# Can a free falling observer localize the event horizon by calculations?

I'm think that in general relativity we can always pass the one curve in one coordinate system for another coordinate system. My intuition say that the free falling observer locate the event horizon in a light-like curve. My intuition is right?

I know that nothing special happens in event horizon for a free-falling observer, so he can measure the event horizon? This implies some issues in the complementarity principle?

• Your question could use some clarification. – lionelbrits Feb 1 '15 at 0:41
• I edited, is better now ? – Nogueira Feb 1 '15 at 1:34
• What do you mean by "pass one curve in one coordinate system for another coordinate system"? Generally you can only locate the event horizon after waiting an infinitely long time to see which trajectories made it out and which didn't. But for static black holes you can use trapped surfaces, which act like event horizons intuitively. – lionelbrits Feb 1 '15 at 1:59

For a freely falling observer the local geometry of spacetime is always flat i.e. described by the Minkowski metric. So the freely falling observer can never observe themself to fall through an event horizon, because that contradicts the requirement that spacetime be locally flat. In fact the freely falling observer will observe an apparent horizon that retreats before them as they fall. They would reach the horizon only at $r = 0$ i.e. the moment they reach the singularity.