the above equation we have also taken the integration constant as zero, from requiring $U_{r} \rightarrow  0$ for $r \rightarrow 0$. Choosing without loss of generality the two characteristic radii $\bar{r}$ and $\bar{r}_{\rho}$ both equal to $GM/c^{2}$, we can now write the full solution as:

$$\rho_{1}=\bar{\rho}_{J}\left( \frac{G M}{c^{2} r} \right)^{1/2} \left(3cos^{2}(\theta)-1 \right),$$

$$V_{1}=\left( \frac{\bar{\rho}_{J}}{\bar{\rho}_{0}} \right)\left( \frac{2c^{4}r}{GM} \right)^{1/2}
\left( \frac{3cos^{2}(\theta)-1 }{g_{\theta}} \right),$$

$$U_{1}=-\left( \frac{12 \bar{\rho}_{J}}{5\bar{\rho}_{0}} \right) \left( \frac{r}{GM} \right)^{3/2}
 \left( \frac{c^{4}cos(\theta)sin(\theta)}{2^{1/2}g_{\theta}} \right),$$

where we have introduced $\bar{\rho}_{J}=\bar{\rho}_{\theta} \bar{\rho}_{r}$ and $g_{\theta}$ as the angular part of $g(\theta)$, $g(\theta)/\bar{\rho}_{0}$. The full solution can be seen to depend only on the two parameters $\bar{\rho}_{0}$ and $\bar{\rho}_{J}$, normalisation constants for the densities of the background state and the perturbation, with the velocities of the perturbation solution depending only, and linearly, on the ratio $Q=2^{1/2} \bar{\rho}_{J} / \bar{\rho}_{0}$, which is expected to be small, beyond natural dependences on the physical parameters of the problem, $M$ and $c$.

We see form eq.(45) that the angular velocity will be zero only for $\theta=\pi/2$, $\theta=0$ and $\theta=\pi$. Thus, movement along the plane of the disk will remain along the plane, but also, along the poles movement will be exclusively radial. This last point, together with the positive sign of 