Riemann curvature in orthonormal frame and Lorentz transformations I have a problem with understanding how the Riemann tensor in an orthonormal frame transforms using Lorentz transformation of frames.
I was reading this paper by Morris and Thorne. The authors have used the metric
$$g_{\mu\nu}={\rm diag}[-e^{2\phi(r)},1/(1-b(r)/r),r^{2},(r\sin\theta)^{2}]
$$
to create an orthonormal (static observer) frame:
\begin{align}
{\bf e}_{\hat{t}} &= e^{-\Phi}{\bf e}_{t}, &
{\bf e}_{\hat{r}} &= (1-b/r)^{1/2}{\bf e}_{r},\\
{\bf e}_{\hat{\theta}} &= r^{-1}{\bf e}_{\theta},&
{\bf e}_{\hat{\phi}} &= (r\sin\theta)^{-1}e_{\phi},
\end{align}
and an orthonormal moving observer frame:
\begin{align}
{\bf e}_{\hat{0}'} &= {\bf u} = \gamma{\bf e}_{\hat{t}}\mp\gamma(v/c){\bf e}_{\hat{r}},&
{\bf e}_{\hat{1}'} &= \mp\gamma {\bf e}_{\hat{r}} + \gamma(v/c){\bf e}_{\hat{t}},\\
{\bf e}_{\hat{2}'} &={\bf e}_{\hat{\theta}},&
{\bf e}_{\hat{3}'} &= {\bf e}_{\hat{\phi}}. 
\end{align}
Then, they calculated the Riemann tensor components for the static observer case — nothing horrifying.
After that they transformed the Riemann tensor from the static frame to the moving one (special relativity transformation). I'm not sure how to do it — I'm guessing that I have to use Lorentz transformation matrices:
$$\Lambda^{\mu}{}_{\nu}= 
   \begin{pmatrix}
    \gamma & \mp\gamma(v/c) &0 &0\\
    \gamma(v/c) & \mp\gamma & 0 &0\\
    0 & 0 & 1 &0\\
    0& 0 & 0 &1 \\
    \end{pmatrix}.
$$
The Riemann tensor is a (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 the right way to do it?
 A: A Lorentz transformation is a special kind of coordinate transformation. Any tensor $T$ is invariant under coordinate transformations. What this means is that, once you         decompose $T$ in terms of some basis of your (co)tangent space, say
$$
T = \sum_{I}T^I{\bf e}_I,
$$
the components $T^I$ change under a (linear) coordinate transformation ${\bf e}_I \mapsto {\bf e}_{\hat{I}} = \sum_{I}\Lambda_{\hat{I}}{}^{I} {\bf e}_I$ in order to keep $T$ as a whole invariant. I have used hats here to indicate the transformed frame. In the particular case of the Riemann tensor ${\rm Riem}$ at a given point $p$, we have
$$
{\rm Riem} = R^\rho{}_{\sigma\mu\nu} {\bf e}_\rho \otimes {\bf e}^\sigma \otimes {\bf e}^\mu \otimes {\bf e}^\nu,
$$
where the Einstein summation convention is implied and ${\bf e}_\rho$ and ${\bf e}^\mu$ are basis elements of the cotangent space $T^\ast_pM$ and tangent space                  $T^\phantom{\ast}_pM$, respectively. Under a coordinate transformation with associated Jacobi matrix $\Lambda^\rho{}_{\hat{\mu}}$, these basis elements transform as
$$
{\bf e}^\rho = \Lambda^\rho{}_{\hat{\alpha}}{\bf e}^{\hat{\alpha}},
\qquad
{\bf e}_\mu = \Lambda^{\hat{\alpha}}{}_\mu{\bf e}_{\hat{\alpha}},
$$
and so on. Therefore, for the Riemann tensor to be invariant, the components transform as
$$
R^\rho{}_{\sigma\mu\nu} =  \Lambda^\mu{}_{\hat{\alpha}} \Lambda^{\hat{\beta}}{}_\sigma  \Lambda^{\hat{\gamma}}{}_\mu \Lambda^{\hat{\delta}}{}_\nu                                R^{\hat{\alpha}}{}_{\hat{\beta}\hat{\gamma}\hat{\delta}}
$$
Notice that this transformation is straightfowardly inverted, which allows one to find the components of the transformed Riemann tensor in terms of the components in the old coordinate system.
This would correspond with the frame of the moving observer in your case.
So, in summary: yes, you can find the components of the Riemann tensor in the frame of the moving observer in terms of the static observer in the way you suspected. In fact, this is how the components of the Riemann tensor in the inertial frame were calculated to begin with; from the transformation of the components in the curved frame. See equation (8) in your reference. Do note that Morris and Thorne perform this final coordinate transformation on the components of the Riemann tensor with all indices down.
