Particle crossing the outer event horizon of a Kerr black hole 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?
 A: 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 $r_+$.
