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How to calculate mass of closed Universe?

  1. Jul 29, 2013 #1
    How does one integrate the mass density over a closed Universe (a 3-sphere?) to obtain the total mass of that Universe?

    Is this the correct integral?

    [tex]
    M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}
    [/tex]

    where [itex]R(t)[/itex] is the radius of the Universe at cosmological time [itex]t[/itex].

    By making the substitution [itex]r=\sin \chi[/itex] one finds that the above integral gives:

    [tex]
    M = \pi^2 R(t)^3 \rho.
    [/tex]

    According to wikipedia the hyperarea of a 3-sphere is [itex]2\pi^2 R^3[/itex] so I'm out by a factor of two.
     
    Last edited: Jul 29, 2013
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  3. Jul 29, 2013 #2

    Simon Bridge

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  4. Jul 29, 2013 #3
  5. Jul 29, 2013 #4
  6. Jul 29, 2013 #5

    Simon Bridge

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    The former applies the cube to the brackets, the latter applies the cube to the function.
    ##R(t)^3=Rt^3##
    ##R^3(t)=\big [ R(t)\big ]^3##

    compare: ##\cos(kx)^3## with ##\cos^3(kx)##
    Since it is also valid to write ##\cos kx## the first version does not make it clear what is intended.
    Is it the cosine of the cube of a product or the cube of the cosine of a product? OP only had one variable in the brackets, which gives notice to the reader.

    ... its really just a notation foible.
    I don't think anyone would get confused in such a small equation but in big long multiline equations it gets easier to be misread.
     
    Last edited: Jul 29, 2013
  7. Jul 29, 2013 #6
    I understand, thanks!
     
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