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One of the Friedman equations is given by $$ H^2\equiv\Big(\frac{\dot{a}}{a}\Big)^2=\frac{8\pi G}{3}\rho-\frac{k}{a^2},\tag{1} $$ which fixes the evolution of $a$ (with given $\rho(t)$) if values for $k$ and $\rho(t_0)$ are given. With $\Omega(t) = \rho(t) \frac{3}{8\pi G H^2(t)}$ this can be expressed as $$|1-\Omega(t)| =\frac{|k|}{(aH)^2}.\tag{2}$$ For matter and radiation we have $$|1-\Omega(t)| \propto t^n,$$ with $n \geq 2/3 $. The quantity decreases to zero for $t \rightarrow 0$, which means that if we today measure $\Omega(t_0) \approx 1$, $|1-\Omega(t)|$ was super tiny for small $t$. This is stated as a fine-tuning problem known as the Flatness-Problem.

However I don't quite see why this poses a problem after all.

  1. No matter which value $\Omega(t_0)$ we would measure, it would be arbitrary close to 1 for small enough $t$

  2. $\rho(t_0)$ and $k$ can be set independently. So increasing the density a little bit doesn't have any influence on the value of $\Omega$ at all (it just changes $H$). So the statement "the matter density has to be precisely fine-tuned to make the universe (that) flat as we measure it today" doesn't make any sense, since matter doesn't affect the spatial curvature

  3. The quantity $1-\Omega(t)$ seems to be arbitrary. Couldn't I also arbitrarily define $b(t) = k e^{-1/t}$? Now $b$ needs to even smaller than $\Omega$ for small $t$ if we measure $k$ to be small nowdays. Creates this then a more severe flatness problem?

So in conclusion: I don't see any problem as long as direct properties of the universe like $a(t)$ do not depend on the initial conditions $\rho(t_0)$ and $k$ in an unstable way.

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  1. No matter which value $\Omega(t_0)$ we would measure, it would be arbitrary close to 1 for small enough $t$

We can't make $t$ arbitrarily small. General relativity breaks down at the Planck scale. When people state the flatness problem, they usually quantify it by talking about fine-tuning at the Planck time in order to produce the currently observed level of flatness.

  1. $\rho(t_0)$ and $k$ can be set independently. So increasing the density a little bit doesn't have any influence on the value of $\Omega$ at all (it just changes $H$). So the statement "the matter density has to be precisely fine-tuned to make the universe (that) flat as we measure it today" doesn't make any sense, since matter doesn't affect the spatial curvature

Matter certainly affects the spatial curvature. The content of the Einstein field equations is precisely that matter causes curvature. In an FLRW spacetime, in the usual coordinates, every nonvanishing component in the Einstein tensor has a term proportional to $k/a^2$, and the Einstein field equations relate this to $\rho$.

It's true that in the equations you can change $\rho$ without changing $\Omega$, provided that you adjust $H$. But we don't have the freedom to adjust $H$. These variables all have specific, observed values.

  1. The quantity $1-\Omega(t)$ seems to be arbitrary. Couldn't I also arbitrarily define $b(t) = k e^{-1/t}$? Now $b$ needs to even smaller than $\Omega$ for small $t$ if we measure $k$ to be small nowdays. Creates this then a more severe flatness problem?

This is a reasonable objection, and others have made it. It's equivalent to complaining that we have no prior probability distribution for the cosmological parameters, since we don't know of any natural principle that should have dictated the parameters. This is a reasonable basis on which to consider the flatness problem not to be a problem at all, and I tend to share this point of view.

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It is easier to think about the flatness problem in the other direction: from the past to the present. First, we need to introduce some quantities: $$E \equiv \frac{H}{H_0},\qquad \bar{\Omega} \equiv \Omega E^2, \qquad \bar{\Omega}_k \equiv \Omega_k E^2, \qquad \Omega_k \equiv -\frac{k}{a^2H^2},$$ where $H_0$ refers to the the Hubble constant today, i.e., $H_0 \equiv H(a_0)$ and $a_0$ representing the scale factor today. In these variables $\bar\Omega + \bar\Omega_k = E^2$, $\bar\Omega(a_0) = \Omega(a_0)$ and $\bar\Omega_k(a_0) = \Omega_k(a_0)$.

It is easy to see that $\Omega \propto \rho$, and therefore, for a fluid with arbitrary constant equation of state $w$, $$\bar\Omega_w = \Omega_{w0}\left(\frac{a_0}{a}\right)^{3(1+w)},\qquad \Omega_{w0} \equiv \Omega_w(a_0).$$ If the only component of the universe is the $w$ fluid, then, $\Omega = \Omega_w$. To simplify it further, lets choose a dust like fluid $\Omega_m$ with $w = 0$ (however, the argument bellow is valid for any combination of components with equations of state larger than $-1/3$).

Now, lets say that initially $\vert\Omega-1\vert$ is very close to one, that is, for an initial value of the scale factor $a_i$, we have $$\Omega_{mi} \equiv \Omega_m(a_i) = 1-10^{-5}, \qquad \Omega_{ki} \equiv \Omega_k(a_i) = 10^{-5}.$$ In terms of these initial densities we have, $$\Omega_m = \frac{\Omega_{mi}(a_i/a)^{3}}{\Omega_{mi}(a_i/a)^{3} + \Omega_{ki}(a_i/a)^{2}} = \frac{1}{1+\alpha a / a_i}, \quad \Omega_k = \frac{\Omega_{ki}(a_i/a)^{2}}{\Omega_{mi}(a_i/a)^{3} + \Omega_{ki}(a_i/a)^{2}} = \frac{\alpha a / a_i}{1+\alpha a / a_i},$$ where we defined $\alpha = \Omega_{ki} / \Omega_{mi}$. Putting these initial conditions around the nucleosynthesis epoch, we have $a_i \approx 10^{-10}a_0$ and consequently $$\Omega_{m0} \approx \frac{1}{1+\alpha 10^{10}} \approx 10^{-5}, \qquad \Omega_{k0} \approx \frac{\alpha 10^{10}}{1+\alpha 10^{10}} \approx 1.$$ The situation is completely reversed, in this case today we would have an universe completely dominated by curvature. To avoid this, we would need $\alpha \ll 10^{-10}$, i.e. $\Omega_{ki} \ll 10^{-10}$.

Now, we can answer your questions:

  1. No matter which value Ω(t0) we would measure, it would be arbitrary close to 1 for small enough t

That's true as long as the matter content of the universe has an equation of state larger than $-1/3$. However, that's exactly the problem, since the matter content tends to dominate in the past, the ratio $\Omega_k/\Omega$ is led dynamically to zero, even if $\vert\Omega_{k0}\vert > \vert\Omega_0\vert$ today. This means that if we want to put initial conditions in the past we need to fine-tune $\Omega_{ki}$ to a very small value in order to obtain the observed value today $\Omega_{k0} \approx 0$ (in the example above, to get $\Omega_{k0} \approx 10^{-5}$ we need $\Omega_{ki} \approx 10^{-15}$!).

Nevertheless, this is not a problem per se. What happens is that we have an implicit expectation that the initial conditions should be natural/common (maybe an implicit use of the Mediocrity principle) and the dynamics above shows that only a very narrow interval of the possible initial conditions for $\Omega_{ki}$ could explain the current state of the universe.

Finally, this has nothing to do with Planck time or quantum gravity, that problem takes place even if you choose to put initial conditions around the nucleosynthesis epoch.

  1. $\rho(t0)$ and $k$ can be set independently. So increasing the density a little bit doesn't have any influence on the value of $\Omega$ at all (it just changes $H$). So the statement "the matter density has to be precisely fine-tuned to make the universe (that) flat as we measure it today" doesn't make any sense, since matter doesn't affect the spatial curvature

That's not true. You are right to say that if one varies $\rho(t0)$ and $k$ independently, a shift in the value of $\rho$ would modify the value of $H$ (due to Friedmann equation, which is actually a constrain equation and not a dynamical one). This certainly would affect $\Omega$, to see this we can write the Friedmann equation in terms of barred densities: $$E^2 = \bar{\Omega} + \bar\Omega_k,$$ remembering that $\bar\Omega \propto \rho$. Thus, a shift in $\rho$ is equivalent to a shift in $\bar\Omega$, then if our new density is given by $\bar\Omega \to \bar\Omega+\Delta$, then $E^2 \to E^2 + \Delta$ and consequently, $$\Omega = \frac{\bar\Omega}{E^2} \to \frac{\bar\Omega+\Delta}{E^2+\Delta}, \qquad \vert1-\Omega\vert \to \left\vert\frac{1-\Omega}{1+\Delta/E^2}\right\vert.$$

The quantity $1−\Omega(t)$ seems to be arbitrary. Couldn't I also arbitrarily define $b(t)=ke^{−1/t}$? Now $b$ needs to even smaller than $\Omega$ for small $t$ if we measure $k$ to be small nowdays. Creates this then a more severe flatness problem?

This quantity is not arbitrary, it is a direct consequence of the Friedmann equation written in terms of $\Omega$ and $\Omega_k$, i.e., $\Omega + \Omega_k = 1$ (yes, this is the Friedmann equation). This expression gives the exact relation between the energy content of the universe and the absolute value of the spatial curvature.

So in conclusion: I don't see any problem as long as direct properties of the universe like $a(t)$ do not depend on the initial conditions $\rho(t0)$ and $k$ in an unstable way.

Yes, that's actually one way to frame the problem. The flat solution is unstable, any small departure from an exact flat solution leads to a curvature dominated evolution at some point in the future (unless you have a matter component with $w<-1/3$ in which case it would dominates instead. The most famous example of this would be the cosmological constant).

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