Vectors, one-forms and gradients

[mathwonk]

in the same vein, of prerequisites, since calculus is the art of approximating non l.inear functions by linear ones, and calculus on manifolds is calculus without coordinates, obvuiously a knowldege of linear Algebra without coordinates is a preprequisite.

the lack of this rpoerequisite explains all confusion displayed so frequentloy, almost ubiquitously here, about upper and lwoer indices, which are a reflection of the fact that in coordinate free linear algebra, it is inescapable that vectors transform differently from scalar valued functions on vectors. i think i have finally put my finger on the problem that i so frequently chastize the physicists here for. namely many have never bothered to learn coordinate free linear algebra, and yet they are attempting to master coordinate free calculus. that is why so many of them are dependent on what i often call "stupid" symbol pushing.

of course it is also possible to find similar inadequate treatments of abstract linear algebra. the point is that to do corrdinate free mathematics it is better and clearer toa ctually throw out the coordinates except when they are needed for calculations. instead these physics books, which use 150 year old mathematics mostly because einstein did so, present the whole subject in coordintes and merel;y show the complicated ways of changing one set of coordinates for another.

how much better to just stop and say what the concepts mean, and then once that is grasped, to show how to compute them in any given coordinate system.

i think i am doomed to keep saying this here until my dying breath, given the huge number of physics books written in the iold style out there, ans that keep proliferating everyday.how ironic too, since it is the physicists who have given us the ideas and concepts that illuminate these symbols in the first place, and who discuss physics itself in such rich and meaningful ways, that they refuse to use ideas when discussing the mathematics.

 

[mathwonk]

i.e. without physics, none of these mathematical constructs would have been produced (calculus, manifolds, tensors), nor have as substantive a meaning or use.

we could always call it geometry, but geometry is more appealing when it describes the universe. and 4 dimensional geometry would not have been introduced so soon without relativity.

 

[mathwonk]

i guess them upshot is that physicists and amtehmaticians need to keep talking to each other, so we math types can learn what our constructs are for, and what they really mean, and physicists can better learn to use them to understand physics.

in that light of course it is at least as stupid of me not to know the physical origins of tensors as for anyone not to know their abstract characterization as multilinear objects.

 

[nrqed]

in the same vein, of prerequisites, since calculus is the art of approximating non l.inear functions by linear ones, and calculus on manifolds is calculus without coordinates, obvuiously a knowldege of linear Algebra without coordinates is a preprequisite.

the lack of this rpoerequisite explains all confusion displayed so frequentloy, almost ubiquitously here, about upper and lwoer indices, which are a reflection of the fact that in coordinate free linear algebra, it is inescapable that vectors transform differently from scalar valued functions on vectors.


i think i have finally put my finger on the problem that i so frequently chastize the physicists here for. namely many have never bothered to learn coordinate free linear algebra, and yet they are attempting to master coordinate free calculus. that is why so many of them are dependent on what i often call "stupid" symbol pushing.

of course it is also possible to find similar inadequate treatments of abstract linear algebra. the point is that to do corrdinate free mathematics it is better and clearer toa ctually throw out the coordinates except when they are needed for calculations. instead these physics books, which use 150 year old mathematics mostly because einstein did so, present the whole subject in coordintes and merel;y show the complicated ways of changing one set of coordinates for another.

But another important factor, I think, is that almost all of us learn things first (and keep learning them for a long time) using specific coordinate systems. It would not make sense to teach intro mechanics (at the level of F= ma an free body diagrams) using equations in general curvilinear coordinates or, even more abstract, using a coordinate free approach! The same for electromagnetism. It is normal then that after years of learning things in specific coordinates it is a challenge to step back and to "undo" some of that learning and to have to relearn everything while distinguishing what results are coordinate free, what depend on a metric, etc. Especially if there is some carelessness in the language used (for example calling both the covector df and th evector nabla f the divergence!

how much better to just stop and say what the concepts mean, and then once that is grasped, to show how to compute them in any given coordinate system.

i think i am doomed to keep saying this here until my dying breath, given the huge number of physics books written in the iold style out there, ans that keep proliferating everyday.


how ironic too, since it is the physicists who have given us the ideas and concepts that illuminate these symbols in the first place, and who discuss physics itself in such rich and meaningful ways, that they refuse to use ideas when discussing the mathematics.

I appreciate all your help, very sincerely.

But I want to say that if it is driving you up the walls and making your blood pressure go up to see all those stupid questions asked again and again, it would be better for your health (mental and physiological) not to spend time answering them. Honestly. It should be *fun* to explain things. If it's a chore and irritates you then you should be spending your time doing more pleasant things for you!
I keep answering questions about lower level mechanics and keep finding myself explaining the importance of free body diagrams and all that extremely basic stuff. And I know that questions like this will keep popping up and will still be asked when I am long dead. On days that I am not in a good mood, I simply don't answer and let others take care of it (even if it means that some students don't necessarily get the answers they are looking for). I prefer not to answer than to get my blood pressure go up and to end up being mad at the student. On *good* days, I look at the Intro homework sections and answer a few, trying to explain things at the level of the students asking them.

I think that those forums should be enjoyable both for the people asking questions and trying to understand new things and for the people answering them!

Regards
Patrick

 

[George Jones]

in that light of course it is at least as stupid of me not to know the physical origins of tensors as for anyone not to know their abstract characterization as multilinear objects.

I kinda like the definition of V tensor W as the free vector space on VxW modded out by the appropriate subspace (very much the same idea as what you wrote https://www.physicsforums.com/showpost.php?p=1029791&postcount=10").

 

[mathwonk]

Patrick, it is a sign of a closed mind, and one that enjoys being closed, to suggest that someone else stop trying to open it for you.

you see that i am not afraid to admit that i do not understand the physics.

as to wanting to enjoy every noble pursuit, just call me the don quixote of math instruction.

george, your favorite definition is of course exactly the object that makes the objects multilinear. i.e. ask yourself why the "appropriate subspace" is appropriate.

 

[nrqed]

Patrick, it is a sign of a closed mind, and one that enjoys being closed, to suggest that someone else stop trying to open it for you.
.

If I am closed minded for not wanting to be looked down on for trying to learn a subject then we agree that I am close minded.

EDIT: Maybe that's narrow-minded but I personally think that learning physics and maths should be fun both for the students and the teachers, especially in a forum like this. I think that there are enough stressful things in life without making it confrontational and stressful to be discussing physics and maths. I have a sister who recently died of multiple sclerosis, just as one example of a stressful situation (and I am sure that everybody here has other stressful things happening in their lives). I think that posting here and discussing things should be done an enjoyable experience for everyone, students and teachers alike.


If that makes me closed minded, then so be it.

 

[mathwonk]

the problem is you are asking someone else who offends you to stop talking, instead of merely not reading their comments. that is not the way an open forum works, or an open society.

and we are getting off topic here. I hope your stresses do subside, and i apologize for apparently adding to them without intending to. i am just teaching math here, and learning physics, nothing deeper. i realize too that your stresses are slopping over into your messages to me, and that is understandable, just as perhaps mine slipped over into my posts you object to.

peace

 

[mathwonk]

Patrick, I know so little physics that i must begin only with a small part of your comments above on coordinate free teaching of it. but i think we can make some progress simply because you have given some examples.

let me go back to this one:

"It would not make sense to teach intro mechanics (at the level of F= ma, an free body diagrams) using equations in general curvilinear coordinates or, even more abstract, using a coordinate free approach!"Now to me F=ma is a perfect example of a coordinate free principle. I.e. without coordinates, it says that the force on a given mass is proportional to acceleration. e.g. if we double the force we double the acceleration.

This idea has nothing at all to do with a specific choice of coordinates. Coordinates come in when we choose to say that a liter of water at sea level, at degrees centigrade is called one unit of mass, and that a force which accelerates that mass by one meter per second, or whatever, shall be denoted by one Newton, or dyne, or whatever.

here the concept takes a back seat and the coordinates enter for purposes of calculation. but they obscure to me at least the concept. my physics instruction semed overly laden with requirements to memorize whAT THE UNITS were INSTEAD OF oops, what the ideas were hence i understand nothing from it.

(my school was famous for good math instruction and poor physics instruction at the time. by the time they put some good lecturers in the physics course i had given up hope and trying to learn.)certainly one would not use curvilinear coordinates in an attempt to discuss the topic in a coordinate free way, rather one would explain what the physical concept was that we were trying to measure in coordiantes, when the time came.

also, please correct me if i am wrong, but since this equation deals with action of a force on a fixed mass, it is "local", i.e. takes place in small region of space. hence global coordinates are inappropriate, and local, i.e. euclidean ones are the right choice.now please explain where i have gone astray here. if you will.

 

[nrqed]

the problem is you are asking someone else who offends you to stop talking, instead of merely not reading their comments. that is not the way an open forum works, or an open society.

No, I was not asking you to stop talking at all.
My point was that if you read my questions and your first reaction is " here we go again having to explain elementary stuff to someone who has not even tried to learn the most basic concepts on his own and who has not even done a google search before asking simple minded questions here.. this is so *annoying*"

then, if this is your feeling upon reading my (admitteddly simple minded) questions, my point was that it was not worth it for you to get worked up in respodning to me if it is annoying and irritating to you which in return makes me feel bad for trying to learn something which is confusing to me.

My point was just that if it makes you feel this way (and it's the impression I was getting from the tone of some replies...if that's totally off then I apologize), then I wanted you to use your time doing something more enjoyable.
But if you feel obliged to reply to my questions while feeling irritated by them then I feel that I should probably stop asking them. I am genuine in my desire to learn and I try to enjoy transmitting to others the limited knowledge I have. I hope that those who are gracious enough to try to help me learn new things are enjoying themselves doing so. If not, I will prefer to stop asking and learn with books alone.

Regards

Patrick

 

[mathwonk]

To get back to having fun, do you have any light you would like to shed on my simple minded question about F = ma?

 

[mathwonk]

maybe i can understand something here. i apoologize in advance for possibly boring the physicists. i want to try to translate physics intio math with a view to understanding the math constructs. sadly i do not know any physics. so please help me

let's start with a space like the 3 dimensional universe we live in.

i am going to suggest that "vectors" in that space are represented by anything whose natural representation involves a function from the real numbers into the space. e.g. a moving particle in time, once we choose some units of time, involves a function from time t to position p(t).

even without a choice of units the path of the motion is a curve. I claim this represents a vector. i.e,. at each instant of time, with units (of time and distance) there is a velocity vector.

even with no units there is a tangent line spanned by the velocity ector, also a vector object, i.e. a vector space.now by a covector i mean any quantity naturally represented bya function from the space to the real numbers, like the heat from a radiant object. with units, each point in space has a temperature assigned to it.

even without units, there are surfaces of constant temperature, which makes sense even without asigning a number to that temperature. this assignment is a covector.

or mroe rpecisely, since no vectors or covectors are yet opresent, i should say these two dual types of phenomena, represent (contra) variant, or covarant quantities.

[ i forget, in physics is it vectors that transfrom "contravariantly"?)

anyway, without any coordinates, the two dual phenomena are distinguished by the dimension of the geometric representatives. in one case, "vectors" are reprented by one dimensional objects namely curves. and covectors, are represented by"codimension one" objects, namely surfaces in three space, i.e. level sets of a function.now my question is, is this a nautural physically meaningful distinction? I.e. is tempertiure naturally represented by level surfaces instead of curves? it would seem so. hence temperature shoud be a (co?) variant quantity, and particle motion a (contra?) variant one.now infinitesimally, we should get vectors representing the first type and covectors representing the second.
i.e. infinitesimal changes in position should be represented by vectors, and infinitesimal chNGES IN TEMPERATURE BY COVECTORS.

NOW WHAT DOES THAT EMAN?

anyway, anybody with me here? or against me? or anything?

 

[mathwonk]

to coninue slightly, if temperature is a function f, then the differntial of that function df, should be the covector corresponding to it.

it pIRS WITH A CURVe to tell us how fast thw temperature is changing along that curve, infinitesimally. i.e. given a velocity vector v to a curve, the number df(v), is the rate of change of temperatiure in the direction of v, in units of temperature per units of time.

If we have a metric and can measure angles, we can pick a curve that is perpendicular to the level set of temperature, and mark off a unit length along that tangent vector to that curve, in the direction of increasing temperature. if we measure the infinitesimal change of temp in that unit norml direction, we get a number we can multiply by that normal vector to get a vector that represents the covector df, under that metric.

hence the covector df has become a gradient "vector".but what are some other impoirtant physical concepts in mechanics say or relativity? and which kind of quantity are they intrinsically?force for example. since acceleration is aderivative of velocity, I will gues that it is a vector ND NOT COVECTOR. then being proprotional to it, LSO FORCWE WOULD BEA VECTOR?
ctice?
how does this compare with physicists intuition or practice?

anyway this is what i mean by discuasing the concepts first without coordinates, and then using that understanding to illuminate how the coordinates are used to measure them.

naturally it would be very helpful if i understood any of the concepts.

jump in anytime here. there is no reason for physics and math to be disjoint. much of math is just an attempt to make physics precise and measurable.

 

[mathwonk]

Patrick, is this ALL NONSENSE? I gues my main question, is does "covariant" quantity have an intrinsic physical meaning?

thanks,

roy

 

[masudr]

How would you know that position is a vector without seeing how it's co-ordinates transform?

 

[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.

?