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Velocity of receding galaxies. Hubble <-> scale factor

  1. Jan 11, 2015 #1
    Hi there,

    This is my first post but I've been a spectator for a long time now. So I've been working on some of the basics of cosmic expansion and there is one contradiction that I came upon that I can't seem to resolve. I've looked around some of the similar threads but I couldn't find anything satisfying so I'll ask myself.

    If the expansion of the universe can be described using the scale factor as d(t)=d0*a(t) then by differentiating I find that d'(t)=d0*a'(t) (I'm just following http://en.wikipedia.org/wiki/Scale_factor_(cosmology)).
    So this tells me that if a'(t)=const (as was thought to be the fact before the discovery of accelerating expansion) then the recession speed of a galaxy d'(t) should be constant, right?
    But if I know look at Hubble's law (which I can even derive from the formula for d(t)) I find that d'(t)=a'(t)/a(t)*d(t)=H*d(t) or simply v=H*D. So doesn't this mean that as the distance becomes greater the speed also becomes greater. So the galaxy is accelerating. Somehow these two expressions must be consistent. What's up?!
    I'd be grateful for any help
     
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  3. Jan 11, 2015 #2

    Orodruin

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    The expressions are consistent. Note that d(t) = d0*a(t). If a'(t) is constant, neither a(t) or d(t) are and since H = a'/a, H is not constant either (it is a constant function divided by one that depends on time).

    Edit: Let me also add that a' never was thought to be constant. Even for a radiation or matter dominated universe, it depends on time. However, in these cases you have a decelerated expansion instead.
     
  4. Jan 11, 2015 #3
    Thanks for the answer. I see what you're saying but I still don't know how to answer the question. The gist of it is: Disregarding accelerated expansion is the recession velocity of a galaxy increasing or not? Does it remain constant? I understand Hubble's law v=H*D to tell me that it is as v will increase as D increases.
    Contrarily the formula d'(t)=v=d0*a'(t) seems to say that the recession speed v of a galaxy remains constant (for the case where we disregard what you said in the edit).
     
  5. Jan 11, 2015 #4

    Orodruin

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    d' will still increase if the comoving distance d0 increases. Also, as I said, a' constant leaves H as a varying quantity, you will end up with the same result for d' regardless of what formula you use. Hubbles law with H constant also only holds for relatively nearby objects. H has been varying throughout the history of the universe. Also note that d' really is not a velocity in the usual sense, nothing is really moving here but space is getting larger.
     
  6. Jan 12, 2015 #5

    bapowell

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    No, a'(t) was never constant. During decelerated expansion, [itex]\ddot{a} < 0[/itex] which means that a'(t) is a decreasing function of time. During radiation-dominated expansion early on, [itex]a \sim t^{1/2}[/itex] so [itex]a'(t) \sim t^{-1/2}[/itex]; during matter-dominated expansion, we have [itex]a \sim t^{2/3}[/itex] and so [itex]a'(t) \sim t^{-1/3}[/itex].
     
  7. Jan 12, 2015 #6

    timmdeeg

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    We compare galaxies at different distances at a certain time. So, according to Hubble's law double distance means double recession velocity. This can be imagined by watching points on a uniformly expanding rubber band.
     
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