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Model for Friedmann Equations

  1. Mar 27, 2015 #1
    You know friedmann equations derived from kinetic energy and potantial energy conservation.I found these shell model for universe.Here I am curious about something. This shell is like a surface of sphere isnt it.I mean it has only surface and that surface mass is m.And we made our equations based on that surface.
    Is that shell has a density ? How can we calculatei it ?
    I made like this m/4πr2surface

    You can look attachment
     

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  2. jcsd
  3. Mar 27, 2015 #2

    marcus

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    The ordinary sphere, denoted S2 is surface of the 3d-solid ball. You have the area formula there. What about the hypersphere?
    Analogous to 2d area, there should be a formula for the 3d volume of the surface of a 4d ball.

    Wikiped has an article on areas and volumes of spheres in higher dimensions. I don't recall what they call it.

    I do recall that the 3d volume of S3 of radius R is 2π2R3
     
    Last edited: Mar 27, 2015
  4. Mar 27, 2015 #3

    Orodruin

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    No it is not. There is a cheap trick often used to heuristically argue this to undergraduate students, but the Friedman equation is only consistently derived in GR with the assumption of a homogeneous and isotropic universe. Misconceptions arising from this trick are generally severe, which is why I would typically argue against using it even for plausibility.
     
  5. Mar 27, 2015 #4
    Are you talking about 4d shells ? I suppose In friedmann eq shell is a 3d sphere and it has a surface 4πr2
     
    Last edited: Mar 27, 2015
  6. Mar 27, 2015 #5

    marcus

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    Our posts crossed. I did not see yours. OK I misunderstood what you were asking about.
    My post here is not relevant to your question and you can ignore it. But I will leave it in case it might help someone else understand the Friedmann equation positive curvature (finite S3 spatial volume) case.

    We don't know that higher dimensions actually exist, may just be abstract conceptual tools we use, not physical.

    The 4d "solid ball" may not be real, maybe only the 3d surface. But that might actually be the 3d space we live in. It is a possibility.

    We do not know that the U is spatially finite. But that is one possible CASE. Let's consider that case. Suppose it is finite and suppose it is an S3 hypersphere, with abstract radius of curvature R (at this time).

    Then the spatial volume at this time is 2π2R3 or about 20 R3

    The radius of curvature can be estimated and a lower bound is estimated about 100 billion LY. At present.

    If you cube that, you get 1000000 billion billion billion cubic light years. Then you multiply by 20. That is a very rough LOWER bound.
    Observations show the curvature is so small that the radius of curvature must be ABOVE 100 billion LY. but we don't have an upper bound estimate, so far.

    The average density is estimated differently, without working with total mass and total volume. One infer it by observing how the Hubble rate is changing. Because density affects both spatial curvature and the change over time of the H(T). If spatial curvature is very small, which it is, one can attribute all the change in H(T) to the density and write a simple equation (derived from Einstein GR equation) which allows to calculate the density just from looking at how H(T) is behaving.
     
    Last edited: Mar 27, 2015
  7. Mar 27, 2015 #6
    I am talking about surface density.Here I understand that change in H(t) affects universe density but I am looking surface I didnt understand
     
  8. Mar 30, 2015 #7
    Is that shell has a density ? How can we calculatei it ?
    Yes, you can use the shell model and derive results that are useful - generally for a surface density engineers use the symbol sigma. And here is a really interesting fact if you use the Hubble parameters which are built into the Friedman equations. Taking the estimate of Hubble mass in the range of 1.5 x 10^53 kgm and the Hubble radius as 1.1 x 10^26 meters, the Hubble surface density is one kgm per square meter.
     
  9. Apr 2, 2015 #8
    Thanks thats really help me
     
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