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We say that the universe has a particular shape (flat, sphere or saddle) depending on what the ratio between the average energy density and the critical energy density is, something we call $\Omega$.

But as dark energy starts to dominate going towards the future, will this also affect the shape of the universe? Suppose that "right now" we observe a flat universe with $\Omega=1$. If that changes even slightly because dark energy starts to dominate, would the Universe's shape as we observe it suddenly "snap" to a saddle or sphere shape?

In other words, as the ratio between the dark energy, dark matter and normal matter contributions change (with dark matter starting to dominate its contribution to the energy density and the other two becoming irrelevant), shouldn't this also change the overall energy density, thus changing the value of $\Omega$?

From what I understand, the proposed solution is that yes, the ratios will change but the shape will remain the same as $\Omega$ itself remains the same. So, instead of our currently observed 70/25/5 dark energy/dark matter/normal matter contributions to the energy density we will have something like 95/4/1 in the future, but $\Omega$ will remain 1.

What I don't understand is why would $\Omega$ remain 1? Aren't the contributions different depending on who is contributing? I am imagining that the dark energy contribution "weight" is different (it is of a different nature, it's not structured like normal matter, for example) than dark matter or normal matter.

What's the mistake in my thinking?

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The sign of the spatial curvature of the Universe is a constant. While current observations are consistent with zero spatial curvature, if there is some small but non-zero amount of curvature, dark energy will actually make that curvature very difficult to see.

This follows from the scaling of different kinds of energy density with the scale factor, $a$. For a perfect fluid with an equation of state relating the pressure $p$ and energy density $\rho$ given by $p=w \rho$, we have that the energy density scales as \begin{equation} \frac{\Omega_w(a)}{\Omega_{w,0}} = a^{-3(1+w)} \end{equation} where $\Omega_w(a)$ is the fractional energy density in the Universe in the component with equation of state parameter $w$ when the scale factor is $a$, and $\Omega_{w,0}$ is the fraction of energy density in that component today (where by convention we take $a=1$).

Spatial curvature effectively behaves like a perfect fluid with $w=-\frac{1}{3}$. Meanwhile, all observations to date show the equation of state parameter of dark energy is very close to $w=-1$. Then the ratio of the energy density in curvature ($w=-\frac{1}{3}$), to the ratio of energy density in dark energy ($w=-1$), as a function of redshift, is \begin{equation} \frac{\Omega_{\rm curvature}(a)}{\Omega_{\rm D.E.}(a)} = a^{-2} \end{equation} This ratio will get smaller and smaller as the Universe expands, meaning it will get harder and harder to detect the spatial curvature.

In this answer I've ignored the other components of the Universe; matter (with $w=0$) and radiation ($w=1/3$). Both of these have a larger $w$ than spatial curvature; flipping this argument around, in the earlier Universe when matter and radiation where the dominant components of the Universe, spatial curvature was also suppressed.

So, if it is there, spatial curvature will be difficult to find. The best hope (if it is there at all) is probably precision measurements of the Cosmic Microwave Background radiation or of the Baryon Acoustic Oscillations, which is currently where the tightest bounds on the spatial curvature come from.

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  • $\begingroup$ "The spatial curvature of the Universe is a constant." I assume you mean uniform throughout space - not constant in time. $\endgroup$
    – D. Halsey
    Aug 24, 2021 at 21:25
  • $\begingroup$ @D.Halsey I mean constant in time; the OP's question included the statement "Suppose that "right now" we observe a flat universe with Ω=1", what I meant was that the sign of $\Omega-1$ doesn't change with time (or if $\Omega=1$ it's always equal to 1). I'll add a note to clarify. $\endgroup$
    – Andrew
    Aug 24, 2021 at 22:03
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    $\begingroup$ That's more clear to me now. $\endgroup$
    – D. Halsey
    Aug 24, 2021 at 23:51

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