Vectors, one-forms and gradients

[mathwonk]

by the way, you see me now functioning as a mathematician, i.e. asking questions, making conjectures, guided by simple plausible examples.

this is what i do at my "best", i.e. make naive guesses. this is what i think i am "good" at.

 

[mathwonk]

that is what i am asking the physicists masudr. you are saying that a quantity is defiend by how we chhjose to represent it, but that is unnatural.

i am asking if there is a inherent physical reason for calling certain quantities covariant and others contravariant.

I.e. whetehr the trnsformation laws are forced by the hpysics.

I think the are. now go back aND READ WHAT ELSE I SAID, and think abut a moving particle, i.e. somethign whose position is changing in time. and ask yourself if that is intrinsically co (or contra) variant or not?

maybe t temperature example is easier.

but if this question has no good answer, then it is hopeless to understand the difference between the two concepts in a physical sense.


what i am saying is: if a quantity is measured naturally bya function from numbers into the space, then it must transform in one way, but if it is natuirlly measured by a function from the space into numbers, then it must transfor te oppoisite way.

if you understand this you will understand the distinction betwen vectors and covectors (on that space).

my whole point s it has nothign t do with arbitrary ways of assigning coordinates, it is intrinsic in the concepts themselves.

but you must take out the coordinates to really see why they were put in in a certain way.

 

[mathwonk]

maybe it is not position per se that is a "vector" quantity, but particle motion in space.

what i am trying to understand is why the tangent space to a manifold is a vector quantity as opposed to a covector quantity. a tangent vector represents an infinitesimal change in position, so i said position was
probably a vector quantity.maybe it is really a relative distinction, i.e. between curves in the space and hypersurfaces in the space, not points in the space at all.in alinear space though like R^n, there is an identification betwen ponts and curves, i.e. a "point" has a relative position wrt the origin, so we can choose the curve starting a te origian and pasing through the point at tme t=1.

i.e. points are position vectors in R^n. maybe that's where it started. or maybe it should be a local concept, and i should fix STARTING POINT, AND THEN join it to other nearby points. but thsi needs a metric to define geodesics.

so i don't understand what is gong on, but it is the main thign want to know, not what is considered a vector or covector in physics, but WHY?

 

[mathwonk]

i.e. unless we know why we are doing something, we cannot know what we are doing.

 

[masudr]

i don't think i can currently answer your question, for i do not have the answer

 

[mathwonk]

heres another one. actually maybve position is more copmplicated since it is measured by assigning n functions i.e. coordinates. so position is measured by the values of n functions from the space to real numbers. but there is nothihng antural about these functions, as there is for tempperature. i.e. the elkvelks ets fo temperature are fixed independent of coordinates, while the level sets of position are only defiend relative to arbitrary coordinate planes.

 

[mathwonk]

lets keep playing.

co means with and contra merans agaoinst. but wiuth or against what? presumably it means the coordinates transform with the position coordinates or against them. so position would be covariant by definiti0on. but of course in physics the words mean the opposite of what they should, so this would be contravariant, i.e. a vector not a covector. (also in math covectoirs transform contravariantly)

but you asked how to tell covariance or contravariance without nkowing how the corodinets tranmsform.

well there are two kinds of constructs, maps into a variable space from a fixed space X, called Map(X,. ), and maps out of a variable space, into a fixed space X called Map (., X).

now in the fiorst situation if we chyange the variable spave say from Y to Z, by map Y-->Z, then composing ewioth that map gives amap

Map(X,Y)-->Map(X,Z), i.e. in the same direction as the map from Y-->Z.

But in the second situation,compising with the map Y-->Z gives amap

Map(Z,X)-->Map(Y,X) i.e. in the opposite doirection from the map Y-->Z. so in algebra we call the second one contravariance and the first one covariance (just the opposite from physics and classical diff geom).now position is determined by a map from a fixed one point space p, into our variable space. so a popint of Y is an element of Map(p,Y), hence is of the covariant type, i.e. behaces like a "vector" as opposed to a "covector", remember the words are backwards.now avelocity vector at p in a space Y is determiend by a curve in Y through p.l i.e. a map from an interval I to Y, so an element of Map(I,Y) hence again behaves liek a vector. i.e. is "covariant" in algebra language, contravriant in diff geom and physics language.temperature ina nay variable space Y is determined by a real valued function on that space, i.e. an element of Map(Y,R) hence contreavariant in algebra wrt Y, or covariant in diff geom and physics wrt Y.the whole point is whether the concept in Y is measured by a map into Y or a map out of Y, and into or out of a fixed object.

particle moption in any space is measured by a map of a fixed intervalk into that space, hence transforms "directly", or the same direction as the map of spaces. (covariant in algebra, contravariant in diff geom,physics language)

 

[garrett]

Here's a simple physicist's take on contravariant vs covariant:

http://deferentialgeometry.org/#[[coordinate change]]

 

[mathwonk]

Patrick, I hope you will feel like returning. Perhaps my ungraciousness is indeed related to stress as you mentioned earlier. my wifes surgery was today and now that it is over i feel more relaxed.

anyway i did not mean to take it out on you.

best regards,

roy

 

[mathwonk]

garret I looked at that link, but it is pretty depressing to me, same old same old, no mathemmatical rigor, all coordinate dependent, no physical insight, strictly symbol pushing without any ideas or concepts at all. the kind of thing i have devoted two years here to trying to eradicate, but it will take another generation or three i guess.

 

[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.