Timeline for Applicability of time-dependent metric for expanding universe
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2018 at 22:45 | history | edited | user4552 | CC BY-SA 4.0 |
added 571 characters in body
|
| Jun 19, 2018 at 22:39 | comment | added | user4552 | @user76568: The latter. See H-J Schmidt, arxiv.org/pdf/gr-qc/9404037v1.pdf and Pravda arxiv.org/abs/gr-qc/0209024 | |
| Jun 19, 2018 at 19:21 | comment | added | user76568 | @BenCrowell Do you mean vanish after gravitational wave passes them or that wave solutions have vanishing scalars? Interesting.. | |
| Jun 19, 2018 at 18:51 | comment | added | user4552 | @user76568: The whole of definitions that are coordinate independent is the scalar R and perhaps another scalar, And whether curvature tensorial qunatities vanish or do not vanish. No? Not true. There are infinitely many curvature scalars that can be constructed, not just R. And knowledge of all curvature scalars does not suffice to describe the spacetime, e.g., all curvature scalars vanish for a gravitational wave. What is the difference between non static and time dependent metric/solution? en.wikipedia.org/wiki/Static_spacetime | |
| Jun 19, 2018 at 18:49 | comment | added | user4552 | @HeisenbergImage: Yes. (Good point, though.) | |
| Jun 19, 2018 at 15:30 | comment | added | user76568 | The whole of definitions that are coordinate independent is the scalar R and perhaps another scalar, And whether curvature tensorial qunatities vanish or do not vanish. No? What is the difference between non static and time dependent metric/solution? | |
| Jun 19, 2018 at 14:24 | comment | added | HeisenbergImage | How about Einstein's static solution? Do we disregard it because of instability? | |
| Jun 19, 2018 at 13:48 | history | answered | user4552 | CC BY-SA 4.0 |