<p>The density parameter of cosmology, <span class="math-container">$\Omega$</span>, is defined as the ratio of the energy density of all forms of matter vs. the critical density.</p>
<p>The forms of matter may include nonrelativistic matter (dust, <span class="math-container">$\Omega_m$</span>, itself perhaps a sum of baryonic matter, <span class="math-container">$\Omega_b$</span> and dark matter, <span class="math-container">$\Omega_{\rm DM}$</span>) with equation of state, that is, the ratio of its pressure to its energy density, <span class="math-container">$w=p/\rho\sim 0$</span>; dark energy, aka., the cosmological constant, <span class="math-container">$\Omega_\Lambda$</span>, with <span class="math-container">$w=-1$</span>; radiation, <span class="math-container">$\Omega_\gamma$</span> (or maybe <span class="math-container">$\Omega_r$</span>) with equation of state <span class="math-container">$w=1/3$</span>; or indeed, anything else. So rather than spelling out all possibilities, let me just denote the sum of it all as <span class="math-container">$\sum\Omega$</span>.</p>
<p>As I mentioned, every one of these <span class="math-container">$\Omega$</span>-s is a ratio of the corresponding energy density to the critical density: <span class="math-container">$\Omega_x=\rho_x/\rho_{\rm crit}$</span>, where <span class="math-container">$\rho_{\rm crit}=3H^2/8\pi G$</span> with <span class="math-container">$H=\dot{a}/a$</span>.</p>
<p>The &quot;critical density&quot; is the density at which the universe is spatially flat, i.e., has no spatial curvature, <span class="math-container">$k=0$</span>. In other words, when <span class="math-container">$\sum\Omega=1$</span> exactly, we have a spatially flat universe.</p>
<p>If <span class="math-container">$\sum\Omega\ne 1$</span>, the universe has (positive or negative) curvature. This is expressed using <span class="math-container">$\Omega_k$</span>, which can be thought of as being defined by</p>
<p><span class="math-container">$$\Omega_k=1-\sum\Omega.$$</span></p>
<p>To offer a little more detail:</p>
<p>This <span class="math-container">$\Omega_k$</span> behaves formally as the density parameter of a perfect fluid with negative pressure, <span class="math-container">$w=-1/3$</span>. To see this, it is instructive to look at the Friedmann equations:</p>
<p><span class="math-container">$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2+\frac{k}{a^2}&amp;=\frac{8\pi G\rho}{3}+\frac{\Lambda}{3},\\ \frac{\ddot{a}}{a}&amp;=-\frac{4\pi G}{3}\left(\rho+3p\right)+\frac{\Lambda}{3},\end{align}$$</span></p>
<p>where <span class="math-container">$a$</span> is the scale parameter, <span class="math-container">$\rho$</span> represents the energy density of everything that is not the cosmological constant or spatial curvature and <span class="math-container">$p$</span> is the corresponding pressure. However, we can also move the <span class="math-container">$k/a^2$</span> term to the right-hand side of the first Friedmann equation, and then re-express both the <span class="math-container">$k$</span> term and <span class="math-container">$\Lambda$</span> in the form of effective densities and pressures, namely <span class="math-container">$\rho_\Lambda=\Lambda/8\pi G$</span>, <span class="math-container">$p_\Lambda=-\rho_\Lambda$</span>, <span class="math-container">$\rho_k=3k/8\pi Ga^2$</span>, <span class="math-container">$p_k=-\rho_k/3$</span>, and get</p>
<p><span class="math-container">$$\begin{align}\left(\frac{\dot{a}}{a}\right)^2&amp;=\frac{8\pi G(\rho+\rho_\Lambda+\rho_k)}{3},\\ \frac{\ddot{a}}{a}&amp;=-\frac{4\pi G}{3}\left[(\rho+\rho_\Lambda+\rho_k)+3(p+p_\Lambda+p_k)\right].\end{align}$$</span></p>
<p>This form makes it explicit that the sum of all <span class="math-container">$\rho$</span>-s is indeed the critical density and thus unavoidably, by definition, <span class="math-container">$\Omega_k=1-\sum\Omega$</span>.</p>

CC BY-SA 4.0
2021-03-31T20:57:19.873Z
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.
John Hobson
CC BY-SA 4.0
2021-03-31T22:03:41.807Z
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.
Viktor Toth
CC BY-SA 4.0
2021-03-31T23:36:01.737Z
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)](https://astronomy.stackexchange.com/q/43230/7982)
uhoh
CC BY-SA 4.0
2021-04-01T16:21:32.797Z
Is there any chance at all that we could let $\Omega_k$ = -(1 - $\Sigma\Omega$)?
John Hobson
CC BY-SA 4.0
2021-04-01T19:23:18.193Z
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.
Viktor Toth
CC BY-SA 4.0
2021-04-01T23:27:20.947Z
I understand what you are saying, as far as my limited understanding allows, but is it mathematically possible to solve/run FLWR with my proposed inputs?
John Hobson
CC BY-SA 4.0
2021-04-02T05:08:05.423Z
No. Your $\Omega$-s do not add up to 1. If this is not due to spatial curvature (that is, if you insist on specifying $\Omega_k$) it has to be due to something else not accounted for. The $\Omega$-s must always add up to 1, by definition.
Viktor Toth
CC BY-SA 4.0
2021-04-02T14:28:37.077Z
Thank you, very helpful. So it seems we can have an FLWR universe which is just a saddle and nothing else but not one which is just a sphere. Intuition suggests otherwise but I suppose that is not always a good guide.
John Hobson
CC BY-SA 4.0
2021-04-02T15:02:51.693Z
That much should be obvious from the first Friedmann equation. In the absence of matter (or dark energy), when $\rho=\Lambda=0$, we have the vacuum field equation in the form $(\dot{a}/a)^2+k/a^2=0$, implying strictly that $k\le 0$ since all the terms are real. Thinking about this in terms of $\Omega$-s just obscures the problem.
Viktor Toth
CC BY-SA 4.0
2021-03-31T16:37:51.143Z
2021-03-31T16:37:51.143Z
Viktor Toth
<p>I am looking for a cosmology calculator that does not have the default that</p>
<p><span class="math-container">$$\Omega_k = 1 - (\Omega_r + \Omega_m + \Omega_{\Lambda}).$$</span></p>
<p>I particularly want to run</p>
<p><span class="math-container">$$\Omega_r = \Omega_m = \Omega_{\Lambda} = 0, \ \text{and} \ \Omega_k = -1.$$</span></p>
<p>Can this be done?</p>

CC BY-SA 4.0
2021-03-31T01:22:26.130Z
I've added [MathJax](https://math.meta.stackexchange.com/q/5020/284619) 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
CC BY-SA 4.0
2021-03-31T09:24:04.970Z
I also tried to improve the title without changing the content, please check.PS: Welcome to astronomy SE :-)
B--rian
CC BY-SA 4.0
2021-03-31T11:24:00.033Z
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
CC BY-SA 4.0
2021-03-31T11:56:19.907Z
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?
John Hobson
CC BY-SA 4.0
2021-03-31T15:59:06.317Z
The mass of a hyperspherical universe increases in proportion to the scale factor. $\Omega_k$ is the only term in FLWR that does that.
John Hobson
CC BY-SA 4.0
2021-03-31T16:16:23.087Z
You can of course set the three values to zero and $\Omega_k$ defaults to 1. This then produces the distance vs redshift relationship exactly as predicted from the geometry of a hyperspherical universe, the same as predicted by the standard model up to double digit redshift values. But of course $\Omega_k$ = 1 is a hyperbolic sphere. This could mean the universe is a hyperbolic sphere but that seems unlikely. That is why I would love to try this with $\Omega_k$ = -1.
John Hobson
<cosmology><mathematics>
Cosmology calculator where curvature density $\Omega_k$ can be independently set?
John Hobson
# Cosmology calculator where curvature density $\Omega_k$ can be independently set?

(Asked by: username_0 on 2021-03-31)

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?

**Tags:** cosmology, mathematics

> **Comments:**
> **username_1**: I've added [MathJax](https://math.meta.stackexchange.com/q/5020/284619) formatting but I'm not a cosmologist so please check it. Should some subscripts have their case modified? Should $\Omega_r$ be  $\Omega_{rel}$?  
> **username_2**: I also tried to improve the title without changing the content, please check.PS: Welcome to astronomy SE :-)  
> **username_3**: 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?  
> **username_0**: 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?  
> **username_0**: The mass of a hyperspherical universe increases in proportion to the scale factor. $\Omega_k$ is the only term in FLWR that does that.  
> **username_0**: You can of course set the three values to zero and $\Omega_k$ defaults to 1. This then produces the distance vs redshift relationship exactly as predicted from the geometry of a hyperspherical universe, the same as predicted by the standard model up to double digit redshift values. But of course $\Omega_k$ = 1 is a hyperbolic sphere. This could mean the universe is a hyperbolic sphere but that seems unlikely. That is why I would love to try this with $\Omega_k$ = -1.  

---

## 1 answers

### Answer by username_4 on 2021-03-31. Score: 1

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

> **Comments:**
> **username_0**: 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.  
> **username_4**: 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.  
> **username_1**: 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)](https://astronomy.stackexchange.com/q/43230/7982)  
> **username_0**: Is there any chance at all that we could let $\Omega_k$ = -(1 - $\Sigma\Omega$)?  
> **username_4**: 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.  
> **username_0**: I understand what you are saying, as far as my limited understanding allows, but is it mathematically possible to solve/run FLWR with my proposed inputs?  
> **username_4**: No. Your $\Omega$-s do not add up to 1. If this is not due to spatial curvature (that is, if you insist on specifying $\Omega_k$) it has to be due to something else not accounted for. The $\Omega$-s must always add up to 1, by definition.  
> **username_0**: Thank you, very helpful. So it seems we can have an FLWR universe which is just a saddle and nothing else but not one which is just a sphere. Intuition suggests otherwise but I suppose that is not always a good guide.  
> **username_4**: That much should be obvious from the first Friedmann equation. In the absence of matter (or dark energy), when $\rho=\Lambda=0$, we have the vacuum field equation in the form $(\dot{a}/a)^2+k/a^2=0$, implying strictly that $k\le 0$ since all the terms are real. Thinking about this in terms of $\Omega$-s just obscures the problem.  

---

