A rotating frame has some similarities with special relativity, but since it is an accelerated frame it is not SR. To describe rotating coordinates consider the transformation between frame $F$ and $F^\prime$
$$
x^{\mu\prime}~=~{\Lambda^{\mu}}_\nu x^\nu,
$$
In cylindrical coordinates the radial and azimuthal coordinates are $r^\prime~=~r$ and $z^\prime~=~z$ and
$$
t^\prime~=~\gamma(t~-~\Omega\chi^t),~\theta^\prime~=~\gamma(\theta~-~\Omega\chi^\theta),
$$
where $\gamma$ is the Lorentz factor. The angular velocity $\Omega$ gives the tangential velocity $v~=~\Omega r$ as measured by a nonrotating observer. The rotation in these coordinates is
$$
\eta~=~{{d\theta}\over{dt}},~\eta^\prime~=~{{d\theta^\prime}\over{dt^\prime}},
$$
where in the rotating frame is given by
$$
\eta^\prime~=~\frac{d\theta~-~\Omega\Big(\frac{\partial\chi^\theta}{\partial t}dt~+~\frac{\partial\chi^\theta}{\partial\theta}d\theta\Big)}
{dt~+~\Omega\Big(\frac{\partial\chi^t}{\partial t}dt~-~\frac{\partial\chi^t}{\partial\theta}d\theta\Big)}.
$$
Define the velocities $v~=~r\eta$ and $v^\prime~=~r\eta^\prime$,
for $\chi^t~=~\theta$ and $\chi^\theta~=~t$. This gives a formula
for the addition of rotation
$$
\eta^\prime~=~{{\eta~-~\Omega}\over{1~-~r^2\Omega\eta/c^2}},
$$
which is analogous to the linear equation for velocity additions.
The metric for a rotating frame is then
$$
ds^2~=~dt^2~-~r^2\Big(d\theta~+~{\Omega\over c} dt\Big)^2~-~dr^2
$$
or
$$
ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2.
$$
The horizon for any observer on the rotating frame exists at $r~=~c/\Omega$. A solution to these coordinates with $dr~=~0$ according to the proper time $s$ is found to be
$$
t~=~ g^{-1}\sinh(gs),~\theta~=~{{g^{-1}}\over r}\cosh(gs)~-~g^{-1}{\Omega\over c} \sinh(gs),
$$
where $g~=~{{\Omega^2r}\over\sqrt{1~-~\Omega^2r^2/c^2}}$ is the acceleration parameter.
A rotating frame has a sort of event horizon, a Rindler wedge horizon $r~=~c/\Omega$. However, this is on the frame, so if you send photons along a radially directed fiber optic they get sort of trapped there a bit like a black hole. If you send them along a free path then there is no horizon. This is also related to the Sagnac effect, where sending photon pulses along a circular fiber optic in both directions results in one group returning before the other, or in an actual measurement a phase shift.
