How to see if a cosmological model is flat?

In summary, if the metric in a particular coordinate system is zero, then the space is said to be flat in that coordinate system.
  • #1
Philip Land
56
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?
 
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  • #2
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.
 
  • #3
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
$$dx^2 + dy^2 + dz^2 = dr^2 + r^2 d\theta^2 + r^2 \sin\left(\theta\right)^2 d\phi^2 .$$
 
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  • #4
Yes this is exactly what I ment and this was very helpful. Thank you very much.
 
  • #5
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.
 

1. What does it mean for a cosmological model to be "flat"?

A flat cosmological model is one in which the universe is geometrically flat, meaning that the curvature of space-time is zero. This is a key concept in modern cosmology and is supported by observations of the cosmic microwave background radiation.

2. How can we determine if a cosmological model is flat?

There are several methods for determining if a cosmological model is flat. One way is to measure the geometry of the universe through observations of the cosmic microwave background radiation. Another way is to study the large-scale structure of the universe through galaxy surveys. Additionally, measurements of the expansion rate of the universe, known as the Hubble constant, can also provide insight into the curvature of space-time.

3. What evidence supports the idea of a flat universe?

The most compelling evidence for a flat universe comes from observations of the cosmic microwave background radiation. These measurements show that the universe is isotropic, meaning that it looks the same in all directions, and that the fluctuations in temperature are consistent with a flat geometry. Other evidence includes observations of the large-scale structure of the universe and the Hubble constant.

4. Are there any competing theories to a flat cosmological model?

Yes, there are alternative theories to a flat cosmological model, such as the closed universe model, in which the universe has a positive curvature, and the open universe model, in which the universe has a negative curvature. These theories are based on different assumptions about the overall geometry of the universe and have been explored and tested by cosmologists.

5. What are the implications of a flat cosmological model?

A flat cosmological model has significant implications for our understanding of the universe. It suggests that the universe is infinite in size and will continue to expand forever. It also supports the idea of dark energy, a mysterious force that is thought to be driving the acceleration of the expansion of the universe. Additionally, a flat universe has implications for the ultimate fate of the universe, as it may continue to expand indefinitely rather than collapsing in a "Big Crunch".

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