Timeline for Inverse metric in linearised gravity
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2023 at 7:53 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
|
| Nov 27, 2016 at 0:38 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
Fixed minor typos.
|
| Nov 26, 2016 at 20:08 | vote | accept | user35305 | ||
| Nov 26, 2016 at 19:56 | comment | added | Qmechanic♦ | $\uparrow$ Yes. | |
| Nov 26, 2016 at 19:26 | comment | added | user35305 | Ok cool. Is the reason why both the metric and its inverse have the form $g_{\mu\nu}=\eta_{\mu\nu}+\delta g_{\mu\nu}$, and $g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$, respectively, simply because one is perturbing both $g_{\mu\nu}$ and $g^{\mu\nu}$ around a Minkowski background (obviously it wouldn't make any sense to expand a metric and its inverse around different backgrounds)? In general, would one have $g_{\mu\nu}=\bar{g}_{\mu\nu}+\delta g_{\mu\nu}$ and $g_{\mu\nu}=\bar{g}_{\mu\nu}+\delta g_{\mu\nu}$, respectively, where $\bar{g}_{\mu\nu}$ is some known background metric?! | |
| Nov 26, 2016 at 19:15 | comment | added | Qmechanic♦ | $\uparrow$ Yes. | |
| Nov 26, 2016 at 19:15 | comment | added | user35305 | ... and since $\delta g_{\mu\nu}= h_{\mu\nu}$, we have from eq.(A) that $$\delta g^{\mu\nu}=-\left(\eta^{\mu\alpha}+\delta g^{\mu\alpha}\right)h_{\alpha\beta}\left(\eta^{\beta\nu}+\delta g^{\beta\nu}\right)=-\eta^{\mu\alpha}h_{\alpha\beta}\eta^{\beta\nu}=-h^{\mu\nu}$$ to first order. Hence, $$g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}$$ Would this be correct? | |
| Nov 26, 2016 at 19:14 | comment | added | user35305 | Ah ok. Is eq.(A) found by noting that $\delta(g^{-1}g)=\delta(g^{-1})g+g^{-1}\delta g=0$, and so $\delta(g^{-1})=-g^{-1}(\delta g)g^{-1}$?! So given this, does one simply use that an infinitesimal perturbation of the inverse metric around a Minkowski background should give: $$g^{\mu\nu}=\overline{g}^{\mu\nu}+\delta g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$$ where $\overline{g}^{\mu\nu}=\eta^{\mu\nu}$ is the background metric one is perturbing around... | |
| Nov 26, 2016 at 18:59 | history | edited | JamalS | CC BY-SA 3.0 |
deleted 1 character in body
|
| Nov 26, 2016 at 18:51 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |