# Linearized gravity: Derivatives of the metric perturbation

In linearized gravity, the metric is given by the Minkowski metric and a small perturbation, $$\begin{equation} g^{\mu\nu} = \eta^{\mu\nu}+h^{\mu\nu},\quad |h^{\mu\nu}|\ll 1. \end{equation}$$ Plugging this expression into the expression for our Christoffel symbols, we obtain $$\begin{equation} \Gamma^{\alpha}_{\mu\nu} = \frac{1}{2} \color{red}{g^{\alpha\lambda}} (\partial_\nu h_{\mu\lambda} + \partial_\mu h_{\nu\lambda} - \partial_\lambda h_{\mu\nu}). \end{equation}$$ Various textbooks and lecture notes like these notes from Carroll (eq. 6.4) simplify this to $$\begin{equation} \Gamma^{\alpha}_{\mu\nu} = \frac{1}{2} \color{red}{\eta^{\alpha\lambda}} (\partial_\nu h_{\mu\lambda} + \partial_\mu h_{\nu\lambda} - \partial_\lambda h_{\mu\nu}) \end{equation}$$ by asserting that $$\partial_\alpha h_{\beta\gamma}$$ is also first order in $$h_{\mu\nu}$$.

How is that justified? Couldn't the perturbation be small, but rapidly oscillating, for example?

• I've been wondering the exact same thing. We know that if a function is small, it does not necessarily follow that its derivative is also small. I think in this case there is no choice but to assume so. Or else it wouldn't be linear in $h$. Nov 12, 2020 at 11:05
• I have seen this in derivations of the gravitational wave equation, though. Given a large enough frequency, the assumption of small derivatives would break down, wouldn't it? Nov 12, 2020 at 11:10
• You are right both assumptions are necessary. Nov 12, 2020 at 11:39
• Is your concern about that the term $\frac{1}{2} h^{\alpha \lambda} \partial_\nu h_{\mu \lambda}$ is not always small if the derivative is large? I think it is tacitly meant, that it is negligible in comparison with the main contribution Nov 12, 2020 at 12:24

This is a common issue that comes up in any classical perturbation theory, not just gravity. At the end of the day, the systematic method for doing any perturbative expansion is to take the would-be true solution for a field, say $$\phi$$, and expand it in powers of some expansion parameter, say $$\alpha$$: $$\phi=\phi_0+\alpha\phi_1+\alpha^2\phi_2+\cdots.$$ The expansion is then in this parameter $$\alpha$$, which since this is a constant is independent of whether we take derivatives or not. The key assumption of perturbation theory after this point is that the equations of motion should hold order by order, meaning we can separate our equations for each order in $$\alpha$$, using the solutions from the lower order equations to simplify the higher order ones.