From: hees@itp.uni-frankfurt.de   
      
   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/   
      
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