Timeline for Gauss' theorem for null boundaries
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Aug 7, 2016 at 22:17 | vote | accept | Bence Racskó | ||
| Aug 5, 2016 at 19:42 | answer | added | Raziel | timeline score: 4 | |
| Aug 5, 2016 at 18:12 | answer | added | Bence Racskó | timeline score: 0 | |
| Aug 4, 2016 at 23:10 | comment | added | Bence Racskó | @Futurologist You are right, however I do not think it is because $N$ is not orthogonal, but because $N$ might not be normed to unity. In fact, it should be possible to take $N$ to be also a null vector, just not the one that is tangent/normal to $\partial M$, which would further complicate things. I am working on this right now, but I am very tired so probably will postpone till tomorrow. | |
| Aug 4, 2016 at 22:52 | comment | added | Futurologist | I am not convinced about the vector field decomposition $X = \langle X, N \rangle N + Y$ because, if I understand correctly, $N$ is not Lorentz orthogonal to your frame of vectors $e_1,..,e_{n-1}$. I would say $X = \lambda N + Y$ where $Y$ is light-like. Then $\langle X, Y \rangle = \lambda \langle N, Y \rangle + \langle Y, Y \rangle = \lambda \langle N, Y \rangle $ and solve for $\lambda$. | |
| Aug 4, 2016 at 17:53 | history | edited | Bence Racskó | CC BY-SA 3.0 |
added 228 characters in body
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| Aug 4, 2016 at 17:18 | history | asked | Bence Racskó | CC BY-SA 3.0 |