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In the context of cosmological perturbation one write the most general perturbed metric as $$ ds^2=-(1+2\alpha)dt^2-2a(t)(\partial_i\beta-S_i)dtdx^i+a^2(t)(\delta_{ij}+2\psi\delta_{ij}+2\partial_i\partial_j\gamma+2\partial_iF_j+h_{ij})dx^idx^j. $$ Then one consider the general gauge transformation as $$ \hat{t}=t+\delta t,\qquad\hat{x}^i=x^i+\delta^{ij}\partial_j\delta x. $$ Looking at the scalar part of the perturbations one find that $\alpha,\beta,\psi,\gamma$ transforms as $$ \hat{\alpha}=\alpha-\dot{\delta t} $$ $$ \hat{\beta}=\beta-a^{-1}\delta t+a\dot{\delta x} $$ $$ \hat{\psi}=\psi-H\delta t $$ $$ \hat{\gamma}=\gamma-\delta x $$ My question is: how to recover this relation? What one must impose?

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closed as unclear what you're asking by ACuriousMind, Kyle Kanos, Martin, Danu, Jim Mar 30 '15 at 14:16

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    $\begingroup$ Have a look at Physical Foundations of Cosmology by V. Mukhanov, he derives similar relations (but in conformal time) on page 293, expression (7.18). You should probably start reading on page 291 (chapter 7.1.1). $\endgroup$ – Photon Mar 24 '15 at 12:23
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    $\begingroup$ I don't understand the question. You have correctly identified the gauge transformation of the scalar perturbations. What relation are you trying to recover? It's fairly obvious how to get these transformations. It should take no more than 30 minutes to do it by hand. And that's allowing time for you to look up the gauge transformation relation and determine the new metric $\endgroup$ – Jim Mar 24 '15 at 12:29
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    $\begingroup$ @Photon thanks so much for the reference. Now i understand $\endgroup$ – yngabl Mar 24 '15 at 13:15