Why can one neglect terms quadratic in derivatives of $h_{\mu\nu}$ in linearised gravity? In the linear approximation, terms quadratic in the Christoffel symbols are all neglected in the Riemann Tensor.
However, these are not quadratic in the $h_{\mu\nu}$ but quadratic in the derivatives of $h_{\mu\nu}$. I don't see how one can therefore neglect them in a first approximation
 A: When linearizing curvature tensors in General Relativity, one assumes that the metric is sufficiently well described by a deviation from some reference metric, which is typically the (flat) Minkowski metric $\eta_{\mu\nu}$.
Mathematically, the metric is written
$$g_{\mu\nu} = \eta_{\mu\nu} + \kappa h_{\mu\nu}$$
It is important to note the presence of the constant $\kappa = \sqrt{\frac{16 \pi G_N}{c^4}}$. Perturbation theory in the linear order is then done by keeping only terms linear in $\kappa$, whether they are found before $h_{\mu\nu}$ or $\partial_\alpha h_{\mu\nu}$.
A: The reason why you don't keep the second derivative terms of $h_{\mu\nu}$ is because those are a total derivative in the Lagrangian. For instance, note that we can only have the following two first order term inside the Lagrangian:
\begin{equation}
\mathcal{L}\subset \alpha_1\partial_\mu \partial_\nu h^{\mu\nu} + \alpha_2\partial_\mu \partial^\mu h,
\end{equation}
as the graviton can only appear in derivative terms.
As the equations of motion are not affected by total derivative terms, they won't contribute to the final state of the system. Another way of seeing it is by noticing that we can integrate by parts, and constraint the boundary terms to vanish at infinity. Therefore, those elements inside the lagrangian will vanish once we integrate by parts (as the differential of $\alpha$ is zero.
Yoy can check that when we have second order terms in $h$, the terms won't vanish. However, we can get rid of the second derivatives as we can always integrate by parts and transform it into the canonically normalisation form. For example, the element:
\begin{equation}
S\subset \int d^4x \sqrt{-g}\ h\alpha_1\partial_\mu \partial_\nu h^{\mu\nu}
\end{equation}
can be transformed into
\begin{equation}
S\subset -\int d^4x \sqrt{-g}\ \alpha_1\partial_\mu h\partial_\nu h^{\mu\nu}.
\end{equation}
I hope you find this helpful :)
