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Ich

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You'll see that even if a(t) increases exponentially with time, H merely stays constant.

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marcus

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... In other words, when we say the universe is accelerating do we really mean that it is decelerating less than before?

Ich answered perfectly. I will just repeat for emphasis. No, we do not mean what you said. When we say the universe is accelerating, we mean that the expansion is accelerating (an increase in the time rate of increase in the scalefactor)

Our handle on the size of the universe is the scalefactor a(t). What plugs into the Friedman metric and makes distances expand.

The scalefactor is the key thing in the Friedman model. The two Friedman equations govern the evolution of the scalefactor.

(Look up Friedman equations on Wiki)

Do not worry about the Hubble rate. It is just a conventional symbol for the ratio a'/a.

It relates in a convenient way to observations, and is extremely useful, but is not fundamental.

The real thing you must think about is a(t). It is the backbone and the guts of the model.

When people say "the universe is expanding" that means nothing else than that

a'(t) > 0

When people say "the expansion is accelerating" that means nothing else than

a"(t) > 0.

There is no reason to be surprised that H(t) is decreasing. H(t) is merely something defined to be equal a'(t)/a(t) and it is quite normal for a function which is increasing at an increasing rate to have that particular ratio be decreasing. Here is an example:

Let f(t) = t

Then f'(t) = 3t

and f"(t) = 6t which is positive so the increase is accelerating.

But what would be analogous to H, namely f'(t)/f(t) = 3t

is obviously decreasing.

==============

Ranku, thanks for posting this question on the forum. It is constructive. Many people do not understand what "accelerating expansion" means and get confused in exactly the way you did. If there is something about this that you now do not understand, please ask more questions. Keep asking until it is clear.

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What I would like to know is how does it relate to the astronomical observation of the value of the Hubble constant, and so I repeat that part of the question: is the Hubble constant decreasing at a lower rate in an accelerating universe than the rate it was decreasing when the universe was decelerating?

Regarding using the scale factor a(t), what is its unit, how do we get the values for it, how is it measured?

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Ich

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We can measure the current value of the Hubble constant directly, not its time derivatives. So there is no relation to the direct measurement of H0.What I would like to know is how does it relate to the astronomical observation of the value of the Hubble constant, and so I repeat that part of the question: is the Hubble constant decreasing at a lower rate in an accelerating universe than the rate it was decreasing when the universe was decelerating?

H is decreasing wth time, but that has nothing to do with acceleration/deceleration. If a(t) were any power funtion, H~1/t would be always true.

It's usualy taken as a dimensionless variable and defined to be 1 in the present epoch. The exact definition does not matter, all you can measure is ratios like a(t1)/a(t2) or a'(t)/a(t) and such.Regarding using the scale factor a(t), what is its unit, how do we get the values for it, how is it measured?

a(t1)/a(t2) is direcly related to the measured redshift of comoving objects: a(now)/a(time of light emission) = 1+z.

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Ich

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Ok, if you could measure H and H', you can construct the deceleration parameter from the ratio H'/H². That means, if H'/H²>-1, the expansion is accelerating.Is the rate of decrease lower in an accelerating universe than in a decelerating universe?

So, yes, if you define "rate of decrease" as -H'/H², your statement is true.

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Thank you Ich for your answer. Until the next time.

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Chronos

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