Timeline for What equation (/solution) predicts the existence of black holes?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2015 at 15:22 | comment | added | Timaeus | @JohnRennie I think you can write a good post, and have comments clearly labelled "for experts" that aren't really for experts in GR but are for people more experienced in physics in general. Because the idea that it is volume density develops early and some people never get rid of it. So if people pick it up here before they formally study GR then it could be permanently crippling to them if they never gets remediated properly. Because they can think they know it right and not listen to people that know better. When a respected person like you can be read as agreeing with them it's a barrier | |
| Sep 11, 2015 at 15:16 | comment | added | John Rennie | @Timaeus: I take your point, and can say only that it's a judgement call how far to simplify an answer for non-GR heads. | |
| Sep 11, 2015 at 15:13 | comment | added | Timaeus | @JohnRennie I think the focus on density can be misleading because people can think it is energy (or mass) per volume that matters instead of energy per area of enclosing surface. You can have a surface density of energy and not form a black hole even though the volume density is infinite. An explanation can be as simple as possible, but don't make it simpler than that. | |
| Sep 11, 2015 at 11:48 | vote | accept | Quantum spaghettification | ||
| Sep 11, 2015 at 11:45 | comment | added | John Rennie | @Joseph: all four metrics predict that a horizon can form for sufficiently high density. The RN, Kerr and KN metrics predict that the horizon will disappear for sufficiently high values of $Q$ and $J$, but these are thought to be unphysical and in practice a horizon will always exist if and only if the density is high enough. So it's really only the density of the spherically symmetric body that matters. Low density (relatively low density that is) objects like the Earth, Sun or even neutron stars won't form a horizon but denser objects will. | |
| Sep 11, 2015 at 10:59 | comment | added | Quantum spaghettification | Can the table on the wiki page for Kerr be interpreted as saying that: The Schwarzschild solution predicts black holes for $J=0$ and $Q=0$, Kerr for $J \ne 0$ and $Q=0$, Reissner-Nordtrom for $J =0$ and $Q \ne 0$ and Kerr-Newman for $J \ne 0$ and $Q \ne 0$. Or is predicts to strong a word? | |
| Sep 11, 2015 at 9:47 | history | answered | John Rennie | CC BY-SA 3.0 |