We start with the metric tensor \begin{equation} g_{\mu\nu}(x) = \eta_{\mu\nu} + h_{\mu\nu}(x) \end{equation} in the linearised theory, or \begin{equation} g_{\mu\nu}(x) = \bar{g}_{\mu\nu}(x) + h_{\mu\nu}(x) \end{equation} in the more general case with a curved background. However, in both cases, we apparently need \begin{equation} \vert h_{\mu\nu}\vert \ll 1. \end{equation} This condition demands that the components of the perturbation are much less than 1, in the coordinate system where the components of the background metric tensor are of order 1. Why do we need this condition?


1 Answer 1


In order to do perturbation the expansion parameter needs to be small. Otherwise the the system will be strongly coupled and you're in the non-perturbative regime. It's the same as for instance in QM: for perturbative calculations the pertubation must be small.

  • $\begingroup$ Additionally, once you calculate the results -- you find that, indeed, h << 1 $\endgroup$ Feb 13, 2013 at 20:58
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    $\begingroup$ Equations for $g$ are non linear and impossible to solve analytically. $h_{\mu\nu}$ is not supposed to be small, but it is normally small (it can be shown numerically) and then linearization reduces equations to solvable ones. $\endgroup$ Feb 13, 2013 at 21:02

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