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Null Energy condition

  1. Mar 5, 2013 #1
    Dear experts,

    I really wonder how to extract the equation of stats [itex]w=\frac P \rho[/itex] from the Friedmann equations and how one can see that dark energy needs to have [itex]w<-\frac13[/itex] and why does [itex]w<-1[/itex] violate the null energy condition.

    Thanks in advance,
  2. jcsd
  3. Mar 5, 2013 #2


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    pressure = w * density (P = w * rho) is not extracted from the Friedmann equation. It's just an assumed equation that gives the equation of state, which is just how pressure changes with density.

    If you look at Baez's "meaning of Einstein's equations" http://math.ucr.edu/home/baez/einstein/, you'll note that if rho+3P is positive, a ball of inert coffee grounds around the matter satisfying that equation will start to contract.

    But dark energy requires the ball of coffee grounds to expand, not contract.

    Thus rho+3P must be negative, hence w must be less than -1/3.
  4. Mar 5, 2013 #3


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    I assume what you mean is how to extract the numerical value of w from observations. One of Friedmann's equations is [itex]\ddot{a}/a=(-4\pi/3)(\rho+3P)[/itex]. Since we observe that [itex]\ddot{a}/a[/itex] is positive, it follows that [itex]\rho+3P[/itex] is negative, and this forces w<-1/3.

    The answer to this would depend on what definition you're using for the NEC. Depending on how you express the NEC, the result could be immediate.
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