# Why is the flatness problem called the "flatness" problem? What is its connection to geometry?

My understanding of the flatness problem is that it says that if we leave out dark energy and inflation, then the density parameter $$\Omega(t)$$ tends to $$\infty$$ or $$0$$ unless we have $$\Omega(t) = 1$$ exactly. Thus, $$\Omega(t) = 1$$ is an unstable equilibrium point, making it very strange to observe $$\Omega(t_{0})\approx 1$$ today.

My question is, why is this called the "flatness problem?" I don't see the connection to geometry or curvature.

I understand that if $$\Omega(t_{0})$$ is close to $$1$$, then $$\Omega_{K}(t_{0})\equiv 1-\Omega(t_{0})$$ would be close to zero, but how does this relate to the actual curvature value $$K$$? In particular, isn't $$K$$ supposed to be constant (so the deviation from flatness is fixed)?

• $K$, as it is usually defined, is not actually the curvature, but rather a dimensionless rescaled quantity that denotes the sign of the curvature. The actual curvature (just the Ricci scalar $R$ of the spatial metric in a homogeneous isotropic spacetime) has units of inverse length squared. Thus, expansion of the universe will change the value of $R$, with inflation, for example, driving it to be nearly zero.
– Buzz
Commented Apr 23, 2022 at 23:58

Relation between curvature $$k$$ and density parameter $$\Omega$$ can be described with 1st Friedmann equation.

$$(\frac{\dot{a}}{a})^2 +\frac{kc^2}{a^2} = \frac{ 8\pi G }{3}\rho$$

Define Hubble parameter be $$H = \dot{a}/a$$, and density parameter be $$\Omega = 8\pi G \rho/3H^2$$, then comparison between $$\Omega$$ and 1 has same meaning with comparison between $$k$$ and 0.

$$\frac{kc^2}{a^2 H^2} = \Omega-1$$

It says $$|\Omega-1| \propto 1/\dot{a}^2$$. If there is no inflation, $$\dot{a}^2$$ will decrease, and $$\Omega$$ increases. Our current observations said $$\Omega \simeq 1$$, so density parameter has to be closer to 1 in the early universe stage. It is called flatness problem.

• I like this answer, but I think you meant to write $k$ instead of $k^{2}$ wherever it appears. Commented Apr 25, 2022 at 4:48
• @MaximalIdeal You're right! Thank you for noticing to me. I edit my typo. Commented Apr 25, 2022 at 7:58

As you pointed out, as we go back in time, $$\Omega_{\rm total}-1$$ needs to be very small (i.e., $$|\Omega_{\rm total}-1| \propto 10^{-61}$$)

This situation brings to mind the following question;

Why should the universe have started from such a unique situation?

In other words, why would the universe initially started from $$|\Omega_{\rm total} -1| = 0.00000...000001$$ when it could have taken different $$\Omega_{\rm total}$$ values?

The situation can be viewed from the following perspective;

If the universe took any other value, we wouldn't be here, so it had to take that value (anthropic principle). But physicists don't like the anthropic principle. So the only solution is to come up with the idea that for any initial $$\Omega_{\rm total}$$ value, the universe would end up with $$\Omega_{\rm total} \approx 1$$. The inflation mechanism provides that.

• I don't think it's so much that physicists dislike the anthropic principle. It's more that it's not obvious that we wouldn't be here if $\Omega$ wasn't nearly $1$. If human life could thrive in a universe with $\Omega=.99$, then the anthropic principle doesn't apply here.
– Chris
Commented Apr 25, 2022 at 19:11

If the universe is finite, then as it gets larger its curvature gets smaller. I am confused why this fact is not discussed in the context of "the flatness problem". The idea that the "density parameter has to be closer to 1 in the early universe stage" seems backward.

• There is no evidence that the universe was finite. Curvature is not related to its "size". So the answer is correct. But in any case, this seems more like a question than an answer...
– rfl
Commented Apr 27, 2022 at 17:32
• There is also no evidence that the universe is infinite (and flat). All of the theoretical possibilities are conjectures.
– Buzz
Commented Apr 27, 2022 at 21:38
• The curvature density, Omega-k, is directly related to the curvature radius. Radius of curvature = R_c = (c/H_0) (1/SQRT( |Omega_k| ).
– Buzz
Commented Apr 27, 2022 at 21:43
• Source of equation: arxiv.org/pdf/1911.02087v1.pdf paragraph beginning on page 2.
– Buzz
Commented Apr 27, 2022 at 22:00