How to see if a cosmological model is flat?

[Philip Land]

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?

 

[PeterDonis]

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.

 

[George Jones]

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
$$dx^2 + dy^2 + dz^2 = dr^2 + r^2 d\theta^2 + r^2 \sin\left(\theta\right)^2 d\phi^2 .$$

 

[Philip Land]

Yes this is exactly what I ment and this was very helpful. Thank you very much.

 

[George Jones]

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.

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.