How does the symbol Λ cosmological constant relate to Ω_Λ or ρ_Λ?

[Buzz Bloom]

TL;DR Summary

From sources (see body below) I recently came across the symbol Λ called the "cosmological constant" with a value 10^-53 m^-2. The cosmological constant which I am familiar with is Ω_Λ with approximate value 0.7 (with no units). I was unable to find in these sources a clear explanation about how these two different concepts called "cosmological constant" are related to each other.

My references are:
https://en.wikipedia.org/wiki/Friedmann_equations#Detailed_derivation
https://en.m.wikipedia.org/wiki/Cosmological_constant

Ω_Λ is a term in the Friedmann equation along with terms for radiation, mass, and curvature.
Λ is the coefficient of the term g_μν in the Einstein field equation with right hand side (8 π G / c^2) T_μν.

I would appreciate someone explaining to me how these two different concepts both called "cosmological constant" are related to each other physically.

The above is a math relationship, but I do not understand the physical implications.

 

[PeterDonis]

I would appreciate someone explaining to me how these two different concepts both called "cosmological constant" are related to each other physically.

They're not two different concepts. They're the same concept expressed in different units. The "math relationship" you refer to is just the conversion between the units. Unfortunately, cosmologists are much less clear about this than they should be.

 

[vela]

From the second link, it says, "Instead of the cosmological constant itself, cosmologists often refer to the ratio between the energy density due to the cosmological constant and the critical density of the universe, the tipping point for a sufficient density to stop the universe from expanding forever. This ratio is usually denoted by Ω_Λ."

 

[LastScattered1090]

@Buzz Bloom PeterDonis is right that these aren't two different concepts, they are just two different ways of expressing the same concept. I don't know if what I'm about to say fully answers your question. But the relationship between the cosmological constant and the dark energy density can be derived from the First Friedmann Equation. This equation describes the relationship between the dynamics of the Universe's expansion, and the density of its mass-energy constituents. In a flat Universe, it is given by:
$$H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho_\mathrm{tot}$$
Here, $a$ is the scale factor, and $\rho_\mathrm{tot}$ is the total energy density of all the constituents making up the "perfect" fluid considered to permeate the Universe. (I think it should actually be $\rho c^2$, but I'm used to using $c=1$ units, and not caring about the distinction between mass density and energy density). So anyway, since I have no curvature term in there (i.e. I'm assuming a flat Universe), it follows also that the total density is equal to the critical density i.e. $\rho_\mathrm{tot} = \rho_\mathrm{crit}$, or $\Omega_\mathrm{tot} = 1$. Substituting in the critical density on the righthand side, we can also derive that, at the present day (where $H = H_0$):
$$\frac{3H_0^2}{8\pi G} = \rho_\mathrm{crit}$$
Let's file that result away for later.

Going back to the First Friedmann equation, we can break up the total density into individual constituents like matter (both baryonic & dark), radiation (photons & relativistic particles), and dark energy: $\rho_\mathrm{tot} = \rho_m + \rho_r + \rho_{de}$. So the Friedmann equation becomes:
$$H^2 = \frac{8\pi G}{3}\left(\rho_m + \rho_r\right)+ \frac{8\pi G}{3}\rho_{de}$$
Now, compare this in your head to a version of the Friedmann equation that only considered a Universe with matter and radiation, but then had the cosmological constant term of $\Lambda/3$ tacked on to the righthand side as a sort of "fudge factor". (I think it's actually $\Lambda c^2/3$. But again, factors of $c^2$ will float around as they do. I care not, for I have set them to unity). The idea here is that if the observed data showing accelerating expansion, which strongly favour non-zero Lambda, can be attributed to some actual substance or field (dubbed "dark energy") permeating space with constant density, then it follows just by comparison of the dark energy term with the Lambda one that:
$$\frac{8\pi G}{3}\rho_{de} = \frac{\Lambda}{3}$$
therefore:
$$\Lambda = 8\pi G \rho_{de}$$
So you can see that $\Lambda$ and $\rho_{de}$ (what you call $\rho_{\Lambda}$) are the same thing, up to a constant of proportionality. And they are both dimensionful constants, it's just that $\Lambda$ absorbs the extra constants of proportionality ($8\pi G$) into the density, giving the cosmo-constant term the same dimensions as the lefthand side: dimensions of "Hubble parameter squared" i.e. 1/time².

We can also relate $\Lambda$ to the dimensionless density parameter for dark energy just by applying the definition of the density parameter for a constituent. It's the ratio of the density of that constituent to the critical density. Here's where we use that result that we filed away. At the present day, this would be given by:
$$\Omega_{\Lambda} \equiv \frac{\rho_{\Lambda}}{\rho_\mathrm{crit}} = \left(\frac{\Lambda}{8\pi G}\right)\left(\frac{8\pi G}{3H_0^2}\right) = \frac{\Lambda}{3H_0^2}$$
This expression for $\Omega_{\Lambda}$ is the same as the one in your OP (up to some factors of $c$ which, again, I don't care about here). And, as we established in my discussion above, $H_0^2$ and $\Lambda$ have the same dimensions, so we know that the units cancel, leaving $\Omega_{\Lambda}$ dimensionless as expected.

Hope that helps, somewhat!

EDIT: note that $\rho_{de}$ may not actually be constant with time. We don't know for sure yet. We need more and better data to be sure, e.g. from cosmology survey telescopes that map out large-scale structure, like WFIRST/Roman and Euclid. If $\rho_{de}$ is constant, then it corresponds to the "Lambda" (cosmo-constant) term in the Friedmann equations. This is the simplest model. But if $\rho_{de}$ varies with time, then it enters into the equation in a scale-factor-dependent way.