Transformations with the Mukhanov variable (Cosmology)

  • Thread starter ergospherical
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  • #1
ergospherical
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Homework Statement
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Relevant Equations
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I've been going around in circles for a while. We have a parameter ##z##, defined through$$z^2 = 2a^2 \epsilon$$where ##a## is the scale factor and ##\epsilon## is the slow-roll parameter. Considering the action$$S = \frac{1}{2} \int d\tau d^3 x \ z^2 \left[(\mathcal{R}')^2 - (\partial_i \mathcal{R})^2 \right]$$with ##\mathcal{R}## the comoving curvature perturbation. We want to write this in terms of a new 'Mukhanov' variable ##v \equiv z\mathcal{R}##. This is where the confusion starts - specifically with ##\mathcal{R}'##. You can find easily that$$\mathcal{R}' = \frac{v'}{z} - \frac{z' v}{z^2}$$We are looking to arrive at $$S = \frac{1}{2} \int d\tau d^3 x \ \left[(v')^2 - (\partial_i v)^2 + \frac{z''}{z}v^2 \right]$$It looks like I need another equation, to get from ##z'##s to ##z''##s. I've previously showed that ##\tfrac{z''}{z} = \mathcal{H}^2(2-\epsilon + \tfrac{3}{2}\eta)##, just from the definition of ##z## and taking some care to keep only first order perturbations, but can't see whether this is useful.
 
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  • #2
Is it taken from some book or paper this question of yours?
 
  • #3
What is ##\partial_i R##?
Derivative w.r.t what?
 
  • #4
##\partial_i = \partial/\partial x^i##
 

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