Conformal Transformations and Hubble Parameter

Here is a simple demonstration that the Hubble effect is compatible
with both Hubble's law and the large value of the Hubble time.

A special conformal transformation is

Xm' = 1/S (Xm + XnXn Am)

where Am is a four-vector and

S = (1 + 2 Am Xm + AnAn XmXm)

If we assume Am is timelike and go to its rest frame where Am =
[0,0,0,a] then

S = (1 + 2 at + (t^2 - r^2) a^2) = (1 + at)^2 - (ar)^2

It turns out that in general

dXm' dXm' = 1/S^2 dXm dXm

This becomes singular when S=0. That happens when

t = r - 1/a

The implication is that a is very small and this t represents the
Hubble time.

For small a,

dXi' = 1/S^2 ( dXi + 2a (t dxi - xi dt) )
dt' = 1/S^2 ( dt + 2a (t dt - xi dxi) )

so

v' = [ v' + 2a (t v - x) ] / [ 1 + 2a(t - x.v) ] = [ v' + 2a
(t v - x) ] * [ 1 - 2a(t - x.v) ]

as long as t - x.v does not become large. To first order

v' = v - 2a x + 2a (x.v) v

= v - 2a (1-v^2) x - 2a v ^ (x ^ v)

If v << 1 then simply

v' = v - 2ax

that is, there is an apparent velocity proportional to the distance.
An object at rest in the distant frame, v' = 0, appears to us to have
an apparent velocity radially outward

v = 2ax

which is just Hubble's law.

-drl

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