• #### The momentum - a cotangent vector?

From Stefan Ram@21:1/5 to All on Wed Aug 7 06:54:34 2024
In mathematical classical mechanics, the momentum is a cotangent
vector, while the velocity is a tangent vector. I don't get this!

A cotangent vector maps a tangent vector to a scalar (real number).
That much I know. But since when is the momentum (in physics)
a function that maps a velocity to a real number, and what is
the physical interpretation (meaning) of that real number?

Using the Lagrange function L, the momentum is p = dL/dq', where

d is the sign for the partial derivative (this newsgroup does
not do Unicode) and
q' is a "q" with a dot above.

How can I see that (given that q' is a tangential vector)
p is a cotangential vector?

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• From Mikko@21:1/5 to Stefan Ram on Wed Aug 7 11:37:02 2024
On 2024-08-07 06:54:34 +0000, Stefan Ram said:

In mathematical classical mechanics, the momentum is a cotangent
vector, while the velocity is a tangent vector. I don't get this!

In the usual formalism a vector is simply a vector. What do you mean
with "tangent" and "cotangent"? Usually they are trigonometric
functions, where cotangent of x is the same as thangent of the
complement of x and also the inverse of the tangent of x. But
those definitions don't apply to vectors.

--
Mikko

[[Mod. note -- I think Stefan is using "tangent vector" and
"cotangent vector" in the sense of differential geometry and tensor
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.

See
https://en.wikipedia.org/wiki/Tensor_calculus
-- jt]]

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• From Stefan Ram@21:1/5 to moderator jt on Thu Aug 8 07:02:29 2024
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.)

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• From Hendrik van Hees@21:1/5 to Stefan Ram on Thu Aug 8 07:49:24 2024
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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• From Mikko@21:1/5 to Stefan Ram on Thu Aug 8 11:00:40 2024
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

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• From Mikko@21:1/5 to the moderator on Thu Aug 8 21:15:43 2024
On 2024-08-07 11:37:02 +0000, the moderator said:

I think Stefan is using "tangent vector" and "cotangent vector"
in the sense of differential geometry and tensor 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.

Thank you. That makes sense.

--=20
Mikko

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• From Hendrik van Hees@21:1/5 to Mikko on Fri Aug 9 13:53:39 2024
It's only that vectors or covectors and in general any tensor of any
rank do not transform at all under coordinate transformations (i.e., diffeomorphisms). What transforms are the basis vectors and the
corresponding dual base and correspondingly the components of the
tensors wrt. these bases. I tried to summarize this briefly for vectors
and covectors in my posting.

Note that in physics you usually have more structure in your manifolds.
E.g., in GR you assume a pseudo-Riemannian manifold, i.e., a manifold
with a fundamental form (of Lorentz signature (1,3) or (3,1) depending
on your sign convention) with the torsion-free compatible affine
connection. Then you can also canonically (i.e., independent of the use
of bases and cobases) identify vectors and covectors, as is usually done
by physicists.

On 09/08/2024 06:15, Mikko wrote:
On 2024-08-07 11:37:02 +0000, the moderator said:

I think Stefan is using "tangent vector" and "cotangent vector"
in the sense of differential geometry and tensor 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.

Thank you. That makes sense.

--=20
Mikko

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/

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• From Stefan Ram@21:1/5 to Stefan Ram on Fri Aug 9 13:54:17 2024
ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
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.

These notions can be somewhat personified by the persons
of /Albert Einstein/ and /Hermann Minkowski/.

Einstein's focus was on the algebraic properties of his theory of
relativity, specifically the equations that express its laws and
their behavior under transformations, known as /covariance/. The
fulfillment of the principle of relativity is demonstrated through
often tedious algebraic manipulations. The equations of the
theory are transformed using Lorentz transformations, showing
that the resulting equations maintain their form.

In contrast, Minkowski emphasized the geometric properties of the
theory, focusing on the geometric units that remain unchanged behind
the transformations, which is referred to as /invariance/. Minkowski
ensures the fulfillment of the principle of relativity through
entirely different means. The only structures allowed in constructing
a theory are the invariants of spacetime. This restriction guarantees
compatibility with the principle of relativity and allows for the
verification of its fulfillment through inspection.

BTW: The word "tensor" for Ricci and Levi-Civita's "contravariant
and covariant systems" was introduced by Einstein and Grossmann.

I pulled this info (about Einstein, Minkowski, and Grossmann) from
"General covariance and the foundations of general relativity:
eight decades of dispute" (1993-03) by John D. Norton.

I also read somewhere - can't remember where right now - that
Einstein didn't actually name his theory "theory of relativity".
That title, like "Big Bang" or "black hole", was thrown out there
by a journalist or a critic. (Maybe Einstein picked it up later on.)
So, it's kind of ironic that some folks today blame Einstein for
giving his "theory of relativity" a name that doesn't really fit.

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• From Stefan Ram@21:1/5 to Stefan Ram on Sat Aug 10 06:16:00 2024
ram@zedat.fu-berlin.de (Stefan Ram) schrieb oder zitierte:
How can I see that (given that q' is a tangential vector)
p is a cotangential vector?

Here's a little calculation I whipped up in the realm of good
old classical mechanics, no relativity involved.

I'm starting with the well-known formula

E = 1/2 m v^2

Using Cartan's calculus, from this, I come up with:

dE = m v dv + 1/2 v^2 dm.

And since dm = 0 (I assume the mass is constant):

dE = p dv.

Now let's write out the implied scalar product as "*":

dE = p * dv.

This "p *" is now a covector acting like a linear function, mapping
changes in velocity (a vector) to changes in energy (a scalar).

BTW, we also can derive the "other" relationship dE = v dp!

Writing "1/2 m v^2" as "1/2 m v v", we can see that

E = 1/2 p v

, so,

dE = 1/2 p dv + 1/2 v dp

. But since we already had established that dE is "p dv" for a
constant mass m, "1/2 p dv" must be "1/2 dE", so that,

dE = 1/2 dE + 1/2 v dp.

Subtracting "1/2 dE" on both sides gives:

1/2 dE = 1/2 v dp,

and multiplication by 2,

dE = v dp.

So, dE is both "v dp" and "p dv" when the mass m is constant!

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