Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Argument that a maximum density Universe expands linearly

  1. Apr 29, 2017 #1
    The Friedmann equation for a spatially flat Universe is given by
    $$\Big(\frac{\dot R}{R}\Big)^2=\frac{8 \pi G}{3}\rho$$
    where ##R(t)## is the proper radius of some spherical volume with us at its center.

    Let us assume that there is a mass ##M## inside this spherical volume of radius ##R##. The density ##\rho## is then given by
    $$\rho=\frac{M}{(4/3)\pi R^3}.$$
    Substituting the above expression for the density ##\rho## into the Friedmann equation gives
    $$\Big(\frac{\dot R}{R}\Big)^2=\frac{2 G M}{R^3}.$$
    Now let us consider a Universe with a maximum density ##\rho##. The maximum density in a spherical volume of radius ##R## is realized by a Black hole whose Schwarzschild radius is equal to ##R##. Therefore we have the relationship
    $$\frac{GM}{R}=\frac{c^2}{2}.$$ If we substitute the above relationship into the Friedmann equation we obtain
    $$\Big(\frac{\dot R}{R}\Big)^2=\frac{c^2}{R^2}$$
    which has the linear solution
    $$R=c\ t.$$
    Therefore it seems that a Universe with a maximum density ##\rho## expands linearly rather than exponentially as would be expected for a de Sitter Universe with a constant Planck scale density.

    Is this reasoning correct?
     
    Last edited: Apr 29, 2017
  2. jcsd
  3. Apr 29, 2017 #2
    Hi jcap:

    I have not yet explored your math, but the title of the thread seems a bit off. The universe you are describing is not "maximum density". It is just flat. Also the word "observable" in an important omission.

    Also, I am curious about where you found a Friedmann equation for the observable universe. The Friedmann equations I have seen are all related to the universe as a whole.

    Regards,
    Buzz
     
  4. Apr 29, 2017 #3
    Hi Buzz,

    I am just assuming a spatially flat Universe for simplicity. The Friedmann equation I am quoting is expressed in terms of some arbitrary proper radius ##R(t)##, with units of length, rather than a dimensionless scale factor ##a(t)##. I think either are correct - it is just a matter of convention. I have chosen to use a proper radius ##R(t)## because I want to argue that the density of the Universe around us up to a radius ##R## cannot be larger than the density of a Black hole with Schwarzschild radius ##R##. When I use this condition I find that the Universe must be scaling linearly rather than exponentially as would be expected of a Planck density de Sitter Universe.

    John
     
  5. Apr 29, 2017 #4
    Hi jcap:

    The R(t) observable universe radius you are using is not a "proper radius". It changes independently of the scale factor a. A proper radius scales with a.

    Regards,
    Buzz
     
  6. Apr 29, 2017 #5
    I have recast my argument entirely in terms of a spherical volume of proper radius ##R(t)## with us at its center. My maximum-density argument does not depend on the notion of an observable Universe.
     
  7. Apr 29, 2017 #6
    Hi jcap:

    I think you have some misunderstandings about the first equation you are using. ρ is not a free variable. It is a specific value, ρc, the critical density, which makes the universe flat.

    [1] ρc = 3 H2 / 8πG​
    If you put that into your first equation you get
    [2] (R'/R)2/H2 = 1​
    (I used apostrophe rather than dot because I do not have convenient use of the dot notation.)
    Since H is defined as
    [3] H = (R'/R)​
    [2] is the identity 1 = 1.

    I suggest that to explore the behavior of a (or R) as a function of time it is simpler to use the other form of the Friedmann equation in
    at the end of the Density parameter section.

    Good luck.

    Regards,
    Buzz
     
  8. Apr 29, 2017 #7

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No. A black hole is a different spacetime geometry from an expanding FRW spacetime. You can't use formulas from one in the other.
     
  9. Apr 29, 2017 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    A de Sitter universe has zero density of ordinary matter. It has a positive cosmological constant. That's not the same thing.
     
  10. Apr 29, 2017 #9

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Aside from my previous comment, Buzz Bloom's point that the density is not a free variable is also correct. If you have zero cosmological constant (which your formulas in the OP assume), and a spatially flat universe, then the density as a function of time and ##R## as a function of time are both determined; there are no free parameters left at all.
     
  11. Apr 30, 2017 #10

    timmdeeg

    User Avatar
    Gold Member

    In addition to was said already it seems worthwhile to check any conclusion regarding the dynamics of the universe if it is consistent with the Friedmann equations.
    The universe expands linearly under the condition ##\dot{a}=const.## which requires ##\rho{a^2}=const.## This doesn't make sense however, because ##\rho## is proportional to ##a^{-3}##. Adding ##\Lambda## doesn't make it better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Argument that a maximum density Universe expands linearly
  1. The expanding universe (Replies: 43)

Loading...