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After discovering the existence of hyperbolic (Rindler) motion, I started wondering if the equations of a circular motion of a relativistic particle are different from those of a classical particle, so I did the calculations.
I started from the Lorentz transformations (where the position $r'$ was decomposed in components parallel and perpendicular to the relative velocity): $$ \begin{cases} r=r'_{\perp}+\gamma(r'_{||}+V_Rt)\\ t=\gamma(t+V_R\cdot r/c^2) \end{cases} $$ I am assuming that $r$ and $t$ are the coordinates in the lab (inertial) frame, whereas $r'$ and $t'$ are the coordinates in an instantaneous inertial frame with relative velocity $V_R$ and comoving with an accelerated object.
After doing so I calculated the ratio $dr/dt$, and after finding the velocity I calculated $dv/dt$ and, considering that $S'$ is the instantaneous comoving inertial frame (i.e. $v'=0$ and $V_R=v$), I eventually found the following formula: $$ \frac{dv}{dt}=\gamma^{-2}a_\perp+\gamma^{-1}(1-v^2/c^2)a_{||} $$ At this point, as a check, I considered the case of an acceleration parallel to velocity, that is $a=a_{||}$ (or $a_\perp=0$). In this case you can see that we have $\frac{dv}{dt}=(1-v^2/c^2)^{3/2}a$, which is the equation for the velocity of a Rindler observer (you can integrate it to find the usual hyperbolic equations $t=\frac{c}{a}\mathrm{sinh}(\frac{a\tau}{c})$ and $x=\frac{c^2}{a}\mathrm{cosh}(\frac{a\tau}{c})$).
However, considering the case of a centripetal acceleration ($a=a_\perp$), and solving the resulting differential equation, I found $$ v(t)=c*\mathrm{tanh}(\frac{at}{c}) $$ which means that, in spite of what happens in the non-relativistic case, the speed of a relativistic particle subjected to a centripetal force increases with time. This seems very weird to me, so I would like to ask you if it is actually the case or if I am wrong.

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