Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Galaxy expansion & cosmological constant

  1. Feb 12, 2012 #1
    Hi Astro forum... I have got something to clarify which came across my mind as I was trying to answer few problems.So without any further delay , let's start :

    For simplicity we will assume that dark matter doesn't influence the center of mass of galaxy A. Now let's say galaxy A has mass 'M' concentrated at the center with a separation distance of 'r' from the edge. In this case with simple arithmetic manipulation we get :

    [itex]V_{escape}[/itex] = [itex]\sqrt{GM/r}[/itex]

    Now I know that at shorter separation distances gravitational forces overwhelm ( i.e are greater) than cosmological constant , and then using Hubble's law v = [itex]H_{0}[/itex] r( assuming redshift is inteperted as doppler shift) .. so far it all makes sense...
    But if I am asked to prove that Vesc >> v ... I was thinking of something as following:

    [itex]V_{escape}[/itex] = [itex]\sqrt{GM/r}[/itex]

    Squaring both sides to get:
    [itex]V_{escape}^2[/itex] = [tex]2GM/r[/tex]
    which gives r = [itex]2GM/ V_{escape}^2[/itex]

    Now as said previously v = [itex]H_{0} r[/itex]

    so v = [itex]H_{0} 2 GM / V_{escp^2} [/itex]

    [itex]V_{escp^2} = H_{0} *2 GM / v [/itex]

    Gives us [itex]V_{escp} \propto[/itex] [itex]v^{-1/2}[/itex]

    Is this suffice enough to prove Vesc >> v ? am I on the dot on this one ?

    Any input is appreciated.


    EDIT: Oh why are my latex command showing up.. hm..
    edit2: latex error fixed
  2. jcsd
  3. Feb 12, 2012 #2


    User Avatar
    Science Advisor

    I have no idea what it is you're trying to prove here, but whatever it is, you're going about it wrong. First off, you state that you want to prove [itex] V_{esc} \gg v[/itex], what is [itex]v[/itex]? The characteristic velocity of stars in the galaxy? I have absolutely no idea what it is you're trying to derive, it just looks like mindless symbol manipulation and combination of equations which leads me to the second point...

    Doing anything on a galactic scale, you cannot use Hubble's velocity relation. It's not merely that the recessional velocities predicted by it will be much smaller than typical velocities of galactic motion (is this what you're trying to prove???), hubble expansion DOESN'T HAPPEN in gravitationally bound clusters. Furthermore, this is completely separate from the cosmological constant! In a traditional Lambda-CDM model, the density of dark energy is constant, and gravitationally bound objects (like our galaxy) will stay gravitationally bound indefinitely. (I say traditional because there are some models in which Lambda is somehow a function of time, and grows, leading to 'big rip' scenarios).

    As a final point, your equation for escape velocity is incorrect:
    Where this [itex]v_{orbit}[/itex] is the velocity of a circular orbit of radius r.
  4. Feb 12, 2012 #3
    Hi Nabeshin .
    I think it would be helpful if I post the actual question than wrongly restating it.

    When you say:
    I have few follow up questions to ask..
    1) Does that refer to a cluster of galaxies ?
    2) if dark energy density stays constant does that mean it's homogeneous , as in not being concentrated at a point ?
    Here's the question:


    P.S : 'v' is the recessional speed of the galaxy.
    Last edited: Feb 12, 2012
  5. Feb 12, 2012 #4


    User Avatar
    Science Advisor

    Yes, clusters are gravitationally bound.

    As far as we know, yes.

    This is just a lousy question. I understand what the author of it is trying to do (explain that expansion is insignificant compared to galactic motions), but this naive way of doing things is simply wrong. As I've said, the Hubble expansion equation doesn't even apply on these scales, so one should take such a 'proof' as heuristic at best. The second part about galactic clusters will probably yield a result that v_esc~v_hubble, but a conclusion here that hubble expansion is important on cluster scales in incorrect.

    Now that I understand the question though, your calculation in post 1 is incorrect. There is a scaling relationship between the two equations: on some scales, one velocity dominates, while on others, the opposite is true. Therefore, the way to do it is to calculate the characteristic velocities of both, and then compare them. (i.e. assume typical scales for M and r).
  6. Feb 12, 2012 #5
    I see. When you say "assume typical scales for M and r" , does that mean any arbitrary value ?
    Do I find a set of equations relating recessional speed using Hubble's constant and v_esc in the above equation ?
    Apart from the escape velocity given ~ 10-3 c (300km/s) and Hubble's constant ~ 75km/s/Mpc there isn't much given.


    EDIT: I took some arbitrary points and the results which i ended up with are: Hubble speed of 757*10^-3 km/s for separation distance of 2.6*1-^17 km... when I plug the same distance in v_escapre equation.. it gives 300km/s ...
    Last edited: Feb 12, 2012
  7. Feb 12, 2012 #6


    User Avatar
    Science Advisor

    I mean you know about how massive a galaxy is, and about how big they are. So using those values, you get some characteristic velocity scale for both equations.
  8. Feb 12, 2012 #7
    I have found some values ( they're on the previous post at the end).
  9. Feb 13, 2012 #8
    So does anyone think my values are sensible ?
  10. Feb 13, 2012 #9


    User Avatar
    Science Advisor

    Seems about right. Our sun's rotational velocity about the galactic center is something like 220km/s, and the Hubble recession velocity is about what I get.
  11. Feb 13, 2012 #10
    Thanks for all your help !
  12. Feb 22, 2012 #11
    Galaxy expansion??
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook