# Why must the gravitational wave components be much less than unity?

We start with the metric tensor $$g_{\mu\nu}(x) = \eta_{\mu\nu} + h_{\mu\nu}(x)$$ in the linearised theory, or $$g_{\mu\nu}(x) = \bar{g}_{\mu\nu}(x) + h_{\mu\nu}(x)$$ in the more general case with a curved background. However, in both cases, we apparently need $$\vert h_{\mu\nu}\vert \ll 1.$$ 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?

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Additionally, once you calculate the results -- you find that, indeed, h << 1 –  DilithiumMatrix Feb 13 '13 at 20:58
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. –  Vladimir Kalitvianski Feb 13 '13 at 21:02