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From: Hendrik van Hees <hees@itp.uni-frankfurt.de>
Newsgroups: sci.physics.research
Subject: Re: The momentum - a cotangent vector?
Date: 8 Aug 2024 07:49:24 GMT
Organization: Goethe University Frankfurt (ITP)
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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/