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Help with Friedman equations and density of energy

  1. Oct 29, 2013 #1
    Hi there,



    I really hope someone can help me with my stupid but urgent problem of understanding something crucial about the Friedman equations.



    So; one of them looks like this (forget about the constants; it is about the principles):



    change of the scale factor with time - density - cosmol. constant = -k (curvature term)


    Then this is sometimes rewritten in terms of densities, which gives:



    change of the scale factor with time - (density of matter + vacuum energy density) = -k (curvature term)




    Now; here is what I do not get.



    Generally the density of the vacuum (or equivalently the cosmol. constant) are treated just like the density of matter; so they have the same effect on the curvature, which somehow should be understandable as energy=matter and so both curve the spacetime.

    But then, it is usually stated that the cosmol. constant, and so the vacuum energy density, are working against gravitation (repulsive).

    However; I do not understand, how to see this in the equations above. I mean; both seem to have the same effect: energy=matter -> attraction (simplified).



    Where is my error of thinking?

    I really hope that someone here can enlighten me.



    Thx
     
  2. jcsd
  3. Oct 29, 2013 #2

    George Jones

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    Cosmological gravitational repulsion means that ##\dot{a}##, the rate of change of the scale factor, is increasing. i.e., that ##\ddot{a}## is positive. The Friedmann equation about which you wrote is

    $$\dot{a}^2 = \frac{8\pi}{3} \rho a^2.$$

    Differentiating this equation with respect to time gives

    $$2\dot{a}\ddot{a} = \frac{8\pi}{3} \left( \dot{\rho} a^2 +2\rho a \dot{a} \right).$$

    Consequently, there is cosmological repulsion when ##0 < \dot{\rho} a^2 +2\rho a \dot{a}##.

    It is somewhat difficult to see what is going on from this, but a couple of things can be noted:

    1) in an expanding universe, there is no gravitational repulsion only when ##\dot{\rho} a^2## is sufficiently negative (as it is for normal matter);

    2) for an expanding universe that consists solely of vacuum energy, which has ##\dot{\rho} = 0##, there is repulsion, since then ##\ddot{a} > 0##.

    Things become a little clearer when the other Friedmann equation is considered,

    $$\ddot{a} = -\frac{4\pi}{3} \left( \rho + 3p \right) R.$$

    Clearly, ##\ddot{a} > 0## when ##w = p / \rho## is less than -1/3. Vacuum energy/cosmological constant has w = -1.

    Roughly, (for non-exotic matter that has positive energy density) repulsion happens when pressure is sufficiently negative.
     
  4. Oct 31, 2013 #3
    Hi;

    thank you for the answers.
    I totally missed (or have overseen) the second more important equation in this context (the second derivative), which shows the dependencies on the pressure terms.

    Will have a deeper look into now that the examination is over. :) Was a little confused the day before.

    Thx
     
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