Dark Energy: Negative Pressure or Density?

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SUMMARY

The discussion centers on the relationship between the cosmological constant and the properties of dark energy, specifically its pressure and density. It is established that for a positive cosmological constant, the density (ρ) is positive while the pressure (p) is negative, leading to the equation of state w = ρ/p = -1. Conversely, a negative cosmological constant results in negative density (ρ < 0) and positive pressure (p > 0). The Friedmann equation indicates that the spatial curvature parameter k can be -1, 0, or 1, which is compatible with both positive and negative cosmological constants, as k describes spatial curvature rather than the properties of dark energy.

PREREQUISITES
  • Understanding of the Friedmann-Robertson-Walker (FRW) metric
  • Familiarity with the equation of state for cosmological fluids
  • Knowledge of the cosmological constant and its implications
  • Basic grasp of spatial curvature in cosmology
NEXT STEPS
  • Study the implications of the Friedmann equations in cosmology
  • Explore the role of the cosmological constant in dark energy models
  • Investigate the relationship between spatial curvature (k) and cosmological constants
  • Learn about the dynamics of perfect fluid cosmology
USEFUL FOR

Astronomers, cosmologists, and physics students interested in the dynamics of dark energy and the implications of the cosmological constant on the universe's expansion.

binbagsss
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<<Mentor note: Edited for readability.>>
<<Follow-up: futher edited to fix LaTeX tags>>

Context: FRW Metric /universe, perfect comological fluid , dark/vacuum energy

So the equation of state ## w \rho=p, w=-1, \rho=-p##
so this clealy implies either ##p## or ##\rho## is negative.
Am I correct in thinking which it is, depends on the cosmological constant?
So ##cosmo constant >0, \rho>0, p<0##
##cosmo constant <0, \rho <0, p>0?##

Now I look at a Friedmann equation given by: ##\Omega -1 = \frac{k}{H^2a^2} ##, where ## \Omega = \frac{\rho}{\rho_{c}} ##, ##H=\frac{\dot{a^{2}}}{a^{2}} ##

and solving for ##a## as a function of ##t## for ##cosmo constant >0##, my book says that in this case ##\Omega <0 ## ( which I understand if my above reasoning is correct) and so from the Friedmann equation this is only possible if ##k=-1 ##. So this is fine, I agree , but the denominator needs to be greater than zero.

Anyway, it than solves for ## cosmo constant >0 ##, and says all ## k=-1,0,1 ## are fine. This is my QUESTION. I don't understand how ##k=-1## can be okay for both ##cosmo constant >0, <0 ## which in turn say different things above which of ## \rho ## and ##p## are negative and positive.

Thanks very much for your help.
 
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You have to fix those Latex tags or else this is pretty unreadable...

I see a note by a mentor saying it's been edited for readability, but this still looks pretty unreadable to me.
 
binbagsss said:
So ##cosmoconstant>0,\rho>0,p<0##

##cosmoconstant <0, \rho<0, p>0##?

Yes.

binbagsss said:
solving for ##a## as a function of ##t## for ##cosmo constant >0##, my book says that in this case ##\Omega<0##

I think you mean for ##cosmo constant < 0##, correct? That's the case for which ##\rho < 0##, which makes ##\Omega < 0##.

binbagsss said:
I don't understand how ##k=-1## can be okay for both ##cosmo constant >0, < 0##

##k## describes the spatial curvature; it doesn't describe either the density or the pressure associated with the cosmological constant. All this is saying is that "open" spatial slices (i.e., negative spatial curvature) are compatible with both a positive and negative cosmological constant.
 

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