tag:blogger.com,1999:blog-4005750306801645234.post8866727297359223032..comments2024-02-02T20:17:24.676+01:00Comments on Spacetime and Geometry: Exercise 3.01 Consequences of metric compatibilityUnknownnoreply@blogger.comBlogger5125tag:blogger.com,1999:blog-4005750306801645234.post-89641614545553195932023-01-03T17:48:49.946+01:002023-01-03T17:48:49.946+01:00Here's [my solution to the exercise](https://p...Here's [my solution to the exercise](https://physics.stackexchange.com/a/743720/269697). It's too long to be posted as a comment.Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-27075491388787280002023-01-03T03:49:05.378+01:002023-01-03T03:49:05.378+01:00The solution to (b) in the link provided by JSBach...The solution to (b) in the link provided by JSBach1801 is simply wrong. It is very important to keep in mind that covariant derivatives are meaningful only when operating on a tensor field. (A tensor field may be a field of scalars, vectors, one-forms, etc.) |g|, or its square root, does not form a scalar field, because its value at each point is coordinate dependent. Idem for the Levi-Civita symbol.Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-9134331375694320402021-06-20T06:05:41.481+02:002021-06-20T06:05:41.481+02:00the fact that the covariant derivative vanishes fo...the fact that the covariant derivative vanishes follows directly from part (a) XD one way of thinking about it is that g is 'constant' wrt covariant derivatives. Other than that I suppose that since the Levi-Civita symbol is not a tensor the covariant derivative of the Levi-Civita symbol is just a partial derivative, so everything vanishes. I wasn't quite sure about this argument so I did things a bit more explicitly. <br /><br />My fully typed-up solution can be found here: https://drive.google.com/file/d/13dBHxlh8CIBFfUIrf18mFPERDQ7IWkHJ/view?usp=sharingAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-39162293576474236922020-03-17T11:07:47.951+01:002020-03-17T11:07:47.951+01:00The proof for part a is great. But I don't fol...The proof for part a is great. But I don't follow the proof for part b. It is true that the determinant will only depend on the metric entries, but why would that mean that the covariant derivative will vanish?Georgehttps://www.blogger.com/profile/04824865122846470839noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-37586509239481108172020-03-16T06:00:59.031+01:002020-03-16T06:00:59.031+01:00Check out this link for some really clever solutio...Check out this link for some really clever solutions to the porblem from Carroll's book. The Part B is so easier than you realize.<br /><br />https://indexguy.wordpress.com/2007/07/13/some-identities-regarding-metric-compatibility/JSBach1801https://www.blogger.com/profile/04780337176436834748noreply@blogger.com