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I Hubble term versus inflaton

  1. Jan 10, 2017 #1
    From cosmology,

    ##H^2 = \frac{ρ}{3M_p^2} = \frac{1}{3M_p^2}(½\dot φ^2 + ½m^2φ^2)##

    Suppose ##V(φ) = ½m^2φ^2##

    where
    ##ρ## = density
    ##M_p## = planck mass

    I want to graph ##H## vs. ##φ## but there is a ##\dot φ## and I know this is a differential equation, can somebody help me what to do here?
     
  2. jcsd
  3. Jan 10, 2017 #2

    Chalnoth

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    The only way is to solve the differential equation so that you know ##\phi(t)##. For that, you'll need more than just this equation, as you have an unknown function of time on both sides of the equation. You'd need to make use of a second equation, possibly the second Friedmann equation, to resolve the discrepancy.
     
  4. Jan 11, 2017 #3
    I was hoping to get rid of ##\dot φ## but it seems I can't find any relationship for that. If I'm to use the second Friedmann equation,
    ##\frac{\ddot a}{a} = -\frac{1}{6M_p^2}(ρ+3p)~~~~~~~~~ ^*~H = \frac{\dot a}{a}~~→~~\dot H = \frac{\ddot a}{a} - (\frac{\dot a}{a})^2~~→~~\dot H = \frac{\ddot a}{a} - H^2##

    ##\dot H + H = -\frac{1}{6M_p^2}(ρ+3p)##

    The problem is the form of ##ρ## and ##p##. For warm inflation, should I consider ##ρ = ρ_λ + ρ_r## and ##p = p_λ + p_r##? Given that ##p_λ = -ρ_λ## and ##p_r = \frac{1}{3}ρ_r##
     
  5. Jan 11, 2017 #4

    Orodruin

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    You cannot get rid of the time derivative in favour of other quantities. That would remove the dynamics of the field itself.

    In some cases, like slow-roll inflation, you can neglect the kinetic term in the energy, but it is still there.
     
  6. Jan 11, 2017 #5
    Yes, that's why I'd like to solve the DE exactly but there is an ##H^2## in front which is also a variable.
     
  7. Jan 11, 2017 #6
    If I define ##t_H = H^{-1}## (Hubble time) then it would be just an ODE so I could use the typical numerical calculation in Mathematica?
     
    Last edited: Jan 11, 2017
  8. Jan 11, 2017 #7
    Have you considered the Klein-Gordon equation?
    $$ \ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0$$
     
  9. Jan 11, 2017 #8
    That would be the case in typical inflationary scenario but in warm inflation KG equation would be modified to

    ##\ddot{\phi}+(3H + Γ)\dot{\phi}+\dfrac{dV}{d\phi}=0##

    There is an extra dissipation term ##Γ##, which I would also need later, so that is also a problem.
    Basically, I want to find the relationship of the tensor to scalar ratio ##r## with the ##Γ## dissipation term by the theoretical result ##r = 16ε## where ##ε## is the Hubble slow roll parameter, but from the equations I can see, ##H##, ##φ##, and ##\dot φ## are in the way since ##ε = -\frac{\dot H}{H^2}## so I think I can numerically calculate ##H## in terms of ##φ## in order to get different values of H to again numerically calculate ##r## in terms of ##H##.
     
    Last edited: Jan 11, 2017
  10. Jan 13, 2017 #9
    I can only think of two equations that I can use to get the behavior of ##H## in terms of ##\phi## for various dissipation term ##\Gamma##,

    $$\ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0 ,\quad H^2 = \frac{1}{6M_p^2}(\dot φ^2 + m^2φ^2)$$

    Can anyone help me figure out how can I use mathematica to solve ##H## for different ##\Gamma##? I only have basic knowledge of mathematica.
     
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