Timeline for Non-inertial frames of reference in empty space
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 14 at 12:56 | comment | added | Ben McKay | I think it was Hegel who said "Nonbeing cannot be", but it might have just been nobody. | |
| Jul 14 at 12:43 | comment | added | Марат Рамазанов | Maybe empty space has a non-zero curvature by default. It makes sence, because there must be something in space, at least gravity. Otherwise it'd be space with nothing in it, but space is expected to contain something | |
| Jul 14 at 12:36 | comment | added | Ben McKay | @МаратРамазанов: I am not a physicist, so I am not the best person to ask about what has a physical meaning. There are physicists who take an interest in locally de Sitter and also in locally anti de Sitter spacetimes, and even in de Sitter and anti de Sitter globally. If you have a frame of reference moving with zero acceleration in all directions, i.e. forming a parallel framing, then clearly the curvature vanishes, so the metric is flat, i.e. locally isometric to Minkowski space. | |
| Jul 14 at 11:59 | comment | added | Марат Рамазанов | Yes, there are non-Euclidean metrics that allow straight geodesic world lines. For example Klein model. But does this have a phisical meaning? Can a frame of reference moving with zero acceleration have a non-Euclidean metrics, have a non-zero curvature tensor? | |
| Jul 13 at 9:14 | vote | accept | Марат Рамазанов | ||
| Jul 13 at 9:07 | comment | added | Ben McKay | It is stronger than vanishing of the stress energy tensor. In a spacetime with no other energy sources, just gravity, the stress energy tensor is the traceless Ricci. But in fact the traceless Weyl projective curvature also vanishes, so the only curvature left is the scalar curvature, which is forced to be constant. The spacetime has constant curvature, and is a space form. | |
| Jul 13 at 8:50 | comment | added | Марат Рамазанов | But can I derive Einstein's equations for zero stress–energy tensor only using this condition? Or they will be more general? | |
| Jul 13 at 8:40 | history | answered | Ben McKay | CC BY-SA 4.0 |