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johne1618
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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?
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?
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