Difference between omega 0 and omega k

[windy miller]

As I try and read cosmology literature with the eyes of layman , I find two different terms that seem to represent the curvature ( or lack of) of the universe .
These are omega sub zero and omega k. But I can't see the difference between the two. Can anyone explain this?

 

[mathman]

you may be confusing terms. \Omega is usually to represent the density of the universe, normalized so that = 1 means flat.

https://en.wikipedia.org/wiki/Shape_of_the_universe

 

[George Jones]

you may be confusing terms. \Omega is usually to represent the density of the universe, normalized so that = 1 means flat.

Yes, but $\Omega_k$ (not an actual mass/energy density) often is defined as $\Omega_k = 1 - \Omega$, so that $\Omega_k = 0$ for a spatially flat universe.

I though that $\Omega_0$ usually means $\Omega_0 = \Omega \left(t_0\right)$, i.e., $\Omega_0$ is the value of the total relative density right now, but I could be wrong. Also, different references can use different notations.

 

[windy miller]

Interestingly in the

Yes, but $\Omega_k$ (not an actual mass/energy density) often is defined as $\Omega_k = 1 - \Omega$, so that $\Omega_k = 0$ for a spatially flat universe.

I though that $\Omega_0$ usually means $\Omega_0 = \Omega \left(t_0\right)$, i.e., $\Omega_0$ is the value of the total relative density right now, but I could be wrong. Also, different references can use different notations.

I get the impression that omega k is the curvature so 0 means zero curvature, whereas omega sub zero is the ratio of the critical density to the actual density and so with zero curvature omega sub zero should be 1. Is that correct?

 

[George Jones]

I get the impression that omega k is the curvature so 0 means zero curvature, whereas omega sub zero is the ratio of the critical density to the actual density and so with zero curvature omega sub zero should be 1. Is that correct?

The ratio is the other way around, .i.e., $\Omega$ is the ratio of the actual density to the critical density. If the present time is $t_0$, then $\Omega_0 = \Omega\left(t_0\right)$, and if at time $t_0$ the universe is spatially flat, then, yes, $\Omega_0 = 1$.
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