Re: "Tired" light

From: N:dlzc D:aol T:com \(dlzc\) (net_at_nospam.com)
Date: 07/02/04 

Date: Fri, 2 Jul 2004 07:27:14 -0700

Dear Marcel Luttgens:

"Marcel Luttgens" <mluttgens@wanadoo.fr> wrote in message
news:86996cba.0407020130.1ba15081@posting.google.com...
> "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> wrote in
message news:<lRUEc.3894$nc.2802@fed1read03>...
> > Dear Marcel Luttgens:
>
> <snip>
>
> > > In the case of a negative acceleration, light is redshifted, meaning
that
> > > it loses energy, like a stone thrown vertically from the ground
> > > (where g is the negative acceleration). In the universe, a_H plays
the
> > > role of g.
> >
> > But in static Euclidian space, there is no such gradient that light
must
> > "climb" against. There is as much mass ahead as behind, with a similar
> > distribution. Therefore you have just proven that your supposition is
in
> > error.
>
> a_H has been observed. Where does it come from?
> Well, only part of the static *infinite* Euclidian space can be observed,
> and that part is a sphere of radius c/H. At some distance from the center
> of the sphere, there is not as much mass "ahead as behind".

We have a straight line view through up to 13 Gy of light's travel, in all
directions, except along the plane of the Milky Way. It is apparently all
uphill to here. It is roughly homogenous to here. We must be tremendously
non-dense, and the path uniformly more non-dense, compared to the point of
emission. Therefore, you have succeeded in shooting your own theory yet
again. Only expansion, and curvature of space, can do this correctly and
simply.

David A. Smith