B Question about calculating the density of the Universe

[tovisonnenberg]

TL;DR Summary

I came across a website whose equation for determining the density of the universe produced a strange result.

Hello! I was reading up on methods for determining the density of the universe and I came across this page: https://hypertextbook.com/facts/2000/ChristinaCheng.shtml

I tried using equation stated, Ω=(2/3Λ)(c^2/H^2), with SI unit versions of both variables:
Λ=1.1056 * 10^-52 m^-2
H=2.1927 * 10^-18 s^-1
And I got an obviously false result: (~10^104). Am I doing something wrong, or is this equation incorrect?

 

[Bandersnatch]

You're doing an algebra error, possibly stemming from misreading the lambda as being in the denominator. That's why the 10^-52s add up instead of cancelling out.

Having said that, it looks wrong. The density should evaluate to 1, but this doesn't. Likely because what's written there is just twice the dark energy density. And why does this equal twice the deceleration parameter? The sign doesn't even match.

 

[Arman777]

The correct equations should be

$$q_0 = \frac{1}{2}\Omega_{m,0} - \Omega_{\Lambda,0}$$

To calculate the density of the universe you can just use the $\rho_{crictical,0}$. Because we are living in a flat universe (at least that is what we are assuming). In this case the total density of the universe should be equal to the critical density.

$$1 = \Omega_{tot,0} \equiv \frac{\rho_{tot,0}}{\rho_{cric,0}} $$So $$\text{Total density of the universe today} \equiv \frac{3H_0^2}{8\pi G} = 8.60 \times 10^{-27}kg/m^3$$