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A Hubble parameter vs Scale factor's derivative

  1. Oct 2, 2016 #1
    How does Hubble parameter and scale factor's derivative differ geometrically? I am reading S. Caroll's GR book. But I cannot get the full representation of these two parameters. On the book, it says
    How can [itex]\dot{H}[/itex] and [itex]\ddot{a}[/itex] be opposite of each other on the same instance if both are representing the speed of expansion?
     
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  3. Oct 2, 2016 #2

    vanhees71

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    The Hubble parameter is defined as ##H=\dot{a}/a##, where ##a## is the scale parameter in the fundamental observer's reference frame, where the FLRW metric takes the form
    $$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left (\frac{\mathrm{d} r^2}{1-k r^2} + r^2 \mathrm{d} \theta^2 + r^2 \sin \theta \mathrm{d} \phi^2 \right).$$
     
  4. Oct 2, 2016 #3
    I know it. I need an explanation from a geometrical perspective.
     
  5. Oct 2, 2016 #4

    PeterDonis

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    Because they're not the same quantity. We have

    $$
    \dot{H} = \frac{d}{dt} \frac{\dot{a}}{a} = \frac{\ddot{a}}{a} - \left( \frac{\dot{a}}{a} \right)^2 = \frac{\ddot{a}}{a} - H^2
    $$

    They aren't; they are representing the acceleration (or deceleration) of expansion, but in different ways. The way I find it easiest to understand the difference is to consider exponential expansion, i.e., ##a = e^t## (in the simplest case). Then we have ##\dot{a} = \ddot{a} = e^t## as well. But we have ##H = 1## and ##\dot{H} = 0##. In other words, ##H## is measuring the (inverse) time constant of exponentiation, and ##\dot{H}## is measuring how that (inverse) time constant changes.

    Contrast this with, for example, a pure matter-dominated critical density universe where ##a = t^{2/3}##. Then ##\dot{a} = (2/3) t^{-1/3}## and ##\ddot{a} = - (2/9) t^{-4/3}##. So we have ##H = (2/3) t^{-1}## and ##\dot{H} = - (2/3) t^{-2}##. So here ##H## and ##\dot{H}## both tend to zero, ##H## from above and ##\dot{H}## from below, and ##\ddot{a}## is negative the whole time.

    In our current universe, which is dark energy dominated but still has significant matter density affecting the expansion rate, ##\dot{H}## is negative even though ##\ddot{a}## is positive, because the expansion is accelerating but the matter density is slowing down the acceleration somewhat; the function ##a(t)## is basically a combination of the two cases I gave above. As the universe continues to expand, the matter density will decrease while the dark energy density remains constant, and ##\dot{H}## will tend to zero from below, while ##H## will tend to a constant value determined by the dark energy density. But ##\ddot{a}## will be positive during all of this.
     
  6. Oct 4, 2016 #5
    Thanks for your detailed answer. But I would like to see somewhat generalization in the explanation of this question. I mean, generally, how does Hubble parameter and derivative of the scale factor or derivative of Hubble parameter and second derivative of the scale factor differ from each other. I understand your examples related to these concepts, but still struggle to imagine them in the geometrical viewpoint.
     
  7. Oct 4, 2016 #6

    phyzguy

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    It's fairly easy to see why [itex] H = \frac{\dot a}{a}[/itex]. Suppose some object is at a distance from you of x0 at time t=0 and all distances are growing by a factor of a(t). Then the distance to the object is given by x = x0 * a(t). So the velocity at which you see the object moving away from you is [itex] \dot x=x_0 \dot a(t)[/itex]. The Hubble constant is defined to be the ratio of the velocity of the object divided by the distance to the object, so [itex] H = \frac{\dot x(t)}{x(t)} = \frac{x_0 \dot a(t)}{x_0 a(t)} = \frac{\dot a}{a}[/itex].
     
  8. Oct 4, 2016 #7

    PeterDonis

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    I'm not sure I understand what you mean by "in the geometrical viewpoint". ##H## and ##\dot{H}## are dimensionless as far as distance is concerned (their only dimensions are time, more precisely time^-1 for ##H## and time^-2 for ##\dot{H}##), so they can be thought of as fractional change. That is the key difference between them and ##\dot{a}## and ##\ddot{a}## conceptually, since the latter two have a dimension of distance in the numerator.
     
  9. Oct 5, 2016 #8
    I mean, if both of these quantities(##\dot{H}##,##\ddot{a}##) represent acceleration of the expansion, how can they differ in ther signs? Today's universe ##\dot{H}## is negative, while ##\ddot{a}## is positive. What do they mean separately? If the expansion of the universe is accelerating, how can ##\dot{H}## be negative?
     
  10. Oct 5, 2016 #9

    PeterDonis

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    Because they are representing "acceleration of the expansion" in different ways.

    In the case of ##\ddot{a}##, the representation is simple: positive ##\ddot{a}## means acceleration, negative ##\ddot{a}## means deceleration. So you only need to look at the sign to see which is happening.

    In the case of ##H## and ##\dot{H}##, the representation is more complex: here acceleration means that ##H## is tending towards some positive value as ##t \rightarrow \infty##, whereas "deceleration" means that ##H## is tending towards zero as ##t \rightarrow \infty## (or, if we include the possibility of a closed universe, we will have ##H \rightarrow 0## as ##t \rightarrow T##, where ##T## is some finite time).
     
  11. Oct 15, 2016 #10
    Yeah, but what does it mean, H is tending towards zero or not? This is the part I have been confusing. Because it is mathematical description, and I have a difficulty in implementing it to real world.
     
  12. Oct 15, 2016 #11

    PeterDonis

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    If ##H## is tending towards zero, there is no limit to how slowly the universe can be expanding; it can even stop expanding and start contracting (if ##H## reaches zero in a finite time).

    If ##H## is tending towards some nonzero positive value, then the universe will end up expanding exponentially forever (since a constant ##H## corresponds to exponential expansion).

    A key difference between these two scenarios (if we leave out the possibility of the universe starting to contract, and only consider ##H \rightarrow 0## vs. ##H \rightarrow H_{\infty} > 0## as ##t \rightarrow \infty##) is that in the former scenario, there is no event horizon in the universe. But in the latter scenario (exponential expansion), there is.

    (NOTE: Edited to correct misstatement about event horizon.)
     
    Last edited: Oct 15, 2016
  13. Oct 15, 2016 #12
    So in our universe, both ##\ddot{a}## and ##\dot{H}## are positive, yes?
     
  14. Oct 15, 2016 #13

    George Jones

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    No, in our universe, ##\ddot{a}## is positive and ##\dot{H}## is negative, i.e., the Hubble parameter is decreasing,
     
  15. Oct 15, 2016 #14
    And this is the confusing part for me. Peter has said, if H is tending towards some positiv value(##\dot{H}## is positive), two comoving observers will eventually be disconnected. I knew, particle horizon is shrinking over time, which means everything near us will be causally disconnected from us one day. Isn't it true?
     
    Last edited: Oct 15, 2016
  16. Oct 15, 2016 #15

    PeterDonis

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    Yes.

    No. ##\dot{H}## is negative, because the current value of ##H## is larger (more positive) than the positive value towards which ##H## is tending. I explained this in post #4.
     
  17. Oct 15, 2016 #16

    PeterDonis

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    Yes. But that has nothing to do with the sign of ##\dot{H}##. It has to do with the fact that ##H## is tending (from above) towards a nonzero positive value instead of towards zero.
     
  18. Oct 15, 2016 #17

    George Jones

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    This is impossible.
     
  19. Oct 15, 2016 #18
    Yeah. Now it has become more clear to me. Thanks for sparing your time to write detailly answer the question.
     
  20. Oct 15, 2016 #19

    PeterDonis

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    Oops, you're right, I misstated it. I should have said there is an event horizon in a dark energy dominated universe and left it at that. :redface:
     
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