Timeline for To which extent is general relativity a gauge theory?
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
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| May 1, 2022 at 20:26 | answer | added | Kregnach | timeline score: 1 | |
| May 1, 2022 at 20:00 | answer | added | ACuriousMind♦ | timeline score: 19 | |
| Dec 10, 2012 at 10:00 | answer | added | Christoph | timeline score: 2 | |
| Dec 10, 2012 at 9:40 | comment | added | Christoph | @Columbia: by using both metric and connection as primary fields, GR can be formulated as a first-order field theory (with second class constraints); this approach is called Palatini gravity | |
| Dec 10, 2012 at 3:46 | comment | added | Columbia | In general the rule of thumb is that you CAN make the analogy with relatively robust, modulo some small details.. Like the fact that the primary variables are first order in the YM gauge fields, whereas in GR the primary variables are 2nd order. | |
| Dec 10, 2012 at 3:45 | comment | added | Columbia | One of the confusing things here is that there are several somewhat related but different formalisms of GR whereby you can sort of make the analogy between YM gauge fields and GR. The exact gauge group is sometimes GL(n,R) sometimes SL(2,C) and sometimes Diff(M). The earliest version I know was written by Utiyama, and later by Sciama-Kibble and others. Meanwhile particle physicists work in the linearized approximation, where the equivalence is manifest | |
| Dec 9, 2012 at 17:23 | vote | accept | FraSchelle | ||
| Dec 9, 2012 at 17:23 | vote | accept | FraSchelle | ||
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| Dec 9, 2012 at 17:23 | vote | accept | FraSchelle | ||
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| Dec 9, 2012 at 17:12 | answer | added | user1504 | timeline score: 9 | |
| Dec 9, 2012 at 15:43 | comment | added | jdm | @namewhere: No, what I meant was more along the lines of what Murod Abdukhakimov posted, namely that the metric connection shows up in the covariant derivative in the same way as the gauge field would. | |
| Dec 9, 2012 at 15:36 | answer | added | Murod Abdukhakimov | timeline score: 15 | |
| Dec 9, 2012 at 15:30 | comment | added | twistor59 | Related discussion here physics.stackexchange.com/q/12461 | |
| Dec 9, 2012 at 13:53 | comment | added | resgh | @jdm A quick search on google reveals what you speak of as the spin connection. (puny)I can't relate it to this question(yet?) though. | |
| Dec 9, 2012 at 13:12 | vote | accept | FraSchelle | ||
| Dec 9, 2012 at 17:23 | |||||
| Dec 9, 2012 at 12:10 | comment | added | jdm | I faintly remember that there is a nice way to think about GR as a gauge theory (or gauge theories as geometry), and it had to do with viewing the Levi-Civita connection as a gauge field. Unfortunately I don't know enough GR to write down the argument. | |
| Dec 9, 2012 at 10:37 | review | Close votes | |||
| Dec 11, 2012 at 14:33 | |||||
| Dec 9, 2012 at 10:08 | comment | added | FraSchelle | @Namehere Thanks again Namehere, you're perfectly right. Tensors are invariant of course (that's the reason of their creation). Sorry, I'm always thinking in terms of components. I think part of my question (or the question I would I've asked) overlaps the one I found in this post. Thanks again. | |
| S Dec 9, 2012 at 10:08 | history | suggested | resgh | CC BY-SA 3.0 |
corrected minor grammar mistakes, changed title to be more relevant
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| Dec 9, 2012 at 10:06 | review | Suggested edits | |||
| S Dec 9, 2012 at 10:08 | |||||
| Dec 9, 2012 at 9:59 | comment | added | resgh | @Oaoa Tensors are not covariant, they are invariant. It is their components that are covariant. And by the way there are many scalars that can be constructed from tensors, such as norms of vectors, traces of matrices and others. Oh, and was there anything else you think I missed or could improve in my answer(I didn't edit my question to include Chris White's comment since I felt it would be like stealing)? | |
| Dec 9, 2012 at 9:43 | history | edited | FraSchelle | CC BY-SA 3.0 |
added 323 characters in body
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| Dec 9, 2012 at 9:40 | comment | added | FraSchelle | Thanks Chris White, that's partly my problem: I see the parallel transport as a change of frame with one extra condition (the moving frame remains as much as possible parallel to the original one). My second problem was that in QM, the invariant quantity is an extra structure (the wave function modulus, whereas the wave function is displaced / covariant). This you partly answered with Namehere: only scalar are invariant, tensor are covariant. I was asking if it exists an extra structure kept invariant when changing frame of reference. Thanks to both of you. | |
| Dec 9, 2012 at 2:13 | comment | added | user10851 | Parallel transport involves moving around in a manifold, while a change of frame is, well just that. The title of the question is a bit misleading. | |
| S Dec 9, 2012 at 1:48 | history | suggested | resgh | CC BY-SA 3.0 |
Modified Title, cancelled excessive self criticism, changed tags
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| Dec 9, 2012 at 1:31 | review | Suggested edits | |||
| S Dec 9, 2012 at 1:48 | |||||
| Dec 9, 2012 at 1:26 | answer | added | resgh | timeline score: 4 | |
| Dec 8, 2012 at 23:38 | history | asked | FraSchelle | CC BY-SA 3.0 |