3 ms·

I tentatively don't agree with your Rindler horizon analysis. Inside an inertial frame, you don't see the Rindler horizon, because only the accelerating observe

I tentatively don't agree with your Rindler horizon analysis. Inside an inertial frame, you don't see the Rindler horizon, because only the accelerating observer can see it. If you are in the accelerating frame of the observer, the apparent position of the Rindler horizon will vary, depending on your position, so you can't actually cross it; but in any case you are not in an inertial frame. Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points). At this point, my understanding of physics gets hazy so I'm not sure how much this extrapolates to the black hole case.

pdonis 14y ago
Inside an inertial frame, you don't see the Rindler horizon, because only the accelerating observer can see it. Actually, the accelerating observer can't "see" the Rindler horizon; light emitted at the Rindler horizon never reaches the accelerating observer (that's the definition of the Rindler horizon). The inertial observer is the one who can actually "see" the horizon, because he passes it. Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points). Any two points within the "accelerating frame" of the accelerating observer can signal to each other. The Rindler horizon is not, strictly speaking, "within" that frame; Rindler coordinates become singular at the Rindler horizon. That said, I don't see what the properties of the accelerating frame have to do with the statement I made that any inertial frame in flat spacetime will include the Rindler horizon, since all inertial frames in flat spacetime are global.