Re: v = cz ?

On Oct 2, 8:35 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
<mluttg...@xxxxxxxxxx> wrote in messagenews:1191321246.585913.37920@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Oct 1, 11:46 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Oct 1, 5:21 am, mluttg...@xxxxxxxxxx wrote:

Wich observations lead to v = cz, but not
to v = cz / (1+z) ?

Marcel Luttgens

"A Relation Between Distance And Radial Velocity Among Extra-Galactic
Nebulae", Proc. Natl. Acad. Sci., Vol. 15, No. 3, pp. 168-173, March
1929.

Might want to look at it, and the entire field of cosmology as it has
developed since then.

Thank you.

According to Hubble, "The results establish a roughly linear relation
between velocities and distances".
Now, we would say v = Hd, and also, if one assumes
that d = (c/H) * z, v = c * z.

Then, from z = 1 on, v > c !

I think that everybody will agree that v = Hd,
but when d = (c/H) * z/(1+z), v = c * z/(1+z), and
v is always less than c.

That's why I consider that the relation
d = (c/H) * z/(1+z) is the correct one.

In the past, people have pointed you hundreds of times to
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.astro.ucla.edu/~wright/cosmo_01.htm

Redshift in relativity:
1 + z = sqrt((1+v/c)/(1-v/c))
and solved for v:
v = c z ( z+2 ) / ( z^2+ 2 z + 2 )
have 2nd order Taylor:
z ~ v/c + 1/2 (v/c)^2 + ...
and
v ~ c z - 1/2 c z^2 + ...

Your
v = c z / ( 1 + z )
has 2nd order Taylor
v ~ c z - c z^2

but don't let that disturb you.

Dirk Vdm



:-) :-)

.

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