special relativity - How do I transform onto a relativistic rotating frame of reference?

In classical mechanics, the usual formula to translate the evolution of a quantity as seen from an inertial frame of reference to a rotational frame is: $$\frac{d \textbf{A} }{dt} \vert_{Inertial} = \frac{d \textbf{A} }{dt} \vert_{Rotational} + \boldsymbol{\omega}\times\textbf{A};$$

How do we make this relativistic?

Answer

The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates $\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates $\{t', x', y', z'\} $ by making an appropriate substitution. For example, if $\vec{\omega} = \omega \hat{z}$, then we have \begin{align} t &= t' \\ x &= x' \cos \omega t' - y' \sin \omega t' \\ y &= x' \sin \omega t' + y' \cos \omega t' \\ z &= z' \end{align}

What's different is that now the metric is not as simple any more. If we take the differentials of all the above terms and substitute them into the Minkowski metric, we get \begin{align} ds^2 &= - dt^2 + dx^2 + dy^2 + dz^2 \\ &= -\left(1 - \omega^2 ({x'}^2 + {y'}^2) \right){dt'}^2 + 2 \omega (- y' dx' dt' + x' dy' dt') + {dx'}^2 + {dy'}^2 + {dz'}^2. \end{align} In particular, this means that an massive object's trajectory in these coordinates can easily have $|d\vec{x}/dt| > 1$; in other words, its coordinate velocity will be greater than $c = 1$. But if you calculate the four-velocity $\eta^\mu$ of this object, it will still be a timelike vector (or, equivalently, $ds^2 < 0$ for nearby points along its worldline.)

As far as the vector transformation law goes, that's a little more problematic. One of the main problems with the derivation of the result you cite is that splits up the derivative in the inertial frame into time derivatives of the coordinate components of the vector with respect to a set of rotating basis vectors, and derivatives of the rotating basis vectors themselves. Moreover, we assume that these basis vectors are constant with respect to space and time; in other words, a unit vector in the $x'$-direction at point A is the same as a unit vector in the $x'$-direction at point B. But you can see that defining a set of basis vectors is going to be problematic in a rotating reference frame; a vector that points in the $t'$-direction will change from timelike to spacelike as we cross the boundary $\omega^2 ({x'}^2 + {y'}^2) = 1$, so there's no way we could possibly make that into a constant vector.

This is not to say that we can't do particle dynamics in a rotating reference system, even for relativistic motion. The better way is to take the Lagrangian approach, in which a particle moving between two events in spacetime will extremize the proper time along its trajectory: $$ \tau = \int \sqrt{ - ds^2}. $$ In the rotating reference frame, this can be written as $$ \tau = \int \sqrt{ \left(1 - \omega^2 ({x'}^2 + {y'}^2) \right) + 2 \omega (- y' \dot{x}' + x' \dot{y}') - (\dot{\vec{r}'})^2 } dt' $$ and we can find a set of Euler-Lagrange equations for $\vec{r}'(t')$ that extremize this integral in the usual way. This would be the easiest way to generalize the Coriolis force and the centrifugal force into a relativistic context.