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No, the observer that is freely falling can determine that there is curvature by measuring second derivatives of the metric tensor, which would include calculations of the first derivatives of the connection, and obtain a Riemann tensor. If there is no curvature they will all be zero. If there is at least some of them will not be zero. And he can compute various scalars from them, where at least one will not be zero.

The error in the thinking is that it ignores that that 'inertial' frame where the observer is freely falling is only local. The observer will measure deviations from flatness as he goes away from his local area, or equivalently, the connection may be zero at his point, but its derivatives, and with them used to construct the curvature tensor and scalars, which will have some nonzero component. If the observer has a good enough accelerometer, some inches or feet or more from his reference point for his local inertial frame, he will measure an acceleration then. If falling into a black hole, for instance, his body will elongate as the gravitational field increases, even if the observer thought his body was local and was in his inertial frame. He simply was wrong.

Both thenWeak and the Einstein Equivalence Principle state that those inertial frames are only in 'small enough regions of spacetime'. I forget what the Strong Equivalence Principle adds to the Weak, but I think it includes the 'small enough regions' statement or idea. Those inertial frames cannot be extended and stay inertial.

So, no, there is no preferred frame. The local free falling inertial frame is just a good initial approximation where he can ignore the gravitational filed, because near that the accelerations are approximately the same. But it is only an approximation, and when he goes a distance away it fails being inertial.

Einstein's General Relativity captured that Relativity principle perfectly. It's held up to all the conceptual and real physical experiments, like your thought experiment.

No, the observer that is freely falling can determine that there is curvature by measuring second derivatives of the metric tensor, which would include calculations of the first derivatives of the connection, and obtain a Riemann tensor. If there is no curvature they will all be zero. If there is at least some of them will not be zero. And he can compute various scalars from them, where at least one will not be zero.

The error in the thinking is that it ignores that that 'inertial' frame where the observer is freely falling is only local. The observer will measure deviations from flatness as he goes away from his local area, or equivalently, the connection may be zero at his point, but its derivatives, and with them used to construct the curvature tensor and scalars, which will have some nonzero component. If the observer has a good enough accelerometer, some inches or feet or more from his reference point for his local inertial frame, he will measure an acceleration then. If falling into a black hole, for instance, his body will elongate as the gravitational field increases, even if the observer thought his body was local and was in his inertial frame. He simply was wrong.

So, no, there is no preferred frame. The local free falling inertial frame is just a good initial approximation where he can ignore the gravitational filed, because near that the accelerations are approximately the same. But it is only an approximation, and when he goes a distance away it fails being inertial.

Einstein's General Relativity captured that Relativity principle perfectly. It's held up to all the conceptual and real physical experiments, like your thought experiment.

No, the observer that is freely falling can determine that there is curvature by measuring second derivatives of the metric tensor, which would include calculations of the first derivatives of the connection, and obtain a Riemann tensor. If there is no curvature they will all be zero. If there is at least some of them will not be zero. And he can compute various scalars from them, where at least one will not be zero.

The error in the thinking is that it ignores that that 'inertial' frame where the observer is freely falling is only local. The observer will measure deviations from flatness as he goes away from his local area, or equivalently, the connection may be zero at his point, but its derivatives, and with them used to construct the curvature tensor and scalars, which will have some nonzero component. If the observer has a good enough accelerometer, some inches or feet or more from his reference point for his local inertial frame, he will measure an acceleration then. If falling into a black hole, for instance, his body will elongate as the gravitational field increases, even if the observer thought his body was local and was in his inertial frame. He simply was wrong.

Both thenWeak and the Einstein Equivalence Principle state that those inertial frames are only in 'small enough regions of spacetime'. I forget what the Strong Equivalence Principle adds to the Weak, but I think it includes the 'small enough regions' statement or idea. Those inertial frames cannot be extended and stay inertial.

So, no, there is no preferred frame. The local free falling inertial frame is just a good initial approximation where he can ignore the gravitational filed, because near that the accelerations are approximately the same. But it is only an approximation, and when he goes a distance away it fails being inertial.

Einstein's General Relativity captured that Relativity principle perfectly. It's held up to all the conceptual and real physical experiments, like your thought experiment.

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No, the observer that is freely falling can determine that there is curvature by measuring second derivatives of the metric tensor, which would include calculations of the first derivatives of the connection, and obtain a Riemann tensor. If there is no curvature they will all be zero. If there is at least some of them will not be zero. And he can compute various scalars from them, where at least one will not be zero.

The error in the thinking is that it ignores that that 'inertial' frame where the observer is freely falling is only local. The observer will measure deviations from flatness as he goes away from his local area, or equivalently, the connection may be zero at his point, but its derivatives, and with them used to construct the curvature tensor and scalars, which will have some nonzero component. If the observer has a good enough accelerometer, some inches or feet or more from his reference point for his local inertial frame, he will measure an acceleration then. If falling into a black hole, for instance, his body will elongate as the gravitational field increases, even if the observer thought his body was local and was in his inertial frame. He simply was wrong.

So, no, there is no preferred frame. The local free falling inertial frame is just a good initial approximation where he can ignore the gravitational filed, because near that the accelerations are approximately the same. But it is only an approximation, and when he goes a distance away it fails being inertial.

Einstein's General Relativity captured that Relativity principle perfectly. It's held up to all the conceptual and real physical experiments, like your thought experiment.