Path: ...!weretis.net!feeder8.news.weretis.net!fu-berlin.de!uni-berlin.de!individual.net!not-for-mail From: Hendrik van Hees Newsgroups: sci.physics.research Subject: Re: The momentum - a cotangent vector? Date: Fri, 09 Aug 2024 13:53:39 PDT Organization: Goethe University Frankfurt (ITP) Lines: 37 Approved: Jonathan Thornburg [remove -color to reply]" References: X-Trace: individual.net WiZBloeQI8QodTg+BROnmQbgPZQIVmHEZb4mmP1u0QSrHbjXFexmMU54l9 Cancel-Lock: sha1:/qSnrFGoudOktUxv6n4R6AoHXt8= sha256:reVVFzSLfOLHD0ojY9RkBA+mBILY2Z76SqO4nr/Tkm8= X-Forwarded-Encrypted: i=2; AJvYcCX3CUDXyXvFLsRzLGgRC2M3qP7OE1bQR5TYHeShkawV4zz8s4LKwKFlF5FtY+dNAgOoIQi2O3VauJLt0kzcwG9QBALTDNIP7iw= X-Orig-X-Trace: news.dfncis.de V4/qSpSjMlZUp1/6ALckqQ1LHlPqOUT2fKJQO6Gm8WjUCu X-ZEDAT-Hint: RO Bytes: 2696 It's only that vectors or covectors and in general any tensor of any rank do not transform at all under coordinate transformations (i.e., diffeomorphisms). What transforms are the basis vectors and the corresponding dual base and correspondingly the components of the tensors wrt. these bases. I tried to summarize this briefly for vectors and covectors in my posting. Note that in physics you usually have more structure in your manifolds. E.g., in GR you assume a pseudo-Riemannian manifold, i.e., a manifold with a fundamental form (of Lorentz signature (1,3) or (3,1) depending on your sign convention) with the torsion-free compatible affine connection. Then you can also canonically (i.e., independent of the use of bases and cobases) identify vectors and covectors, as is usually done by physicists. On 09/08/2024 06:15, Mikko wrote: > On 2024-08-07 11:37:02 +0000, the moderator said: > >> I think Stefan is using "tangent vector" and "cotangent vector" >> in the sense of differential geometry and tensor calculus. In >> this usage, these phrases describe how a vector (a.k.a a rank-1 >> tensor) transforms under a change of coordintes: a tangent vector >> (a.k.a a "contravariant vector") is a vector which transforms the >> same way a coordinate position $x^i$ does, while a cotangent vector >> (a.k.a a "covariant vector") is a vector which transforms the same >> way a partial derivative operator $\partial / \partial x^i$ does. > > Thank you. That makes sense. > > --=20 > Mikko -- Hendrik van Hees Goethe University (Institute for Theoretical Physics) D-60438 Frankfurt am Main http://itp.uni-frankfurt.de/~hees/