Curved or flat

1. May 16, 2013

Hi there.
I have a dump question for you guys.
I really wonder about curvature of spacetime.

I read that due to Omega_tot=1 the Universe is assumed to be flat. But on the other hand something like the curvature of the universe is mentioned... I also thought that the energy stress tensor always induces curvature. Can someone please explain to me.
Do I confuse two different curvatures?
thx

2. May 16, 2013

WannabeNewton

Indeed you are confusing space-time curvature with curvature of the constant time spatial slices. When we speak of a flat universe we mean that the constant time spatial slices are flat, not space-time itself.

3. May 16, 2013

so I suppose spacetime curvature is given by the metric and influences the spacetime distance ds.

But what are constant time spatial slices?
thx

4. May 16, 2013

yenchin

Simply put, at any instant of cosmic time, we have a 3-dimensional space, there is a Riemannian metric defined on this space. You calculate the curvature from that metric. This is different from the curvature calculated for the 4-dimensional spacetime.

5. May 16, 2013

WannabeNewton

I'll try not to make things too mathematical (so this is going to be hand-wavy). For the particular case of the FLRW universe, basically if our space-time is given by $M$ we can separate it up into spatial slices $\Sigma_{t}$ which are "snapshots" of the universe at each instant of time (more precisely the proper time read on a clock carried by the congruence of isotropic observers). The curvature tensor of these $\Sigma_{t}$ is given by an induced spatial metric and it turns out that isotropy of the universe leads to a relation between the spatial metric and the associated curvature tensor that includes a constant factor $K$ which we call the sectional curvature of $\Sigma_{t}$. $K$ is what determines if we are dealing with a flat universe or closed universe etc.

6. May 16, 2013

Naty1

http://en.wikipedia.org/wiki/Shape_of_the_Universe#Flat_universe

7. May 16, 2013

Naty1

The stress energy tensor is the source of gravity...the metric tensor describes the curvature...problem is there are a lot of measures of curvature...Ricci, Riemann, Weyl, Christoffel....They each capture a different aspect of curvature....

http://en.wikipedia.org/wiki/Weyl_tensor

8. May 16, 2013

Popper

Yes. You're confusing spatial curvature with spacetime curvature.

9. May 24, 2013

backward

"WannabeNewton

Indeed you are confusing space-time curvature with curvature of the constant time spatial slices. When we speak of a flat universe we mean that the constant time spatial slices are flat, not space-time itself. "

But surely, the path of a satellite near a star is curved, being the geodesic in the local space. It means that space is curved near a gravitating body.
I think the space is flat except in the vicinity of matter; and spacetime, of course, is curved in general. Am I right?

10. May 24, 2013

WannabeNewton

That is again space-time curvature. Also, you are no longer talking about a solution for the universe but rather a solution for an isolated gravitating body - it is a different scenario (cosmological metrics consider fluid models "in the large" i.e. averaged over large distances - not local distributions at the solar system level). "Space curvature" is not always well-defined but for the FRW universe it is well defined if we speak of the sectional curvature of the spatial slices of the universe at different cosmic times and under the FRW model, the sectional curvature is a constant.

EDIT: Again space-time is flat except when there is some energy-momentum distribution around. This is what Einstein's equations tell us. "Space curvature" is completely different.

Last edited: May 24, 2013
11. May 24, 2013

Naty1

For a good discussion on curvature, try here:

Is Spacetime Curvature Absolute

Spacetime Curvature Observer and/or Coordinate Dependent?

In that discussion :

My attempt at a Summary:
[I don’t think I’ve ever seen one, so I thought I’d give it a try..]

[#19] Is spacetime curvature observer dependent ?

None of the answers to the OP are precise…none can be unless an agreement is reached on exactly what component of 'curvature' measurement will be used...and that would provide only a partial answer; the reason is given at the end of this post with some of the great insights along the way summarized here first.…

Originally Posted by Mentz114

If we take 'spacetime curvature' to mean gravity then the answer must be 'yes'.

passionflower asked several questions that I really liked:

[1]
and [2]

I’ll try some explanations; these, too will be imprecise, but hopefully again illustrate
The problem; feel free of course pick them apart if you don’t like them:

[1] Locally, everything stays the same....spacetime stays flat….but as a differential instead of infinitesimal observation is made, curvature becomes apparent. Time dilation enables the observer to make a comparison with a ‘distant’ observer to see their clocks run at different speeds; such an observer may also see a distant inertial observer apparently moving in an arc...with some acceleration, that is, some ‘curvature’ is apparent resulting from the acceleration of either. docAl might say: "yes, that's the effect of different grid lines but drawn on the original curved graph paper.…the original gravitational field. [This is another imprecise explanation.]

[2] Jonathan Scott provided insights to this one in posts #13,14; I want to try a complementary approach: I think he said different curvatures, accelerations and velocities are all related in curved spacetime.

I am reminded that ‘light follows null geodesics’ is an APPROXIMATION; a good one, but what really follow null geodesics are test particles...ideals with no mass, no energy, no disturbance of the initial gravitational field. So we see that photons of different energy follow slightly different paths. [I got this idea elsewhere from pervect.] So these ‘bodies’ are affected by energy….and you [OP] were worried about just speed affecting curvature’!!

Next, suppose we have a particle with some mass at some velocity relative to an otherwise stationary gravitational field. Send a second particle at the same initial trajectory but, say, double the velocity of the first; or impart it with some spin [angular momentum] but at the original velocity: All the particles will follow different trajectories. In docAl's language, the original gravitational field maintains it's 'curved graph paper' shape, unless it’s evolving, but there is an overlay of different grid lines [a/w each particle] on that curved graph paper representing ‘spacetime curvature…...due to different energies; hence different paths result. The ‘gravitational curvature’ may remain invariant [depending on how we define that] but the effect of the now differently curved grid lines results in a different overall ‘curvature’ and hence different worldline paths.

We seem to have agreed that: An invariant would be something that has exactly the same value in whatever frame it is calculated; And Dalespam and Mentz seem to agree that: Invariant refers to scalars (tensors of rank 0), but Tensors are 'covariant' not invariant.” And Dalespam notes:
It also depends on what means by "observer dependent". Certainly the numerical values of the components of a tensor can vary when you change frames….[
but the observers WILL agree on the total..] No controversy nor obstacles here.

Is that progress?? It sure helps me, but I don’t know about the OP…where is he??

So even if you all were to agree with all those descriptions, which would be a miracle, and the ones Pallen described in post #23, we are still faced with the problem of ascribing those descriptions of ‘grid lines’ and ’curved graph paper’ [or some other equivalents] to some mathematical component of the EFE.

That seems impossible because if my understanding is correct MTW lays out an absolute impass: [from my post #27, quote courtesy of pervect from elsewhere]

MTW:

12. May 24, 2013

pervect

Staff Emeritus
By convention, "time" here is cosmological time, the sort of time measured by the clocks of an observer to whom the universe appears isotropic and the cosmic microwave background appears isotropic.

Using an array of such isotropic observers, one can define a particular spatial slice, for instance "1 million years after the big bang". These are the constant cosmological time spatial slices.