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Friedmann equations for critical density is:

$$\rho_c = \frac{3H^2}{8\pi G}$$

Is there any other way to write this equation? For example:

$$\rho_c = \frac{3}{8\pi GH^2}$$

I saw the above form on another website, and was wondering if it was right?

I think he used Hubble time instead of Hubble's parameter.

Does this work? Even though it isn't like the original equation of Critical Density?

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  • $\begingroup$ Comment to the question (v2): OP's second equation doesn't seem to accurately reproduce the equation mentioned on the linked website. $\endgroup$ – Qmechanic Sep 16 '14 at 12:42
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The web site you link is using the expression:

$$ \rho_c = \frac{3}{8\pi G \theta^2} $$

where $\theta$ is the Hubble time and is equal to $1/H$. So your second equation should be:

$$ \rho_c = \frac{3}{8\pi G \theta^2} = \frac{3}{8\pi G \left(1/H\right)^2} = \frac{3H^2}{8\pi G} $$

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  • $\begingroup$ So it is true that both forms mentioned are correct and will result in the same answer, after unit conversions? $\endgroup$ – Hatmix5 Sep 16 '14 at 9:23
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    $\begingroup$ ...I don't think this should be an answer, John. Then again, the whole question (as you seem to have interpreted it) is so trivial that I don't think it really deserves anything more than a comment (I already left a couple). Anyways, I interpreted the question as: Are there any more (meaningful) ways to express $\rho_c$? This answer does not actually answer that question. $\endgroup$ – Danu Sep 16 '14 at 9:26
  • $\begingroup$ @NickI: yes, it's the same equation written in two different ways. $\endgroup$ – John Rennie Sep 16 '14 at 9:31
  • $\begingroup$ great! @Danu yes the question is now answered. $\endgroup$ – Hatmix5 Sep 16 '14 at 9:35

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