Transformations with the Mukhanov variable (Cosmology)

• ergospherical
ergospherical
Homework Statement
See below
Relevant Equations
N/A
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.

Is it taken from some book or paper this question of yours?

What is ##\partial_i R##?
Derivative w.r.t what?

##\partial_i = \partial/\partial x^i##

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