[antognini]

If the spacetime is static, then yes, the trajectory of a light beam (or anything else) is reversible. The question you pose is an excellent one and is one of the reasons that many physicists believed that black holes could not exist after Karl Schwarzschild discovered the Schwarzschild metric that describes spacetime around a spherically symmetric black hole.

So what happens if you direct a beam of light radially into a black hole? A distant observer will calculate that as the light beam descends into the black hole it will be (according to a distant observer) progressively blueshifted until it reaches the event horizon, at which point it will have blueshifted to infinite energy. In fact, according to the calculations of a distant observer, the light beam will never cross the event horizon, so the statement that the light beam is reversible remains true. The reason that the light beam (according to the distant observer) doesn't cross the horizon is that if you look at the metric, it states that ds^2 = (1 - R_Schw / r) c^2 dt^2 - 1 / (1 - R_Schw / r) dr^2, where R_Schw is the radius of the event horizon. Notice that at the event horizon r = R_Schw and the dr^2 term diverges. The distant observer therefore concludes that there is a singularity at the event horizon. This analysis lead physicists to initially conclude that black holes were unphysical.

A more careful analysis reveals that this isn't the case, however. We have been using this metric which describes the shape of spacetime according to a distant observer. But what we're really concerned with is not what a distant observer calculates, but how a local observer perceives spacetime. If we now switch to a coordinate system in which we are free falling into the black hole along with the beam of light, we find that nothing remarkable happens when we cross the event horizon. The light is blueshifted relative to its initial energy, but it doesn't have an infinite amount of energy. Thus there is no true singularity at the event horizon. This singularity that the distant observer found must therefore have been an artifact of the particular coordinate system that the distant observer was using. The free falling observer can then hold up a mirror and send off a light beam and have it return back to him even after they have crossed the event horizon.

[lisper]

You have this backwards. To a distant observer, light going into a black hole is red-shifted, not blue-shifted. If this were not the case, sending even a single photon into a black hole would produce infinite energy.

[antognini]

No, any photons that are emitted from matter close to the event horizon and travel toward the distant observer will be redshifted. Photons falling into the potential well of the black hole will be blueshifted. Of course, the distant observer is not able to directly detect photons falling into the black hole because he is far away from it. All he can do is calculate what he would expect the energy of the photons to be. This calculation gives an infinite energy as the photon approaches the event horizon, but it's not a "real" energy, so the black hole doesn't gain an infinite amount of energy.

An analogous effect occurs for an observer stationed at rest, hovering above the event horizon. Such an observer will observe the photons from afar to be blueshifed in agreement with the distant observer. Moreover, as this stationary observer gets closer and closer to the event horizon, the photons will be blueshifted to infinitely large energies.

[lisper]

Ah. OK, I misunderstood what you were saying. This is right.