Re: The momentum - a cotangent vector?

On 2024-08-08 07:02:29 +0000, Stefan Ram said:

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.

For physicists' purposes this definition looks more asymmetric that
necessary. It is simple to postulate that there are vectros and
covectors and there is a multiplication of a covector and a vector,
and there is a relation that connects directions of vectors to
directions of curves at a point, and likewise drections of covectors
and directions of gradients of scalar fields. Of course the exact
details depend on what kind of space one wants to have and that
may depend on why one wants to have it.

  When one defines the "canonical" (or "generalized") momentum as
  the derivative of a Lagrange function, it points toward being a
  covector.

Derivative of scalar with respect to position is a covector.
Canonical momentum is the derivative of Lagrangian with
respect to one of its arguments and that argument is not a position
but a velocity, which is a vector.

--
Mikko