# Equation of relativistic circular motion

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.