvature" separately. The "space-time
curvature" is always well defined, while you have to be really
specific to give proper meaning to the other two terms.

Finally, to answer your question about about the effects of curvature
on the large scale, the most obvious effect is the observed redshift
from remote galaxies. Without curvature, there would be no redshift.

Hope this helps.

Igor

[Jonathan Thornburg [remove -animal to reply]]

Chalky <chalkyspam@bleachboys.co.uk> asked
# However, the new side issue which intrigues me right now, is this:
# To what extent can we say that the effects of spatial curvature
# [as opposed to spacetime curvature] correspond to the effects of
# Newtonian gravity.....or would such a suggestion be completely
# bonkers?

Not bonkers, just not meaningful. Until you choose a cosmological time
coordinate, you can't even define the "space" implicit in the phrase
"spatial curvature". As Steve Carlip put it,

| To talk about "spatial curvature [as opposed to spacetime curvature],"
| you have to specify what you mean by "space," or more precisely, "space
| at a fixed time." In a theory with an absolute tiome, this makes sense.
| In relativity, it doesn't: there are an infinite number of time
| coordinates, that is, an infinite number of ways of synchronizing
| clocks, and none of them is better than any other.

Chalky went on to say:
> I don't see that this need be a problem in any system of coordinates
> constructed around the observer at the (thus defined) spatial origin.
> You then only need one clock, the observer's, to describe the dynamism
> of the observable universe, relative to that observer. Space at a
> fixed time difference from that clock is then space at a fixed light
> travel time, relative to that observer.

I think you're trying to define a cosmological time coordinate, but you
don't have a usable definition yet. That is, I don't think you've
specified what either of the phrases "space at a fixed time difference
from that clock" or "space at a fixed light travel time" means in an
operational sense. To make things clearer, let's put our reference
observer at the center of the Earth. What do you mean by the phrases
"space at a 1 year time difference from the center-of-the-earth clock"
and "space at a 1 year light travel time, relatve to the center of the
earth"?

The problem is that the phrases "time difference" and "light travel
time" suggests you're comparing two clock readings... but the only
clocks we have to start with are at the center of the earth, so as soon
as your light signals propagate away from the earth, we don't have any
clocks at the signals' reception events.

[We started this discussion in the context of cosmology, but I think the
same basic problems are already present for special relativity (i.e. the
physics of (flat) Minkowski spacetime).]

Also, don't forget that in general relativity, the light travel time
between two observers need not be uniquely defined. That is, given two
observers A and B, if A sends out a flash of light in all directions, B
might see some of that light at a certain time, but some more of it
1.23456 year later (as measured by B's local proper time), and more of
it 6.54321 B's-local-proper-time-years after that. What does your
definition say about how to synchronize A and B's clocks in this case?

You might find
http://en.wikipedia.org/wiki/Einstein_synchronisation
instructive, particularly the passage
# In general relativity frames, most importantly in rotating ones,
# the non-transitivity of Einstein synchronisation diminishes its
# usefulness. If clock 1 and clock 2 are not synchronised directly,
# but by using a chain of intermediate clocks, the synchronisation
# depends on the path chosen.

There's a somewhat more extensive discussion at
http://en.wikipedia.org/wiki/Relativity_of_simultaneity
and in instructive example at
http://en.wikipedia.org/wiki/Andromeda_paradox

Chalky also write
> I have just looked up "Painleve-Gullstrand coordinates" and this
> appears to be one (limiting) example of such a coordinate system.
>
> That would seem to confirm that space is, indeed, flat relative to all
> such free falling observers.

It confirms that *if* you define "space" to be the level sets of the
Painleve-Gullstrand coordinates in Schwarzschild spacetime, then space
is 3-flat. However, that statement is specific to that choice of
coordinates. It's also true that "Schwarzschild space" (defined as the
level sets of the Schwarzschild time coordinate in that same
Schwarzschild spacetime) is not 3-flat. And the "ingoing
Eddington-Finkelstein space" (defined as the level sets of the ingoing
Eddington-Finkelstein time coordinate in that same Schwarzschild
spacetime) also isn't 3-flat (and its 3-curvature at a given position is
probably a different nonzero number from the 3-curvature of
"Schwarzschild space" there).

All of these time coordinates (and their corresponding definitions of
what "space" is) are equally valid -- none of them have any intrinsic
superiority over the others, they're just different ways of describing
the same physics, or more precisely different ways of splitting
4-dimensional spacetime into a 1-dimensional family of 3-dimensional
spacelike "spaces".

--
-- From: "Jonathan Thornburg [remove -animal to reply]" <jthorn@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[Igor Khavkine]

On Jan 10, 5:22 pm, cma...@yahoo.com wrote:

> Assuming that the universe is truly spatially flat (in the common
> astrophysical sense that corresponds to the empirical observations
> discussed above in this thread), in what GR coordinate(s) would this
> be true?

The common astrophysical (or rather cosmological) sense refers to
spatial slices at constant cosmological time. This cosmological time
is measured from the Big Bang along the world lines of observers that
observe the universe to be isotropic, that is they are co-moving with
the background matter/energy density. Take the cosmological time as
one coordinate, and any set of coordinates on these spatial slices and
you've got yourself a coordinate system for the whole space-time.

> Chalky wrote "On this large scale we know that space is flat, or
> nearly so. On the large scale, however, we also know that spacetime is
> strongly curved" and nobody argued with that. So I guess this is a
> correct statement (assuming the empirical evidence is 100% correct).
> Now I wonder if it is too naive of me to conclude that---on the large
> scale---only time is strongly curved. Also I understand that local
> time curvature manifests itself as gravitational effects, but what is
> its manifestation on the large scale? [I guess it must be observable
> in the so-called big-bang, but I'm not quite sure I really get it.]

Unless you really know what you are doing, it's best not to invoke
"space curvature" and "time curvature" separately. The "space-time
curvature" is always well defined, while you have to be really
specific to give proper meaning to the other two terms.

Finally, to answer your question about about the effects of curvature
on the large scale, the most obvious effect is the observed redshift
from remote galaxies. Without curvature, there would be no redshift.

Hope this helps.

Igor

[cmaj10@yahoo.com]

On Jan 6, 9:05 am, carlip-nos...@physics.ucdavis.edu wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [...]
>
> > However, the new side issue which intrigues me right now, is this:
> > To what extent can we say that the effects of spatial curvature
> > [as opposed to spacetime curvature] correspond to the effects of
> > Newtonian gravity.....or would such a suggestion be completely
> > bonkers?
>
> As others have said, in almost this context, the question doesn't make
> sense. It runs into the standard problem that lies behind most
> relativity "paradoxes," the relativity of simultaneity.
>
> To talk about "spatial curvature [as opposed to spacetime curvature],"
> you have to specify what you mean by "space," or more precisely, "space
> at a fixed time." In a theory with an absolute tiome, this makes sense.
> In relativity, it doesn't: there are an infinite number of time
> coordinates, that is, an infinite number of ways of synchronizing
> clocks, and none of them is better than any other.
>
> For the Schwarzschild metric, for example -- the metric that describes
> the spacetime outside a static, spherical mass -- one can use standard
> "Schwarzschild coordinates," the ones you are most likely to see in a
> textbook. In these coordinates, there is a sense in which the curvature
> is half "purely spatial" and half "spacetime." (To be technical, the
> square of the spatial curvature tensor is, I believe, half of the square
> of the fiull spacetime curvature tensor, where by "square " I mean
> R_{abcd}R^{abcd}.) On the other hand, one can instead choose to use
> Painleve-Gullstrand coordinates. In these coordinates, space at a fixed
> time is flat.
>
> Now, if you want to forget general relativity, it is also possible to
> formulate Newtonian gravity -- with its absolute time -- in terms of
> spacetime curvature. This was first done by Cartan, and is described in
> detail in chapter 12 of Misner, Thorne, and Wheeler. In Cartan- Newton
> gravity, "space at a fixed time" makes sense, since there is an absolute
> time, and it turns out that the relevant geometry is spatially flat --
> all the curvature is in spacetime [as opposed to space]." But this is
> *not* general relativity, and is also not our universe.
>
> Steve Carlip

I have a few questions.

Assuming that the universe is truly spatially flat (in the common
astrophysical sense that corresponds to the empirical observations
discussed above in this thread), in what GR coordinate(s) would this
be true?

Chalky wrote "On this large scale we know that space is flat, or
nearly so. On the large scale, however, we also know that spacetime is
strongly curved" and nobody argued with that. So I guess this is a
correct statement (assuming the empirical evidence is 100% correct).
Now I wonder if it is too naive of me to conclude that---on the large
scale---only time is strongly curved. Also I understand that local
time curvature manifests itself as gravitational effects, but what is
its manifestation on the large scale? [I guess it must be observable
in the so-called big-bang, but I'm not quite sure I really get it.]

cmaj10

[Chalky]

On Jan 11, 11:02Â*am, "Jonathan Thornburg [remove -animal to reply]"
<jth...@astro.indiana-zebra.edu> wrote:

> Chalky went on to say:

> > I don't see that this need be a problem in any system of coordinates
> > constructed around the observer at the (thus defined) spatial origin.
> > You then only need one clock, the observer's, to describe the dynamism
> > of the observable universe, relative to that observer. Space at a
> > fixed time difference from that clock is then space at a fixed light
> > travel time, relative to Â*that observer.

> The problem is that the phrases "time difference" and "light travel
> time" suggests you're comparing two clock readings... but the only
> clocks we have to start with are at the center of the earth, so as soon
> as your light signals propagate away from the earth, we don't have any
> clocks at the signals' reception events.

We don't need any clocks there, we just need to specify that the
signal is promptly returned, then use our same local clock to measure
the resultant time difference over the round trip.

Yes, there might be potential practical difficulties at very large
distances but it can be (and has already been) used locally, at least
on the scale of the solar system.
Your method of synchronising clocks at the big bang, however, means
that lunar and Earth observers can never even agree on the time,
because clocks run at different speeds in different gravitational
fields.

[Jonathan Thornburg [remove -animal to reply]]

[[trying to define a cosmological time coordinate]]
Chalky <chalkyspam@bleachboys.co.uk> originally wrote
> I don't see that this need be a problem in any system of coordinates
> constructed around the observer at the (thus defined) spatial origin.
> You then only need one clock, the observer's, to describe the dynamism
> of the observable universe, relative to that observer. Space at a
> fixed time difference from that clock is then space at a fixed light
> travel time, relative to ??that observer.

I questioned just how this is defined operationally:
| The problem is that the phrases "time difference" and "light travel
| time" suggests you're comparing two clock readings... but the only
| clocks we have to start with are at the center of the earth, so as soon
| as your light signals propagate away from the earth, we don't have any
| clocks at the signals' reception events.

Chalky replied:
> We don't need any clocks there, we just need to specify that the
> signal is promptly returned, then use our same local clock to measure
> the resultant time difference over the round trip.

Ok, now you have operationally defined a clock synchronization scheme.
In fact, this is just the standard Einstein clock synchronization,
http://en.wikipedia.org/wiki/Einstein_synchronisation

> Yes, there might be potential practical difficulties at very large
> distances but it can be (and has already been) used locally, at least
> on the scale of the solar system.

The problem I described in an earlier posting (that there could be
multiple paths (with differing light-travel-times) between a given
pair of observers) is indeed absent if we choose a small enough
spatial scale. Since the Sun's gravity bends light by around 1.75
arcseconds, the Sun acts as a gravitational lens with a focal length
of approximately
(1 solar diameter) / (2 * 1.75 arcseconds, expressed in radians)
= about 275 astronomical units (1 AU = mean Earth-Sun distance),
if my arithmetic is correct. So we should be ok within the solar
system, or at least that part of it occupied by the major planets
(i.e. we might not be ok out in the Kuiper belt or Oort cloud).

Unfortunately, in a curved spacetime Einstein-synchronization is
path-dependent, and isn't transitive. That is, for example, consider
3 observers A, B, and C, all in different places in a curved spacetime.
If we synchronize A's clock to B's, then B's to C's, we'll get a
different result than if we synchronize A's directly to C's. In
fact, using the A --> B --> C chain of synchronizations, in general
the result depends on the position of B.

Considering the Sagnac effect makes things even worse: If we
start someplace on the Earth (say on the equator for simplicity),
and Einstein-synchronize a clock somewhat to the west, then use
that clock to Einstein-synchronize a clock somewhat farther to
the west, ... and keep going till we've circumnavigated the Earth's
equator and returned to our starting clock, we'll wind up with
a *different* clock reading than the one we started with! And
if we change the round-the-world path we followed, we'll get a
different answer still! See
http://en.wikipedia.org/wiki/Sagnac_effect
for a nice discussion of this.


> Your method of synchronising clocks at the big bang, however, means
> that lunar and Earth observers can never even agree on the time,
> because clocks run at different speeds in different gravitational
> fields.

Indeed. Curved spacetimes are tricky things!

However, Einstein synchronization doesn't solve this "problem":
In fact, we can see this even for observers at different elevations
on the Earth. For example, suppose we have an observer at sea level
somewhere on the Earth send out a radio pulse once per second (as
measured by her local clock). Due to gravitational redshift, her
colleague on a mountaintop will measure observer will measure these
pulses as arriving every T seconds, where T is a number slightly
larger than 1. This means our two observers right here on the
Earth's surface can't even agree on the time! Curved spacetimes
are *very* tricky things!

In standard general-relativist cosmology we deal with this problem
by assuming a spacetime that's locally smooth, so it doesn't have
an Earth or Moon to introduce gravitational fields/potentials.


The fundamental message to take away from this whole topic is that
in a curved spacetime, in general there is no inherent (unique)
global time coordinate. You can certainly define such a coordinate,
... but it will be just one of an infinite family of time coordinates,
and without some extra physics, there's no obvious reason to prefer
one over another (apart from convenience for some particular calculation).

In observational cosmology, we do indeed have some "extra physics":
in a suitable reference frame at the Earth's position the large-scale
universe looks isotropic (which also implies nonrotating), and
(assuming this holds throughout in the universe, i.e. assuming the
universe is in fact homogeneous and isotropic (& hence also nonrotating)),
we can turn this around to define a suitable reference frame at each
event (one where there's no Doppler anisotropy in the large-scale
appearance of the universe), and thence a "cosmological time coordinate"
(proper time of the preferred-reference-frame observers). This works
fine... but only because the universe is observationally "nice" (i.e.
homogeneous and isotropic and nonrotating to some fairly high degree
of accuracy).


Using the standard homogeneous-and-isotropic-universe time coordinate
(let's call it the HIU time for short), if we compare the cosmic
microwave background radiation (CMBR) temperatures measured by many
different observers throughout the universe, all at the same HIU-time,
then (as we understand cosmology) they'll all get the same answer
(assuming a truly homogeneous & isotropic
Friedmann-Lemaitre-Robertson-Walker universe, i.e. ignoring
the tiny deviations from non-uniformity in the real universe).
In contrast, if you do this same gedanken experiment using observers
all at the same Einstein-synchronized-with-respect-to-some-origin time
(let's call this ESWRTSO time for short), I believe that they would
*not* all measure the same CMBR temperature.

More generally, consider the (3-dimensional spacelike) set ("slice")
of all events at a given HIU-time. This "slice" is homogeneous (i.e.
the universe looks the same from everywhere in it). This statement
would, I believe, also fail to hold for slices of constant ESWRTSO-time.

--
-- From: "Jonathan Thornburg [remove -animal to reply]" <jthorn@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[cmaj10@yahoo.com]

On Jan 10, 8:33*am, Igor Khavkine <igor...@gmail.com> wrote:
> On Jan 10, 5:22 pm, cma...@yahoo.com wrote:
>
> > Assuming that the universe is truly spatially flat (in the common
> > astrophysical sense that corresponds to the empirical observations
> > discussed above in this thread), in what GR coordinate(s) would this
> > be true?
>
> The common astrophysical (or rather cosmological) sense refers to
> spatial slices at constant cosmological time. This cosmological time
> is measured from the Big Bang along the world lines of observers that
> observe the universe to be isotropic, that is they are co-moving with
> the background matter/energy density. Take the cosmological time as
> one coordinate, and any set of coordinates on these spatial slices and
> you've got yourself a coordinate system for the whole space-time.

I have no concrete reasoning behind this, but I somehow imagine that
space would necessarily be curved for all observers accelerating with
respect to the background matter/energy density. And it would be flat
for all other observers. Is that correct?

cmaj10

[Igor Khavkine]

On Jan 30, 5:53 pm, cma...@yahoo.com wrote:
> On Jan 10, 8:33 am, Igor Khavkine <igor...@gmail.com> wrote:
> > On Jan 10, 5:22 pm, cma...@yahoo.com wrote:
>
> > > Assuming that the universe is truly spatially flat (in the common
> > > astrophysical sense that corresponds to the empirical observations
> > > discussed above in this thread), in what GR coordinate(s) would this
> > > be true?
>
> > The common astrophysical (or rather cosmological) sense refers to
> > spatial slices at constant cosmological time. This cosmological time
> > is measured from the Big Bang along the world lines of observers that
> > observe the universe to be isotropic, that is they are co-moving with
> > the background matter/energy density. Take the cosmological time as
> > one coordinate, and any set of coordinates on these spatial slices and
> > you've got yourself a coordinate system for the whole space-time.
>
> I have no concrete reasoning behind this, but I somehow imagine that
> space would necessarily be curved for all observers accelerating with
> respect to the background matter/energy density. And it would be flat
> for all other observers. Is that correct?

Unfortunately, this question is still not precise enough to be
answered concretely. In a previous post, quoted above, I described a
specific way of slicing up space-time into 3-dimensional
hypersurfaces. Each one of those spatial hypersurfaces can then be
said to be flat, according to the best current cosmological models.
You haven't made the description of the observers you have in mind
specific enough to determine what kind of spatial hypersurfaces you
want to foliate space-time with. Hence, a similar statement is
difficult to make.

Here's something that's true for *any* particular observer in a
cosmological space-time. She will see a Hubble redshift when looking
at distant galaxies, which is a manifestation of *space-time*
curvature.

Hope this helps.

[cmaj10@yahoo.com]

On Jan 30, 7:19*pm, Igor Khavkine <igor...@gmail.com> wrote:
> On Jan 30, 5:53 pm, cma...@yahoo.com wrote:
>
>
>
> > On Jan 10, 8:33 am, Igor Khavkine <igor...@gmail.com> wrote:
> > > On Jan 10, 5:22 pm, cma...@yahoo.com wrote:
>
> > > > Assuming that the universe is truly spatially flat (in the common
> > > > astrophysical sense that corresponds to the empirical observations
> > > > discussed above in this thread), in what GR coordinate(s) would this
> > > > be true?
>
> > > The common astrophysical (or rather cosmological) sense refers to
> > > spatial slices at constant cosmological time. This cosmological time
> > > is measured from the Big Bang along the world lines of observers that
> > > observe the universe to be isotropic, that is they are co-moving with
> > > the background matter/energy density. Take the cosmological time as
> > > one coordinate, and any set of coordinates on these spatial slices and
> > > you've got yourself a coordinate system for the whole space-time.
>
> > I have no concrete reasoning behind this, but I somehow imagine that
> > space would necessarily be curved for all observers accelerating with
> > respect to the background matter/energy density. And it would be flat
> > for all other observers. Is that correct?
>
> Unfortunately, this question is still not precise enough to be
> answered concretely. In a previous post, quoted above, I described a
> specific way of slicing up space-time into 3-dimensional
> hypersurfaces. Each one of those spatial hypersurfaces can then be
> said to be flat, according to the best current cosmological models.
> You haven't made the description of the observers you have in mind
> specific enough to determine what kind of spatial hypersurfaces you
> want to foliate space-time with. Hence, a similar statement is
> difficult to make.

I was assuming the observers used your specific description. Let's say
the observer is initially co-moving with the background matter/energy
density, and is using the said description of space. If we are to
believe the empirical evidence, this observer would experience a flat
space on the large scale. Now if that observer subsequently changed
their motion with respect to the background matter/energy density,
would it be true that using the same description of space, they would
experience a curved space on the large scale ** if and only if ** they
are accelerating with respect to the background matter/energy density?

I wrote earlier that I didn't have any concrete reasoning behind this
question. That wasn't really true. I should say that I heuristically
base that on the principle of equivalence, and on the typical
'observer in an elevator' gedanken experiment.

cmaj10

[Igor Khavkine]

On Jan 31, 5:05 pm, cma...@yahoo.com wrote:

> I was assuming the observers used your specific description. Let's say
> the observer is initially co-moving with the background matter/energy
> density, and is using the said description of space. If we are to
> believe the empirical evidence, this observer would experience a flat
> space on the large scale. Now if that observer subsequently changed
> their motion with respect to the background matter/energy density,
> would it be true that using the same description of space, they would
> experience a curved space on the large scale ** if and only if ** they
> are accelerating with respect to the background matter/energy density?

I see, I was confused by the fact that you introduced accelerating
observers, which you've now defined more precisely.

But, there's yet another problem with your question. The problem that
any *one* observer *does not* experience space on the large scale in
any meaningful sense, if space is defined by surfaces of constant
cosmological time. The last statement does not change, whether the
*one* observer of interest is accelerating with respect to the
background matter/energy or not. That's what makes your question
difficult to answer.

Any one observer can only be influenced by (experience, if you will)
the space-time events contained within its past light cone, which
necessarily excludes the spatial slice containing the event where the
observations are made. As such, the experiences of two observers whose
world-lines intersect at a particular event will be the same, up to a
local Lorentz transformation, no matter if either one is accelerating
with respect to the background matter/energy density or not.

I described the cosmological spatial slicing not with one observer,
but with infinitely many. Every space-time event must sit on the world-
line of some observer. Cosmological spatial slices are experienced
only by all of these observers collectively. Unfortunately, these
observers would not be able to communicate their experiences to each
other, except with signals traveling no faster than light. This is one
more reason to discuss cosmology in terms of space-time rather than
space and time. While the latter two can still be defined, as this
thread illustrates, their separate use is rather awkward.

Hope this helps.

Igor

[cmaj10@yahoo.com]

On Jan 10, 8:33*am, Igor Khavkine <igor...@gmail.com> wrote:
> On Jan 10, 5:22 pm, cma...@yahoo.com wrote:
>
> > Assuming that the universe is truly spatially flat (in the common
> > astrophysical sense that corresponds to the empirical observations
> > discussed above in this thread), in what GR coordinate(s) would this
> > be true?
>
> The common astrophysical (or rather cosmological) sense refers to
> spatial slices at constant cosmological time. This cosmological time
> is measured from the Big Bang along the world lines of observers that
> observe the universe to be isotropic, that is they are co-moving with
> the background matter/energy density. Take the cosmological time as
> one coordinate, and any set of coordinates on these spatial slices and
> you've got yourself a coordinate system for the whole space-time.

I have no concrete reasoning behind this, but I somehow imagine that
space would necessarily be curved for all observers accelerating with
respect to the background matter/energy density. And it would be flat
for all other observers. Is that correct?

cmaj10

[Igor Khavkine]

On Jan 30, 5:53 pm, cma...@yahoo.com wrote:
> On Jan 10, 8:33 am, Igor Khavkine <igor...@gmail.com> wrote:
> > On Jan 10, 5:22 pm, cma...@yahoo.com wrote:
>
> > > Assuming that the universe is truly spatially flat (in the common
> > > astrophysical sense that corresponds to the empirical observations
> > > discussed above in this thread), in what GR coordinate(s) would this
> > > be true?
>
> > The common astrophysical (or rather cosmological) sense refers to
> > spatial slices at constant cosmological time. This cosmological time
> > is measured from the Big Bang along the world lines of observers that
> > observe the universe to be isotropic, that is they are co-moving with
> > the background matter/energy density. Take the cosmological time as
> > one coordinate, and any set of coordinates on these spatial slices and
> > you've got yourself a coordinate system for the whole space-time.
>
> I have no concrete reasoning behind this, but I somehow imagine that
> space would necessarily be curved for all observers accelerating with
> respect to the background matter/energy density. And it would be flat
> for all other observers. Is that correct?

Unfortunately, this question is still not precise enough to be
answered concretely. In a previous post, quoted above, I described a
specific way of slicing up space-time into 3-dimensional
hypersurfaces. Each one of those spatial hypersurfaces can then be
said to be flat, according to the best current cosmological models.
You haven't made the description of the observers you have in mind
specific enough to determine what kind of spatial hypersurfaces you
want to foliate space-time with. Hence, a similar statement is
difficult to make.

Here's something that's true for *any* particular observer in a
cosmological space-time. She will see a Hubble redshift when looking
at distant galaxies, which is a manifestation of *space-time*
curvature.

Hope this helps.