Doubt regarding 4-velocity of a particle in Kerr space-time The Kerr metric written in Boyer-Lindquist coordinates is:
$$ds^2=-\frac{\Sigma\Delta}{A}dt^2+\frac{A\sin^2\theta}{\Sigma}(d\phi-\omega dt)^2+\frac{\Sigma}{\Delta}dr^2+\Sigma d\theta^2$$
where $\Sigma=r^2+a^2\cos^2\theta,\quad \Delta=r^2+a^2-2r,\quad A=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=2ar/A.$
The 4-velocity of a particle moving with angular velocity $\Omega (=u^\phi/u^t)$ around a Kerr black hole can be written as
$$\mathbf{u}=u^t\left(\frac{\partial}{\partial t}+\Omega\frac{\partial}{\partial\phi}\right),$$
where the $(\partial/\partial t)$ and $(\partial/\partial\phi)$ are the Killing vectors associated with stationarity and axisymmetry of the Kerr metric.

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*I couldn't understand how the above form of the 4-velocity  $\mathbf{u}$ is derived. I found the above equation in this textbook. The above equation is Eqn. 8.74 of the textbook.


*From the above equation how could I show that $\mathbf{u}\cdot\mathbf{u}=-1$?
I had tried but couldn't arrive at the results. I couldn't understand why only $u^t$ and $u^\phi$ is considered here and not $u^r$ and $u^\theta$.
Can someone help me in this regard? Any references would also  be sufficient.
 A: From what I read, the author wants to show that there are no static observers under ergosurface. So he is not interested in observers moving in the direction of $\partial_r$ and $\partial_\theta$ - he considers only those with circular orbits.
Then he shows that the normalization condition $\mathbf{u}\cdot \mathbf{u} = -1$ creates constraint on possible values of $\Omega$. In particular he shows, that under ergosurface this constraint implies $\Omega \neq 0$ and therefore no static observers are possible, whatever their proper acceleration.


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*I couldn't understand how the above form of the 4-velocity  $\mathbf{u}$ is derived. I found the above equation in [this textbook][2]. The above equation is Eqn. 8.74 of the textbook.


The form is not derived. It is the most general four velocity of observers constrained to move only in the $\partial_\phi$ direction.



*From the above equation how could I show that $\mathbf{u}\cdot\mathbf{u}=-1$?


Again, you are not showing this. The normalization condition is constraint that every real observer must satisfy. You will use this to reduce the space of all possible 4-vectors of the form
$\mathbf{u}=u^t\left(\frac{\partial}{\partial t}+\Omega\frac{\partial}{\partial\phi}\right)$
to space of only those 4-vectors that can describe motion of real, physical observer.
