5
$\begingroup$

From my understanding, the Hubble constant $H_0$ calculates from observed redshifts $z$ of distant galaxys against their proper distance $D$. The current value appears to be 67.80(77) $\frac{km}{s}Mpc^{-1}$

Calculating the Hubble constant via the redshift, I assume one only wants those velocity contributions due to the expansion of the universe, and not those from the real movement of the galaxies within in cluster (peculiar motion) or so. This, I also read in various online resources (which I can't recall right now).

On the other Hand, if your galaxy cluster is far enough away (eq. ~ 1 Gpc), one can neglect peculiar motion, which is in the order of 1000 $\frac{km}{s}$ (1000 $\frac{km}{s}$ / (1 Gpc $\times$ 67.80(77) $\frac{km}{s}Mpc^{-1}$) $\approx$ 1.4%)

Nonetheless, how would you try to correct for the peculiar motion, or is it really just neglected? Calculating all the gravitational components in each cluster? Another idea could be to assume that the galaxies within the cluster move randomly in respect to each other and the peculiar motion cancels out over the average? Any other possibilities?

Note that this question is partly a copy of another question of mine at physics.stackexchange.com, which wasn't fully answered but commented to ask it here.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

I think this might answer your question, I have quoted the important paragraph below.

If peculiar velocities could have any value, then this would make Hubble’s law useless. However, peculiar velocities are typically only about 300 km/sec, and they very rarely exceed 1000 km/sec. Hubble’s law therefore becomes accurate for galaxies that are far away, when H0d is much larger than 1000 km/sec. Furthermore, we can often estimate what a galaxy’s peculiar velocity will be by looking at the nearby structures that will be pulling on it.

So it seems that in answer to your question, if the galaxy is far enough away then the peculiar motion can be neglected, or if its close by, a reasonable approximation found using estimates from nearby galaxies (in a cluster for example).

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Or it simply averages out if you are looking at a a large number of galaxies in a cluster of galaxies (at similar distance) of course. $\endgroup$
    – ProfRob
    Feb 2, 2016 at 12:58
  • $\begingroup$ And for very far away galaxies I suppose one has to take into account that the inflation rate of space has shifted over time. But that's maybe kind of a circular argument? $\endgroup$
    – LocalFluff
    Feb 2, 2016 at 14:53
  • $\begingroup$ @LocalFluff Yeah because even though they're further away they're also further back in time, so the expansion of the universe was slower when the light we see now was emitted. Is that what you mean by circular argument? $\endgroup$
    – Dean
    Feb 2, 2016 at 15:03
  • 1
    $\begingroup$ I just meant that the Hubble constant is also a premise for the early history of inflation, for which the very observations on which the Hubble constant relies has to be adjusted. It's not a logically circular argument, just a complication I suppose. Finding standard candles all the way to the end of space and beginning of time. However, it seems to have been done! $\endgroup$
    – LocalFluff
    Feb 2, 2016 at 15:12
  • $\begingroup$ @LocalFluff The Hubble constant is $H_0$ and measures the expansion rate now. If you look at things very far away, you model that with a Hubble parameter $H(z)$. $\endgroup$
    – ProfRob
    Feb 3, 2016 at 7:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .