Timeline for Falling angle of a photon near the event horizon
Current License: CC BY-SA 4.0
21 events
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| Nov 9, 2023 at 14:09 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 9, 2023 at 9:33 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 9, 2023 at 8:26 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 9, 2023 at 7:24 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 9, 2023 at 5:09 | comment | added | Yukterez | @Daniel P - the worldline will look the same after you transform to Gullstrand Painleve or Eddington Finkelstein coordinates since they only differ in the coordinate time t, while the parameter over which the path is integrated is τ, only on an animation with the coordinate time t as the animation parameter the coordinate speed would be different, but on an image where only r,θ,φ or x,y,z are shown it doesn't make a difference. | |
| Nov 8, 2023 at 10:57 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 8, 2023 at 10:50 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 8, 2023 at 7:27 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 7, 2023 at 18:10 | comment | added | Daniel P | I will take some time to fully digest it. As a first time use of this site, I am not fully sure I posted this answer correctly. Finally for clarification, I did use the Schwarzschild coordinates. I will now practice coordinate changes to fully understand this situation. Thanks again. Daniel | |
| Nov 7, 2023 at 18:07 | comment | added | Daniel P | Thank you very much for this very clear and detailed answer. | |
| Nov 7, 2023 at 17:42 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 7, 2023 at 17:08 | comment | added | ProfRob | @A.V.S. see physics.stackexchange.com/a/787543/43351 | |
| Nov 7, 2023 at 11:37 | comment | added | ProfRob | @A.V.S. I am busy for the next couple of hours but I am going to work it out in both coordinate systems. The OP's plot is correct and I confirmed using the GRorbits software. Looks like 69 degrees to me. It isn't exactly radial until $r=r_s$ but there can't be a shell observer at $r=r_s$. | |
| Nov 7, 2023 at 11:31 | comment | added | A.V.S. | Yes, and I agree with the last sentence of your answer in its current version. But the sentence before that raises doubts in me, since I would think that “being emitted radially” is a coordinate independent property. | |
| Nov 7, 2023 at 11:10 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 7, 2023 at 11:08 | comment | added | ProfRob | @A.V.S. but the plot is in Schwarzschild coordinates (I think), so isn't my angle correct in terms of a trajectory in a Schwarzschild $r, \phi$ diagram? | |
| Nov 7, 2023 at 10:50 | comment | added | A.V.S. | I think that there must be a $(g_{rr}) ^{-1/2}$ multiplier in $\tan^{-1} $ argument in your equation for $\theta$ because the proper radial distance is $(g_{rr}) ^{1/2}dr$. | |
| Nov 7, 2023 at 9:53 | history | undeleted | ProfRob | ||
| Nov 7, 2023 at 9:52 | history | edited | ProfRob | CC BY-SA 4.0 |
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| Nov 7, 2023 at 8:53 | history | deleted | ProfRob | via Vote | |
| Nov 7, 2023 at 8:20 | history | answered | ProfRob | CC BY-SA 4.0 |