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ISW during matter dominant era

  1. Feb 1, 2014 #1
    I've been reading reading an older article on ISW (integrated Sache-wolfe effect). One line struck me as odd in the paper.

    on page two the line " In particular, we know that during the matter dominated era the gravitational potential stays constant "

    I can understand that during the matter dominant era, the cosmological constant effect is greatly reduced. However wouldn't there be an ISW potential due to dynamics of the matter dominant era?
  2. jcsd
  3. Feb 1, 2014 #2


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    The gravitational potential referred to is the analog Newtonian potential associated with the matter perturbations (recall that the Newtonian potential, [itex]\Phi[/itex], vanishes in any homogeneous cosmology). We start with Poisson's Eq.,
    [tex]\nabla ^2 \Phi = 4\pi G \delta \rho_m[/tex]
    where [itex]\delta \rho_m[/itex] is the matter perturbation. Usually a different quantity is actually solved for -- the density contrast: [tex]\delta = \delta \rho/\bar{\rho}[/tex] where the bar denotes an average, so we instead have
    [tex]\nabla^2 \Phi = 4\pi G \bar{\rho}\delta_m[/tex] In order to solve Poisson's equation, we work in Fourier space so that the [itex]\nabla^2[/itex] brings down two factors of the physical wavenumber, [itex](k/a)^2[/itex]. Putting it all together we have
    [tex]\Phi_k = 4 \pi G\left(\frac{a^2}{k^2}\right)\bar{\rho}\delta_{m,k}[/tex]
    Now, during matter domination we know that [itex]\bar{\rho} \sim a^{-3}[/itex], and there is a growing mode [itex]\delta_k \sim a[/itex] (this is found by solving the perturbation equation in Newtonian gauge). We find then
    [tex]\Phi_k = 4\pi G \left(\frac{a^2}{k^2}\right) \frac{a}{a^3} = const[/tex].

    EDIT: The important conceptual point here is that [itex]\Phi[/itex] is not associated with the background density, [itex]\rho[/itex] (or really [itex]\bar{\rho}[/itex]), but with the perturbations. In a homogeneous cosmology, there is no gravitational potential/field.
    Last edited: Feb 1, 2014
  4. Feb 1, 2014 #3
    Thanks a ton that excellent explanation, answered the question beautifully.
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