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Universe expansion + light speed limit -> scale factor linear in time?

  1. Nov 16, 2011 #1
    The Hubble parameter is defined by:

    H(t) = a'(t) / a(t)

    where a is the scale factor which is a function of cosmological time t.

    This definition is equivalent to the Hubble relation:

    v(t) = H(t) r(t)

    where v(t) and r(t) are the proper velocity and distance of an object at cosmological time t.

    I assume that this relationship is exact provided that we acknowledge that the Hubble parameter H(t) is constant only in space and *not* in time. It can be interpreted as simply asserting that space is expanding uniformly at a rate H(t).

    Now let us assert that a universal speed limit exists, namely the velocity of light c. We know this is true observationally as the redshift diverges as v -> c.

    This assumption allows us to define the Hubble radius R(t) by

    c = H(t) R(t).

    Then the Hubble radius, R_0, at the present time t_0 is given by:

    c = H_0 R_0

    where H_0 is the present Hubble parameter value.

    As we have:

    R(t) = R_0 a(t)

    R(t) = (c / H_0) * a(t)

    Therefore

    a'(t) / a(t) = c / R(t)

    a'(t)/ a(t) = H_0 / a(t)

    So that finally

    a(t) = H_0 * t

    or equivalently

    R(t) = c * t

    Thus we seem to have derived a scale factor with a linear time dependence using only two facts:

    1. Space is expanding uniformly.

    2. A cosmological speed limit, namely c, exists.

    Is this correct?

    I expect I'm also assuming that space is flat.

    This very simple scale factor time dependence, the so-called coasting cosmology, is actually close to what is observed.
     
    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2
    It is true that the late time behaviour of an open universe with no cosmological constant is given by:

    a [itex]\propto[/itex] t

    But I think the curvature term: - k / [itex]a^{2}[/itex]

    Does not limit the expansion to c.
     
  4. Nov 16, 2011 #3

    marcus

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    John, you made an error in your derivation about 15 lines from the bottom where you said:

    R(t) = R_0 a(t)

    It does not follow from the preceding stuff.
     
  5. Nov 16, 2011 #4

    marcus

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    Each age has its own Hubble radius R(t). It is not simply equal to the Hubble radius of some other age, subjected to normal scalefactor expansion. If that were true the Hubble volume would always contain the same set of galaxies (expect for changes due to their individual random motions).

    It's good to see all the algebra, almost all steps of which are correct. And you are digging into the actual definitions and exploring their consequences---the right thing to be doing. Just that one mistake.
     
    Last edited: Nov 16, 2011
  6. Nov 17, 2011 #5

    Chalnoth

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    Also, in General Relativity, there cannot be any global speed limit. There is a local speed limit, meaning that objects cannot ever outrun light rays (any observer anywhere always sees light moving past them moving at c). But there is no unique way to define the velocity of far-away objects. So it turns out that the velocity of far-away objects just depends upon the way we write down our coordinates, and we can just as easily say something far away has no motion relative to us, is moving away at the speed of light, or is moving away at twice the speed of light. It makes no difference.

    The only time that relative velocity is well-defined is when you are comparing two velocities at a single point. So it is only then that the speed of light limitation makes sense.

    Thus the idea that the speed of light could possibly impose a limit on the Hubble expansion rate is just incorrect.
     
  7. Nov 17, 2011 #6
    Thanks for the replies - very helpful!
     
  8. Nov 17, 2011 #7
    But if one wanted to assert that the Hubble Radius R(t) is the "radius of the Universe" then one would want it to be constant in the comoving frame (the frame in which the Universe is static with respect to time)

    I realise that the radius of the "observable" universe can be larger than the Hubble radius because objects that we see now have been expanding away from us as the light from them travelled to us. That distance depends on the time dependence of the scale factor.

    In fact if one assumes a scale factor that is linear with time for all times then one finds that the radius of the observable universe is actually infinite whereas the Hubble radius is equal to the speed of light times the age of the Universe. Rather neat I think. The Hubble radius is thus the radius of the lightcone of events since the Big Bang that can effect us now.

    To explain: The radius of the observable universe is given by:

    Observable Radius = Integral [t=t_s:t_0] c dt' / a(t')

    where t_0 is the present age of the universe and t_s is the starting time of the Universe.

    If we assume a linear scale factor time dependence then:

    a(t') = t' / t_0

    where t_0 is the present age of the Universe.

    Thus substituting into the above expression for the observable radius and doing the integration we find:

    Observable radius = c t_0 ln (t_0 / t_s)

    As t_s -> 0 the observable radius -> infinity.

    We seem to get a flat infinite Euclidean space popping out of a point in a finite time which is rather interesting.
     
    Last edited: Nov 17, 2011
  9. Nov 17, 2011 #8

    Chalnoth

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    Except the universe isn't static with respect to time, no matter your frame of reference.
     
  10. Nov 17, 2011 #9
    According to the Wikipedia entry on comoving distance:

    Space in comoving coordinates is usually referred to as being "static", as most bodies on the scale of galaxies or larger are approximately comoving, and comoving bodies have static, unchanging comoving coordinates. So for a given pair of comoving galaxies, while the proper distance between them would have been smaller in the past and will become larger in the future due to the expansion of space, the comoving distance between them remains constant at all times.
     
  11. Nov 17, 2011 #10

    marcus

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    John, I probably was too complimentary and encouraging in my post #4. I picked out an error 15 lines from the end and said "only that one mistake". I should have emphasized that the train ran off the track at that point. Your conclusions in the last 15 lines are wrong as a consequence.

    In your latest post you do begin to sound a bit crackpotty, so I urge you to back off and take stock of the situation. Please do not press on with the mistaken assumptions of time-linear scalefactor and "universal speed limit" on the distance expansion rate.

    I should have pointed out there was some conceptual trouble early in your post #1. The v(t) in the Hubble relation is not an ordinary velocity, and is not governed by the velocity speed limit. But you refer to it as a "proper velocity"

    ==quote John post #1==
    This definition is equivalent to the Hubble relation:

    v(t) = H(t) r(t)

    where v(t) and r(t) are the proper velocity and distance of an object at cosmological time t...

    ...Now let us assert that a universal speed limit exists, namely the velocity of light c. We know this is true observationally as the redshift diverges as v -> c.
    ==endquote==

    If v is the expansion rate in the Hubble relation the redshift certainly does not diverge :biggrin:. It approaches something like 1.4 as v -> c.

    The divergence of the Doppler shift as velocity approaches c is in the fixed non-expanding frame of special relativity, Minkowski space. That does not apply at cosmological distances.
    I dimly suspect that you know this already, but perhaps that is overly pessimistic.
     
    Last edited: Nov 17, 2011
  12. Nov 17, 2011 #11

    Chalnoth

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    There is still the curvature of space-time, in any coordinate system, that means that it just doesn't fit the definition of static. There's also the point to be made that even in a coordinate system where the average coordinate distance between far-away galaxies doesn't change, you still get the formation of galaxy clusters, galaxies, and smaller objects.
     
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