Re: Hubble deep field question

On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzc D:aol T:com \(dlzc\)"
<dlzc1@cox.net> wrote:

No, at red shift 6, they would be traveling at c*6/7 =0.857c, because
in fact v/c = z/1+z.
You are making the common error of assuming that v/c =z.
This is a Doppler effect, in which v(z) is widely misinterpreted, even
on NASA's site. I am preparing a paper on this topic.

John Polasek

Dear Craig Franck:

No,, that assumes the objects look to be 12.7 Gly away "now".

... and hence further away, and with the correct intensity.

No.  A Z of 1 is "receeding" at c.

http://www.astro.ucla.edu/~wright/cosmo_01.htm
... the text below the little sketch with the text "You're
receding"

I think you are trying to get to a number that has little
meaning.

David A. Smith

I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he
states on page 140 that "blobjects" at redshift 6 that were
13,000 light years across would appear 0.2 seconds of arc in
size. But that's assuming the light left when the object was 12.7
billion light years away.

I had thought that the objects would have been much closer when
the light first left and it took 12.7 billion years to reach us because
of cosmic expansion, which would not have made the objects look
smaller.

At redshift 6 they would be traveling at about 0.9c, but how would
you figure out how far away they were when the light first left from
that? I thought it would be 1/6 * 12.7bly, but that's assuming the
author is wrong, which may not be warranted.