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I am confused about the age of the Universe.

If you calculate it by $1 / H_0$, won't the answer be roughly the same today as it will be 30 billion years from now?

I know $H_0$ is a parameter, not a constant, but it doesn't change that much, does it?

And if the expansion is accelerating then $H_0$ is going up, implying the age of the universe $1 / H_0 $ was higher in the past than it is today.

MikeHelland
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    Possible duplicates: http://physics.stackexchange.com/q/69050/ and http://physics.stackexchange.com/q/10400/ – Kyle Kanos Sep 25 '13 at 19:29
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    As I said in the message I know Hubble's parameter is not constant. At the moment it is getting larger, right? Because expansion is accelerating. That would imply that in the future the age of the universe (1 / H_0) is smaller? – MikeHelland Sep 25 '13 at 19:36
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    Have you read my answer here, and other answers to related questions? It seems like that explains what you want to know... – David Z Sep 25 '13 at 19:40
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    It seems to say as the universe gets older, H_0 gets smaller. But H_0 is getting bigger according to the acceleration of the expansion. It seems to me like you are defining H_0 in terms of the age of the universe and then using it to compute the age of the universe. – MikeHelland Sep 25 '13 at 19:47
  • $H_0$ is an empirically determined value. – Kyle Kanos Sep 25 '13 at 19:56
  • And it seems to be smaller in the past, meaning the age of the universe was larger in the past. Surely that has been addressed in a more convincing manner. – MikeHelland Sep 25 '13 at 20:00
  • "But H_0 is getting bigger according to the acceleration of the expansion." Where did you hear that? Perhaps we can explain it if we know what the context was. – David Z Sep 25 '13 at 21:28
  • 30 billion years from now the galaxies are still going to have a redshift consistent with a recessional velocity of $H_0 * D $ . That is how $H_0$ is an empirically determined value, as Kyle says. Is $H_0$ expected to be roughly a third its present value when determined empirically 30 billion (13.8 B x 2 = 27.6 billion) years from now? – MikeHelland Sep 25 '13 at 21:46
  • David, http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/100528a.html – MikeHelland Sep 25 '13 at 22:15
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    The accelerated expansion implies that $\dot{a}$ increases (the derivative of the scale factor). But the Hubble parameter is defined as $H = \dot{a}/a$, which decreases. The age of the universe is calculated here: http://physics.stackexchange.com/a/63673/24142 and here: http://physics.stackexchange.com/a/69052/24142 – Pulsar Sep 27 '13 at 21:44
  • If the a with the dot over it increased, wouldn't H increase too? 28 billion years from now is H really supposed to be 1/3rd its current value? If not, 1 / H won't predict 44 billion years. – MikeHelland Sep 30 '13 at 00:38
  • The Hubble parameter is governed by the Friedmann equation (particularly the first one), which allows you to find the value of $H$ given the energy density and curvature of the universe. You can neglect the curvature term and the radiation term unless you go very early in the universe. So you really just need to keep in mind that the matter density and dark energy density scale differently (as $a^{-3}$ and $a^0$ respectively). – Michael Oct 24 '13 at 00:45

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