Transformations with the Mukhanov variable (Cosmology)

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The discussion focuses on the transformation of the action in cosmology using the Mukhanov variable, defined as v = zR, where z is related to the scale factor and slow-roll parameter. The confusion arises in expressing the comoving curvature perturbation's derivative, R', in terms of the new variable. The goal is to reformulate the action to include the second derivative of z, z'', which is derived from the definition of z and first-order perturbations. The participants explore the necessary equations to connect z' and z'' while clarifying the meaning of the spatial derivative operator. The conversation emphasizes the mathematical intricacies involved in this transformation process.
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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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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##
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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