What does it mean to say that the Universe is flat?

Question

The curvature parameter $k$ in the FRW metric can take three possible values $k=0,\pm1$. The case $k=0$ is called a flat Universe. But observationally $k$ can take fractional value say k=0.001. What is the meaning of this? One can redefine $k\to k^\prime=10^3k$ such that $k^\prime=1$. Since the value of $k$is not reparametrization invariant, does a value $k=0.001$ can really ensure a flat universe?

P.S. The existing answers do not explain that if $k$ is always adjustable, how is it meaningful to measure $k$ and comment on the flatness of the Universe from the measured value of $k$.

Answer 1

The FRW metric is

$$ds^2=a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2d\Omega^2\right)-dt^2,$$

where $a(t)$ is the scale factor, and $d\Omega^2$ is the metric of a unit sphere. Consider the transformation $$k'\equiv \frac{k}{|k|}=\pm 1,\quad r'\equiv\sqrt{|k|}r$$ for nonzero $k$, or equivalently $$k\equiv k'|k|,\quad r\equiv\frac{r'}{\sqrt{|k|}}.$$ With these transformations, the metric then becomes after some simplification

$$ds^2=a'^2(t)\left(\frac{dr'^2}{1-k'r'^2}+r'^2d\Omega^2\right)-dt^2,$$

where $$a'(t)\equiv\frac{a(t)}{|k|}.$$

Dropping the primes we get back the original metric, but with $k=\pm1.$

Answer 2

If I get it straight, you want to know what happens for $k\in R$ instead of $k=0,\pm1$.

The FRW metric has the form $$ds^2=a^2(t)\bigg( \dfrac{dr^2}{1-kr^2}+r^2d\Omega^2\bigg)-dt^2$$ As the user aRockStr showed above, you can reduce all the values $k>0$ to a metric of the same form as for k=1. Similarly for $k<0$ you reduce the metric to that of k=-1. So we only look at the case $k=0,\pm1$ because these are the only cases where the results are qualitatively different.

Another way to go at it, is to say that these -$k\in R$- are all permitted solutions by our assumptions -that is, to have a homogeneous and isotropic universe-. This is a lot like checking the function $f(x)=ax^2$ for $a\in R$. You can categorize it's behaviour using $a=0,\pm1$.

Lets get more hands on and rewrite the FRW metric for k>0 as $$ds^2=a^2(t)\bigg( \dfrac{dr^2}{1-(\dfrac{r}{r_0})^2}+r^2d\Omega^2\bigg)-dt^2$$ where $r_0$ is a free parameter. The argument here is that we are not interested in carring two unknowns, $r_0$ and $a(t)$ so we can just make a transformation and carry them together as one. Of course, different values of $r_0$ describes different universes, but they are qualitatively the same.

Now lets see what happens for $k=0$. The FRW metric becomes $$ds^2=a^2(t)\bigg( dr^2+r^2d\Omega^2\bigg)-dt^2$$ Notice that $ds|_E=dr^2+r^2d\Omega^2$ is the Euclidean metric for space. By definition, we call flat universes the universes with this spatial component, this spatial behaviour.

Finally, a third way to go about it, is to notice that for all the values of k, if we look at $r\rightarrow 0$ , we get locally flat universe by taylor expanding $\dfrac{1}{1-kr^2}\approx1+0(r)$. So don't get confused, that for small k we have flat universe.

I hope you find this a satisfactory answer to both your post and your comments. Cheers.

Answer 3

No it doesn't ensure a flat universe but a universe with the smallest of curvature that only precise measurements can allow it's existence