I was reading some papers on astrophysics, and in several of them, I've encountered velocity being used as distance. Or more precisely, distance being in dimensions of distance over time. For example, a paper referred to a group of galaxies at "roughly $2000\:\mathrm{km\:s^{-1}}$ in distance." Another used the specific phrase "velocity distance." It said there was a super-cluster "an observed concentration of galaxies at a velocity distance of ${\sim}2500$ to $4000\:\mathrm{km\:s^{-1}}$." I searched for what this could mean for quite a bit, and couldn't find it. If anyone has any idea what this means, some insight would be greatly appreciated.
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$\begingroup$ Related: Do we actually measure distances in light years? $\endgroup$– Emilio PisantyCommented Apr 27, 2017 at 14:32
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2 Answers
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It is used because of uncertainty in the Hubble constant. The relationship between recession velocity and distance is given by
$v=H_{0}d$
Where $H_{0}$ is the Hubble constant. Since that isn't known precisely, distances aren't known precisely, so talking about distances doesn't necessarily make sense.
Taking about the distance velocity on the other hand is using something they can be measured pretty precisely from the redshift. And you know that if object B had twice the velocity of object A then it is twice as far away even if you don't know the actual distances.
Since distances depend on the Hubble parameter then, especially back when it was only known to be in the range 50-100 km/s/Mpc there wasn't a great deal of point in reporting distances: to compare distances between two papers you'd have to convert them to velocities using whatever value of the Hubble parameter each paper used, since there is no guarantee they'd use the same value.
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$\begingroup$ So to be clear, is the velocity distance essentially the velocity a galaxy would be moving if no other forces besides the Hubble expansion acted on it? $\endgroup$– RothXCommented Apr 27, 2017 at 14:19
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$\begingroup$ That's a better answer than mine - a very obvious (but never thought about by me) reason as to why it's done this way. $\endgroup$ Commented Apr 27, 2017 at 14:22
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$\begingroup$ @RothX Phill's answer is also, of course, the reason why the same data are often talked about as $z$ (redshift) values; without thinking about it too much, I guess I'd always just assumed that it was just easier to have an everyday sized number instead of some godzillion number of parsecs. $\endgroup$ Commented Apr 27, 2017 at 14:28
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2$\begingroup$ @RothX kind of. For the great majority of things in the universe, the expansion velocity is much much greater than any extra velocity they might have. For nearby objects (such as in the local group of galaxies the local gravitational motion within the cluster is bigger, but then we are measuring the distance by other means (such as supernovae) target than from the redshift. So for galaxies where it is worth measuring the redshift, out is just assumed (quite reasonably) that all of the redshift velocity is due to cosmic expansion $\endgroup$– PhillSCommented Apr 27, 2017 at 17:26
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Hubble's law will convert the velocity to a proper (as opposed to co-moving) distance for you; indeed Wikipedia states one form of Hubble's law as:
"... the observation ... that ... Objects observed in deep space (extragalactic space, 10 megaparsecs (Mpc) or more) are found to have a Doppler shift interpretable as relative velocity away from Earth"
This law comes from the FLRW metric in the special case when the the standard $\Lambda{\rm CDM}$ cosmological equation of state prevails, and is also, of course, confirmed by observations first made by Edwin Hubble (but formerly theoretically predicted by Georges Lemâitre and contemporary theoreticians).
answered Apr 27, 2017 at 14:13
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$\begingroup$ So is the velocity distance essentially the velocity a galaxy would be moving if no other forces besides the Hubble expansion acted on it? $\endgroup$– RothXCommented Apr 27, 2017 at 14:17
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