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In a Kerr Black hole there is a region where the component $g_{t t}$ of the metric changes sign (the ergoregion). The surface where $g_{t t}=0$ is called the ergosphere.

So, if we consider an observer with four-velocity $u^{\mu}=\frac{d x^{\mu}}{d \tau}=\gamma \dot{x}^{\mu}$, and (for simplicity) with $u^{\theta}=0$ and $u^{\phi}=\gamma \dot{\phi} \geq0$, we have:

$$-1= u^{\mu} u_{\mu}= (g_{t t}+g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2)\gamma^2 \\ \Rightarrow \gamma=\sqrt{-\frac{1}{g_{t t}+g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2}}$$.

We can have ${g_{t t}+g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2}=0$ for:

  • $\dot{r}=\dot{\phi}=g_{t t}=0$

This means that the observer is "standing still" on the ergosphere, this case is not particularly interesting because we can demonstrate that for having $\dot{\phi}=0$ at the ergosphere, we actually have to move at the speed of light in the counter rotational direction of the BH, so it is obvious that $\gamma \to \infty$.

  • $\dot{\phi}>0,\,\,\, -g_{t t}=g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2$

The real problem comes here, being $g_{t t}<0$ outside the ergoregion this situation is possible. And in principle, an observer falling towards the ergosphere should ever reach a point where the equation is satisfied. Indeed $g_{t t}$ is negative and $g_{t t} \to 0$, at the moment that the observer "crosses" the ergosphere we have $\gamma=\sqrt{-\frac{1}{g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2}}$, but we said that $\dot{\phi}>0$ so $\gamma$ is an imaginary number that is not possible.

This means that the denominator goes to zero ($\gamma \to \infty$) before the ergoregion.

The physics implication is very strong, an outside observer should never "see" (all photons are redshifted to infinity) the first observer crossing the ergosphere (this is totally analogue with the event horizon of the Schwarzschild Black Hole).

  • The point of all this is:

From the point of view of an observer at infinity, how can a particle escape from the ergoregion if it never entered for him?

(In the case it can't, this means that also the ergoregion is a "black hole region"?, is the Penrose Process not feasible?).

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Di Pinto Andrea asked: "From the point of view of an observer at infinity, how can a particle escape from the ergoregion if it never entered for him?"

The observer at infinity does see the particle cross the ergosphere, $g_{\rm tt}=0$ (covariant) at the ergosphere but $1/g^{\rm tt}$ (contravariant) is not.

$1/\sqrt{g_{\rm tt}}$ in Boyer Lindquist coordinates gives you the time dilation relative to an observer who is stationary with respect to the fixed stars, which as you know is not possible to be stationary at the ergosphere.

$\sqrt{g^{\rm tt}}$ on the other hand gives you the time dilation relative to a locally nonrotating and radially stationary frame (that of the local ZAMO). Outside of the horizon the local velocities relative to a ZAMO are always $\rm v \leq c$ (and inside the horizon radially stationary ZAMOs can not exist), so the total time dilation of a test particle in the frame of the far away coordinate bookkeeper $\dot{\rm t}=\sqrt{g^{\rm tt}/(1-\rm v^2)}$ is always positive and finite outside of the horizon.

Di Pinto Andrea wrote: "The real problem comes here, being $g_{\rm tt}<0$ outside the ergoregion this situation is possible."

This is not possible, $g_{\rm tt}=1-\rm \frac{2 r}{a^2 \cos ^2 \theta +r^2}$ and the ergosphere is at $\rm r=1+\sqrt{1-a^2-a^2 \cos ^2 \theta }$ so there is no way for $g_{\rm tt}$ to be smaller than $0$ outside the ergosphere.

$g_{\rm tt}$ is a component of the metric tensor and therefore independent of the proper time derivatives $\{ \rm \dot{t},\dot{r},\dot{\theta},\dot{\phi} \}$, it depends only on the coordinates $\{ \rm r, \theta \}$ itself.

You seem to have some mess in your calculation, for the correct equation to find $\rm v$ (in Boyer Lindquist coordinates relative to a local ZAMO) and therefore $\gamma$ see here and the links therein. The correct equation for $\gamma$ (in a metric where the only crossterm is $g_{\rm t \phi}$) is

$$\gamma = \sqrt{\frac{1}{1-\rm v^2}}=\frac{\sqrt{g_{\rm tt}+{g_{\rm t\phi}}^2 \ \dot{\phi}^2-g_{\rm tt} \ g_{\rm rr} \ \dot{\rm r}^2-g_{\rm tt} \ g_{\theta \theta} \ \dot{\theta}^2-g_{\rm tt} \ g_{\phi \phi} \ \dot{\phi}^2} - g_{\rm t\phi} \ \dot{\phi}}{\sqrt{g^{\rm tt}} \ g_{\rm tt}}$$

where you also need the contravariant time component of the metric, not only the covariant ones. Right on the ergosphere you get $0/0$ in Boyer Lindquist coordinates (which is better than $1/0$, because you can still take the limit and get a finite value).

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  • $\begingroup$ Can you rephrase your first statement please? I haven't understood the meaning sorry. I have used the signature (-+++) so $g_{tt}$ is always negative outside the ergosphere (and the others are positive). Yeah $g_{tt}$ doesn't depend on $\{\dot{t},\dots\}$, when I wrote: $\,\,$ "- $\dot{\phi}>0, \,\,\, -g_{tt}=g_{\phi \phi}\dot{\phi}^2+2g_{t \phi}\dot{\phi}+g_{r r}\dot{r}^2$ " I meant it as an equation, in the sense that for an observe with $\dot{\phi}>0$ there is a point in the spacetime where the equality is satisfied and so $\gamma \to \infty$. $\endgroup$ – Andrea Di Pinto Jul 7 '20 at 21:29
  • $\begingroup$ The signature should be the least problem, but if you choose your conditions such that $\gamma = \infty$ then your local velocity relative to a ZAMO is $\rm v=c$ and therefore unphysical. If you want to have $\rm v=c$ you need to set $\rm ds^2=0$ when solving for your initial conditions, then you can differentiate by the photon's affine parameter instead of the particle's proper time. $\endgroup$ – Gendergaga Jul 7 '20 at 21:36
  • $\begingroup$ Is this true in general? I mean, in the Schwarzschild BH we have $\gamma \to \infty$ approcing the horizon but the velocity shouldn't be $v=c$. Using the same analogy the formula for $\gamma$ is $\gamma=\sqrt{-\frac{1}{g_{tt}+g_{rr}\dot{r}^2}}$, and so if $\dot{r} \neq 0$ we should have $\gamma \to \infty$ "before" the horizon. I used this analogy because $g_{tt} \to 0$ at the horizon in the Schw. BH as at the ergosphere in Kerr. $\endgroup$ – Andrea Di Pinto Jul 7 '20 at 22:16
  • $\begingroup$ Why should the velocity not be c at the horizon, if you fall in from infinity you have the negative escape velocity which is c at the horizon. Only photons can be stationary at the horizon, so since at the horizons only photons can be the stationary reference particles relative to whom the local velocity is measured (at least in the regular Droste coordinates) there is only one velocity you can have relative to them, and that is c. The same goes for stationary particles at the ergosphere, only a retrograde photon can be stationary there, so relative to it you have v=c and γ=∞. $\endgroup$ – Gendergaga Jul 7 '20 at 22:38
  • $\begingroup$ Ok I didn't think about this. About the formula for $\gamma$ I have searched in all my books and all seems to agree with my formula rater than yours. How have you obtained it? $\endgroup$ – Andrea Di Pinto Jul 11 '20 at 20:14
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A distant observer would see the collapse of the star and all other events at or inside the event horizon stop, when the event horizon was formed. If an observer could exist at or inside the horizon, he would see things occurring at a normal rate.

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