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Derivation of (linearized) perturbation equations

  1. Jun 3, 2009 #1
    Hi, there!

    I'm trying to understand the derivation of (linearized) pertubation equations given in my lecture.

    As usual the density and velocity fields are split into a homogeneous and an inhomogeneous part:

    [tex]\rho(t,x) = \rho_0(t) + \delta \rho(x,t)[/tex]
    [tex]\vec{v}(t,x) = \vec{v_0}(t) + \vec{\delta v}(x,t)[/tex]

    Then (and that is what I don't understand) the claim is made that the background quantrities need to separately fullfill mass conservation. In other words:

    [tex]\frac{\partial \rho_0}{\partial t} + \rho_0 \nabla * \vec{v_0} = 0[/tex]

    Any idea where this comes from?
    Many thanks in advance!
    Angelos
     
  2. jcsd
  3. Jun 4, 2009 #2

    Chalnoth

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    Okay, so first of all, there's something wrong with your equations, because if [tex]\vec{v_0}[/tex] is independent of position, then it implies that [tex]\rho_0(t)[/tex] must be independent of time.

    Instead, [tex]\vec{v_0}[/tex] should have a very simple dependence upon position, one that stems from the simple expansion of the universe, with the peculiar velocity of all particles described by [tex]\vec{\delta v}[/tex]. So you should have:

    [tex]\vec{v_0}(x, t) = \dot{a}\vec{x}[/tex]

    Now, unfortunately I don't have my cosmology texts with me, but I believe that [tex]\rho_0(t)[/tex] and [tex]\vec{v_0}(x,t)[/tex] are specifically chosen to satisfy the equations of motion, as it makes the later equations simpler. I don't think it's something derived, just an arbitrary choice that makes the equations easy to work with (and makes [tex]\rho_0(t)[/tex] and [tex]\vec{v_0}(x,t)[/tex] carry physical significance as the "background" density and expansion).
     
  4. Jun 4, 2009 #3

    Wallace

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    v is usually defined as the peculiar velocity, hence v_0(x,t)=0 everywhere. I'm assuming this derivation is done in the synchronous gauge, in which case the velocity would be defined as above.

    For Angelos, check the derivation in your (or any) textbook. Unfortunately you haven't given us enough information to be able to help you (for example exactly what your velocity is defined as). I'm happy to help if you can provide some more information, i.e. the definitions and all steps leading to the part you don't understand.
     
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