Skip to main content
11 events
when toggle format what by license comment
Aug 16, 2022 at 7:27 vote accept Ryder Rude
Aug 15, 2022 at 21:46 answer added peek-a-boo timeline score: 8
Aug 15, 2022 at 18:37 comment added Kurt G. The point is probably that on an $\color{red}{n}$ dimensional manifold $dx^{\mu_1}\wedge...\wedge dx^{\mu_\color{red}{n}}$ (what you call $n$-form field) is the volume form. The Graviation book by Misner, Thorne & Wheeler spends a lot of time explaining this very pedagogically.
Aug 15, 2022 at 12:17 comment added Ryder Rude @KurtG. I don't understand how its interpretation as a volume integral and as a n-form field can co-exise. If it's an n-form field, then what does it mean to apply $\int$ on it?
Aug 15, 2022 at 9:52 comment added Kurt G. Carroll writes on the previous page : that "the naive volume element $d^nx$ transforms as a density, not as a tensor...". Furthermore, his $\epsilon$ on the LHS of $\equiv$ in (2.96) is not the Levi-Civita tensor. The integral is a volume integral of a scalar function $\phi$. That's it.
Aug 15, 2022 at 9:48 comment added Ryder Rude @KurtG. $d^n x$ is still a tensor
Aug 15, 2022 at 9:46 comment added Ryder Rude @KurtG. Yes, but that's still the same tensor field, even when expressed in terms of $d^n x$. Integrals are supposed to be numbers.
Aug 15, 2022 at 9:42 comment added Kurt G. ... and Carroll showed $$\tag{2.96} \epsilon\equiv\epsilon_{\mu_1...\mu_n}dx^{\mu_1}\otimes...\otimes dx^{\mu_n} =...=\sqrt{|g|}d^nx. $$
Aug 15, 2022 at 9:30 comment added Kurt G. Carroll writes explicitly that (2.98) is an abstract notation for $$\tag{2.97} \boxed{I=\int \phi(x)\sqrt{|g|}d^nx}. $$ Imho: we can do it because nothing stops us from doing it.
Aug 15, 2022 at 8:08 history edited Qmechanic CC BY-SA 4.0
added 10 characters in body; edited tags
Aug 15, 2022 at 7:17 history asked Ryder Rude CC BY-SA 4.0