even in the black-hole spacetimes; moreover, it is not possible to define the co-rotating (counter-rotating) orbits in relation to stationary observers at infinity, as can be done in the Kerr spacetimes, since the KdS spacetimes are not asymptotically flat.

Orientation of the circular orbits in the KdS spacetimes must be related to locally non-rotating frames (LNRF), similarly to the case of asymptotically flat Kerr spacetimes. In the KdS spacetimes, the tetrad of 1-forms corresponding to the LNRF is given by Ref. :

$$\begin{aligned}
\omega^{(t)}&\equiv&\left
(\frac{\Delta_{r}\Delta_{\theta}\varrho^{2}}{I^{2}A}\right )^{1/2}{\rm d}t,
\nonumber \\
\omega^{(r)}&\equiv&\left (\frac{\varrho^{2}}{\Delta_{r}}\right )^{1/2}{\rm
  d}r, \nonumber\end{aligned}$$

$$\begin{aligned}
\omega^{(\phi)}&\equiv&\left (\frac{A\sin^{2}\theta}{I^{2}\varrho^{2}}\right
)^{1/2}({\rm d}\phi-\Omega {\rm d}t), \\ 
\omega^{(\theta)}&\equiv&\left (\frac{\varrho^{2}}{\Delta_{\theta}}\right
)^{1/2}{\rm d}\theta, \end{aligned}$$

2exwith the angular velocity of the LNRF being given by $$\Omega\equiv\frac{{\rm d}\phi}{{\rm d}t}=\frac{a}{A}\left
                 [-\Delta_{r}+(r^{2}+a^{2})\Delta_{\theta}\right ];\quad 
A\equiv (r^{2}+a^{2})^{2}-a^{2}\Delta_{r}.$$

Locally measured components of 4-momentum in the LNRF are given by the projection of a particle’s 4-momentum onto the tetrad $$p^{(\alpha)}=p^{\mu}\omega^{(\alpha)}_{\mu}, \quad
p^{\mu}=m\frac{{\rm d}x^{\mu}}{{\rm d}\tau}\equiv m\dot{x}^{\mu}=\frac{{\rm
    d}x^{\mu}}{{\rm d}\lambda}.$$ A simple calculation reveals the intuitively anticipated relation $$p^{(\phi)}=\frac{mr}{A^{1/2}}L.$$ We can see that the sign of the azimuthal component of the 4-momentum measured in the LNRF is given by the sign of the specific angular momentum of a particle on the orbit of interest. Therefore the circular orbits 