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In Carroll's online Lecture Notes on General Relativity (p112) he says, "the weakness of the gravitational field allows us to decompose the metric into the Minkowski form plus a small perturbation"

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He then says that, to first order in h

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How does that happen? I've tried juggling round with various inverse metrics but I just can't see where that minus sign comes from.

Nice and simple please.

Thank you.

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    $\begingroup$ Hint: Think about what you might want $g_{\mu\nu}g^{\mu\nu}$ to come out to be $\endgroup$
    – twistor59
    Sep 10, 2011 at 14:09
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    $\begingroup$ Hint 2: if $h_{\mu \nu}$ is small, how does $h^{\alpha \nu}h_{\mu \nu}$ compare to the other terms? $\endgroup$
    – Zo the Relativist
    Sep 10, 2011 at 15:39

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The inverse matrix of (I+A) is (I-A) when you can ignore $A^2$. This is just like the fact that the reciprocal of .999 is 1.001.

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  • $\begingroup$ Ron Maimon - Thank you. Does that mean when Carroll says $g^{\mu\nu}g_{\mu\sigma}$ equals some form of the kronecker delta ($\delta^\mu_\sigma $) he is also saying $g_{\mu\nu}g^{\mu\nu}$ equals the identity matrix? I didn't realize that. $\endgroup$
    – Peter4075
    Sep 11, 2011 at 6:41
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    $\begingroup$ As a matrix $g^{\mu\nu}$ is the inverse matrix of $g_{\mu\nu}$, when you write them out as a square box of numbers in a coordinate system. They aren't matrices exactly, because they have two upper indices, they transform differently under coordinate transformations. A matrix, strictly speaking, has one upper and one lower, $A^\alpha_\beta$. But index-contraction is just the most general form of matrix multiplication for multiple index objects. You should learn to see how to recast an index contraction as the multiplication of the appropriate matrix (although matrix notation is less useful). $\endgroup$
    – Ron Maimon
    Sep 11, 2011 at 7:22

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