- #1
johne1618
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An effort to understand the physical implications of the Hubble radius.
I start by assuming a flat radial FRW metric given by:
[tex]
ds^2 = - c^2 dt^2 + a(t)^2 dr^2
[/tex]
Let us assume that we are holding one end of a rigid ruler with the other end out in space at a fixed proper distance [itex]R[/itex] away from us.
[tex]
R = a(t) r
[/tex]
Therefore the comoving radial co-ordinate [itex]r[/itex] of the end of the ruler is given by
[tex]
r = \frac{R}{a(t)}\ \ \ \ \ \ \ \ \ \ (1)
[/tex]
Now using the relation for proper time [itex]ds = -c\ d\tau[/itex] in the expression for the metric and dividing through by [itex]d\tau^2[/itex] we obtain the differential relation
[tex]
c^2 \left(\frac{dt}{d\tau}\right)^2 - a^2\left(\frac{dr}{d\tau}\right)^2 = c^2. \ \ \ \ \ \ \ \ \ \ \ \ (2)
[/tex]
By differentiating Equation (1) by proper time [itex]\tau[/itex] we find
[tex]
\frac{dr}{d\tau} = -\frac{R}{a^2}\frac{da}{dt}\frac{dt}{d\tau}
[/tex]
Substituting the above expression into Equation (2) we obtain
[tex]
c^2\left(\frac{dt}{d\tau}\right)^2 - R^2 \left(\frac{\dot a}{a}\right)^2\left(\frac{dt}{d\tau}\right)^2 = c^2
[/tex]
Rearranging, and substituting the Hubble parameter [itex]H=\dot{a}/a[/itex], we obtain:
[tex]
\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{R^2 \ H^2}{c^2}}} \ \ \ \ \ \ \ \ (3)
[/tex]
Now the Hubble law, for the proper recession velocity [itex]v[/itex] of a galaxy at proper distance [itex]R[/itex], is given by
[tex]
v = H \ R
[/tex]
Substituting into Equation (3) we finally obtain
[tex]
\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.
[/tex]
This implies that as one approaches the Hubble radius [itex]R_H=c/H[/itex], where the galaxy recession velocities approach [itex]c[/itex], we find that the local proper time [itex]\tau[/itex] slows down to a standstill compared to coordinate time [itex]t[/itex]. (It is interesting that the time dilation formula is exactly as one would expect from a special relativity viewpoint.)
This argument seems to show that there is a cosmological event horizon at the Hubble radius. The Hubble radius really is the radius of our Universe, at the present cosmological time [itex]t[/itex], and the event horizon is the edge of our Universe.
I would say that the space beyond the Hubble radius event horizon is not part of our Universe just as the space beyond a black hole's event horizon is not part of our Universe.
P.S. Implication for Universal expansion rate
As the Hubble radius really is the radius of the Universe then it must expand with the Universe therefore
[tex]
R_H = a(t) r_h
[/tex]
where [itex]r_h[/itex] is a constant.
Substituting into Hubble's law with [itex]v=c[/itex] we have
[tex]
\frac{\dot a}{a} = \frac{c}{R_H} \\
\frac{\dot a}{a} = \frac{c}{a \ r_h}
[/tex]
Thus [itex]\dot a[/itex] is a constant so that [itex]a[/itex] must increase linearly with time.
I start by assuming a flat radial FRW metric given by:
[tex]
ds^2 = - c^2 dt^2 + a(t)^2 dr^2
[/tex]
Let us assume that we are holding one end of a rigid ruler with the other end out in space at a fixed proper distance [itex]R[/itex] away from us.
[tex]
R = a(t) r
[/tex]
Therefore the comoving radial co-ordinate [itex]r[/itex] of the end of the ruler is given by
[tex]
r = \frac{R}{a(t)}\ \ \ \ \ \ \ \ \ \ (1)
[/tex]
Now using the relation for proper time [itex]ds = -c\ d\tau[/itex] in the expression for the metric and dividing through by [itex]d\tau^2[/itex] we obtain the differential relation
[tex]
c^2 \left(\frac{dt}{d\tau}\right)^2 - a^2\left(\frac{dr}{d\tau}\right)^2 = c^2. \ \ \ \ \ \ \ \ \ \ \ \ (2)
[/tex]
By differentiating Equation (1) by proper time [itex]\tau[/itex] we find
[tex]
\frac{dr}{d\tau} = -\frac{R}{a^2}\frac{da}{dt}\frac{dt}{d\tau}
[/tex]
Substituting the above expression into Equation (2) we obtain
[tex]
c^2\left(\frac{dt}{d\tau}\right)^2 - R^2 \left(\frac{\dot a}{a}\right)^2\left(\frac{dt}{d\tau}\right)^2 = c^2
[/tex]
Rearranging, and substituting the Hubble parameter [itex]H=\dot{a}/a[/itex], we obtain:
[tex]
\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{R^2 \ H^2}{c^2}}} \ \ \ \ \ \ \ \ (3)
[/tex]
Now the Hubble law, for the proper recession velocity [itex]v[/itex] of a galaxy at proper distance [itex]R[/itex], is given by
[tex]
v = H \ R
[/tex]
Substituting into Equation (3) we finally obtain
[tex]
\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.
[/tex]
This implies that as one approaches the Hubble radius [itex]R_H=c/H[/itex], where the galaxy recession velocities approach [itex]c[/itex], we find that the local proper time [itex]\tau[/itex] slows down to a standstill compared to coordinate time [itex]t[/itex]. (It is interesting that the time dilation formula is exactly as one would expect from a special relativity viewpoint.)
This argument seems to show that there is a cosmological event horizon at the Hubble radius. The Hubble radius really is the radius of our Universe, at the present cosmological time [itex]t[/itex], and the event horizon is the edge of our Universe.
I would say that the space beyond the Hubble radius event horizon is not part of our Universe just as the space beyond a black hole's event horizon is not part of our Universe.
P.S. Implication for Universal expansion rate
As the Hubble radius really is the radius of the Universe then it must expand with the Universe therefore
[tex]
R_H = a(t) r_h
[/tex]
where [itex]r_h[/itex] is a constant.
Substituting into Hubble's law with [itex]v=c[/itex] we have
[tex]
\frac{\dot a}{a} = \frac{c}{R_H} \\
\frac{\dot a}{a} = \frac{c}{a \ r_h}
[/tex]
Thus [itex]\dot a[/itex] is a constant so that [itex]a[/itex] must increase linearly with time.
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