Path: ...!weretis.net!feeder8.news.weretis.net!fu-berlin.de!uni-berlin.de!news.dfncis.de!not-for-mail From: Hendrik van Hees Newsgroups: sci.physics.research Subject: Re: The momentum - a cotangent vector? Date: 8 Aug 2024 07:49:24 GMT Organization: Goethe University Frankfurt (ITP) Lines: 135 Approved: hees@itp.uni-frankfurt.de (sci.physics.research) Message-ID: References: X-Trace: news.dfncis.de +6tSlhXs+NjHQZ4fY50icwFdZlcue6Lxi13TsXh3YLY6E8rzQ0bPmYUHTEYq89Oc3H Cancel-Lock: sha1:uxyNfYKJjF9znWBzS/iDOgpruUM= sha256:GQ1ITIZQA155lDuIeRDxlC0n4lS6rP0no1Sr43KCEPA= Bytes: 6868 The confusion is due to the physicists' sloppy language. They usually call components of a vector or a dual vector vector or dual vector. When they say "a quantity is a vector" they mean the components and call these components "a vector", because they transform as components of a vector do under some class of transformations (general basis transformations, orthogonal, special-orthogonal transformations etc., i.e., it's also important to know from the context which transformations are considered). If you have a plain differentiable manifold, you have a set of points forming a topological (Hausdorff) space and an atlas with maps defining (locally, i.e., around some neighborhood of a point) coordinates x^j (with a upper index by convention). The physical quantities are defined as fields, starting with scalar fields, phi(x)=phi(x^1,x^2,\ldots,x^n). Then you can define curves x^j(t) and define tangent vectors at any point in the neighborhood by taking the "directional derivative, using the Einstein summation convention (over an pair of equal indices, one an upper, one a lower you have to sum) d_t phi[x(t)]=dx^j \partial_j \phi, and tangent vectors are defined as the differential operators V^j=V^j \partial_j Now under coordinate transformations (i.e., arbitrary local diffeomorphisms) a scalar field transforms by definition as phi'(x')=phi(x) It's easy to prove with the chain rule that dx'^j \partial_j'=dx^j \partial_j Now dx'^j=dx^k \partial_k x'^j, and since a vector should be a coordinate-independent object its components should transform as these coordinate differentials, V'^j = V^k \partial_k x'^j The partial derivatives transform like \partial_j' phi'=\partial_j' x^k \partial_k phi, i.e., \partial_j'=\partial_j' x^k \partial_k, i.e. contragrediently to the coordinate differentials. They form components of dual vectors of the tangent vectors, also called cotangent vectors. In the Lagrange formalism you deal with curves x^j(t) and d_t x^j(t)=\dot{x}^j obviously transform like vector components, and the Lagrangian should be a scalar. since the \dot{x}^j are vector components, and thus p_j = \partial L/\partial \dot{x}^j are the components of a co-vector. On 08/08/2024 09:02, Stefan Ram wrote: > moderator jt wrote or quoted: >> 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. > > Yeah, that explanation is on the right track, but I got to add > a couple of things. > > Explaining objects by their transformation behavior is > classic physicist stuff. A mathematician, on the other hand, > defines what an object /is/ first, and then the transformation > behavior follows from that definition. > > You got to give it to the physicists---they often spot weird > structures in the world before mathematicians do. They measure > coordinates and see transformation behaviors, so it makes sense > they use those terms. Mathematicians then come along later, trying > to define mathematical objects that fit those transformation > behaviors. But in some areas of quantum field theory, they still > haven't nailed down a mathematical description. Using mathematical > objects in physics is super elegant, but if mathematicians can't > find those objects, physicists just keep doing their thing anyway! > > A differentiable manifold looks locally like R^n, and a tangent > vector at a point x on the manifold is an equivalence class v of > curves (in R^3, these are all worldlines passing through a point > at the same speed). So, the tangent vector v transforms like > a velocity at a location, not like the location x itself. (When > one rotates the world around the location x, x is not changed, > but tangent vectors at x change their direction.) > > A /cotangent vector/ at x is a linear function that assigns a > real number to a tangent vector v at the same point x. The total > differential of a function f at x is actually a covector that > linearly approximates f at that point by telling us how much the > function value changes with the change represented by vector v. > > When one defines the "canonical" (or "generalized") momentum as > the derivative of a Lagrange function, it points toward being a > covector. But I was confused because I saw a partial derivative > instead of a total differential. But possibly this is just a > coordinate representation of a total differential. So, broadly, > it's plausible that momentum is a covector, but I struggle > with the technical details and physical interpretation. What > physical sense does it make for momentum to take a velocity > and return a number? (Maybe that number is energy or action). > > (In the world of Falk/Ruppel ["Energie und Entropie", Springer, > Berlin] it's just the other way round. There, they write > "dE = v dp". So, here, the speed v is something that maps > changes of momentum dp to changes of the energy dE. This > immediately makes sense because when the speed is higher > a force field is traveled through more quickly, so the same > difference in energy results in a reduced transfer of momentum. > So, transferring the same momentum takes more energy when the > speed is higher. Which, after all, explains while the energy > grows quadratic with the speed and the momentum only linearly.) -- Hendrik van Hees Goethe University (Institute for Theoretical Physics) D-60438 Frankfurt am Main http://itp.uni-frankfurt.de/~hees/