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

Equation involving H and a re Accerlerating Universe

  1. Mar 26, 2015 #1
    In the Wikipedia article https://en.wikipedia.org/wiki/Accelerating_universe#Evidence_for_acceleration the following equation
    (1)
    Equation2.png
    [where the four currently hypothesized contributors to the energy density of the universe are curvature, matter, radiation and dark energy]

    is given without any derivation from the previous equations
    (2)
    Equation.png
    (3)
    Equation3.png
    (4)
    Equation4.png


    I have several question about these equations.

    (a) I assume that in (2) K is -1, 0, +1 respectively for the space being hyperbolic, flat, or spherical. Then R would be the radius of curvature of the universe at time t. Is that correct?
    (b) How is (1) derived form (2), (3), and (4)?
    (c) Is the sum of the four Omegas in (1) supposed to equal the Omega in (4)?
    (d) In the Einstein equations, isn't Lambda a constant density, independent of a?
    (e) Isn't Omega[DE] = Lambda/rho[sub-c] and Lambda a constant independent of a?
    (f) If e is correct, then
    Omega[DE] = Lambda * (8*pi*G/3) * (a/adot)^2
    OK, if that's right, where does the exponent with w in the coefficient of Omega[DE] come from?
     
  2. jcsd
  3. Mar 26, 2015 #2

    wabbit

    User Avatar
    Gold Member

    I found this document very useful as a thorough introduction to flrw/lcdm equations:
    http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Mar 26, 2015 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes.

    If K = 1, yes. If K = 0 or -1, R doesn't have that simple interpretation; it's just a "scale factor" than can be used to track the universe's expansion.

    No. (4) is just a definition; each individual ##\Omega## in (1) is obtained from the corresponding energy density by dividing by the critical density.

    If it's a cosmological constant, yes. But the term "dark energy" is also used more generally to denote anything that can cause accelerating expansion; for example, a scalar field.

    ##w## is the parameter in the equation of state that relates pressure to density; in general, ##p = w \rho##. For non-relativistic matter, ##w = 0##; for radiation or highly relativistic matter, ##w = 1/3##; for a cosmological constant, ##w = -1##; and for other more general types of "dark energy" such as a scalar field, ##w## can vary.

    If you look at the second Friedmann equation, and note the factor ##\rho + 3p## on the RHS (where we include a cosmological constant and dark energy in ##\rho## and ##p##), you can see that, if we know ##w##, the factor becomes ##\rho \left( 1 + 3 w \right)##, so any ##w < - 1/3## will make the factor negative and therefore the acceleration ##\ddot{a} / a## will be positive. Combining the second Friedmann equation with the first leads to the factor ##- 3 \left( 1 + w \right)## in (1); note that for a cosmological constant, ##1 + w = 0## and the dark energy density term is constant, as desired. But any ##w < - 1/3## can in principle appear in that term, so it is not necessarily constant in general.
     
    Last edited: Mar 26, 2015
  5. Mar 26, 2015 #4

    Chalnoth

    User Avatar
    Science Advisor

    Not quite. The way the equations above are written, R is a constant (the current radius of curvature), and the radius of curvature at any given time would be Ra.

    (PeterDonis' answer above is for the more common notation where R(t) replaces a(t).)

    By using stress-energy conservation. Stress-energy conservation determines how the energy density of each component scales with the expansion.

    The sum of the four ##\Omega## parameters is equal to one, by definition. You can see this by considering the situation where ##a = 1## (now), as by definition ##H(a=1) = H_0##.

    Yes.

    As PeterDonis mentioned, Lambda is the special case for dark energy where ##w = -1##. You should be able to show that if ##w = -1##, the dark energy in equation (1) is a constant.

    The way that equation (1) is written, ##\Omega_{DE}## is the current density fraction of dark energy. That is to say, equation (4) should be written as ##\Omega_x = \rho_x(now) / \rho_c(now)##, where ##x## is whatever component of the energy density we're referring to.
     
    Last edited: Mar 26, 2015
  6. Mar 26, 2015 #5
    Thanks wabbit. The document looks like I will find a lot of useful stuff in it. I will probably have more questions after I study it.
     
    Last edited by a moderator: May 7, 2017
  7. Mar 26, 2015 #6
    Thanks PeterDonis and Chalnoth.

    It seems like the notation is a bit confusing. The clarification in Chalnoth's post helps to clarify it.
     
  8. Mar 26, 2015 #7
    I think what confused me was seeing in several places that the sum of the contributing Omegas added to 1.

    Ah, I see now that Chalnoth explained that the sum always equals 1 by definition.
     
    Last edited: Mar 26, 2015
  9. Mar 26, 2015 #8
    Thanks once again to PeterDonis. Your answers to my questions (e) and (f) were very clear and helpful.answers to my questions (e) and (f) were very clear and helpful.
     
  10. Apr 10, 2015 #9
    I have been thinking some more about the 1st equation of my original question.

    equation2-png.81011.png

    (a) What would be the value of Omega[k] be if |radius of curvature| = R[0], and k = +1 or -1?

    (b) Also, since a is a relative distance with a[0] arbitraily set = 1, as I now undestand it, the equation would need to be modified, replaceing a with R/R[0]. Would an alternative to this be to modify H[0] by replacing its standard distance 1 Mpc with R[0], that is, dividing H[0] by R[0] and/or by reinterperting the unit of time from 1 sec to R[0]/c?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook