tag:blogger.com,1999:blog-4005750306801645234.post5933643939843822003..comments2024-04-20T13:45:15.715+02:00Comments on Spacetime and Geometry: Exercise 3.02 Spherical gradient divergence curl as covariant derivativesUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4005750306801645234.post-16539807326814668552023-01-04T07:01:20.480+01:002023-01-04T07:01:20.480+01:00A comment or two on the curl. (a) when we move to ...A comment or two on the curl. (a) when we move to a general coordinate system, we probably want to replace the Levi-Civita symbol by the Levi-Civita tensor. (b) curl is in fact a two-form disguised as a vector. A more compact way to write it is simply $$dV_\mu$$.Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-80546207107751686852021-03-07T11:21:48.861+01:002021-03-07T11:21:48.861+01:00Hello, I wonder if you are still working on the ex...Hello, I wonder if you are still working on the exercises of Carroll's book. Continue the good work, I think it's very helpful.<br /><br />First of all, all the Christoffell symbols are correct, but a few things,<br /><br />1. When working with Christoffell symbols, it's best not to contract, lower or raise indices because they are not tensors. So when you calculate the covariant form of the divergence, you should do g^{ij}\nabla_i V_j first, instead of carrying the contraction of i into the calculation. And then do the covariant derivative of V_j (there'll be a minus sign in front of the \Gamma 's).<br /><br />2. The covariant form of curl should be \epsilon^{ijk}\nabla_j V_k \partial_i and the whole thing divided by the square root of the determinant of the metric. The way you wrote in the pdf will give you a number, not a vector. And the square root of det(g) is because \epsilon is not a tensor but a tensor density.<br /><br />3. As you said at the end of the pdf, you'll need to convert everything to an orthonormal basis in order to get the form in Jackson's book. Basically, the new basis is \hat{\partial_i} = \partial_i/\sqrt{g_{ii}} (no summation) and correspondingly, the new component is \hat{A^i} = A^i * sqrt{g_{ii}} (no summation).<br /><br />And there will be places where you need to change back and forth between V_i and V^i. It's not straightforward in any means. <br /><br />Thank you for putting in the time and effort to try to work out the exercises. I really appreciate it. I'm sure many other people do too.<br /><br />Analyse Fonctionnellehttps://www.blogger.com/profile/09152584146670985770noreply@blogger.com