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Meaning of setting the scale factor equal to 1 at present epoch

  1. Mar 15, 2008 #1

    kdv

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    EDIT: I corrected a typo below: I had forgotten to put my entire expression for a(t) to the power 1/3. It's corrected now



    This may sound like a silly question but it's bugging me quite a bit.

    Consider a cosmological model, let's say the simple case of no pressure, no cosmological constant, no curvature so that the solution is simply of the form

    [tex] a(t) = (6 \pi G \rho_0 t^2 + C)^{1/3} [/tex]

    where rho_0 is the density at the present epoch. Now, this specifies completely how the scale factor evolves with time up to the integration constant C. We may impose th einitial condition a(0) = 0 (Big Bang) which sets C=0.

    Fine so far. But then we also impose that a_now is equal to one and this leads to an equation giving the age of the universe in terms of H_0. It's this step that baffles me. I am not sure what it means. This seems to correspond to a rescaling of the time axis so that a_now is equal to one. But then I don't know what this time means at all. We may use this equation to then show that in this universe the age of the universe is 2/(3 H_0).

    The question is then: how is this time related to our physical time (that we measure on our watch? We measure a certain value of H_0 experimentally so we have an actual number for H_0. How can we see that this experimental value is actually consistent with this calculation that assigns a value of 2/(3H_0) to the age of the universe?

    It's not clear to me at all. Putting it another way, what if I decided that the scale factor at the present epoch was 2 instead of 1? Wouldn't that lead to a different estimae of th eage of the universe or a different estimate of the density required to have a flat universe?
     
    Last edited: Mar 16, 2008
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  3. Mar 16, 2008 #2

    cristo

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    I'm not really sure where your equation has come from. For pressureless matter in a flat universe, the scale factor goes as t^2/3. More precisely, we can write this as [tex]a(t)=a_0\left(\frac{t}{t_0}\right)^{2/3}[/tex]. Now, by writing this in terms of these constants, we see that when the scale factor takes its present day value, time takes its present day value also. Now, one can normalise this and take a_0=1; and yes, I guess this would be a change of units (much like setting c=1), but it won't physically change anything. Recall the definition of the scale factor: y=ax, where y is the physical coordinate and x is the comoving coordinate. If you set the present day value of the scale factor to 1, then this is equivalent to saying that physical distance scales today are the same as comoving distance scales.

    I've probably not answered the right question, but I hope that slightly helps!
     
  4. Mar 16, 2008 #3

    George Jones

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    Actually, changing the scale factor is just a rescaling of spatial coodinates, but isn't a change in anything physical. Consider the spatially flat metric

    [tex]g = -c^2dt^2 + a\left(t\right)^2 \left( dx^2 + dy^2 + dz^2),[/tex]

    and introduce the change of coordinates [itex]x' = Ax[/itex], [itex]y' = Ay[/itex], and [itex]z' = Az[/itex], where [itex]A[/itex] is a constant. Then, [itex]dx = dx'/A[/itex], etc., or in terms of transformations of tensor components,

    [tex]g_{\mu' \nu'} = \frac{\partial x^\alpha}{\partial x^\mu'} \frac{\partial x^\beta}{\partial \x^nu'} g_{\alpha \beta}.[/tex]

    In particular (no sum),

    [tex]g_{a'a'} = g_{aa}/A^2[/tex]

    and thus

    [tex]g = -c^2dt^2 + \left(\frac{a\left(t\right)}{A}\right)^2 \left( dx'^2 + dy'^2 +dz'^2),[/tex]

    and thus we can define [itex]a'\left(t\right) = a\left(t\right)/A[/itex] such that [itex]a'\left(t_0\right) = 1[/itex] by changing spatial coordinates.

    Note that this doesn't change anything physical; proper spatial distances, the forms of the equation that cristo gave, and of the Hubble parameter [itex]H\left(t\right)[/itex] and the age of the universe remain unchanged.
     
    Last edited: Mar 16, 2008
  5. Mar 16, 2008 #4

    kdv

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    Thanks to both of you.

    Cristo, yes, I had forgotten to type in an overall exponent of 1/3 in my answer (now corrected). Sorry about that typo.


    Ok, it all makes sense now. I had made a mistake in my calculations. Now I see that if I leave a_now completely arbitrary it cancels out completely in my expression for the age of the universe (in terms of H_0) and in the expression for the critical density. My mistake is that in an earlier equation I had set a_now=1 and then in later equations I had kept a_now arbitrary. When I solved for the critical density of the universe (to get k=0 in a dust-filled universe with no cosmological constant) I ended up with an equation for rho_critical which contained a_now. So that bothered me. Now I went back and ketp a_now arbitrary everywhere and it does cancel out from all physical results.

    Thanks to both of you for claryfying this up for me!
     
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