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View Full Version : Does Hubble's law implies accelerated expansion?


LedPhoton
Oct27-09, 03:31 PM
I`ve read that the universes expansion is accelerating. To explain this we have introduced concept of Dark matter and energy.
However, Hubble's law: v = H*d means that a speed that separates two points is proportional to their distance. But since they are moving apart the distance becomes greater and so the speed must become greater. I think this means the speed of expansion is increasing. So the expansion of the universe is accelerating.
I`m sure I`ve got something wrong here, is the acceleration we explain with Dark energy a "surplus"?
Hope I made myself clear.

ZikZak
Oct27-09, 04:01 PM
No, you are confusing yourself by thinking of the rate of expansion as a speed. It's not. It's a percentage per unit time, approximately H = 0.007% per million years.

S.Vasojevic
Oct27-09, 04:08 PM
No. It is just saying that universe is expanding. It says nothing about universes acceleration or deacceleration of expansion throughout time.

Naty1
Oct27-09, 04:33 PM
There are a number of subtlies underlying the answer to your question.

Observations and measurements in 1998 surprised most scientists, including those doing the work, finding the expansion of the universe is currently accelerating , although the Hubble factor is still decreasing over time.

The Hubble "constant" is not really constant varing with time according to your choice of cosmological model.

Try as a starter: http://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=2413312
and New Wrights tutorial and FAQ: http://www.astro.ucla.edu/~wright/cosmology_faq.html

Ich
Oct28-09, 04:10 AM
But since they are moving apart the distance becomes greater and so the speed must become greater. I think this means the speed of expansion is increasing.
Exactly. What you describe here is the case of a constant Hubble parameter, which is equivalent to exponential acceleration.
In a non-accelerating universe the Hubble parameter will be proportinal to 1/t. Distance d = v*t, v=const -> H = d/v ~ 1/t.