Timeline for answer to On the definition of rest frame in special & general relativity by Markus Klyver
Current License: CC BY-SA 4.0
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| Mar 22 at 21:29 | history | edited | Markus Klyver | CC BY-SA 4.0 |
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| Mar 19 at 22:12 | history | edited | Markus Klyver | CC BY-SA 4.0 |
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| Mar 19 at 22:04 | history | edited | Markus Klyver | CC BY-SA 4.0 |
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| Mar 19 at 22:03 | comment | added | Markus Klyver | Frames are per definition sections of the frame bundle, a basis for $T_pM$ at one $p\in M$ is not a frame. I also think you confuse and intertwine the differential structure and the projective structure of the manifold $M$, both are needed for any identification to happen. Only when the concepts of both velocity and geodesics exist, and the exponential map is built from them. Special relativity really do not have way to talk about acceleration, so the geodesics are kind of ad hoc defined. I expand on this in my edit above. | |
| Mar 19 at 0:50 | comment | added | Markus Klyver | Charts can be inertial in special relativity. | |
| Mar 19 at 0:48 | comment | added | WillO | Third, identifying points in the tangent space with points on the manifold has absolutely nothing to do with frames, which are bases for the tangent space and have nothing to do with the rest of the manifold. If you had a disembodied vector space with no manifold in sight, you could still talk about frames, and if you had an inner product you could still talk about orthonormal (i.e. inertial) frames. Anything that mentions the rest of the manifold is irrelevant. (It becomes relevant if you want to talk about sections of the frame bundle, but that's a different matter.) | |
| Mar 19 at 0:44 | comment | added | WillO | "The reason for why you can pretend that your chart is inertial in special relativity is that, in special relativity, you can identify the tangent space TpM with points on the manifold. This is absolutely not true in general relativity..." First, this is just wrong. You can use the exponential map to identify points in the tangent space with points on the manifold in general relativity just as well as you can in special relativity. Second, even if it were right, I think you misspoke when you wrote "pretend your chart is inertial; it's frames, not charts that are inertial. | |
| Mar 18 at 19:12 | history | answered | Markus Klyver | CC BY-SA 4.0 |