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Just for fun: recast the Friedmann eqn. to be about volume instead

  1. May 23, 2015 #1

    marcus

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    What if the Friedmann equation told how the (square of) fractional growth rate of spatial volume relates to the energy density, instead of describing how the (square of) fractional growth rate of spatial distance relates to energy density?

    Just two slightly different ways of describing the same physical relationship--how scale and energy density interact and change over time.

    Maybe someone here has seen the Friedmann formulated that way, and can give us a pointer to it. I don't recall seeing it. And I think it's really just on the level of a fun exercise--by which one might get to know the Friedmann a little better. I'll just consider the spatially flat case, with matter dominant over radiation. To keep it simple and manageable, think of radiation as a negligible part of the overall energy density. The cosmological curvature constant Λ is just that, here, (not related to any imagined form of "dark energy".) The ordinary Friedmann:

    H(t)2 - Λ/3 = [const]ρ(t)

    where H = a'/a is the fractional growth rate of the (linear) scale factor, and [const] is some constant which relates the energy density ρ to the excess square expansion rate (eser) over and above the residual amount inherent in spacetime geometry. You can see that as density goes to zero the excess square expansion rate (eser) goes to zero and the growth rate H must decline and approach Λ/3 in the limit.
    H2 = Λ/3
    H(t)2 - H2 = [const]ρ(t)

    The present energy density ρnow determines the present excess square expansion rate
    Hnow2 - H2 = [const]ρnow
    Since by assumption:
    H(t)2 - H2 = [const]ρnow/a(t)3
    we have:
    H(t)2 - H2 = (Hnow2 - H2 )/a(t)3

    this is the kind of thing we could "transpose into a different key". We'd see how it would work when, instead of being about the linear scale factor a(t) and its fractional growth rate a'/a it is adapted to be the equivalent equation about a volumetric scale factor f(t) and its fractional growth rate f'/f.

    Suppose we have a function f(t) that tracks volume as the U expands. What "Friedmann" equation would it satisfy? We'd better not call the fractional growth rate f'(t)/f(t) of volume by the letter H(t).
    Might cause confusion. Maybe we can call it J(t) the "Jubble" instead of the "Hubble"? Or perhaps we don't need to. Maybe the volume growth can be described by 3H(t) using the same growth rate function H(t) that we use for distances. The growth rate might not care whether it is used to describe growth of distances or volumes---except for a factor of 3.

    I'll leave this here for the time being. Nice to get some response, if any be forthcoming.
     
    Last edited: May 24, 2015
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  3. May 24, 2015 #2

    marcus

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    To recapitulate, here is a version of the Friedmann equation (spatial flat, post radiation era)
    H(t)2 - H2 = (Hnow2 - H2 )/a(t)3

    It's the kind of thing we could "transpose into a different key". We'd see how it would work when, instead of being about the linear scale factor a(t) and its fractional growth rate Hubble H(t) = a'/a it is adapted to be the equivalent equation about a volumetric scale factor f(t) and its fractional growth rate (call it "Jubble") J(t) = f'/f.

    And BTW we know that a nice hypertrig f(t) = sinh2(1.5 t) tracks volume. Is this going to work?

    What I have in mind is something like this:
    J(t)2 - J2 = (Jnow2 - J2 )/f(t)
     
    Last edited: May 24, 2015
  4. May 24, 2015 #3

    marcus

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    First let's apply a little easy calculus to make sure that we have J(t) = f'(t)/f(t)
    IOW that we really do have the fractional growth rate of volume
    f(t) = sinh2(1.5 t)
    f'(t) = 2sinh(1.5 t)cosh(1.5 t)3/2 = 3sinh(1.5 t)cosh(1.5 t)
    f'(t)/f(t) = 3 sinh(1.5 t)cosh(1.5 t)/sinh2(1.5 t) = 3 cosh(1.5 t)/sinh(1.5 t) = 3 coth(1.5 t)
    That looks right. The fractional growth rate of volume should be 3 times that of length. But I'm still worried. Is this going to work out?

    We still have to multiply f(t) by 0.443 to normalize it, so that our volume tracker equals one at the present.
    that won't change the f'/f fractional growth rate.

    Let's divide through by J2 the way we did earlier with H2
     
    Last edited: May 24, 2015
  5. May 24, 2015 #4

    marcus

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    $$\frac{J(t)^2}{J_\infty^2} - 1 = (\frac{J_{now}^2}{J_\infty^2} - 1)/f(t)$$

    where f(t) = 0.443sinh2(1.5t)
    and J(t) = f'(t)/f(t) = 3coth(1.5t)

    f(t) is the volumetric scale factor---the size of a generic volume, normalized to equal one at the present.
    J(t) is its log derivative, IOW fractional growth rate.
    The present is t = 0.797
     
  6. May 24, 2015 #5

    Chalnoth

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    Seems like it'd be rather more complicated to me. For example, the spatially-flat metric would change from:

    [tex]ds^2 = dt^2 - a(t)^2\left(dx^2 + dy_2 + dz^2\right)[/tex]

    To:

    [tex]ds^2 = dt^2 - f(t)^{2 \over 3}\left(dx^2 + dy_2 + dz^2\right)[/tex]

    The asymptotic form of the result that you wrote down isn't so horrible, but most calculations would be made significantly more complicated.
     
  7. May 24, 2015 #6

    marcus

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    Why would you change that to that? I would keep
    [tex]ds^2 = dt^2 - a(t)^2\left(dx^2 + dy^2 + dz^2\right)[/tex]
    for the metric.
    I like the linear scale factor a(t) for most purposes.
     
  8. May 24, 2015 #7

    marcus

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    So, to recap, here (for fun and because we might learn something by looking at it from a new angle) is the Friedmann equation transposed to be about volume scale instead of distance scale
    $$\frac{J(t)^2}{J_\infty^2} - 1 = (\frac{J_{now}^2}{J_\infty^2} - 1)/f(t)$$

    where f(t) = 0.443sinh2(1.5t)
    and J(t) = f'(t)/f(t) = 3coth(1.5t)

    f(t) is the volumetric scale factor---the size of a generic volume, normalized to equal one at the present.
    J(t) is its log derivative, IOW fractional growth rate.
    The present is t = 0.797

    Notice the simple cancelation on the righthand side. If you looked at one or more of the other threads (from Aeon to Zeon...) you know that 0.443 is a number that introduces a handle on the present into the model. And it keeps coming up over and over again: otherwise the model has no idea where the present is.
    $$ \frac{J_{now}^2}{J_\infty^2} - 1 = \frac{H_{now}^2}{H_\infty^2} - 1 = 0.443$$

    Furthermore 0.443 is what you have to multiply the volumetric scale factor sinh2(1.5t) by to normalize it so that it equals one at present.

    So this form of the Friedmann, when you introduce the functions which SOLVE the equation, boils down to a familiar hypertrig identity. The 0.443 in the numerator and denominator cancel.

    $$\frac{J(t)^2}{J_\infty^2} - 1 = (\frac{J_{now}^2}{J_\infty^2} - 1)/f(t)$$
    $$\frac{J(t)^2}{J_\infty^2} - 1 = \frac{0.443}{0.443\sinh^2(1.5t)}$$
    $$\coth^2(1.5t) - 1 = \frac{1}{\sinh^2(1.5t)}$$
     
    Last edited: May 24, 2015
  9. May 25, 2015 #8

    ChrisVer

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    Why not taking the [itex]a^3(t)[/itex] and look at its time derivative?

    [itex] J(t) \equiv \frac{d a^3(t)/dt }{a^3(t)} = 3 a^2 (t) \frac{\dot{a}(t)}{a^3(t)} = 3 H(t)[/itex]
     
  10. May 25, 2015 #9

    marcus

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    No reason not to! I did that too, several times :smile:
    What you get is a useful equation. I use that J = 3H a lot in this thread, in fact.

    But it does not solve the problem, which is to find an analogue of the usual version of the Friedmann equation:
    to see what the Friedmann equation might look like if it were about volume instead of distance.
     
  11. May 25, 2015 #10

    ChrisVer

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    Where in the Friedmann equation do you use the distance? The 1st Friedmann equation tells you how the scale factor scales with time due to the existence of energy densities...
    So I would replace H in it with J/3 (it would be the same equation). The equation itself is not supposed to change, at least not without changing the metric... It's a solution of your current metric's EFE afterall.
     
  12. May 25, 2015 #11

    marcus

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    Here's a problem for you, Chris!

    Suppose you go into suspended animation (as in cryonic storage) and wake up some time in the future.
    The people who re-animate you are very good at measuring temperature but they have lost all the records of history and do not know about cosmic expansion either. How do you discover the age of the universe?
     
  13. May 25, 2015 #12

    marcus

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    Our posts crossed. Sure that is essentially what I did. then I divided thru by J.
    And because I like it that way, I expressed the DENSITY on the righthand as the presentday density divided by the volume factor (so if the volume is half of today's the density will be twice).

    Friedmann equation has a variety of versions. If you would like to, how about you write the form you prefer and transpose that to be about volume rather than distance. In a way it's completely trivial. You could, as you suggest, simply replace H by J, take care of the factor of 3, and leave it like that.

    I was interested in going into it a little deeper and trying out a volumetric scale factor f(t) which is analogous to a(t) but tracks volume instead of distance.
     
    Last edited: May 25, 2015
  14. May 25, 2015 #13

    marcus

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    In case anyone wants to think about it, here's a simple problem. These people have discovered the microwave background and measured its temperature but they don't know what it is. they have no idea of the age of the universe.

    You have just woken up, they can't tell you how long you slept, you want to know what time it is.

    Then you get an idea. You ask them what is the temperature of the microwave background!

    Somehow they explain their temperature scale to you (the melting point of ice, and so on) so you can translate into Kelvin. You learn from them that the CMB temperature is 1.3625 Kelvin.
     
  15. May 25, 2015 #14

    ChrisVer

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    What I am saying is that this [itex]f(t)[/itex] will have to be [itex]a^3 (t)[/itex] ...because it's the cubic power of the scale factor that gives the scaling of the volume. I don't think this inserts anything interesting (?).

    As for the problem, I guess you can measure the redshift of the CMB temperature, and answer your question by the known integral of the U age...
     
  16. May 25, 2015 #15

    marcus

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    Thanks for responding! I'm interested in how our methods would compare. You would say, I guess, that the redshift is about twice 1090 because the temperature is half---1.3625 is half of 2.725.
    And then you would use an integral (?) to find the U age?

    I would not use an integral but rather the two formulas given on page one "Aeon to Zeon" thread.
    In this case the s factor going from the present to the time in the future is s = 0.5 (our distances are half the ones in the future. So the calculation would go:
    $$ H = \sqrt{.443*0.5^3 +1} = \sqrt{.443/8 +1} =\sqrt{1.0554} = 1.027326$$
    $$T = \ln(\frac{2.027326}{0.027326})/3 = 1.4355$$

    Multiply that times 17.3 billion years to get the answer in terms of the usual billion years.
    It comes to year 24.83 billion. Agrees with Jorrie's Lightcone to 4 significant figures.
     
    Last edited: May 25, 2015
  17. May 26, 2015 #16

    ChrisVer

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    I mean the solution to the age of the universe can be obtained from the Friedman equation:

    [itex] \dot{a}/a = H[/itex]
    [itex] dt = \frac{da}{\dot{a}}= \frac{da}{a H^{1/2} } = \frac{da}{aH_0 (\Omega_{m0} a^{-3} + \Omega_{r0} a^{-4} + \Omega_{\Lambda} )^{1/2} }[/itex]

    Or better written:

    [itex] dt =\frac{da}{H_0(\Omega_{m0} a^{-1} + \Omega_{r0} a^{-2} + \Omega_{\Lambda} a^{2} )^{1/2} }[/itex]

    You can neglect the radiation density:

    [itex] T_{uni}= \int dt = \int \frac{da}{H_0(\Omega_{m0} a^{-1} + \Omega_{\Lambda} a^{2} )^{1/2} }[/itex]

    Then you only need to replace the ##a## with the redshift and do the appropriate integration.
     
  18. May 26, 2015 #17

    marcus

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    I assume that if you worked out the integral you would get the same answer I did, namely 24.83 billion years. My method is based on the Friedmann equation, and as you point out yours is also based on the Friedmann equation. Aside from round-off error they should give the same answer.

    I'm curious to know if you see any pedagogical value in introducing this kind of problem and showing a quick arithmetical path to the answer. Is anything useful learned?
     
  19. May 26, 2015 #18

    marcus

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    For example, what will the expansion age be when the CMB temperature is 2.18 kelvin instead of present 2.725 kelvin?
    Here is what you's paste into google to get the answer in H units:
    ln(((.4433*.8^3+1)^(1/2)+1)/((.4433*.8^3+1)^(1/2)-1))/3

    Or, if you want it in terms of billions of years, you would multiply by 17.3, or simply paste in:
    17.3*ln(((.4433*.8^3+1)^(1/2)+1)/((.4433*.8^3+1)^(1/2)-1))/3

    The point is 2.18/2.7525 = 0.8, so our distances now are 80% of those in future, so as you can see I put the cube of 0.8 into the calculation.

    In case you want to work it out with your integral and see if you get the same answer, I'll paste my expression in and tell you want it works out to.

    Google calculator says 17.15 billion years. (That would have been about 0.99 zeon if we wanted it in H terms.)
     
  20. May 26, 2015 #19

    marcus

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    In case anyone new to the topic happens to read the thread here, the s factor is the size of distances and wavelengths now compared with those in the past or future. s = 1/a, the reciprocal of the normalized scale factor.
    $$H = \sqrt{0.4433s^3 + 1}$$$$t = \ln(\frac{H+1}{H-1})$$
     
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