Timeline for How do you calculate the black hole diameter of M87*?
Current License: CC BY-SA 4.0
14 events
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| Nov 19, 2020 at 22:51 | comment | added | ProfRob | @MarkBeadles your physical diameter is too big by a factor of $10^6$. | |
| Apr 13, 2019 at 15:04 | vote | accept | not2qubit | ||
| Apr 12, 2019 at 2:50 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 1 character in body; edited title
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| Apr 12, 2019 at 1:10 | history | edited | Qmechanic♦ |
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| Apr 12, 2019 at 0:16 | answer | added | ProfRob | timeline score: 6 | |
| Apr 11, 2019 at 23:17 | comment | added | not2qubit | Never mind. I think I found the answer in the 5th paper. | |
| Apr 11, 2019 at 23:07 | comment | added | not2qubit | I see. But what exactly is the definition of the emission ring in this case, and how far is it from the (ideal Schwarzchild) event ($r=2M$) or photon ($r=3M$) horizon? | |
| Apr 11, 2019 at 23:02 | vote | accept | not2qubit | ||
| Apr 13, 2019 at 15:04 | |||||
| Apr 11, 2019 at 23:02 | history | edited | not2qubit | CC BY-SA 4.0 |
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| Apr 11, 2019 at 15:03 | comment | added | Mark Beadles | Note that it's the emission ring that has the angular diameter of 42±3 μas, not the central black hole. If you calculate the actual diameter of the emission ring from 2*R*tan(α/2) with R=16.8 Mpc and α=42 μas, you get ~1e20 m, much larger than the Swarzchild radius of the central black hole | |
| Apr 11, 2019 at 15:00 | history | edited | Kyle Kanos | CC BY-SA 4.0 |
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| Apr 11, 2019 at 14:47 | answer | added | alfred | timeline score: 3 | |
| Apr 11, 2019 at 14:30 | review | First posts | |||
| Apr 11, 2019 at 14:32 | |||||
| Apr 11, 2019 at 14:27 | history | asked | not2qubit | CC BY-SA 4.0 |