Does the Equation 1=Ω_k+Ω_m+Ω_Λ Hold for All Universe Types?

[Mysteryciel]

I was searching a cosmology calculator and In one calculator I saw this equation.
$1=Ω_k+Ω_m+Ω_Λ$
is that true true for all type universe's (open,flat,closed) ?
and is here $Ω_k={-k\over H^2_0 /a^2_0}$ isn't it.
(And If you can give me an article about this issue I will be very happy,I need exact proof)
Thanks

 

[bapowell]

It's true for globally isotropic universes with matter and cosmological constant. For proof, consult any introductory cosmology text.

 

[Chalnoth]

I was searching a cosmology calculator and In one calculator I saw this equation.
$1=Ω_k+Ω_m+Ω_Λ$
is that true true for all type universe's (open,flat,closed) ?
and is here $Ω_k={-k\over H^2_0 /a^2_0}$ isn't it.
(And If you can give me an article about this issue I will be very happy,I need exact proof)
Thanks

That equation is part of the definition of the $\Omega$ parameters. It's true as long as there is no other component of the energy density (e.g. it's true as long as the radiation density is small enough to be ignored). If you have some other component, such as $\Omega_r$ for radiation, then you have to add that to the equation:

1 = \Omega_k + \Omega_m + \Omega_\Lambda + \Omega_r

And yes, that equation for $\Omega_k$ is correct.

 

[mathman]

Sum = 1 implies a flat universe. > 1 implies closed, < 1 implies open.

 

[Garth]

Sum = 1 implies a flat universe. > 1 implies closed, < 1 implies open.

Not when \Omega_k is included, as Chalnoth said the sum is unity; \Omega_k absorbs the k term.

\Omega_k > 0 implies open, \Omega_k = 0 implies flat and \Omega_k< 0 implies closed, (see the OP definition of \Omega_k).

Garth