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

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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]]

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]]

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...

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. :-)

>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.

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

Dec 14, 2021, 4:47:04 PM12/14/21

to

In article <61b59704....@news.aioe.org>, er...@flesch.org (Eric

> 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)

> http://adsabs.harvard.edu/abs/1993ApJ...403...28H

>

> -- jt]]

Indeed. See also the corresponding chapter in his textbook Cosmology:

The Science of the Universe. For that matter, everything Harrison wrote

is worth reading.

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.
> 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.

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.

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.

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)

> http://adsabs.harvard.edu/abs/1993ApJ...403...28H

>

> -- jt]]

The Science of the Universe. For that matter, everything Harrison wrote

is worth reading.

Dec 14, 2021, 5:31:25 PM12/14/21

to

In article <61b83d2c....@news.aioe.org>, er...@flesch.org (Eric

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.

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.)

> > 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!?

Yep, that's it!
> 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.)

> > 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!?

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.

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.

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.

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
>er...@flesch.org (Eric Flesch) writes:

>> z = v / (c-v)

>> Simple & easy.

>But where did it come from?

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?

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.

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

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.

Dec 16, 2021, 8:09:12 AM12/16/21

to

In article <61b92469....@news.aioe.org>, er...@flesch.org (Eric

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)?

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
> 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.

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.

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.

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.

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