# 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.

1. 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.

2. 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.

• In which page of which textbook did you find this equation? Context often matters and might help answering the question Commented Jun 29, 2022 at 0:32
• I don't get your question. $u^t\partial_t + u^\phi \partial_{\phi}$ is the most general 4-velocity for particle moving in the direction of $\partial_{\phi}$. What is there to derive? Or is $u^t$ and $u^\phi$ something specific and you forgot to write what it is? Also saying $\partial_t$ and $\partial_\phi$ are killing vector fields does not, as far as I know, specify which killing vector fields are you talking about. Maybe you could also mention in what coordinates are you working? Commented Jun 29, 2022 at 3:54
• in general though, you could derive what $u^t$ and $u^\phi$ is by demanding 4-velocity to be normalized and tangent vector to a geodesic you seek. Commented Jun 29, 2022 at 4:04
• I had edited the question with the relevant details. Commented Jun 29, 2022 at 6:53
• @Umaxo The question is based on the Kerr metric written in Boyer-Lindquist coordinates and the Killing vectors are associated with the stationarity and axisymmetry of the Kerr space-time. My concern is the reason why only $u^t$ and $u^\phi$ is considered. Since the particle can also have motion along $r$ and $\theta$ coordinates, shouldn't we need to include them to show that $u^\mu u_\mu=-1?$ Commented Jun 29, 2022 at 6:58

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.

1. 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.

1. 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.

• Thank you for the detailed answer. I have a question that if I consider motion also in the $\theta$ direction, how would the 4-velocity expression change? Commented Jun 29, 2022 at 19:51
• the same way as in geometry in your town. if $\partial_x$ is some basis vector from east to west and $\partial_y$ is some basis vector from south to north and $\partial_z$ from ground to sky, then the most general velocity vector for moving in the east-west direction will be $v^x\partial_x$, with $v^x$ being your speed. If you also want to move in the north-east direction, i.e. beable to move anywhere within your town (west, north, north-west or in any other such direction) your will just add the given term $v^x\partial_x + v^y\partial_y$. (cont.) Commented Jun 29, 2022 at 20:46
• @Richard ... The linear combination of these two basis vectors generate any possible velocity with the plane of the town. If you also like Moon and would like to visit there, you also need to be able to travel in ground-sky direction $v_x\partial^x + v^y\partial_y + v^z\partial_z$. In this case you are unconstrained to move anywhere in the universe since linear combination of all three basis vectors generate any possible direction and velocity of your movement Commented Jun 29, 2022 at 20:48
• ... so the answer is $u^t\partial_t+u^\phi\partial_\phi +u^\theta\partial_\theta$ Commented Jun 29, 2022 at 20:50
• nonzero component in direction $\partial_\theta$ will produce even more stringent constrained on Ω. Simply put, the normalization condition is there to make sure you will never move faster than the speed of light (and to properly dilate your clocks, since it is appropriate in STR for anyone moving with nonzero speed to have their clocks slowed down). So if you have nonzero speed in the direction of $\partial_\theta$, i.e. nonzero component $u^\theta$, then there is even less freedom on the choice of $u^\phi$ that will keep your speed below the speed of light. Commented Jun 29, 2022 at 21:02