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Big Bang singularity

  1. Oct 4, 2013 #1
    Does a FRW universe with the equation of state:

    $$p = -\frac{\rho c^2}{3}$$

    have a singularity at the Big Bang?

    I was looking at:

    http://en.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

    and trying to decide if such a Universe obeys the "dominant energy condition" for the Penrose-Hawking theorem (energy greater than pressure) or not.
     
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 5, 2013 #2

    Simon Bridge

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    What would it mean to have "a singularity at the big bang"?
    Crunch the numbers and see.
     
  4. Oct 5, 2013 #3

    Chalnoth

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    The dominant energy condition basically states that mass-energy can never be observed to be flowing faster than light:
    http://en.wikipedia.org/wiki/Energy_condition#Dominant_energy_condition

    So yes, it holds necessarily.

    But it's also pretty easy to show it using the Friedmann equations.

    First, we can get a(t) using the second derivative equation with [itex]\Lambda=0[/itex]:

    [tex]{\ddot{a} \over a} = -{4 \pi G \over 3} \left( \rho + {3p \over c^2}\right) = 0[/tex]

    Thus, we can write:
    [tex]a(t) = a(t=0) + H_0 t[/tex]

    This is important because it shows that [itex]a = 0[/itex] at some finite time.

    Second, energy conservation shows that the energy density of this type of matter scales as [itex]\rho \propto 1/a^2[/itex]. Thus, when [itex]a=0[/itex], the energy density is infinte, so there's your singularity.

    The only universe in which there doesn't appear, at first glance, to be a singularity is the de Sitter universe, where [itex]\rho[/itex] is a constant. But this doesn't quite work out because even a single photon causes a singularity to happen somewhere in the finite past, and a de Sitter universe produces Hawking radiation at its horizon.
     
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