Re: Hubble deep field question

On Thu, 29 Jan 2009 14:49:13 -0800 (PST), dlzc <dlzc1@xxxxxxx> wrote:

Dear John Polasek:

On Jan 29, 1:59 pm, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Thu, 29 Jan 2009 09:45:47 -0800 (PST),dlzc<dl...@xxxxxxx> wrote:
On Jan 28, 5:57 pm, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Wed, 28 Jan 2009 11:43:07 -0800 (PST),dlzc<dl...@xxxxxxx> wrote:
On Jan 28, 8:51 am, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Tue, 27 Jan 2009 06:45:37 -0800 (PST),dlzc<dl...@xxxxxxx> wrote:
On Jan 26, 3:14 pm, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Mon, 26 Jan 2009 10:33:28 -0800 (PST),dlzc<dl...@xxxxxxx> wrote:
On Jan 23, 9:51 pm, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)"

<dl...@xxxxxxx> wrote:
Dear Craig Franck:

"Craig Franck" <craig.fra...@xxxxxxxxxxx> wrote in message
snip

No.  A Z of 1 is "receeding" at c.

I beg to differ. This is a common and grievous
error, if we attribute redshift to straight Doppler.

The Universe has no unique center, therefore the
expansion is not from an explosion in space.  So
there is no theoretical support to your
"manipulations".

I did not discuss 'center'.

But you did invoke straight classical Doppler shift,

which, I think you will agree, is merely the most
plausible *explanation* for a universally
acknowledged phenomenon, redshift, and which
involves mere kinematics.

It is the first baby step to be taken in exploring a
cosmology.  Essentially, it requires that we are the
center of the Universe.
Again.

No, I don't find that "plausible".

You are confusing redshift with the Hubble precept
that the radial velocity is proportional to range (an
extrapolation by others, as Hubble didnt go that far).

He actually did, but Hubble is not the problem...

At first glance that makes us the center, as you
seem to presume, but that is easily debunked.

Actually required by your mathematics.  

If you would please be specific, what is required?

Motion in a Euclidian Universe, with all objects visible moving away
from us, with speed (more or less) proportional to distance.

You simply say "well that is puerile", but you
do not follow your mathematics to their
inescapable conclusion.

What inescapable conclusion? Be specific.

I have been. You do not acknowledge the cosmology required by your
"simple solution".

This has all the earmarks of a debate that
could continue far into the spring.
Just for starters, accept or reject:  
        f' = f(1-dS/cddt)
        where S is the separation.

No, the "debate" ends now.

David A. Smith
You must think my logic requires a Hubble-expanding universe to have a
center, (why didn't you say so?). Not at all.
There is little science that can be extracted from Hubbles constant,
since cosmology cannot put a definite value on it, nor is it possible
to implicate it as a possible cause of expansion (for which cosmology
presently has no raison). But since it and z are about the only real
data available, we analyze the Hubble principle:
There is no center, as follows, using stars A(us), B and X:

A sees X with vector distance and its colinear velocity = AX
Clearly, B equally sees a different distance and velocity BX
which, given large separation of A and B, the vectors mismatch by a
wide angle yet BX can claim legitimacy also. How? the 3d leg AB.
So A can view B and get redshift range and velocity = AB.
And B can view A and get BA. Simply combine:
Now it's obvious that AB + BX = AX so B is like A, and v.v.
BA + AX = BX so A is like B.
There's only one conclusion: they are separating without the semblance
of a center.
So to cover this contretemps, cosmology uses the anomalous coefficient
a, which for us now is a0, and its value, fortuitously is unity: We
are just the right size now!
In fact that's just what H is: H = 100%/age of our star 13Gyr.
So, what do you think, is
    f' = f(1-dS/cdt)
or not?
John Polasek
.

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