In the <Levi-Civita connection>, the background metric is constant and replacing $g^{\alpha\rho}$ by its $O(\epsilon)$ correction would multiply a derivative of $h$ and produce an $O(\epsilon^2)$ term. Hence the <linearized Levi-Civita connection> is
$$
\boxed{
\Gamma^\alpha{}_{\beta\gamma}
=\frac12\eta^{\alpha\rho}
\left(
\partial_\beta h_{\gamma\rho}
+\partial_\gamma h_{\beta\rho}
-\partial_\rho h_{\beta\gamma}
\right)}.
$$
The products of two connection coefficients in the <Riemann curvature tensor> are also quadratic and may be discarded. Lowering its first index with $\eta_{\mu\nu}$ gives
$$
\boxed{
R_{\mu\rho\alpha\beta}^{(1)}
=\frac12\left(
\partial_\alpha\partial_\rho h_{\mu\beta}
-\partial_\alpha\partial_\mu h_{\rho\beta}
-\partial_\beta\partial_\rho h_{\mu\alpha}
+\partial_\beta\partial_\mu h_{\rho\alpha}
\right)}.
$$

Because the coordinate change is $\widetilde x^\alpha=x^\alpha-\xi^\alpha$, the perturbation changes at linear order by
$$
\widetilde h_{\mu\nu}
=h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu.
$$
Substitution into $R_{\mu\rho\alpha\beta}^{(1)}$ produces terms containing three partial derivatives of $\xi$. Since partial derivatives commute, every term cancels another with the opposite sign. Therefore
$$
\boxed{\widetilde R_{\mu\rho\alpha\beta}^{(1)}
=R_{\mu\rho\alpha\beta}^{(1)}}.
$$
This is the <gauge invariance of the linearized Riemann tensor>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/1/b/solution