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I am looking for a cosmology calculator that does not have the default that

$$\Omega_k = 1 - (\Omega_r + \Omega_m + \Omega_{\Lambda}).$$

I particularly want to run

$$\Omega_r = \Omega_m = \Omega_{\Lambda} = 0, \ \text{and} \ \Omega_k = -1.$$

Can this be done?

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  • $\begingroup$ I've added MathJax formatting but I'm not a cosmologist so please check it. Should some subscripts have their case modified? Should $\Omega_r$ be $\Omega_{rel}$? $\endgroup$
    – uhoh
    Mar 31 at 1:22
  • $\begingroup$ I also tried to improve the title without changing the content, please check.PS: Welcome to astronomy SE :-) $\endgroup$
    – B--rian
    Mar 31 at 9:24
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    $\begingroup$ What does such a universe mean, physically? A negative value of Ωk means a closed space, but how will you close space if it's empty? $\endgroup$
    – pela
    Mar 31 at 11:24
  • $\begingroup$ Thanks for the formatting. 'r' stands for radiation. I believe this will model a hyperspherical universe. A hypersphere is a surface or boundary. If such a boundary contains the mass of the universe then this should appear as k in FLWR. After all k is energy density which is mass. Surely it must be worth a try? $\endgroup$ Mar 31 at 11:56
  • $\begingroup$ The mass of a hyperspherical universe increases in proportion to the scale factor. $\Omega_k$ is the only term in FLWR that does that. $\endgroup$ Mar 31 at 15:59
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The density parameter of cosmology, $\Omega$, is defined as the ratio of the energy density of all forms of matter vs. the critical density.

The forms of matter may include nonrelativistic matter (dust, $\Omega_m$, itself perhaps a sum of baryonic matter, $\Omega_b$ and dark matter, $\Omega_{\rm DM}$) with equation of state, that is, the ratio of its pressure to its energy density, $w=p/\rho\sim 0$; dark energy, aka., the cosmological constant, $\Omega_\Lambda$, with $w=-1$; radiation, $\Omega_\gamma$ (or maybe $\Omega_r$) with equation of state $w=1/3$; or indeed, anything else. So rather than spelling out all possibilities, let me just denote the sum of it all as $\sum\Omega$.

As I mentioned, every one of these $\Omega$-s is a ratio of the corresponding energy density to the critical density: $\Omega_x=\rho_x/\rho_{\rm crit}$, where $\rho_{\rm crit}=3H^2/8\pi G$ with $H=\dot{a}/a$.

The "critical density" is the density at which the universe is spatially flat, i.e., has no spatial curvature, $k=0$. In other words, when $\sum\Omega=1$ exactly, we have a spatially flat universe.

If $\sum\Omega\ne 1$, the universe has (positive or negative) curvature. This is expressed using $\Omega_k$, which can be thought of as being defined by

$$\Omega_k=1-\sum\Omega.$$

To offer a little more detail:

This $\Omega_k$ behaves formally as the density parameter of a perfect fluid with negative pressure, $w=-1/3$. To see this, it is instructive to look at the Friedmann equations:

$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2+\frac{k}{a^2}&=\frac{8\pi G\rho}{3}+\frac{\Lambda}{3},\\ \frac{\ddot{a}}{a}&=-\frac{4\pi G}{3}\left(\rho+3p\right)+\frac{\Lambda}{3},\end{align}$$

where $a$ is the scale parameter, $\rho$ represents the energy density of everything that is not the cosmological constant or spatial curvature and $p$ is the corresponding pressure. However, we can also move the $k/a^2$ term to the right-hand side of the first Friedmann equation, and then re-express both the $k$ term and $\Lambda$ in the form of effective densities and pressures, namely $\rho_\Lambda=\Lambda/8\pi G$, $p_\Lambda=-\rho_\Lambda$, $\rho_k=3k/8\pi Ga^2$, $p_k=-\rho_k/3$, and get

$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2&=\frac{8\pi G(\rho+\rho_\Lambda+\rho_k)}{3},\\ \frac{\ddot{a}}{a}&=-\frac{4\pi G}{3}\left[(\rho+\rho_\Lambda+\rho_k)+3(p+p_\Lambda+p_k)\right].\end{align}$$

This form makes it explicit that the sum of all $\rho$-s is indeed the critical density and thus unavoidably, by definition, $\Omega_k=1-\sum\Omega$.

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    $\begingroup$ Thank you. I understand there are good reasons not to enter $\Omega_k$ = -1 into these equations but are you saying it is impossible? Perhaps the equations would break down. But I would still love to see what happens if the density parameters were entered as in the question. It almost seems trivial to me, but perhaps it isn't. $\endgroup$ Mar 31 at 20:57
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    $\begingroup$ You can of course have $\Omega_k=-1$ but the books need to be balanced: In this case, the remaining $\Omega$-s must add up to +2 because the sum of all $\Omega$-s (including $\Omega_k$) is always 1, by definition. $\endgroup$ Mar 31 at 22:03
  • $\begingroup$ hope you don't mind my linking in What's the largest angle that light has been “seen to bend” by gravity? (of one object by a separate object) $\endgroup$
    – uhoh
    Mar 31 at 23:36
  • $\begingroup$ Is there any chance at all that we could let $\Omega_k$ = -(1 - $\Sigma\Omega$)? $\endgroup$ Apr 1 at 16:21
  • $\begingroup$ No. The very definition of $\Omega_k$ is that it is what's "left over" after all other contributions are accounted for. Whatever you call it, if $1-\sum\Omega_{\rm other~stuff}$ is not zero, it characterizes spatial curvature; and you cannot willy-nilly flip the meaning of positive vs. negative curvature. $\endgroup$ Apr 1 at 19:23

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