# Cosmology calculator where curvature density $\Omega_k$ can be independently set?

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?

• 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 $\Omega_{rel}$?
– uhoh
Mar 31 at 1:22
• I also tried to improve the title without changing the content, please check.PS: Welcome to astronomy SE :-) Mar 31 at 9:24
• 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?
– pela
Mar 31 at 11:24
• 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? Mar 31 at 11:56
• The mass of a hyperspherical universe increases in proportion to the scale factor. $\Omega_k$ is the only term in FLWR that does that. Mar 31 at 15:59

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

• 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. Mar 31 at 20:57
• 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. Mar 31 at 22:03
• 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)
– uhoh
Mar 31 at 23:36
• Is there any chance at all that we could let $\Omega_k$ = -(1 - $\Sigma\Omega$)? Apr 1 at 16:21
• 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. Apr 1 at 19:23