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Gravitational waves

  1. Mar 12, 2013 #1
    Hi not sure if this is GR or cosmology,
    does anyone know what the gravitational wave equation is for GW's produced during cosmic inflation is it just [itex] \ddot{h}+2H\dot{h} +k^{2}h=0 [/itex] because this is derivable using the FRW metric which isn't valid during inflastion, does this govern the dynamics as the waves re-enter the horizon?
     
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  3. Mar 12, 2013 #2

    bapowell

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    The equation governing the evolution of tensor perturbations (gravitational waves) generated during inflation is
    [tex]h'' + 2\frac{a'}{a}h' + k^2 h = 0[/tex]
    where the primes denote derivatives wrt conformal time, [itex]d\tau = dt/a[/itex] (note that [itex]a'/a[/itex] is not the Hubble parameter, which is defined this way in terms of coordinate time: [itex]H = \dot{a}/a[/itex].)

    I'm not clear on your claim about the FRW metric not being valid during inflation -- what do you mean by this? The above EOM is derived assuming FRW expansion, and it is valid at all length scales (including after the mode re-enters the horizon).
     
  4. Mar 12, 2013 #3

    George Jones

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    It is also unclear to me why you say
     
  5. Mar 12, 2013 #4
    well surely inflation is quasi de sitter, so the metric is [itex] dt^2-d\underline{X}^2 [/itex], whereas after infltaion in is FRLW so[itex] dt^2-a^{2}d\underline{X}^2 [/itex].

    Isn't that tensor perturbation eqn derived from the action [itex] S^{(2)}=\frac{a^{2}(t)}{2}d^{4}x [/itex] which surely can't occur during inflation.
     
  6. Mar 12, 2013 #5

    George Jones

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    de Sitter is a vacuum FRLW spacetime.
     
  7. Mar 12, 2013 #6
    Yeah but it doesn't have the scale factor so why in here for example does it derive that equation using an action that contains an a(t)?
     
  8. Mar 12, 2013 #7

    bapowell

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    The scale factor for de Sitter is [itex]a(t) \propto \exp(Ht)[/itex]. Why do you think there is no scale factor? (Keep in mind that while the full topological de Sitter space is static, inflationary spacetimes are described by only half of this space).
     
  9. Mar 12, 2013 #8
    yeah but the metric describing de-sitter space is dt^2 - dx^2 isn't it?

    Or during inflation is it dt^2 - a^2 dx^2 ?
     
  10. Mar 12, 2013 #9

    George Jones

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    The same spacetime can look very different in two different coordinate systems. I don't think that I am familiar with a coordinate system for which de Sitter looks like dt^2 - dx^2. Can you give a reference?
     
  11. Mar 12, 2013 #10

    bapowell

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    de Sitter space looks like [itex]ds^2 = dt^2 - \exp(2Ht)dx^2[/itex], with [itex]H = {\rm const}[/itex]. It has the FRW form with [itex]a(t) = \exp(Ht)[/itex], and is inflationary since [itex]H = {\rm const} \rightarrow \ddot{a}>0[/itex].
     
  12. Mar 12, 2013 #11
  13. Mar 12, 2013 #12

    George Jones

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    No!!!

    Minkowski spacetime has zero spacetime curvature; de Sitter spacetime has non-zero spacetime curvature.

    de Sitter spacetime can be considered to be a curved 4-dimensional hypersurface of a 5-dimensional space that has zero curvature.
     
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