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It's a rate (not a constant) so let's call it that!

  1. Apr 5, 2013 #1

    marcus

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    One of the main points of cosmology is that the Hubble rate of distance expansion is not constant. In the standard cosmic model it evolves according to the Friedman equation. Recently I've been noticing more professional cosmologists referring to it as a rate, like these people (Roy Maartens is a prominent world-class figure, Clarkson is also a leader in the field)

    http://arxiv.org/abs/1205.3431
    Using H(z) data as a probe of the concordance model
    Marina Seikel, Sahba Yahya, Roy Maartens, Chris Clarkson (Cape Town and Western Cape)
    (Submitted on 15 May 2012)
    Direct observations of the Hubble rate, from cosmic chronometers and the radial baryon acoustic oscillation scale, can out-perform supernovae observations in understanding the expansion history, because supernovae observations need to be differentiated to extract H(z). We use existing H(z) data and smooth the data using a new Gaussian Processes package, GaPP, from which we can also estimate derivatives. The obtained Hubble rate and its derivatives are used to reconstruct the equation of state of dark energy and to perform consistency tests of the LCDM model, some of which are newly devised here. Current data is consistent with the concordance model, but is rather sparse. Future observations will provide a dramatic improvement in our ability to constrain or refute the concordance model of cosmology. We produce simulated data to illustrate how effective H(z) data will be in combination with Gaussian Processes.
    Comments: 9 pages, 8 figures.

    Or take this earlier paper by Chris Clarkson and friends.

    http://arxiv.org/abs/1011.3959
    The Hubble rate in averaged cosmology
    Obinna Umeh, Julien Larena, Chris Clarkson (ACGC, University of Cape Town)
    (Submitted on 17 Nov 2010)
    The calculation of the averaged Hubble expansion rate in an averaged perturbed Friedmann-Lemaitre-Robertson-Walker cosmology leads to small corrections to the background value of the expansion rate, which could be important for measuring the Hubble constant from local observations. It also predicts an intrinsic variance associated with the finite scale of any measurement of H0, the Hubble rate today. Both the mean Hubble rate and its variance depend on both the definition of the Hubble rate and the spatial surface on which the average is performed. We quantitatively study different definitions of the averaged Hubble rate encountered in the literature by consistently calculating the backreaction effect at second order in perturbation theory, and compare the results. We employ for the first time a recently developed gauge-invariant definition of an averaged scalar. We also discuss the variance of the Hubble rate for the different definitions.
    Comments: 12 pages, 25 figures.
     
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2

    marcus

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    The use of misleading traditional terminology (established by historical accident) can confuse newcomers, so I'm glad to see the older term "Hubble constant" starting to be replaced in professional research writing.

    One of the impressive things about cosmology is how rapidly the Hubble rate changed in the past and how slowly we expect it to change in the distant future! According to standard model by year 60 billion the rate will be almost constant at 1/173 percent per million years.

    The RECIPROCAL of the Hubble rate, namely 1/H (for the moment let's denote it R for reciprocal, R = 1/H) has of course been increasing as the rate H declines. And as the rate gradually levels out and becomes essentially constant at 1/173% per My, the reciprocal, called the "Hubble time" will essentially stop increasing and level off at 17.3 billion years. I'm using the latest parameters here--from the Planck mission report.
     
  4. Apr 5, 2013 #3

    marcus

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    To illustrate the rapid past change in R = 1/H and, by contrast, slow change in future.

    [tex]{\begin{array}{|c|c|c|c|c|c|c|}\hline R_{now} (Gy) & R_{inf} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline14.4&17.3&3400&67.92&0.693&0.307\\ \hline\end{array}}[/tex] [tex]{\begin{array}{|r|r|r|r|r|r|r|} \hline S=z+1&a=1/S&T (Gy)&R=T_{Hub}(Gy)&D (Gly)&D_{then}(Gly)&D_{hor}(Gly)&D_{par}(Gly)\\ \hline1090.000&0.000917&0.000373&0.000628&45.332&0.042&0.057&0.001\\ \hline1.000&1.000000&13.787206&14.399932&0.000&0.000&16.472&46.279\\ \hline0.040&25.000000&68.366857&17.299636&-15.809&-395.235&17.300&1551.841\\ \hline\end{array}}[/tex]

    Back at S=1090 when the background radiation was emitted, that we are now receiving, a generic distance was expanding at the rate of 159% per million years.
    That's because 1/0.00628 = 159

    By contrast at the present S=1, the rate is 1/144% per million years. To take an example, a sample distance of 1 billion lightyears is expanding at 1/14.4 c. A bit less than 1/14 of the speed of light.

    In the distant future, around year 68 billion, the rate is projected to be 1/173% which is to say that a sample distance of 1 billion lightyears will be expanding at 1/17.3 c, or around 1/17 of the speed of light.

    That is not much different from what it is now, only slightly smaller in fact, and it is what the rate is converging to according to the standard picture.

    However notice that our sample 1 billion lightyears distance would at that S=1090 epoch have been expanding at 1590 c, or almost 1600 times the speed of light. Between the time the CMB radiation was emitted and the present day there has, in other words, been a big slowdown in the speed a given-size distance is expanding.
     
    Last edited: Apr 5, 2013
  5. Apr 5, 2013 #4
    This raises a valid point one I hadn't considered thank you for pointing it out the term Hubble rate would prevent confusion. by the way good articles
     
  6. Apr 6, 2013 #5
    Thank you Marcus. It's actually a good point and something that I did experience last week. I discussed the hubble "constant" with someone and I said that the hubble "constant" is not constant through time :confused::confused: ??? and noticed the very bad logic in the words that stumbled out of my mouth... From now on I'll call it the hubble rate - it will probably not confuse as much, just as you have mentioned.

    Best regards,
    Robin Andersson
     
  7. Apr 6, 2013 #6

    cepheid

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    I'm used to seeing H(t) referred to as "the Hubble parameter", and H(t0) = H0 as "the Hubble constant." However, I agree that it seems strange to cherry pick a particular value of a time-variable quantity and refer to that value as being a "constant." I think the term is a result of the history of its discovery. When Hubble's law was discovered, it was found that no matter how far away a galaxy was, the ratio of its rate of recession (as given by the proxy of redshift) to its distance (as given by the proxy of magnitude) was constant. This was indicative of a uniform expansion. So, in that sense, at the present epoch, there is a quantity that is constant (spatially) that characterizes the uniform expansion that the universe is presently undergoing, and this is the context in which the term "constant" makes sense. What do you think?
     
  8. Apr 6, 2013 #7

    marcus

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    I think you are right. In fact that was the "historical accident" I was referring to in post #2.

    For Hubble, as he was measuring it, the expansion rate was indeed a constant. He was measuring over a comparatively short time range and distance range. Presumably he was not, at that point, giving any thought to the longterm evolution of that "constant" rate of expansion.

    So "constant" was the natural word to use at the time, and the name stuck.
     
  9. Apr 7, 2013 #8

    Jorrie

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    One must then also be on guard not to confuse the changing 'Hubble rate' with the changing 'expansion rate', because they are two related, but different things: the Hubble rate is [itex]\dot{a}(t)/a(t)[/itex] while expansion rate is just [itex]\dot{a}(t)[/itex].
     
  10. Apr 7, 2013 #9
    Thanks I already realized that however it is an important distinction

    As interesting as the second article is its rather tricky to go through mind I dont fully understand enough on perturbation theory. In particular what distinquishes between first order and second order perturbations. Answering that however is best done in another thread
     
    Last edited: Apr 7, 2013
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