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?
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 &quot;critical density&quot; 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:
$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2+\frac{k}{a^2}&amp;=\frac{8\pi G\rho}{3}+\frac{\Lambda}{3},\\ \frac{\ddot{a}}{a}&amp;=-\frac{4\pi G}{3}\left(\rho+3p\right)+\frac{\Lambda}{3},\end{align}$$
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
$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2&amp;=\frac{8\pi G(\rho+\rho_\Lambda+\rho_k)}{3},\\ \frac{\ddot{a}}{a}&amp;=-\frac{4\pi G}{3}\left[(\rho+\rho_\Lambda+\rho_k)+3(p+p_\Lambda+p_k)\right].\end{align}$$
This form makes it explicit that the sum of all $\rho$-s is indeed the critical density and thus unavoidably, by definition, $\Omega_k=1-\sum\Omega$.