Linear model uniquely implies a boundary to our Universe?

1. Apr 25, 2012

johne1618

The Hubble radius R is defined by:

$R(t) = c / H(t)$

where $H(t)$ is the Hubble parameter which is a function of time.

Objects beyond the Hubble radius are receeding from us faster than the velocity of light.

At first glance one would think that light from those objects can never reach us. However the Hubble radius generally moves relative to the Universal expansion so that objects that were inside or outside the Hubble radius at a particular time move outside or inside at a later time.

If the Hubble radius was stationary in co-moving cordinates then there would be a true cosmological event horizon at that distance separating objects within our Universe from those outside it for all time.

For this to be true

$R(t) \propto a(t)$

where $a(t)$ is the scale factor.

Thus

$\frac{1}{H(t)} \propto a(t)$

Now we have

$H(t) = \frac{\dot{a}}{a}$

Therefore we get

$\frac{a(t)}{\dot{a}(t)} \propto a(t)$

This implies

$\dot{a}(t) \propto 1$

Therefore

$a(t) \propto t$

So a linearly expanding Universe is unique because it has a true "impermeable" boundary at its Hubble radius.

Have I got this right?

Last edited: Apr 25, 2012
2. Apr 25, 2012

Ich

No, this would be the case if it was at a constant cosmological proper distance, 1/H=const.
What you're describing is a freely coasting universe, with the empty universe as a special case.

3. Apr 25, 2012

Calimero

As Ich said in de Sitter expansion horizon is constant in time. You are describing empty universe, where H=1/t, therefore Rh=ct, so Hubble radius is growing at the speed of light and eventually all objects in the universe will be in causal contact. Cosmological event horizon is a feature of accelerated spacetimes only!

Last edited: Apr 25, 2012
4. Apr 26, 2012

johne1618

I guess I shouldn't use the term "cosmological event horizon".

But I still think that the linearly expanding Universe is unique in that it has a true boundary at the Hubble radius acting as an "edge" to the Universe. In the linear model no matter ever crosses this boundary in either direction.

This is in contrast to, say, the de Sitter model where matter is constantly crossing out of the boundary at the (constant) Hubble radius.

5. Apr 26, 2012

Ich

Great. But:
In every (ideal) FRW-spacetime, no matter ever crosses any comoving sphere. The Hubble-sphere in the linearly expanding model is just one example.
Google "shell crossing", that's something you don't want to have for your model to be well-behaved.

6. Apr 29, 2012

johne1618

Hi,

Sorry everyone!

I now understand that one can only have an event horizon in an accelerating Universe.

John