# velocity -> redshift

9 views

### Eric Flesch

Dec 14, 2021, 12:48:54 AM12/14/21
to
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)

-- jt]]

### Eric Flesch

Dec 14, 2021, 4:46:26 PM12/14/21
to
On Mon, 13 Dec 2021 21:48:52 PST, er...@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.)

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.

Harrison did not mention my little pretty equation at all. :-)

### Phillip Helbig (undress to reply)

Dec 14, 2021, 4:47:04 PM12/14/21
to
In article <61b59704....@news.aioe.org>, er...@flesch.org (Eric
Flesch) writes:

> 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.

They just multiply the speed of light by z.

> 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.

But where did it come from? The non-relativistic Doppler formula is
v = cz, which holds for small v (i.e. v/c << 1). Your equation is
equivalent to v = cz-vz. So, if v/c << 1 is small, then the second term
on the r.h.s. is much smaller than the first, and thus your formula is
approximately correct and the error is smaller than the error of using
the non-relativistic Doppler formula in the first place (strictly
speaking valid only in the limit of vanishing v). But is there any
justification for your formula?

> Should work quite adequately for z<0.1 which is where
> one encounters such given velocities.

Yes. Everything is linear to first order. :-) But that doesn't mean
that all approximations are equally valid logically (even if the
difference, mathematically, is negligible).

> 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.

Define "work". Can you plug in a number and get another number? Yes.
Does it mean anything useful? No.

The equation for arbitrarily high redshift is very simple: v=HD, where H
is the Hubble constant and D is the proper distance. That is the
definition of the Hubble constant. However, the proper distance cannot
be directly measured, but can be calculated given the cosmological model
(which can be inferred from the dependence of other distances on
redshift). Yes, v can become arbitrarily large. No, no conflict with
special relativity.

Not even wrong is using the relativistic Doppler formula for high
redshift. Simple proof: it contains no cosmological parameters (not
even the Hubble constant), so using it implies that the velocity at high
redshift is independent of the cosmological model, which is absurd.

> [[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)
>
> -- jt]]

Indeed. See also the corresponding chapter in his textbook Cosmology:
The Science of the Universe. For that matter, everything Harrison wrote

### Phillip Helbig (undress to reply)

Dec 14, 2021, 5:31:25 PM12/14/21
to
In article <61b83d2c....@news.aioe.org>, er...@flesch.org (Eric
Flesch) writes:

> On Mon, 13 Dec 2021 21:48:52 PST, er...@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.)
>
> 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!?

Yep, that's it!

In the old days, when 0.1 was a huge redshift, it sort of made sense: at
low redshift, the Doppler formula does give the recession velocity (in
the limit of 0 redshift), and differences of hundreds of km/s are easier
to visualize than the difference between 0.001 and 0.0025 or whatever.
Of course, although theoretically predicted by de Sitter and Lema=EEtre,
Hubble (whether or not he knew about their work) was very empirically
minded and used the standard astronomer conversion of redshift into
velocity (familiar from motions of double stars or whatever).

When redshift became larger, most (but not all---as Harrison points out,
even some professional astronomers at least seemed confused) realized
that it was just a placeholder for redshift, which also aided comparison
with older data.

> I avoided
> that as too simple, not to mention grossly wrong at z=1.

Right; it's a low-redshift approximation. But your formula, as far as I
know, has no justification.

> 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.

The literature you have almost certainly has v=cz, so just convert back.

Where it is more difficult is where people observe something like the
distribution in redshift and flux, i.e. the luminosity-dependent
redshift distribution or, equivalently, the redshift-dependent
luminosity function. (Some astronomers call those relations "Hubble
diagrams"---redshift plotted against apparent magnitude or vice
versa---even for objects (QSOs, say) which are not standard candles and
hence no (approximately) limited relationship is even expected; again,
are you an empiricist or concerned with interpretation?) The
observational data, apparent magnitudes and redshifts, are clear. But
sometimes results are presented in terms of absolute magnitude (or
luminosity) and (co-moving) volume, which necessarily implies conversion
via some cosmological model, which they might have failed to specify.
Even if known, converting back to the original data is non-trivial and,
if binning is involved, impossible. Best for observers to (at least
also) report their data (not necessarily all the raw data) in terms of
observable quantities.

### Eric Flesch

Dec 14, 2021, 7:42:35 PM12/14/21
to
On Tue, 14 Dec 2021 13:47:01 PST, Phillip Helbig wrote:
>er...@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.

>Define "work". Can you plug in a number and get another number? Yes.
>Does it mean anything useful? No.

It has only so much meaning as "non-relativistic cosmological
recession" has a meaning. No less, no more. Unless I'm missing
something basic.

> However, the proper distance cannot
>be directly measured, but can be calculated given the cosmological model

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.

### Phillip Helbig (undress to reply)

Dec 16, 2021, 8:09:12 AM12/16/21
to
In article <61b92469....@news.aioe.org>, er...@flesch.org (Eric
Flesch) writes:

> On Tue, 14 Dec 2021 13:47:01 PST, Phillip Helbig wrote:
> >er...@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's wrong in the sense that at v=c/2 one can't say anything about the
velocity without knowing the cosmological model. It is right in the
sense that at z=1 frequencies are halved. But the frequency is the
rest-frame frequency divided by (1+z). For z=1 both formulae give the
same answer, but not in general.

Interestingly, some people (such as Zel'dovich) used to use Delta rather
than z to characterize redshifts, with Delta = 1 - 1/(1+z) = z/(1+z),
which is similar to your formula. That has the advantage that it ranges
between 0 and 1 rather than 0 and infinity. But nothing to do with
velocity.

> It has only so much meaning as "non-relativistic cosmological
> recession" has a meaning. No less, no more. Unless I'm missing
> something basic.

That has a meaning only at low redshift, in which case it would be
easier to use v=cz.

> 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.

Right.

I'm not sure what you mean by cosmological recession velocity being
non-relativistic. It's true that one can't calculate it with the
relativistic Doppler formula, but that doesn't mean that the
non-relativistic Doppler formula (or your formula) is correct, except in
the limit of low redshift (in which case the relativistic Doppler
formula is also correct).

> 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.

What could have been the motivation to use V = cz/(1+z)?