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Cosmology Questions

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data
    (a)
    From the continuity equation show that if ##P=\omega \rho## and ##\omega > -1/3##, show that an expanding universe will eventually reach a maximum size. Take ##k=1## (closed universe).

    (b) Show that if ##\omega <-1##, the energy density ##\rho## will increase as the universe expands.

    2. Relevant equations

    Continuity equation: ##\dot{\rho}+3 \frac{\dot{a}}{a} (\rho +P)=0##

    First Friedmann equation:

    ##\left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G}{3} \rho - \frac{k}{a^2}##

    3. The attempt at a solution

    (a)
    The hint says I must first solve for ##a## when ##\dot{a}=0## the first Friedmann equation and then consider ##\ddot{a}##. So solving that equation yields:

    ##a=\sqrt{\frac{3k}{8 \pi G \rho}}=\sqrt{\frac{3}{8 \pi G \rho}}##

    To find ##\ddot{a}##, I substitute this into the second Friedmann equation:

    ##\frac{\ddot{a}}{a}= - \frac{4 \pi G}{3} (\rho + 3P) \implies \ddot{a} = \frac{-4 \pi G}{3} (\rho +3P). \sqrt{\frac{3}{8\pi G \rho}}##

    Using the equation of state ##\rho=P/\omega##

    ##\ddot{a} = \frac{-4 \pi G}{3} ((P/\omega)+3P). \sqrt{\frac{3}{8\pi G (P/\omega)}}##

    Do I need to show that we have positive acceleration (i.e. ##\ddot{a}>0##)? I think the equation above shows that acceleration is reduced as the universe expands (##\ddot{a}## decreases as P decreases).

    Also, how can I use the continuity equation here? :confused:

    (b) Do I need to be looking at the expression I found previously with negative P?

    Any corrections or explanation is greatly appreciated.
     
  2. jcsd
  3. Oct 23, 2014 #2

    Orodruin

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    I would use the equation of state to remove P instead of ##\rho##. Pressure can be negative, energy density is not.
     
  4. Oct 23, 2014 #3
    Thank you. I got

    ##\ddot{a}=\frac{-4\pi G}{3}(\rho+3\rho \omega) \sqrt{\frac{3}{8 \pi G \rho}}##

    So, for the case ##\omega = -1/3##, we have ##\ddot{a}=\sqrt{\frac{3}{8 \pi G \rho}}>0##. So the acceleration is always positive.

    But how could we use the continuity equation in this case?
     
  5. Oct 23, 2014 #4

    Orodruin

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    I assume you mean for ##\omega < -1/3##. Why would you not expect the continuity equation to hold? A component with negative pressure simply behaves very differently from ordinary matter or radiation. In particular, the case of ##\omega = -1## corresponds to a cosmological constant.

    Also note that you have assumed that ##\dot a = 0## in your expression. This is not necessary to know the sign of the acceleration, but it will be true in particular when this holds.
     
  6. Oct 23, 2014 #5
    No, I meant for the case when ##\omega >-1/3##. How does one explain that such a universe reaches a maximum size with reference to the continuity equation?
     
  7. Oct 23, 2014 #6

    Orodruin

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    The second Friedmann equation is derived by inserting the continuity equation into the first Friedmann equation so when you are applying the second Friedmann equation you are implicitly assuming the continuity equation.

    If ##\omega > -1/3## you have ##\ddot a \propto - \rho (1 + 3\omega) < 0## so deceleration. In order to have ##\ddot a## positive, ##1+3\omega## must be negative, i.e., ##1 + 3\omega < 0 \Rightarrow \omega < -1/3##. What is the nature of the local extrema (where ##\dot a## is zero) if the second derivative is negative?
     
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