Timeline for How can I show that the inverse of the induced metric $h_{\alpha \beta}$ is $h^{\alpha \beta}$?
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| Dec 5 at 5:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 25 at 16:25 | answer | added | Zo the Relativist | timeline score: 0 | |
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| May 11, 2022 at 9:26 | answer | added | Mozibur Ullah | timeline score: -2 | |
| May 11, 2022 at 9:23 | comment | added | Mozibur Ullah | In fact, if I recall correctly $h$ is simply the pullback of spacetime to the worldsheet. | |
| May 11, 2022 at 9:21 | comment | added | Mozibur Ullah | I don't see the problem, so I might be missing something. Surely once you note that $h$ is a metric then the rest follows? | |
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| Oct 2, 2019 at 18:10 | comment | added | Rahul S. | @Brick Thank you so much! Your explanation makes a lot of sense! I will look into the lapse and shift! | |
| Oct 2, 2019 at 18:06 | answer | added | JamalS | timeline score: 0 | |
| Oct 2, 2019 at 18:04 | comment | added | Brick | You are essentially projecting down to a lower-dimensional subspace. The compatibility should be with the elements of the original metric that "have legs" pointing out of that subspace, e.g. the lapse and the shift. Write $g_{ab} g^{ac}$ in terms of $h_{ab}$ and $h^{bc}$ plus lapse and shift. | |
| Oct 2, 2019 at 18:03 | comment | added | Rahul S. | @Qmechanic then can you make a comment on the compatibility condition that Brick was talking about? | |
| Oct 2, 2019 at 17:57 | comment | added | Qmechanic♦ | Yes, that would be circular logic. | |
| Oct 2, 2019 at 17:52 | comment | added | Rahul S. | @Brick Yeah I considered the fact that it was a definition. But I figured that the compatibility condition would make it so that arbitrary diffeomorphisms of my worldsheet were not allowed. I suspect I am wrong though... | |
| Oct 2, 2019 at 17:48 | comment | added | Rahul S. | @Qmechanic I was thinking about this as well. Intuitively, I would assume that $\sigma_{\alpha} = h^{\alpha}_{\beta} \sigma^{\beta}$ but this logic seems a bit circular. Is there a better way to define it? | |
| Oct 2, 2019 at 17:41 | comment | added | Brick | This a definition. The question that perhaps you should ask instead is what compatibility condition allows this as a free-choice so that everything remains consistent between the two metrics. | |
| Oct 2, 2019 at 17:39 | comment | added | user1379857 | How do you define $\sigma_a$? I think $h^{ab}$ is simply defined to be the inverse of $h_{ab}$, no? | |
| Oct 2, 2019 at 17:39 | comment | added | Qmechanic♦ | What's the definition of the coordinates $\sigma_{\alpha}$ with sub-indices (rather than super-indices)?? | |
| Oct 2, 2019 at 17:35 | history | edited | Qmechanic♦ |
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| Oct 2, 2019 at 17:13 | comment | added | AGML | I think it's only the inverse once you impose ADM coordinates? I'm a bit busy to answer specifically, but this question is answered in most numerical relativity textbooks, which tend to be a bit gentler with this sort of thing than is otherwise typical. See Baumgarte and Shapiro for example. | |
| Oct 2, 2019 at 17:01 | history | asked | Rahul S. | CC BY-SA 4.0 |