The metric directly gives
$$
g(e_0,e_0)=-1,\qquad
g(e_i,e_j)=t^{-p_i}t^{-p_j}t^{2p_i}\delta_{ij}
=\delta_{ij},
$$
with no mixed products. Hence
$$
\boxed{e_0=\partial_t,\qquad e_i=t^{-p_i}\partial_{x^i}}
$$
is an orthonormal frame.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-309/3/a/solution