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#9
Sep5-08, 08:29 AM
PF Gold
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P: 522
Hi,
Could anybody help clarify the assumed nature of dark energy in terms of its energy density and pressure? In the standard form of Friedmann's equation the implication of the cosmological constant [tex][\Lambda][/tex] is that it overpowers any contraction due to gravity, as implied by [tex]\rho[/tex].

However, it seems possible to make the following assumptions about energy density associated with the cosmological constant based on the equivalence of units in the Friedmann equation, i.e.

[1] [tex]\rho_\Lambda \equiv \frac {\Lambda c^2}{8 \pi G}[/tex]

As an aside, the units of [tex][\Lambda=1/metres^2][/tex] and the numbers suggest that this is equivalent to [tex][1/R^2][/tex], where [R] seems to be in the order of the radius of the visible universe?

[2] [tex]\rho = \frac {E}{V} = \frac {mc^2}{V}[/tex] such that [tex]m= \frac {\rho V}{c^2}[/tex]

This suggests that dark energy also has an effective mass and therefore a gravitational effect that would slow down the expansion of universe in opposition to the expansion assumed by the following equation:

[3] [tex]P = \omega \rho c^2[/tex] where [tex]\omega_\Lambda=-1[/tex]

Equation [1] suggests that the use of the cosmological constant in Friedmann’s equation can be replaced by an equivalent energy density. While I have shown a transposition into pressure, I am not sure how this ‘overpowers’ the effective gravitational mass given that [tex]P_\Lambda = -1 \rho_\Lambda c^2[/tex]?

A similar problem seems to occur with this substitution into the Fluid equation, while the Acceleration equation is offset by the factor [tex][\rho + 3P][/tex] that allows [-P] to overcome [tex][\rho][/tex] and give a net positive acceleration.

Appreciate any insights. Thanks.