Vectors, one-forms and gradients

[mathwonk]

but i could only read the first page. maybe it got better later.

 

[garrett]

Well, yah, "contravariant" and "covariant" tensors are an anachronism. The goemetric objects they're meant to describe are better understood today as vectors and differential forms.

 

[mathwonk]

does he say anywhere in there why certain things are covariant or contravariant, in physical terms? or in any terms?

 

[George Jones]

does he say anywhere in there why certain things are covariant or contravariant, in physical terms? or in any terms?

Velocity is a vector; momentum is a covector.

 

[mathwonk]

now were talking! why is momentum a covector? it is ovious that velocity is a vector. but of course, momentum is a number assigned to a velocity and a mass, hence dual to velocity.

thank you! more...more,...

 

[mathwonk]

but momentum is something like mv^2 right? so momentum seems quadratic i.e. bilinear not linear in velocity, and hence a 2nd order [co]tensor?>??

 

[George Jones]

but momentum is something like mv^2 right? so momentum seems quadratic i.e. bilinear not linear in velocity, and hence a 2nd order [co]tensor?>??

No, this is (proportional to) kinetic energy. In elementary treatments of mechanics, momentum is mass times velocity. In more advanced treatments of mechanics, i.e., in Lagrangian and Hamiltonian mechanics, velocities lie in the tangent bundle of a differentiable manifold (not necessarily R^n), while momenta lie in the cotangent bundle. Kinetic energy is used as the metric for tangent vectors, and so is used also to identify vectors and covectors.

All physics students are taught Lagrangian and Hamiltonian mechanics, but few are taught this using the language of differential geometry. As a student, I was never taught it this way, and I have never taught mechanics this way, so I am a bit shaky on the details, but, in any case, l will try to give some of these details. At times, these details rely heavily on coodinates (charts), but there some ideas behind the "symbol pushing".

 

[mathwonk]

since i presume mass is scalar, if momentum were mass times velocity it would just be proportional to velocity, hence also a vector.

is momentum rather a number assigned to a velocity? if we have a moving point, its velocities give a family of velocity bectors, i.e. a curve in the tangent bundle.

then a momentum does what? asigns a curve in the cotangent bundle? or is the momentum a covector field which then asigns a family of numbers to the curve of velocity vectors?

i.e. what does it take to define m omentum, and what type of quantity is momentume, and to what is it assigned?

a moving object has at any instant a momentum i presume. but does that momentum have a direction? or simply a magntude?

when you say momentum lives in tyhe cotangent bundle, do you mean the momentum ofa given moving object does so, or that the f=unction which assign momentum to moving objects does so?

 

[nrqed]

since i presume mass is scalar, if momentum were mass times velocity it would just be proportional to velocity, hence also a vector.

is momentum rather a number assigned to a velocity? if we have a moving point, its velocities give a family of velocity bectors, i.e. a curve in the tangent bundle.

then a momentum does what? asigns a curve in the cotangent bundle? or is the momentum a covector field which then asigns a family of numbers to the curve of velocity vectors?

i.e. what does it take to define m omentum, and what type of quantity is momentume, and to what is it assigned?

a moving object has at any instant a momentum i presume. but does that momentum have a direction? or simply a magntude?

when you say momentum lives in tyhe cotangent bundle, do you mean the momentum ofa given moving object does so, or that the function which assign momentum to moving objects does so?

I am sure that I won't use the right terminology and that all of that is way too basic for you guys but I will just make a few comments about the way I understand it.

One can think of the generalized coordinates q^i as forming a manifold. Then the generalized velocities {\dot q^i } are vectors. So the pair q^i, {\dot q^i} form a tangent bundle which is what we, physicists, call the configuration space.

Now, a mapping from the vectors to the one-forms is not present in general for an arbitrary manifold, such a mapping requires some extra structure. The way I understand it, in mechanics this extra structure is provided by the lagrangian. Basically, the Lagrangian introduces a mapping from two vectors to a scalar (the lagrangian itself is the name we give to the resulting scalar) so it introduces a metric and hence, a mapping from vectors to covectors.

For potentials which are velocity independent, the Lagrangian, which is a scalar, takes the form
L = T( {\dot q^i } , {\dot q^j}) - V(q)
where the potential is a scalar function of the coordinates only (for the simple case of velocity independent potentials) and T, the kinetic energy, is something which assigns a number to a pair of vectors, so we may write
T \equiv {1 \over 2} g_{ij} {\dot q^i} {\dot q^j}

This is where a metric enters. (the use of the conventional factor of 1/2 will be clear below)

Now, we can use this metric to go from the tangent bundle associated to (q, {\dot q}) to the cotangent bundle (p, q) where p_i is the covector associated to the vector {\dot q^i} (I am sure I am not using the correct terminology here , sorry) . The covectors are what we, physicists, call the generalized momenta.

And the cotangent bundle is what we, physicists, call the phase space.

From the above definition of L, it is clear how to get p_i from {\dot q^j}. We simply have
p_i = g_{ij} { \dot q^j}

But, obviously from the definition of L, this can be written as
p_i = {\partial L \over \partial {\dot q^i} }
which is the way we, physicists, first learn how to get the generalized momenta from the lagrangian. The reason for the 1/2 in T is obviously so that there is no factor of 2 in the derivative in the previous formula.

In the simplest case of a particle moving in 3D and working in cartesian coordinates, the metric g_{i,j} is simply the mass times the identity matrix.


So, the momentum is a covector (one-form) and "feeding" a velocity vector at a point spits out a number which is essentially twice the kinetic energy.

Things get really interesting when there is, say, an electromagnetic interaction, though,



I hope I haven't said anything *too* stupid.

Personally, I find this all nice except that it raises much more questions than it answers. The problem is that I am pretty sure all the questions that I want to ask are probably quite dumb so I won't ask them. I will go back to the more elementary boards now. I just hope this was not too trivial.

Regards

 

[mathwonk]

well if you have to use a metric to view momentum as a covector, that suggests it is really a vector. as it seems it should be if it is proportional to the velocity vector.

?

 

[nrqed]

well if you have to use a metric to view momentum as a covector, that suggests it is really a vector. as it seems it should be if it is proportional to the velocity vector.

?

I don't understand. If there is a metric, there is a natural correspondence between vectors and covectors, no? And if we use the metric to obtain a covector from a vector, the covector is a bona fide covector, no?

And since the metric is not in general a constant, the generalized momenta will not be proportional to the generalized velocities in general (the generalized momenta do not even have the for of m {\vec v} in general, where here v is the velocity vector in the usual high school sense.

By the way, the genralized velocities and momenta are not necessarily the velocities and momenta we learned in elementary physics. They don't even have the correct dimensions, in general.

You see how confusing things are for a physicist trying to learn the stuff?
It's more difficult than learning things from scratch especially given that books rarely show explicitly the correspondence between vectors, forms, metric and everything we have learned before. That may sound like "symbol pushing" ot you but some of that is required in order to really understand things, imho. (adding to the the whole business of integrating over forms does not help)

Going back to an example. For a particle moving in two dimensions in a central potential (let's say),

L = {m \over 2} ({\dot r}^2 + r^2 {\dot \theta}^2) - V(r)
The generalized velocities have components {\dot r} and {\dot \theta}. So the metric is diagonal(m, m r^2).

The generalized momenta (covectors) have components p_r = m {\dot r} and p_\theta =m r^2 {\dot \theta}.

with those definitions, "feeding" the generalized vector field {\dot q}_r \partial_r + {\dot q}_\theta \partial_\theta to the covector p_r dr + p_\theta d\theta gives half the kinetic energy.

 

[mathwonk]

i guess if you can argue that the metric is somehow natural then a vector that uses the metric t be viewed as a covector is somehow naturllly a covector.

but if you have ametric there is essentillyno difference between a vector and a covector. but metrics are not so natural.