z = v / (c-v) Simple & easy. Should work quite adequately for z<0.1... >[[Mod. note -- (I suspect the author knows this, but others may not.)
http://adsabs.harvard.edu/abs/1993ApJ...403...28H
The occasional paper reports AGN redshift as a velocity, e.g., 5000
km/sec. Since the original measurement was that of a redshift, i.e., wavelength displacement of spectral lines, the authors thus used some equation to convert that to velocity. But the equation is not given.
So, treating this generically, I convert the velocity back to a
redshift. But I'm not interested to use a cosmological model with
various parameter values, instead I use a simple cosmology-free
equation, to wit:
z = v / (c-v)
Simple & easy.
Should work quite adequately for z<0.1 which is where
one encounters such given velocities.
In principle, it could work all the way to z=infinity, not that I want
to. But I wonder if anyone has any thoughts on this.
[[Mod. note -- (I suspect the author knows this, but others may not.)
There is a superb discussion of this & many related issues in
Edward R Harrison
"The Redshift-Distance and Velocity-Distance Laws"
Astrophysical Journal 403(1), 28-31 (Jan 1993)
http://adsabs.harvard.edu/abs/1993ApJ...403...28H
-- jt]]
On Mon, 13 Dec 2021 21:48:52 PST, eric@flesch.org (Eric Flesch) wrote:
z = v / (c-v) Simple & easy. Should work quite adequately for z<0.1... >[[Mod. note -- (I suspect the author knows this, but others may not.)
http://adsabs.harvard.edu/abs/1993ApJ...403...28H
Thanks for that, it is a great discussion. The part relevant to me
comes at the very end where Harrison describes the "habit of
converting redshifts into radial velocities by means of the Doppler approximation V=cz" as being "convenient astronomically".
Is *that* all that is used to produce the velocity figure!?
I avoided
that as too simple, not to mention grossly wrong at z=1.
Well, if
that's what they do, then my reverse equation z=v/(c-v) will show a
10% discrepancy at z=0.1, so I'd better go back and fix those.
eric@flesch.org (Eric Flesch) writes:
z = v / (c-v)
Simple & easy.
But where did it come from?
But is there any justification for your formula?
Define "work". Can you plug in a number and get another number? Yes.
Does it mean anything useful? No.
However, the proper distance cannot
be directly measured, but can be calculated given the cosmological model
On Tue, 14 Dec 2021 13:47:01 PST, Phillip Helbig wrote:
eric@flesch.org (Eric Flesch) writes:
z = v / (c-v)
Simple & easy.
But where did it come from?
It's just an isomorphic mapping of non-relativistic cosmological
recession inferred from the spectral line shifts. So non-relativistic recessional velocity could be written as
V = c * z/(1+z)
Where did it come from? As far as I know, this was how redshift was originally quantified as a measure (using spectral displacement as a placeholder for velocity), but I have no citation.
But is there any justification for your formula?
Only that it's accurate and trivial, e.g., at v=c/2, z=1 and light frequencies are halved. Tell me that's wrong.
It has only so much meaning as "non-relativistic cosmological
recession" has a meaning. No less, no more. Unless I'm missing
something basic.
I was specifically avoiding cosmological models and cosmological
distances. I was only looking for a conversion between redshift and cosmological recession velocity (which I understand to be
non-relativistic). As it turns out, all I needed was z=V/c, silly as
it may be, because V=cz is what is used to calculate the recessional velocities presented in some papers.
They could have used
V = cz/(1+z) , but they did not. I'm surprised that they did not, and
that's the end of it, I guess.
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