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So we have one of Friedmann's equation:

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

Using This website, resources where gathered for specific times in the universe. The resources being the Hubble constant at the specific times (i.e. 3.38By). The critical density was worked out for the specific times using the Hubble constant. The critical density was in Kg/m-3 units and was converted into amu/m-3 units, by timing the value by 6.02214129 × 10^26. I then graphed the values:

enter image description here

I could then conclude from this data that (equation of the line): $$\rho_c = E H^2$$ $$H^2= \frac{\rho_c }{E}$$ Where E = 1081.6, isn't this a much simpler equation to work out the critical density or Hubble constant? Not saying one is better than the other, but just an idea. For further investigation i will use the equation to find the Hubble constant at 13.79By where the critical density is 5.2 amu/m-3: $$H^2= \frac{\rho_c }{E}$$ $$H^2= \frac{5.216128}{1081.6}$$ $$H= \sqrt{0.00482}$$ $$H= 0.069445$$ And we convert this from s-1 mpc to km/s mpc, through google, or times by 977.6 (google is better though). The final awnser we get is: H = 67.9km/s mpc. And if we refer to the Planck Mission where they observed the Hubble constant to be within 67.13 and 68.57km/s mpc, which corresponds to the answer we have through the new equation. Does this 'new' equation have any relevance in physics at all?

Selene Routley
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Hatmix5
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1 Answers1

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If we start with the equation you quote:

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

and rewrite it as:

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

then it's the same as your equation:

$$ \rho_c = E H^2 $$

because all you've done is to replace the constant factor of $3/8\pi G$ with the symbol $E$. This is of couse a perfectly reasonable thing to do, but it isn't new in any sense.

The reason we prefer the original form is that $G$ is an important constant of nature - it's Newton's constant. It's the same constant as used to calculate the (non-relativistic) gravitational force between two masses:

$$ F = \frac{Gm_1m_2}{r^2} $$

John Rennie
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  • @NickI: you posted then deleted a question about calculating the Hubble time. I had almost finished writing an answer when you deleted your question, so rather than waste the answer I've posted it here. – John Rennie Sep 18 '14 at 15:23
  • Sorry. I got a -1 on the question, so i was guessing i was wrong so i deleted it to save my account reputation you see. Sorry about that! – Hatmix5 Sep 19 '14 at 11:52
  • Also is it fine if i post the question again under different contexts, still have a question about your working with Hubble parameter. – Hatmix5 Sep 19 '14 at 11:58
  • hopefully you get these notifications, the @ sign doesn't appear when the comment is posted. – Hatmix5 Sep 19 '14 at 12:07
  • @NickI: why not add a comment to my question explaining what you would like clarified. I might be able to extend my answer to address it. – John Rennie Sep 19 '14 at 13:56
  • Okay, i will have to wait until i get 50 reputation though. Also did you think the question that i deleted was worth the -1 vote? Should i worry about them when asking questions? – Hatmix5 Sep 19 '14 at 22:34