Timeline for Is the angular size of the black hole in the movie "interstellar" completely overblown?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 19, 2020 at 0:20 | comment | added | hirschme | @simonp2207 I am still not sure if this answer defends the sizes seen in the movie/image. According to the graph, even for an angular size of 20 degrees (much less than the image) we would still be around 10 AU far away. For an angular size of ~90 degrees we would be maybe even < 1 AU which seems unrealistically close. Also in the answer you give the example fo standing on Pluto, a BH on the sun would be 4 x larger than the moon, which is still a very small angular size. Does this all hint towards the black hole in the image being way too large for a realistic distance? | |
| S Oct 16, 2020 at 9:56 | history | suggested | Nilay Ghosh | CC BY-SA 4.0 |
mathematical expressions written in $...$ format
|
| Oct 16, 2020 at 4:07 | review | Suggested edits | |||
| S Oct 16, 2020 at 9:56 | |||||
| Oct 16, 2020 at 1:54 | comment | added | PM 2Ring | The spin of the BH in Interstellar is inconsistent. According to physics.stackexchange.com/q/148567/123208 Kip Thorne used a spin parameter of $a=1-10^{-14}$ (where 1 is the maximal spin), with the planet very close to the event horizon, to get the desired time dilation. (At such an extreme spin, the photon sphere is almost at the event horizon, and there are stable orbits for massive bodies that are almost as close). However, for visualisation purposes, they used a much less extreme spin of $a=0.6$. | |
| Oct 15, 2020 at 21:07 | comment | added | user24157 | @DavidHammen - the black hole in the movie is supposed to have a spin parameter close to 1, which enables stable orbits much closer to the horizon. | |
| Oct 15, 2020 at 20:55 | comment | added | David Hammen | Re Sub-light speed orbits can still be stable outside the event horizon: Even photons can't orbit just outside the event horizon. The photon sphere for a non-spinning back hole is 1.5 times the Schwarzschild radius. For a particle with mass, the innermost stable circular orbit for a non-spinning back hole is 3 times the Schwarzschild radius. | |
| Oct 15, 2020 at 17:05 | comment | added | simonp2207 | The safe distance would be defined as any distance outside the event horizon, since inside the event horizon not even light can escape and therefore escape is impossible! I've edited the answer to include this. However, this doesn't include the safe distance from the accretion disc which is heated to millions of degrees! I'd have to do a few more calculations to establish the safe distance from that... Though accretion discs around black holes may be transient, appearing and disappearing as matter falls within the black hole's gravitational influence. | |
| Oct 15, 2020 at 17:03 | history | edited | simonp2207 | CC BY-SA 4.0 |
Added extra information in response to the question
|
| Oct 15, 2020 at 16:55 | comment | added | hirschme | Thank you for an excellent answer. I am still missing the "safe distance" aspect to it. This answer implies then that the distance the observers are from the supermassive black-hole is nearer, than the distance from Pluto to a supermassive black-hole in place of the Sun. Is this a realistic distance to orbit? (your comments about a stable planetary system inside makes me think it is not) | |
| Oct 15, 2020 at 16:47 | history | answered | simonp2207 | CC BY-SA 4.0 |