# Why can't a particle rotate opposite to the central mass within the ergosphere?

Wiki says about the Kerr metric:

A moving particle experiences a positive proper time along its worldline, its path through spacetime. However, this is impossible within the ergosphere, where $g_{tt}$ is negative, unless the particle is co-rotating with the interior mass $M$ with an angular speed at least of Ω. Thus, no particle can rotate opposite to the central mass within the ergosphere.

I don't get it. Can it be somehow seen that $g_{tt}$ is negative from

\begin{align} g^{\mu\nu}\frac{\partial}{\partial{x^{\mu}}}\frac{\partial}{\partial{x^{\nu}}} = & \frac{1}{c^{2}\Delta}\left(r^{2} + \alpha^{2} + \frac{r_{s}r\alpha^{2}}{\rho^{2}}\sin^{2}\theta\right)\left(\frac{\partial}{\partial{t}}\right)^{2} + \frac{2r_{s}r\alpha}{c\rho^{2}\Delta}\frac{\partial}{\partial{\phi}}\frac{\partial}{\partial{t}} \\ & - \frac{1}{\Delta\sin^{2}\theta}\left(1 - \frac{r_{s}r}{\rho^{2}}\right)\left(\frac{\partial}{\partial{\phi}}\right)^{2} - \frac{\Delta}{\rho^{2}}\left(\frac{\partial}{\partial{r}}\right)^{2} - \frac{1}{\rho^{2}}\left(\frac{\partial}{\partial{\theta}}\right)^{2} \color{red}{?} \end{align}

And why can't no particle can rotate opposite to the central mass within the ergosphere?

## 2 Answers

I am writing my Ph.D. thesis and wanted to address the same question.
The best demonstration I found is at page 133 of "A First Course in General Relativity (Second Edition)" by Bernard Schutz. It goes like this:

Let us consider a photon at a radius $$r$$, moving in an equatorial orbit ($$\theta=\pi/2$$ and $${\rm d}\theta=0$$) around the Black Hole (BH). The photon can go in the directions $$\pm \phi$$ tangent to the orbit. The metric reads in this case (we have a photon so $${\rm d}s^2 = 0$$): $$0 = g_{tt}\,{\rm d}t^2 + 2g_{t\phi}\,{\rm d}t{\rm d}\phi + g_{\phi \phi}\,{\rm d}\phi^2.$$ Solving for the tangential velocity $${\rm d}\phi / {\rm d}t$$: $$\frac{{\rm d}\phi}{{\rm d}t} = -\frac{g_{t\phi}}{g_{\phi\phi}} \pm \left[ \left( \frac{g_{t\phi}}{g_{\phi\phi}} \right)^2 - \frac{g_{tt}}{g_{\phi\phi}} \right]^{1/2}$$ Now, in the words of Schutz, a remarkable thing happens if $$g_{tt}=0$$, which is where the ergosphere occurs. There are two solutions: $$\begin{equation} \begin{split} \frac{{\rm d}\phi}{{\rm d}t} &= 0, \\ \frac{{\rm d}\phi}{{\rm d}t} &= -2 \frac{g_{t\phi}}{g_{\phi\phi}}. \end{split} \end{equation}$$ One can prove that the second solution, using the $$g_{t\phi}$$ and $$g_{\phi\phi}$$ from the Kerr metric, gives the same sign of $$a$$, i.e. the Black Hole spin ($$\alpha$$ in the Original Poster's metric notation).
$${\rm d}\phi / {\rm d}t = 0$$ represents the photon that is trying to move in the direction opposite to the BH rotation and cannot do it (the dragging of the orbits prevents it); $${\rm d}\phi / {\rm d}t$$ with the same sign as $$a$$ (or $$\alpha$$) represents the photon moving in the same direction of the BH rotation. Any (massive) particle slower than a photon will be forced to co-rotate with the BH.

Cheers!

If you require that the particle's trajectory is timelike, i.e. $u^2 = -1$, as it must be for a massive particle, then you can see that this equation cannot be satisfied unless the particle is rotating with the black hole (once it passes inside the ergoregion).

If you artificially forced the particle to not rotate, and you dipped it behind the ergoregion, then if would be a spacelike trajectory, not appropriate for a massive particle.

• I think I got something you say, but I don't see that this equation cannot be satisfied unless the particle is rotating with the black hole (once it passes inside the ergoregion)... – draks ... Feb 12 '15 at 1:30
• Work it out! Here's a quick way to see that some sort of rotation is needed: a static trajectory (not geodesic) would have $u^{\mu} = (\partial_t )^{\mu}$, and the norm of this is $u^2 = g_{tt}$ which is positive in the ergoregion by definition (with the mostly plus convention). – Surgical Commander Feb 12 '15 at 1:35
• No, sorry I need more detail...if you don't mind! Is this related to Frame dragging? – draks ... Feb 12 '15 at 1:46