Re: The momentum - a cotangent vector?

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/