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We know that this density results in a flat universe. $$\rho_c=\frac{3 H^2}{8 \pi G}$$ And we know that if the universe isn't flat, the density as a proportion of critical density can be expressed as this: $$\frac{\rho}{\rho_c}=\frac{8 \pi G\rho}{3 H^2}$$ The FRW metric tells us that that spacetime is shaped like this (I'm ignoring the flat and hyperbolic versions of the formula here): $$ds^2=(ict)^2dt^2+a(t)^2\left(dr^2+\left(\frac{sin(r\sqrt{k})}{\sqrt{k}}\right)^2\left(d\theta^2+sin^2\theta\space d\phi^2 \right) \right)$$ I've seen many, many articles talk about how something less than or greater than critical density causes spacetime to be curved, but I've never seen an article put it all together. If $\frac{\rho}{\rho_c}$ is less than 1 (that is, a closed universe), then what value do I have for $k$? How is $k$ related to the sum total of all the mass/energy components: $\rho=\rho_c\left(\Omega_M+\Omega_\gamma+\Omega_\lambda\right)$?

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This is just the first Friedmann equation (including the cosmological constant within $\rho$):

$$H^2 + \frac{k}{a^2} = \frac{8\pi G \rho}{3},$$

which you can rearrange and evaluate at the present time (so $a=1$) to get

$$k = H^2 \left( \frac{\rho}{\rho_c} - 1 \right) .$$

This gives you the present curvature as a function of the quantities you have. Note that this is not just $k=1$; it has dimensions of inverse length squared, so it is the actual curvature of space, equal to the inverse of the squared radius of curvature of the 3-sphere.

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  • $\begingroup$ I don't understand why you cosmologists use 1 instead of c. The previous equation doesn't produce something with inverse length squared. This equation does: $$H^2 + \frac{kc^2}{a^2} = \frac{8\pi G \rho}{3}$$ $$kc^2 = \frac{8\pi G \rho}{3}-H^2$$ $$\frac{kc^2}{H^2} = \frac{8\pi G \rho}{3H^2}-\frac{H^2}{H^2}$$ $$kc^2 = H^2\left(\frac{\rho}{\rho_c}-1\right)$$ $$k = \frac{H^2}{c^2}\left(\frac{\rho}{\rho_c}-1\right)$$But thank you for sending me in the right direction. $\endgroup$
    – Gluon Soup
    Jun 21 '20 at 17:50
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    $\begingroup$ @GluonSoup we set $c=1$ because it's convenient, because we know how to convert back to normal units, but fundamentally because the way it simplifies equations reflects fundamental symmetries of the laws of physics, namely the unification of space and time. We find that this has conceptual and practical advantages. $\endgroup$
    – Javier
    Jun 21 '20 at 18:10

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