# Conservation equation of a perfect fluid in Kerr geometry

I am reading a paper in which a perfect fluid in the Kerr geometry is studied. The stress-energy tensor is

$$T_{\mu \nu} = (\epsilon + P) u_\mu u_\nu + P g_{\mu \nu},$$

where $P$ is the fluid pressure and $\epsilon$ the fluid proper energy density. The four-velocity is $u^\mu = u^t (1,0,0,\Omega)$ (circular motion).

They say that in this case, the energy-momentum conservation equation, $\nabla_\nu T^\nu_\mu = 0$, takes the form

$$\frac{\nabla_\mu P}{\epsilon + P} = - \nabla_\mu \ln(-u_t) + \frac{\Omega \nabla_\mu \mathcal{l}}{1-\Omega \mathcal{l}},$$

with $l \equiv -u_\phi/u_t$, and $\Omega \equiv u^\phi / u^t$.

So, my questions are

1. Which are the intermediate steps to reach the second form for the energy-momentum conservation equation?

2. Why the derivatives of the scalars $P$ and $l$ are kept in a covariant-form?

Thank you very much!