<p>Equation (34) in this <a href="https://articles.adsabs.harvard.edu/pdf/1937ApJ....86..217Z" rel="nofollow noreferrer">paper</a> shows an average radial velocity. The author wrote that the value was obtained by measuring the radial velocities of many galaxies in the cluster of galaxy and by averaging the velocities over time and again over the different galaxies.</p>
<p>How do you obtain the &quot;time average&quot; of a radial velocity? When you measure the doppler shift of an object, does it give only the time average, or would you have to measure velocity over a long time? If the latter is the case, what would be the common time scale for averaging?</p>

CC BY-SA 4.0
2023-06-08T10:58:26.927Z
@planetmaker, thank you for the reply! Then (1) are the galaxies (dots in the sky) tracked over the period? and (2) Do the time-averaged velocities over a few years sufficiently account for the virial of the cluster of galaxies?
Nownuri
CC BY-SA 4.0
2023-06-08T11:34:37.833Z
It would actually take _millions_ of years to see meaningful changes in the individual velocities of galaxies in a cluster (using modern-day technology, anyway). Eqn. 34 of that paper is an average _over multiple galaxies within the cluster_; it is not an "average over time".
Peter Erwin
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2023-06-08T11:39:56.697Z
I don't think Earth's annual motion is the issue, since that is well known and can be accounted for even in a single exposure. The average should _in principle_ be taken over time, and are indeed earlier in the paper, as a "theoretical" answer. That's of course impossible given the timescales (~100 Myr), but using the galaxies' instantaneous velocities is a good approximation.
pela
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2023-06-08T13:12:10.510Z
@pela Thank you. Is the approximation justified by the assumption of dynamical equilibrium? I'm sorry for asking many questions...
Nownuri
CC BY-SA 4.0
2023-06-08T13:34:13.993Z
No worries :) Yes, clusters are in general assumed to have virialized (at least the low-z ones that Zwicky was able to see).
pela
<radial-velocity><doppler-effect>
Time average of radial velocities of multiple galaxies
Nownuri
# Time average of radial velocities of multiple galaxies

(Asked by: username_0 on 2023-06-08)

Equation (34) in this [paper](https://articles.adsabs.harvard.edu/pdf/1937ApJ....86..217Z) shows an average radial velocity. The author wrote that the value was obtained by measuring the radial velocities of many galaxies in the cluster of galaxy and by averaging the velocities over time and again over the different galaxies.

How do you obtain the "time average" of a radial velocity? When you measure the doppler shift of an object, does it give only the time average, or would you have to measure velocity over a long time? If the latter is the case, what would be the common time scale for averaging?

**Tags:** radial-velocity, doppler-effect
> **Comments:**
> **username_0**: @planetmaker, thank you for the reply! Then (1) are the galaxies (dots in the sky) tracked over the period? and (2) Do the time-averaged velocities over a few years sufficiently account for the virial of the cluster of galaxies?  
> **username_1**: It would actually take _millions_ of years to see meaningful changes in the individual velocities of galaxies in a cluster (using modern-day technology, anyway). Eqn. 34 of that paper is an average _over multiple galaxies within the cluster_; it is not an "average over time".  
> **username_2**: I don't think Earth's annual motion is the issue, since that is well known and can be accounted for even in a single exposure. The average should _in principle_ be taken over time, and are indeed earlier in the paper, as a "theoretical" answer. That's of course impossible given the timescales (~100 Myr), but using the galaxies' instantaneous velocities is a good approximation.  
> **username_0**: @pela Thank you. Is the approximation justified by the assumption of dynamical equilibrium? I'm sorry for asking many questions...  
> **username_2**: No worries :) Yes, clusters are in general assumed to have virialized (at least the low-z ones that Zwicky was able to see).  

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## 0 answers

