"Jaxtraw" wrote in message
...
"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift
which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.
http://home.i-zoom.net/~wgilmour/Redshift.html

Interesting. It seems to me though that there is a flaw; if, wherever a
photon is created, it sees itself as in the centre of a gravitational
well,
then wherever it moves to it will continue to see itself in the centre of
the gravitational well since, wherever it moves to, it is still at the
centre of the universe (from its own POV). Therefore it never climbs out
of
the well, so this won't cause a redshift.
Ian

Yes I have thought of this argument before and it's a good one.
However in the climbing out from the center of the earth analogy there
clearly is a redshift which is measurable! And I am attempting to argue that
the two situations are identical.

Another poster in another forum states

Boundary conditions. There's no boundary to the universe,
so there is no center from which to measure the distance

This is true, when we speak of the universe as a whole, but surrounding each
point within the universe there is a visible horizon which is determined by
the value of C. There is of course more universe beyond this boundary but
this can never be communicated with. I am suggesting that this horizon
imposes a boundary condition [artificial if you will, but there none the
less] which makes the analogy between the earth example and universe analogy
valid. I tried to emphasis that due to Gausses law, the value of R [the
radius of the earth or radius to the visible horizon] is always unknown and
unknowable, yet in the climbing from the center of the earth analogy we do
indeed get a gravitational redshift even though there would appear to be no
known boundary at this point or at least one that can never be known. [A
photon climbing from the center can never know where the surface is before
it gets there.]

traveled towards any boundary beyond which there is no
mass.

And again I would suggest any point beyond the visible horizon would
represent this no mass area, since any mass there has yet to communicate its
presence.

Cheers