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Calculating critical density of mass, from Hubble time, for exercise

  1. Apr 5, 2014 #1

    marcus

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    Jorrie's calculator (Lightcone) makes cosmic history tables which tell you among other things the Hubble times in past years. For convenience let's temporarily use greek Theta Θ to stand for THubble so we don't have to write so much.
    Basic facts (definitions actually) are that Θ = 1/H and the Hubble radius R = cΘ.

    Someone asked me this recently: if they want to work out critical density of mass they use:
    ρ = 3/(8πG Θ^2)
    And replace Θ with Hubble time.
    What mass unit would this be in? And would they have to convert it at all?

    I think that's a good simple exercise in quantitative cosmology so I want to reply to the question in open thread. I use google calculator:

    If you open Lightcone it tells you the current Hubble radius R = 14.4 billion lightyears. So that means Θ = 14.4 billion years.

    So you can type this into the google window:
    3/(8 pi G (14.4e9 years)^2)

    If you paste that into google window, you get:
    8.66146267 × 10-27 kg / m3

    So that is the answer to the question "what mass unit would it be in?" Google tends to give you answers expressed in standard metric units, like kilogram. If you want the answer in grams per cubic meter then you should type this in:
    3/(8 pi G (14.4e9 years)^2) in g per m^3

    If you paste that in, it will realize you want the answer in terms of g per m^3 and it will say:
    8.66146267 × 10-24 g per (m^3)

    Or you can paste in:
    3/(8 pi G (14.4e9 years)^2) in g per cubic kilometer

    and then it will tell you:
    8.66146267 × 10-15 g per (cubic kilometer)

    Basically you get the answer in whatever units you want and specify to your calculator.
     
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  3. Apr 5, 2014 #2

    marcus

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    I got to thinking about what units would be more meaningful and tried typing in this:
    3/(8 pi G (14.4e9 years)^2) in amu per cubic meter

    If you paste that into google you get:
    5.2160552 amu per (cubic meter)

    So that is the mass of about 5 hydrogen atoms. In a cubic meter.
    An "amu" (atomic mass unit) is officially 1/12 of the mass of a carbon atom which consists of 6 protons and 6 neutrons surround by 6 electrons---the common isotope of carbon.

    So a hydrogen atom masses about 1 amu, and an oxygen atom masses about 16 amu, and a water molecule masses approximately 18 amu.

    So you can see that the critical mass density of the U as a whole is on average about 5 hydrogen atoms per cubic meter. Only a small fraction of that is ordinary matter!

    The conventional critical density needs to be discounted for the cosmological constant by this amount
    3/(8 pi G (17.3e9 years)^2)

    If you just want MATTER (dark and ordinary) then paste in

    3/(8 pi G (14.4e9 years)^2) - 3/(8 pi G (17.3e9 years)^2)

    And THAT makes more sense to evaluate in atomic mass units, so try:

    3/(8 pi G (14.4e9 years)^2) - 3/(8 pi G (17.3e9 years)^2) in amu per cubic meter

    When I put that in, I get:
    1.60216497 amu per (cubic meter)

    so the mass of about 1.6 hydrogens in a cubic meter, but 85% of that is dark matter so only 15% is ordinary, and 15% is 0.24 amu per cubic meter.

    So only about a QUARTER of a hydrogen atom's worth of ordinary matter in a cubic meter
     
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