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Why is the age of the Universe the reciprocal of the Hubble constant?

  1. Sep 10, 2012 #1
    According to the wikipedia entry, the latest values for the Lambda-CDM model parameters for the age of the Universe, [itex]t_0[/itex], and the Hubble constant, [itex]H_0[/itex] are

    [itex]t_0 = 13.75 \pm 0.11 \times 10^9 \mbox{ years}[/itex]
    [itex]H_0 = 70.4 \pm 1.3 \mbox{ km s}^{-1} \mbox{Mpc}^{-1}[/itex]

    If you combine the errors this implies the following relationship

    [itex]t_0 H_0 = 0.99 \pm 0.02[/itex]

    Why is the age of the universe the reciprocal of the Hubble constant to within experimental error?

    Is this just a coincidence?

    It almost seems that the entire Lambda-CDM model could simply be summarized by

    [itex] a(t) = H_0 t [/itex].
     
  2. jcsd
  3. Sep 10, 2012 #2

    cepheid

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    Short answer: it's not in general, but in the standard LCDM model, they're pretty close.

    Yeah, as far as I know.

    Not even remotely. This relationship doesn't hold true in any reasonable Friedmann world model. You need to solve the Friedmann equation to get the dependence of scale factor with time, and it's certainly not linear.
     
  4. Sep 10, 2012 #3
    As pointed out above, it is just a good approximation to the age of the universe.
    In simple terms this is due to the Hubble parameter not being constant, varying depending on whether the universe expansion accelerates, decelerates or neither one thing nor the other.
    Simplifying, this is also the main reason the L-CDM cannot be summarized as you propose, and the scale factor is not a linear function of time, because Ho is not a constant.
     
  5. Sep 11, 2012 #4
    The linear relationship

    [itex] a(t) = H_0 t [/itex]

    solves the Friedmann equations for [itex]k = 0[/itex] and [itex]\Lambda=0[/itex]

    provided one has the equation of state

    [itex] p = -\frac{\rho c^2}{3} [/itex]

    Thus when applied to the Friedmann equations the linear expansion solution requires the existence of a kind of dark energy with repulsive gravity.
     
    Last edited: Sep 11, 2012
  6. Sep 11, 2012 #5
    But the linear scale factor equation

    [itex] a(t) = H_0 t [/itex]

    does not assume that the Hubble parameter is constant.

    In this model the Hubble parameter is given by

    [itex] H = \frac{\dot{a}}{a} = \frac{1}{t} [/itex]

    which is not constant.
     
  7. Sep 11, 2012 #6

    Garth

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    Actually H0 is constant. It is H, the Hubble parameter that is not constant, H0 is the present value of that parameter.

    The coincidence referred to in the OP is the equality within observational error bars of the present value of the inverse of H, 1/H0 or present Hubble Time, TH, and the present age of the universe, t0.

    Without DE and with Ω =1 the age of the universe t0 would be the 2/3TH, if Ω < 1 then t0 would lie between 2/3TH and TH, depending on the value of the actual density parameter and the coincidence would not be remarkable.

    However if we add DE to the 'cosmic mix' then t0 could be anything from 2/3TH → ∞, depending on the amount of DE you add.

    Why is it then that in the observed universe there is just enough DE for t0 = TH in the present epoch?

    Some consider this might point to any alternative cosmology which does linearly expand so that t1 = TH1 in every epoch. (Where the subscript 1 refers to the values of H, TH and t at that epoch), so in the last post by johne1618 should read:

    The linear relationship is:

    [itex] a(t) = Kc t[/itex] where K is some constant that can be normalised by choice of units to 1.

    i.e. [itex] a(t) = c t[/itex]

    and then

    [tex] H_{1} = \frac{\dot{a}}{a} = \frac{1}{t_{1} } [/tex]


    Such a cosmology is generally called A coasting cosmology (Astrophysical Journal, Part 1 (ISSN 0004-637X), vol. 344, Sept. 15, 1989, p. 543-550.) and some who have looked at it find that it might be concordant with observation. ( A Concordant “Freely Coasting” Cosmology)

    Also you may enjoy The Cosmic Spacetime.

    One advantage of such a model (in which the equation of state ω = -1/3) would be that it would not require Inflation to resolve the horizon, flatness, or smoothness problems of the standard model as they would not exist in the first place.

    Just a thought...
    Garth
     
    Last edited: Sep 11, 2012
  8. Sep 11, 2012 #7

    cepheid

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    Hold on a minute. You said, and I quote:

    LCDM refers to the standard or "concordance" model of cosmology, which is so-named because its parameter values (Ωm = 0.27, Ωλ = 0.73) are consistent with multiple sets of observations, and it is the one that we think most closely corresponds to reality. Therefore, that is what everyone in this thread has assumed you have been referring to. The statement quoted directly above is patently false for Lambda-CDM, which is why I objected. Now you try to justify it by proposing a totally different cosmological model that has no Lambda, and no matter! How is that even remotely the "Lambda Cold Dark Matter" model?

    I will grant you that if P = -ρ/3, then the Friedmann acceleration equation says that ##\ddot{a} = 0##, which implies that ##\dot{a} = \textrm{const.}## But from this, one CAN'T presume that H = const., since ##H = \dot{a} / a##. In any case, what you have done is tried to assume a priori a Friedmann world model in which ##\dot{a}## = const. and therefore a is linear. What we have been trying to tell you in this thread is that this is NOT true in general for any Friedmann world model, and it is not true for the one corresponding to LCDM. Here's a link that shows how the scale factor evolves with time for various models, including LCDM (credit to marcus, another member here, for posting this): http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure14.jpg

    Note: I severely edited my post, because I saw that I made and error and that some of what you were saying in (esp in #5) is right, but again, it bears no relation to LCDM.
     
    Last edited: Sep 11, 2012
  9. Sep 11, 2012 #8
    Thanks for the reference - very interesting!

    I'm definitely a fan of Fulvio Melia's R = ct cosmology.
     
    Last edited: Sep 11, 2012
  10. Sep 11, 2012 #9
    I agree that I shouldn't have asserted that a linear expansion model "summarized" the Lambda CDM model.

    Basically I'm saying that if one accepts Fulvio Melia's justification for a linear expansion model then the fact that the age of the Universe is the reciprocal of the Hubble parameter is not surprising.
     
    Last edited: Sep 11, 2012
  11. Sep 11, 2012 #10
    Fulvio Melia's R = ct model

    What do people think of Fulvio Melia's R = ct model described in The Cosmic Spacetime ?

    By arguing that the Hubble radius is equivalent to the radius of a gravitational horizon he derives a very simple linear model of the Universal expansion.

    He has demonstrated in a number of arXiv papers that such a model could explain many cosmological observations on its own without the need to assume inflation etc.
     
  12. Sep 12, 2012 #11
    Re: Fulvio Melia's R = ct model

    I don't think people take it very seriously. One thing about some theories is that they fall into the category of "so broken no one bothers arguing against it". I think this falls into that category. The papers are publishable because he manages to come up with some "smoking gun" observational signatures of his models. I'd bet a substantial amount of money that those observations will come up the other way.

    Yes. The trouble is that you then have to explain why gravity works in such a way to set up that rate of expansion. If expansion is linear then gravity doesn't exist.

    Yes. If you slow the expansion rate of the universe, you no longer have the horizon problem. You've thrown away general relativity in the process.

    But for every cosmological thing that he does explain, there are a dozen big ones that he doesn't. The two big ones are big-bang nucleosynthesis and the galaxy power spectrum. Those calculations are not difficult to do, and if you put in a(t) is proportional to t, then everything falls apart. The fact that he hand waves those away leads me to strongly suspect that he's done the calculations himself and he can't make it work. If he runs a(t) proportional to t, and gets the right amount of He4, I'm pretty sure that he would use that as evidence for his model.

    But the papers are publishable because he's come up with a "smoking gun" test for his theories, and if God is in a good mood, then when people see that smoking gun, they'll take his ideas seriously and give him a Nobel prize.
     
    Last edited: Sep 12, 2012
  13. Sep 12, 2012 #12
    Re: Fulvio Melia's R = ct model

    Also, this is a good example of how to present an "nutty" idea in a way that people will take you seriously. One thing that you get from "cranks" is that the idea that anything that goes against the prevailing wisdom won't get a fair hearing. In this situation, the idea that R=ct is "nutty" but it's presented in a way that passes peer review.

    The reason it's a useful paper is that its more than "I've got this crazy new idea for how to set up the universe" but rather "I've got this crazy new idea for how to set up the universe and here are the consequences of that idea, one of which is a "smoking gun" that shows that I'm right or wrong."
     
    Last edited: Sep 12, 2012
  14. Sep 12, 2012 #13

    Haelfix

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    Re: Fulvio Melia's R = ct model

    Sorry if this is getting OT, but it is actually quite common that crackpot papers contains falsifiable predictions (usually by some experiment that is conveniently dated a good interval of time into the future).

    The cranks then bang the drum, produce a conspiracy theory about how the establishment is keeping them down and violating their own principles and how their own theories are more 'scientific' than some of the non falsifiable papers that the establishment produces etc

    What is typically lost in translation and what many well intentioned laymen don't understand about physics (and especially theoretical physics), is that the falsifiable part really only comes at the end of the journey, not the beginning. Theoretical physicists aren't interested in theory A b/c it makes some random prediction in an experiment: eg "my trinity particle has mass x". People are interested b/c the detailed arguments seem to follow from preexisting mathematic lore in a novel, selfconsistent and interesting way.

    Once that interest is established, only then do people look for ways to scythe the theory in such a way as to be able to rule it out. Sometimes this isn't possible even in principle, for instance "the interpretations of quantum mechanics", but that doesn't stop scientists from exploring their consequences.
     
  15. Sep 12, 2012 #14

    Garth

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    Re: Fulvio Melia's R = ct model

    Who are you calling a crank?

    Fulvio Melia is not a crank, he is Professor of Physics and Astronomy at the University of Arizona, and is a member of the Theoretical Astrophysics Program. He is also the Series Editor for Theoretical Astrophysics with the University of Chicago Press.

    He has 186 refereed publications to his name, this year's being:

    The Astrophysical Journal (Letters), submitted (2012): "The Inevitability of Omega_m=0.27 in LCDM," F. Melia

    The Astrophysical Journal (Letters), submitted (2012): "High-z Quasars in the R_h=ct Universe," F. Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2011): "Two-Body Dark Matter Decay and the Cusp Problem," K. McKinnon and F. Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2011): "Anisotropic Electron-Positron Annihilation at the Galactic Centre," T. M. McClintock and F. Melia

    The Astrophysical Journal (Letters), submitted (2006): "Periodic Modulations in an X-ray Flare from Sagittarius A*," G. Belanger, R. Terrier, O. de Jager, A. Goldwurm, and Fulvio Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2012): "Proper Size of the Visible Universe in FRW Metrics with Constant Spacetime Curvature," Fulvio Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2012): "CMB Multipole Alignment in the R_h=ct Universe," Fulvio Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2012): "Angular Correlation of the CMB in the R_h=ct Universe," Fulvio Melia

    Monthly Notices of the Royal Astronomical Society, submitted (2011): "The Rh=ct Universe Without Inflation," Fulvio Melia

    Journal of Cosmology and Astroparticle Physics (JCAP), in press (2012): "The Cosmic Horizon for a Universe with Phantom Energy," Fulvio Melia

    The Astrophysical Journal (Letters), 757, L16 (2012): "Diffusive Cosmic-ray Acceleration in Sagittarius A*," Marco Fatuzzo and Fulvio Melia

    The Astronomical Journal, 144, 110 (2012): "Analysis of the Union2.1 SN Sample with the R_h=ct Universe," Fulvio Melia

    Australian Physics, 49, 83 (2012): "The Cosmic Spacetime," Fulvio Melia

    Monthly Notices of the Royal Astronomical Society, 422, 1418 (2012): "Cosmological Redshift in FRW Metrics with Constant Spacetime Curvature," Fulvio Melia

    The Astrophysical Journal, 750, article id. 21 (2012): "Assessing the Feasibility of Cosmic-Ray Acceleration by Magnetic Turbulence at the Galactic Center," M. Fatuzzo and F. Melia

    Monthly Notices of the Royal Astronomical Society, 421, 3356 (2012): "Photon Geodesics in FRW Cosmologies," Ojeh Bikwa, Fulvio Melia, and Andrew Shevchuk

    Monthly Notices of the Royal Astronomical Society, 419, 2579 (2012): "The Rh=ct Universe," Fulvio Melia and Andrew Shevchuk

    Monthly Notices of the Royal Astronomical Society, 419, 2489 (2012): "Polarimetric Imaging of Sgr A* in its Flaring State," Fulvio Melia, Maurizio Falanga, and Andrea Goldwurm

    You might also want to discuss his paper Angular Correlation of the CMB in the Rh = ct Universe published in Monthly Notices of the Royal Astronomical Society, submitted (2011). The linearly expanding model cannot be dismissed as a "nutty idea", if you are going to do that then some people might want to argue that the standard LCDM model is "nutty" as it depends on Inflation, Dark Matter and Dark Energy, all hypothetical concepts that have no laboratory confirmation whatsoever.

    Garth
     
  16. Sep 12, 2012 #15
    First you need a constant to have a linear equation.
    Also see Garth's correction, H is not a constant, Ho is H at the present time, you cannot simply assume that H at the present time has always been and will always be the same. The L-CDM model certainly doesn't assume it and that's one reason it can't be summarized with a linear equation.
     
  17. Sep 12, 2012 #16
    I know H is not constant.

    In the linear model it is given by

    [itex] H(t) = \frac{1}{t} [/itex]
     
  18. Sep 12, 2012 #17
    Question about Fulvia Melia's gravitational horizon

    I have a question about Fulvio Melia's derivation of a gravitational radius around any observer in a homogeneous and isotropic Universe in The Cosmic Spacetime.

    He says that if you consider a spherical region of mass centered on the observer A then an observer B on the surface of this mass will experience a gravitational acceleration relative to and towards observer A.

    But if the Universe is homogeneous and isotropic then the gravitational potential at each point should be the same.

    Thus there is no difference between the gravitational potentials at A and B and thus there should not be any relative acceleration between them.

    I think Fulvio Melia's argument only works if the Universe is spherically symmetric about observer A. Of course this would go against the Copernican principle as it would put observer A in a priveleged position. The Universe is homogeneous and isotropic rather than being spherically symmetric about a unique point.

    (I personally think his R = ct Universe is still valid but that it has to be derived by assuming that the gravitational potential has the value [itex]-c^2[/itex] at every point. Thus every particle's rest mass energy [itex]m c^2[/itex] is balanced by its gravitational potential energy [itex]-m c^2[/itex] giving a zero-energy Universe. The radius of the spherical mass around each point required to produce this potential is what Melia calls the gravitational radius.)
     
    Last edited: Sep 12, 2012
  19. Sep 12, 2012 #18

    Garth

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    Re: Question about Fulvia Melia's gravitational horizon

    Two points johne1618:

    First you have started a second thread on a subject currently being discussed in Fulvio Melia's R = ct model , I wonder if a Moderator would like to combine the two as the discussion will get crossed?

    In response to your point here; I think Fulvio Meilia's argument point was that if the universe is homogeneous and isotropic then it could also be considered spherically symmetric about every point.

    Garth
     
  20. Sep 12, 2012 #19
    Re: Question about Fulvia Melia's gravitational horizon

    I agree.

    I agree with that definition.

    But surely in practice if observer B releases a particle, observer A is not going to see it accelerate towards himself (or vice-versa)?

    Both observers assume exactly the same spherically symmetric mass distributions around each other. They know that if they release a particle it has zero acceleration relative to themselves. Thus each will expect that particles released by the other will have zero acceleration relative to themselves.
     
    Last edited: Sep 12, 2012
  21. Sep 12, 2012 #20

    Garth

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    Re: Fulvio Melia's R = ct model

    To get the discussion off the ropes and onto a good footing a contrary view has been expressed in
    We do not live in the Rh = ct universe.

    This alternative Rh = ct model certainly has many problems to face in order for it to be concordant with all the cosmological observations. But I would point out that back in the 1970's the standard BB model had problems, which would be far greater with today's 'precision cosmology' observations. After forty years of work with the invention of inflation, Dark Matter and Dark Energy - all unverified in the laboratory, the LCDM model works well, but would getting the Rh = ct universe model to work be any easier? We won't know unless it is given a fair chance.

    After reviewing Melia's and Shevchuk's paper The Rh = ct Universe which basically makes the same case as The Cosmic Spacetime, Bilicki and Seikel argue
    I would point out that 1). Getting the w = -1 EOS to work in the standard model results in a ~10-120 mismatch with QM expectation and 2) The cosmological solution of Self Creation Cosmology An Alternative Gravitational Theory yields an EOS
    w = -1/3 as a matter of course.


    Just a thought,
    Garth
     
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