Does the Hubble Parameter Go to Zero as the Universe Expands Forever?

[palmer eldtrich]

I recently came across this claim:
"On the other hand, in certain cases even an expanding cosmology may have 0 average expansion, due to the fact that we are averaging over an infinite amount of time. "
Is this correct? can someone explain how averaging over infinity time leads to zero ?
http://www.wall.org/~aron/blog/did-the-universe-begin-iii-bgv-theorem/

 

[palmer eldtrich]

I get that when you divide by inifnity you get zero, but would the volume of space also be infinity?

 

[Chalnoth]

"On the other hand, in certain cases even an expanding cosmology may have 0 average expansion, due to the fact that we are averaging over an infinite amount of time."

I'm pretty sure this statement is utterly incoherent. You're right to be confused. It just doesn't make sense.

 

[DrStupid]

"On the other hand, in certain cases even an expanding cosmology may have 0 average expansion, due to the fact that we are averaging over an infinite amount of time."

Maybe it means something like this:

In an Einstein–de Sitter universe the scale factor has the time dependence

a = a_0 \cdot \sqrt[3]{{\frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot t^2 }}{3}}}

That results in

\mathop {\lim }\limits_{t \to \infty } \frac{a}{t} = 0

 

[Chalnoth]

Maybe it means something like this:

In an Einstein–de Sitter universe the scale factor has the time dependence

a = a_0 \cdot \sqrt[3]{{\frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot t^2 }}{3}}}

That results in

\mathop {\lim }\limits_{t \to \infty } \frac{a}{t} = 0

That's not an average of the expansion rate, however.

Also, this requires perfect flatness, which is essentially impossible.

 

[DrStupid]

Also, this requires perfect flatness, which is essentially impossible.

The average of the corresponding expansion rate

H = \frac{1}{t}

is

\mathop {\lim }\limits_{t \to \infty } \frac{{\int\limits_{t_0 }^t {H \cdot dt} }}{{t - t_0 }} = 0

 

[Chalnoth]

The average of the corresponding expansion rate

H = \frac{1}{t}

is

\mathop {\lim }\limits_{t \to \infty } \frac{{\int\limits_{t_0 }^t {H \cdot dt} }}{{t - t_0 }} = 0

Fair enough, provided you don't consider t=0 (which, I suppose, is pretty reasonable). But still, put any cosmological constant or positive curvature into this equation and it changes dramatically.

 

[DrStupid]

But still, put any cosmological constant or positive curvature into this equation and it changes dramatically.

That's correct, but it doesn't contradict the statement that "in certain cases even an expanding cosmology may have 0 average expansion".

 

[Chalnoth]

Still not a helpful statement, though. A universe that expands forever also has an average matter/radiation density of zero. So in effect the reason this occurs is because you're averaging over an infinity of time where the universe is empty.

 

[palmer eldtrich]

So assuming the universe expands forever then, would you be correct to say that the HUbble parameter does go to zero?

 

[marcus]

So assuming the universe expands forever then, would you be correct to say that the HUbble parameter does go to zero?

That's what the Friedmann equation says will happen in the spatial flat case, with zero curvature constant Lambda

Until 1998 it was generally assumed that Lambda was zero and so it was commonly thought H would go to zero as the density thinned out.
After 1998, it was generally accepted that Lambda was a small positive spacetime curvature implying a leveling out of H at some asymptotic growth rate H_∞ like 1/173 of a percent per million years.