# Riemann curvature in orthonormal frame and Lorentz transformations

I have problem with understading how Riemann tensor in orthonormal frame transforms using Lorentz transformation of frames. I was reading Morris Thorne paper from 1988 (American Journal of Physics 56, 395 (1988); https://doi.org/10.1119/1.15620). Authors have used metric: $$g_{\mu\nu}=diag[-e^{2\phi(r)},1/(1-b(r)/r),r^{2},(rsin\theta)^{2}]$$ to create orthonormal (static observer) frame:

and orthonormal moving observer frame:

Then, they've calculated riemann tensor components for static observer case - nothing horrifying. Afther that they transformed Riemann tensor from static frame to moving one (special relativity transformation). I'm not sure how to do it -i'm guessing that i have to use lorentz transformation matrices:

$$\Lambda^{\mu}_{\nu}= \begin{pmatrix} \gamma & ^-_+\gamma(v/c) &0 &0\\ \gamma(v/c) & ^-_+\gamma & 0 &0\\ 0 & 0 & 1 &0\\ 0& 0 & 0 &1 \\ \end{pmatrix}$$

Riemann tensor is (1,3) rank tensor so it will be 3 inverse lorentz matrices ($$\Lambda^{\nu}_{\mu}$$) and one standard $$R^a_{bcd}=\Lambda^{a}_{\mu}\Lambda^{\nu}_{b}\Lambda^{\sigma}_{c}\Lambda^{\zeta}_{d}R^\mu_{\nu\sigma\zeta}$$

Is this right way to do it?