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I'm trying to do a problem in which I am given the Einstein tensor for the following metric:

$$ ds^2 = -e^{2\Phi}(d{x^0})^2 + e^{2\Psi}\delta_{ij}dx^idx^j, $$ And then asked to find the Einstein tensor for the perturbed cosmological metric:

$$ ds^2 = -(1 +2\phi)dt^2 + a(t)^2(1+2\psi)(dx^2 + dy^2 + dz^2). $$ (Here, $\Phi, \phi, \Psi, \text{ and } \psi$ are all scalars.)

I begin by equating $e^{2\Phi} = (1+2\phi)$, and: $e^{2\Phi} = a(t)^2(1 +2\psi)$.

And I end up with products involving derivatives; for instance, $G_{00}$ contains a term: $$ \frac{1}{a}(1 + 2\phi - 2\psi)\left[ \partial_k\psi\partial^k\psi + 2\partial^k\partial_k \psi \right]. $$

My question is: We assume that $\phi, \psi \ll 1$, so we can ignore second order terms. What do we do about products of derivatives? Do we also assume that $\partial_\mu \phi, \partial_\mu \psi \ll 1$? I don't really see any justification for making this assumption, except in the special case of $2\psi\partial_k\psi = \partial_k(\psi^2) \cong 0,$ by the product rule.

Of course, if we don't make this assumption, then our theory is no longer linear, so what can we even do?

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  • $\begingroup$ as usuall: don't think about it and just skip this terms (although even if magnitudes $\psi$ are small, their variations $\partial_k\psi$ could be large), or give your problem to a mathematican, who, unlike physicists, knows how to get approximations correctly. $\endgroup$ – image Apr 3 '15 at 11:27
  • $\begingroup$ So, what would the correct sort of approximation look like? $\endgroup$ – user109527 Apr 3 '15 at 11:38
  • $\begingroup$ I'm not sure. I've also only been thaught the usuall nonsense, and have been trying to get some mathematical foundations ever since. For first, one should check if there exist error terms for your way of approximation, that is if the numerical approximation advances to the realy solution (of the geodesic equations) if one increases the order. This is in general not clear, since it involves the change of two limiting processes. There's a whole branch of maths about it. Don't expect to get an easy answer. $\endgroup$ – image Apr 3 '15 at 11:56

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