# I Difference between omega 0 and omega k

1. Sep 2, 2016

### 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?

2. Sep 2, 2016

3. Sep 2, 2016

### George Jones

Staff Emeritus
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.

4. Sep 4, 2016

### windy miller

Interestingly in the
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?

5. Sep 4, 2016

### George Jones

Staff Emeritus
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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