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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.