I am quite puzzled by the following statement in Sean Carroll's 'Spacetime and geometry' (formula 6.100).

A particle with momentum $p^\mu$ crossing the outer event horizon of a Kerr black hole $r=r_+$ "moving forward in time" satisfies $$p^\mu\chi_\mu \lt 0. $$ $\chi = \partial_t +\frac {a}{r_+^2+a^2}\partial_{\phi}$ is the killing vector that is null on the outer horizon, with $a $ being the ratio between Komar angular momentum and the Komar energy of the black hole.

Using the components of the Kerr metric tensor $g_{\mu\nu}$ and evaluating the inner product at $r=r_+$, I get $$p^\mu\chi_\mu = 0 $$ for any value of $p^\mu$. Can somebody explain me how to prove the inequality and what I am doing wrong?


That is a statement about the energy, as seen by a particular observer.

Remember that the energy is an observer dependent quantity. In special relativity we defined the energy of a particle with 4-momentum $p^{\mu}$ measured by an observer with 4-velocity $u^{\mu}$ as:

$$E^{(u)} = - \eta_{\mu \nu} u^{\mu} p^{\nu} > 0$$

that in general relativity generalizes to


For instance for a static observer in special relativity, that is $u^{\mu} = (1,0,0,0)$:

$$E^{(static)} = - p_{0}$$

For the particle to be moving forward in time, the energy must be positive. Notice that this is a tensorial statement, so it's true in every coordinate frame.

Now in the kerr spacetime

$$E^{(static)} = E$$

where $E$ is the constant of motion $-(\partial_t)^{\mu} u_{\mu} = -u_0 = -p_0$ (the last equality can always be satisfied, using the reparametrization freedom of the geodesic) associated to the timelike Killing vector $\partial_t = (1,0,0,0)$, therefore $E$ can be interpreted as the energy seen by a static observer at infinity, and must be positive. If we are inside the ergoregion, there are not static observers, since the black holes is dragging us. A convenient observer that is corotating with the hole has four velocity $u^{\mu} \propto (1,0,0,\Omega_H)$, therefore:

$$E^{(rotating)} \propto (E-\Omega_H L)$$

where again $L$ is the constant of motion associated to the rotational Killing vector $\partial_\phi = (0,0,0,1)$. The statement that the energy seen by such an observer is positive implies the statement $p^{\mu}\chi_{\mu} < 0$.

The Kerr spacetime is peculiar since following a process of particle decay $E^{(0)} = E^{(1)} + E^{(2)}$ certain particles can have $E^{(2)} < 0$, but there is no contradiction with what I said before, since this happen only if these particles are unable to escape to infinity, therefore there isn't an interpretation as energy seen by a static observer at infinity.

Notice that all the above reasonings are done before crossing $r_+$.

  • $\begingroup$ I understand that saying that moving forward in time, means that the energy measured by a timelike moving observer should be <0. As static observers dont't have a timelike velocity fourvector, we need to find another one. The obvous choice is one that is rotating around the BH just outside the horizon and with the angular speed Omega_H. In that case the U fourvector is equal to the killing vector chi defined above. I can expect that such a chi is timelike by construction. Is there a way to show directly and without too much calculations that chi is timelike just outside the horizon? $\endgroup$ – jac Nov 28 '17 at 19:51
  • $\begingroup$ I don't think it's possible to do this without plugging in the metric and computing the norm of chi. $\endgroup$ – Rexcirus Nov 28 '17 at 21:01

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