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Hubble lookback time based on energy density.

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    If the energy density of the vacuum were the value 10^133 eV/m^3 , what would the value of the Hubble lookback time be for such a universe with no curvature and no other matter?

    2. Relevant equations
    Possible equation...

    ...Lookback time = ln(1+z)/H

    3. The attempt at a solution
    I am not sure what to assume for z? And should I convert the energy density to kg/m^3. I am very confused, any direction for this question would be IMMENSELY appreciated.
     
  2. jcsd
  3. Oct 3, 2015 #2

    marcus

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    Tell us what you mean by Hubble lookback time.

    Do you mean how far back in time a specific telescope e.g. the HST can see?
     
  4. Oct 3, 2015 #3

    marcus

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    There is also a well-defined conventional quantity known as the Hubble time
    it is not a "lookback" time and it is not associated with any particular telescope. It is just 1/H where H is the usual Hubble expansion rate "constant" (which is actually not constant over time)
    The current value of the Hubble time is about 14.4 billion years.
    Not to be confused with any "lookback" time.
     
    Last edited: Oct 3, 2015
  5. Oct 4, 2015 #4
    R
    Right. So what my professor wants to know is what is the lookback time for a flat universe, with no matter, and a density of 10^133 eV/m . Would you know how to go about starting this? OmegaM would be 0 and OmegaK would be 0 right?
     
  6. Oct 4, 2015 #5

    marcus

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    I'm not the right person to help you. In my universe "lookback" time doesn't mean anything unless you specify a telescope or an object you are looking at. If your professor were asking about the Hubble time then I would know how to proceed.

    I would just multiply the energydensity by 8piG/3c^2 and that would give me H^2 and then the Hubble time is just 1/H and we're done.
    It would be a simple application of the Friedmann equation and the Hubble time is constant in that situation.
     
  7. Oct 4, 2015 #6
    I think I will just do that, as that is what some of my peers are doing. Thanks so much for the assistance.
     
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