A How to see if a cosmological model is flat?

AI Thread Summary
To determine if a cosmological model is flat, open, or closed, one can analyze the metric's structure, particularly focusing on the curvature parameter k in the RW metric. If the model lacks explicit k-dependence, transforming it into a known form or computing the Riemann tensor for the spatial metric can help assess its curvature. For instance, a metric expressed in Cartesian coordinates can reveal flatness if all components of the curvature tensor are zero. The discussion emphasizes that while the overall spacetime may not be flat, specific three-dimensional hypersurfaces can still exhibit flatness. Ultimately, calculating the curvature tensor is a reliable method for determining the model's geometric properties.
Philip Land
Messages
56
Reaction score
3
The RW metric reads
$$ds^2 = -dt^2 + a^2(t) \Big( \frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 sin(\theta)^2 d\phi^2 \Big)$$

The value of k determines the model is flat/open/closed.

But say if we have a model on a completely different form, with no explicit k-dependence. How would I determine if it's closed(open/flat?

Take a very simple example: $$ ds^2 = -dt^2 a(t)^2 \Big( dx^2 + dy^2 + dz^2 \Big)$$

How would you see what model this lin-element would correspond to?
 
Space news on Phys.org
Philip Land said:
How would I determine if it's closed(open/flat?

One way would be to transform it to the coordinates used in the first line element.

A second way would be to compute the Riemann tensor for the 3-d spatial metric inside the parentheses.
 
Philip Land said:
The RW metric reads
$$ds^2 = -dt^2 + a^2(t) \Big( \frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 sin(\theta)^2 d\phi^2 \Big)$$

The value of k determines the model is flat/open/closed.

But say if we have a model on a completely different form, with no explicit k-dependence. How would I determine if it's closed(open/flat?

Take a very simple example: $$ ds^2 = -dt^2 a(t)^2 \Big( dx^2 + dy^2 + dz^2 \Big)$$

How would you see what model this lin-element would correspond to?

Did you mean to type
$$ds^2 = -dt^2 + a\left(t\right)^2 \left( dx^2 + dy^2 + dz^2 \right) ?$$
If you didn't mean to type this, then your metric doesn't make sense, as its fourth-order in differentials. If you meant what I wrote, then ##k=0##. To see this, set ##x = r\sin\theta\cos\phi##, ##y = r\sin\theta\sin\phi##, ##z = r\cos\theta##. Then,
$$\begin{align}
dx &= \frac{\partial x}{\partial r} dr +\frac{\partial x}{\partial \theta} d\theta + \frac{\partial x}{\partial \phi} d\phi \\
&= \sin\theta\cos\phi dr + r\cos\theta\cos\phi d\theta - r\sin\theta \sin\phi d\phi .
\end{align}$$
Similarly,
$$dy = \sin\theta\sin\phi dr + r\cos\theta\sin\phi d\theta - r\sin\theta \cos\phi d\phi $$
and
$$dz = \cos\theta dr + r\sin\theta d\theta .$$
A little computation gives
$$dx^2 + dy^2 + dz^2 = dr^2 + r^2 d\theta^2 + r^2 \sin\left(\theta\right)^2 d\phi^2 .$$
 
  • Like
Likes Philip Land and PeroK
Yes this is exactly what I ment and this was very helpful. Thank you very much.
 
PeterDonis said:
One way would be to transform it to the coordinates used in the first line element.

A second way would be to compute the Riemann tensor for the 3-d spatial metric inside the parentheses.
Philip Land said:
Yes this is exactly what I ment and this was very helpful. Thank you very much.
I explicitly carried through Peter's first suggestion.

Peter's second suggestion is more general. The idea is that while the above spacetime is not flat, space (specific 3-dimensional hypersufaces in FLRW spacetimes) can be flat. These 3-dimensional hypersufaces are picked out by setting ##t = \mathrm{constant}##, so that ##dt =0##, and ##a\left(t\right)##, and the metric for space is
$$dr^2 + r^2 d\theta^2 + r^2 \sin\left(\theta\right)^2 d\phi^2$$
in spherical coordinates, and
$$dx^2 + dy^2 + dz^2$$
in Cartesian coordinates. Because all the components of this spatial metric are constant in Cartesian coordinates, it is easily seen that, in this coordinate system, all the components of the curvature tensor are zero. If all the components of a tensor are zero in a particular coordinate system, then all the components of a tensor are zero in every coordinate system. Consequently, putting the components of this spatial metirc in spherical coordinates in the standard express for the Riemann curvature tensor will give zero for all components. If I had to verify this, I would you a computer package, e.g., Maple Mathematica, Macsyma.

It might not be easy (or possible if the space is not flat) to see a transformation to Cartesian coordinates, but calculation of the curvature tensor in any coordinate system is enough to indicate flatness/non-flatness.
 
https://en.wikipedia.org/wiki/Recombination_(cosmology) Was a matter density right after the decoupling low enough to consider the vacuum as the actual vacuum, and not the medium through which the light propagates with the speed lower than ##({\epsilon_0\mu_0})^{-1/2}##? I'm asking this in context of the calculation of the observable universe radius, where the time integral of the inverse of the scale factor is multiplied by the constant speed of light ##c##.
The formal paper is here. The Rutgers University news has published a story about an image being closely examined at their New Brunswick campus. Here is an excerpt: Computer modeling of the gravitational lens by Keeton and Eid showed that the four visible foreground galaxies causing the gravitational bending couldn’t explain the details of the five-image pattern. Only with the addition of a large, invisible mass, in this case, a dark matter halo, could the model match the observations...
Hi, I’m pretty new to cosmology and I’m trying to get my head around the Big Bang and the potential infinite extent of the universe as a whole. There’s lots of misleading info out there but this forum and a few others have helped me and I just wanted to check I have the right idea. The Big Bang was the creation of space and time. At this instant t=0 space was infinite in size but the scale factor was zero. I’m picturing it (hopefully correctly) like an excel spreadsheet with infinite...
Back
Top