Assume we had an elevator shaft long enough for a free-falling elevator to reach terminal velocity.
As I understand it, when the elevator begins to fall a person inside would experiences weightlessness because they would be accelerating downward at the same rate as the elevator. The rate of downward acceleration for the elevator should reach a constant when air resistance equals gravity (right?). However, the person in the elevator would not be exposed to air resistance and should continue to accelerate downward.
So would they gradually sink back to the floor and experience gravity as though the elevator were at rest… at least until the inevitable termination of the experiment?
Best Answer
That is exactly right. A fundamental tenet of physics is that all inertial reference frames are equivalent and indistinguishable.1 Furthermore, given one inertial frame (standing at rest2), any other frame moving with respect to it with a constant velocity is also inertial. The frame "moving at terminal velocity" is just as inertial as "sitting still" and so you would not even be able to tell you were moving.
By definition you feel no acceleration at constant velocity. Thus the acceleration due to gravity must be exactly balanced by some other force. By construction that force is not air resistance for you (as would be the case of a sky diver at terminal velocity) but simply the normal force of the elevator floor, which would make the experience feel exactly like standing in a non-moving elevator in the same gravitational field.
1 At least locally, meaning that any experimental apparatus and things you measure are confined to objects also in that frame.
2 To be pedantic, standing "still" in a gravitational field is considered inertial in Newtonian mechanics but not general relativity. I am speaking in Newtonian terms here, but the conclusion would be just the same if analyzed with the machinery of GR.