Why does one form need to be used over the other?

[nrqed]

In differential geometry, the usual curl operation that we are familiar with from elementary calculus is generalized to \,^*dA (where A is a one-form). In three-dimensions, this gives back a one-form.

Now, the components of this one-form are \sqrt{g} \epsilon_{ijk} \partial^j A^k.

however, the corresponding contravariant components are \frac{1}{\sqrt{g}} \epsilon^{ijk} \partial_j A_k.

Now, to obtain the formula that we learned in elementary calculus, it is the second form that must be used. Why is that the case?

On the other hand, if one looks at the generalization of the gradient, it's the formula for the covariant components of d \phi = \partial_i \phi that one must use to get the usual formula we have learned for the gradient.

So what is the rationale behind choosing one form over another? Maybe one must pick the form that differentiates with respect to the x^i so that one must have the index on the partial derivative downstairs?

I am sure there is something fundamental going on here that I am obviosuly completely missing.

Thanks

Patrick

 

[mathwonk]

learn differential forms.

 

[HallsofIvy]

It is "contravarient" components that we use in "normal" Rⁿ vector. Strictly speaking the "covarient" components are the components of the dual space. In Rⁿ ("Euclidean Tensors") there is a natural isomorphism with its dual so that we never need to mention covariant components.

 

[nrqed]

learn differential forms.

I thought that from my post it was clear that this is what I am trying to do.

Thank you for your help. When students ask questions about inclined planes, friction problems and circular motion in the Homework forum I will be sure to reply by telling them "Learn Mechanics"!

 

[nrqed]

It is "contravarient" components that we use in "normal" Rⁿ vector. Strictly speaking the "covarient" components are the components of the dual space. In Rⁿ ("Euclidean Tensors") there is a natural isomorphism with its dual so that we never need to mention covariant components.

Thanks for your help.

This makes sense to me and that was my first inclination. But when I look at the generalization of the gradient through the exterior derivative of a scalar function d \phi with components \partial_i \phi, it seems that I now have to consider the covariant components in that case. I mean, if I wanted the contravariant components, I would need to consider g^{ji} \partial_i \phi but this does not give the components of the gradient, let's say in spherical coordinates, that we learn in elementary calculus.

 

[mathwonk]

well sorry it wasnt helpful - i was a little puzzled by the very coordinate dependent approach you were using, so maybe i should have said, try to learn a more intrinsic version of differential forms, like the one in david bachmans little book, once read here communally.

curl is just "d" of a one form, and the result is a 2 form, not a one form, even in three space. so the first thing to learn is that a one looks like

fdx +gdy + hdz and that d of it, i.e. the curl,

looks like df^dx + dg^dy + dh^dz

= oops i need a curly d now,

maybe ill use a question mark

?f/?y dy^dx + ?f/?z ^dz^dx

+ ?g/?x dx^dy + ?g/?z dz^dx

+ ?h/?x dx^dz +?h/?y dy^dz

now rearrange all these terms by the rule dy^dx = -dx^dy etc...and note i have already deleted the terms dx^dx, dy^dy, dz^dz.

try it.

 

[mathwonk]

if i may make another suggestion, when asking for help, a comment like "in fact that's what I am trying to do" is more productive than "if you weren't so stupid youd realize that was what I am trying to do." just a suggestion.

you see what you wrote was so far from differential forms as i view them, that indeed i did not realize that was what you were trying to do.

 

[nrqed]

if i may make another suggestion, when asking for help, a comment like "in fact that's what I am trying to do" is more productive than "if you weren't so stupid youd realize that was what I am trying to do." just a suggestion.

you see what you wrote was so far from differential forms as i view them, that indeed i did not realize that was what you were trying to do.

I apologize sincerely.
Since I started my post by mentioning the intrinsic expression \,^* dA and then I talked A being a one-form, I had mistakenly thought that you had seen that and that your remark was derogatory since I was (in my mind) clearly trying to make sense of the meaning of this form.

My sincere apologies.

Patrick

 

[nrqed]

well sorry it wasnt helpful - i was a little puzzled by the very coordinate dependent approach you were using, so maybe i should have said, try to learn a more intrinsic version of differential forms, like the one in david bachmans little book, once read here communally.

curl is just "d" of a one form, and the result is a 2 form, not a one form, even in three space. so the first thing to learn is that a one looks like

fdx +gdy + hdz and that d of it, i.e. the curl,

looks like df^dx + dg^dy + dh^dz

= oops i need a curly d now,

maybe ill use a question mark

?f/?y dy^dx + ?f/?z ^dz^dx

+ ?g/?x dx^dy + ?g/?z dz^dx

+ ?h/?x dx^dz +?h/?y dy^dz

now rearrange all these terms by the rule dy^dx = -dx^dy etc...


and note i have already deleted the terms dx^dx, dy^dy, dz^dz.

try it.

Yes, indeed, I know that this works. But I am trying to get the general expression for the curl in arbitrary coordinates, not just in cartesian coordinates (which is where the power of the differential form approach should become more clear to me). And then it seems that to really get the general expression, it's not enough to apply the exterior derivative, one must then take the Hodge dual. This is why I am looking at \,^* dA (which is obviously a one-form if I am in 3 dimensions) and not just dA. My goal is to obtain from the differential form approach all the expressions we learn in elementary calculus for the curl in spherical or cylindrical coordinates as well as the gradient and the divergence. But then I run into the question I mentioned in my first post about having to make a certain choice of using the components of the one-form I get or if I have to use the metric to get the contravariant components. I am trying to understand what motivates one choice over another.

I find that often, if one concentrates purely on the mathematical definitions, things are not too difficult. But when trying to connect with the equations used in physics, I often encounter some difficulty.

Thank you for your help, it is appreciated.

 

[mathwonk]

thank you for your kind response. my sincere apologies as well.

and i could not tell the hodge star from a small smudge!

but differential forms are so suited that they work the same in all

systems of coordinates.

(and no need for a metric, which hodge thery requires.)for instance if x and y are functions of u and v, then an expression in dx and dy is simply replaced by one in terms of du and dv by means of

dx = ?dx/?u du + ?x/?v dv, and so on for dy.

so i am still unable to understand why it makes any difference what coordinates we use? probably i am being slow, but i seem to succeed in using differential forms without ever caring which coodinates i am using.

 

[mathwonk]

i look at them this way: suppose we have a space, with no coordinates at all.

[for purposes of deriving component expressions, ignore this blah blah and see example below.]

then to every parametrized curve in the space we can assign a velocity vector at every point, using the coordinates on the time interval, or domain of the curve.

this gives us a family of vectors in our space.dually, if we have a function f on our space, then at each point we have a one form df, i.e. a linear function on vectors, which assigns to a given vector v, the derivative at t=0 of any curve through v, composed with f.

similarly, there are families of parallelograms in space, tangent plkanes to pieces of surface, and 2 forms which are functions on parallelograms.

then justa s one can take the boundary of a piece of surface, getting a curve, dually one can take d of a one form, which acts on a surface by letting the original one form act on the boundary of th surface.

i.e. curl of a one form is just adjoint to the action of the boun dary of a surface. then in varous coordinates this geometry and calculus gets computed numerically, via the fundamental theorem of calculus.

i.e. greens and stokes and gauss' theorems say that the goofy formula for d of a one form, i.e,. curl, grad, etc,, is in fact just a comoputation of this adjoint action.

 

[mathwonk]

letme see if i can do an example, in 2 dimensions. saypollar coordinates, which will at least suffice for cylindrical ones.

so then x = x(r,t) = rcost, y = rsint, so dx = dx/dr dr + dx/dt dt, etc..


then write also f for
f(x(r,t), y(r,t)),

so if we have a one form in x,y coords, say w = f(x,y)dx + g(x,y)dy,

then you can either put it in r,t coordinates and then take d, or vice versa, and you get the same result, which is nice.

i.e. dw = (after canceling and rearranging) [dg/dx - df/dy] dxdy

and then replacing dx by dx = dx/dr dr + dx/dt dt = (i hope)

cost dr - rsint dt, and replacing dy by let's see, dy/dr dr + dy/dt dt

= sint dr + rcost dt, we get, omigosh, what was i doing??

oh yeah, dw = [dg/dx - df/dy] dxdy

= [dg/dx - df/dy] [cost dr - rsint dt] [sint dr + rcost dt]

= [dg/dx - df/dy] [rcos^2 t+ rsin^2 t] drdt

= r [dg/dx - df/dy] drdt .

good heavens, no wonder i dislike coordinates, they are so messy.

ok anyway, i am claiming this is the same as if we first rtansformed into r,t coordinates and then took d.

this i usually leave as exercise for the hapless student (not really knowing if it is true or not, but believing somehow it is.)

well let's make a stab at it.

 

[mathwonk]

to be continued...

ok back, again we have w = f(x,y)dx + g(x,y)dy,

so tranforming to polar coordinates first gives,

w = w = f(x,y)[dx/dr dr + dx/dt dt] + g(x,y)[dy/dr dr + dy/dt dt]

= [f dx/dr g dy/dr] dr + [f dx/dt + g dy/dt] dt

= [fcost + g sint]dr + [f rsint - g rcost]dt

now take d, getting oh boyy...

d/dt of that first mess times dtdr, then d/dr of that second mess times drdt.

hopefully that simplifies.

i am needed on the home front, i.e. my wife won't let me complete this mess, if you believe that. but i assure you it works really, if done right...

 

[nrqed]

to be continued...

ok back, again we have w = f(x,y)dx + g(x,y)dy,

so tranforming to polar coordinates first gives,

w = w = f(x,y)[dx/dr dr + dx/dt dt] + g(x,y)[dy/dr dr + dy/dt dt]

= [f dx/dr g dy/dr] dr + [f dx/dt + g dy/dt] dt

= [fcost + g sint]dr + [f rsint - g rcost]dt

now take d, getting oh boyy...

d/dt of that first mess times dtdr, then d/dr of that second mess times drdt.

hopefully that simplifies.

i am needed on the home front, i.e. my wife won't let me complete this mess, if you believe that. but i assure you it works really, if done right...

I hope you did not spend too much of your time writing all that! I do appreciate very much! I have to spend some time going through the details of your posts.

Just a comment: for example, in spherical coordinates, the expression we learn in elementary calculus contains a factor of 1/(r sin \theta) which looks like 1/\sqrt{g}. In fact, the expression \sqrt{g} \epsilon_{ijk} \partial^j A^k that I gave in my first post is exactly the expression given for the generalization of the curl given in a book I am looking at (by Felsager. I could give the complete reference when I get back to my office).

But let me spend a bit more time digesting yoru posts and working through the calculation myself.

Thank you very much for the food for thoughts!

Regards


Patrick

 

[mathwonk]

you are very welcome. the whole point is thaT THE operation d haS an intrinsic meaning, as explained in post 11. hence if it has a meaning independent of coordinates, then all coordinate expressions are derivable from any one.

i do recall from my courses of fourty yeARS AGo or so that the actual calculations are tedious.

the short version of the calculation i was trying to do above, is that if w is a form and f a change of variables, then f*dw = df*w.

best regards.

 

[mathwonk]

ok let's try this again, simplifying it. the whole point is that to find the curl in polar coordinates, one can either transform to those coordinates and then use the same curl procedure as in cartesian coordinates, or vice versa, namely compute the curl in cartesian coordinates and then transform that to polar ones.

the comoputation in full has been seen to be lengthy, so let's just do the simpelst possible one, where the given one form is just dx.

then we know the curl of dx is zero, namely dx = 1dx so the curl equals

d1^dx = 0^dx = 0. and zero of course transforms to zero.

now let's do it the other way around, namely transform dx to polar coordinates, and then see if the curl of the polar transform is also zero.

but in polar coordinates, dx = dx/dr dr + dx/dt dt

= cost dr - rsint dt. then curl of that

is [d(-rsint)/dr - d(cost)/dt] drdt

= [-sint +sint]drdt = 0.

indeed there is no need to restrict to polar coordinates, for let x be any smooth function of r,t.

then we have again ddx = 0, and now

curl [dx/dr dr + dx/dt dt]

= [d^2 x/dtdr - d^2 x/drdt] drdt, which equals zero by the principle of equality of mixed partials.


next we do a more complicated one, but still simpler than those above.

 

[mathwonk]

of all the frustrating,... i finished the calculation and the biorwser won't accept it.

advanced mode doeswn t work either, so ill break it uo in pieces.

ok, now notice that since the curl of a sum is the sum of the curls, we can restrict to things just of form fdx.

then we have curl(fdx) = df^dx = df/dy dy^dx, in cartesian coordinates.

which transforms to the polar form:

df/dy [dy/dr dr + dy/dt dt] ^ [dx/dr dr + dx/dt dt]

= df/dy [dy/dr dx/dt - dy/dt dx/dr]drdt

which will probably need some transformation later.

 

[mathwonk]

but now going the other way around,

f dx = f [dx/dr dr + dx/xt dt],

and curl of that is, oh boy, maybe ill try to use the leibniz product rule,curl {f [dx/dr dr + dx/xt dt]}

=? df ^ [dx/dr dr + dx/xt dt] + f d[dx/dr dr + dx/xt dt] ?

= [df/dt dx/dr] dt^dr + df/dr dx/dt dr^dt

+ f [d^2 x/drdt dt^dr + d^2 x/dtdr dr^dt]

(notice i started writing wedges again because i have a repeated derivative in the denominator which looks similar to a 2 form otherwise.)

 

[mathwonk]

man this is frustrating.

 

[mathwonk]

and again the second part is zero because of equality of mixed partials and the fact that dr^dt = -dt^dr.

so we get just [df/dt dx/dr] dt^dr + df/dr dx/dt dr^dt]

= [df/dr dx/dt - df/dt dx/dr] dr^dt,

 

[mathwonk]

so does [df/dt dx/dr] dt^dr + df/dr dx/dt dr^dt]

=df/dy [dy/dr dx/dt - dy/dt dx/dr]dr^dt ?

well they would be if df/dr = df/dy dy/dr, and so on, but that looks false to me, so where did i screw up? i.e. these are partials, so shouldn't we have

df/dr = df/dy dy/dr + df/dx dx/dr?

well anyway, let's hope for some canceling,

 

[mathwonk]

i.e. [df/dr dx/dt - df/dt dx/dr] dr^dt
=
{ [df/dy dy/dr + df/dx dx/dr] dx/dt - [df/dy dy/dt + df/dx dx/dt] dx/dr} dr^dt= df/dy dy/dr dx/dt - df/dy dy/dt dx/dr

= df/dy [dy/dr dx/dt - dy/dt dx/dr]

hooray! i think we got it!.
and we did it in general coords, not just for polar coords.

 

[mathwonk]

moral, it is easier to take the curl in cartesian coords and then transform that, than to transform first and take curl afterwards.

so the polar curl of fdx seems to be:

df/dy [-rsin^2 t -rcos^2 t] drdt = -r df/dy dr^dt.

so i am guesing the polar curl of gdy is r dg/dx dr^dt.

which would make the polar curl of fdx + gdy, equal to r[dg/dx -df/dy]dr^dt

 

[mathwonk]

if you compare the previous result with post 12 you will see they agree.

 

[mathwonk]

another moral is that one should learn the machinery of differential forms once for all, intrinsically, and how they transform, and then matters like this begin to look just like F*(dw) = d(F*w), and one never ever does these messy explicit calculations again, except in teaching them i guess.

i.e. the messy one was d(F*w), but the equality shows we could have done just F*(dw) instead.

peace.

 

[mathwonk]

oh and just to make a point, notice that the sort of specific "components" you began by writing have never appeared at all. which is why i didn't even recognize your original question as really being in the same subject, or at least not in the same spirit. i.e. to me, using differential forms is about manipulating d and F*, not writing out "a upper ij" and so on.

 

[mathwonk]

all you really need to know is how to use d, ^ and F*.

i.e.
heres the scoop as i see it: smooth differential forms are a differential graded exterior algebra over the smooth functions, locally freely generated by the gradients of the coordinate functions, i.e. they look like sums of products like fdx, and are graded by the number of such factors.

i.e. one has an alternating multiplication called ^, and a differentiation called d,

and for any smooth function F, there is a pullback operation F* which is a homomorphism both for ^ and for d.

I.e. F*(dw) = d(F*w), and F*(w^v) = F*w^F*v, and d obeys a graded leibniz rule, i.e. d(w^v) = dw^v +(-1)^deg(w)w^ dv.

 

[mathwonk]

with just those rules, just knowing the rukle for taking gradients in cartesian coordintaes determiens everything, curl, div, etc,,, in all possible cordinate systems.

one never needs to write out "components" unless a particular specific calculation is needed, and if so, they should disappear immediately afterwards.

 

[nrqed]

with just those rules, just knowing the rukle for taking gradients in cartesian coordintaes determiens everything, curl, div, etc,,, in all possible cordinate systems.

one never needs to write out "components" unless a particular specific calculation is needed, and if so, they should disappear immediately afterwards.

Thanks for all those posts. It will take me some time to absorb that!

I see what you are saying about starting from Cartesian and then transforming,but to me it seemed interesting (and efficient) to have a formula that would directly give the components of the curl in any dimensions and in any coordinate system. Something that would not require making a change of coordinate. A master formula if you will. Do you see what I mean? For example, I wanted that if working in spherical coordinates in 3 dimensions, I could simply *directly* get the components of the culr using only the metric. Basically I was looking for a general formula. It does seem like the formula \sqrt{g} \epsilon^{ijk} \partial_j A_k does work but I was looking to understand why it was the correct one.


Or I would also be happy to start with the exterior derivative of a one form dA and then to have an algorithm to generate the curl out of this by working directly in the coordinate system used (spherical for example).
So I was trying to get an expression that would give directly the answer in any coordinate system without requiring to do a change of coordinate.

But I will try to reproduce all your work in details because that will teach me a different approach and I will learn from that.

Thank you!


Patrick

 

[mathwonk]

d of a one form IS the "curl" of that one form.

i.e. no matter WHAT coordinates u,v are,

the curl of fdu + gdv is ALWAYS df^du + dg^dv, in THOSE coordinates.

e.g. the curl of the angle form dtheta, in polar coords, is ddtheta = 0.d of a one form, i.e. the curl, is always computed the same way, in all coordinates. there is no need to transform anything.

all i was doing was proving this.there is nothing else to memorize. by introducing hodge duals you are just making life needlessly more complicated.

the only time a metric is needed is when one wants to study harmonic functions, i.e.laplacians.

somehow i feel we are still not communicating. you are trying to express something which is natural in all coordinate systems, i.e. exterior derivatives d, in terms of something less natural, i.e. components gij using a metric.

perhaps you have a need for this, but i do not see it.

 

[nrqed]

d of a one form IS the "curl" of that one form.

i.e. no matter WHAT coordinates u,v are,

the curl of fdu + gdv is ALWAYS df^du + dg^dv, in THOSE coordinates.

e.g. the curl of the angle form dtheta, in polar coords, is ddtheta = 0.


d of a one form, i.e. the curl, is always computed the same way, in all coordinates. there is no need to transform anything.

all i was doing was proving this.


there is nothing else to memorize. by introducing hodge duals you are just making life needlessly more complicated.

the only time a metric is needed is when one wants to study harmonic functions, i.e.laplacians.

somehow i feel we are still not communicating. you are trying to express something which is natural in all coordinate systems, i.e. exterior derivatives d, in terms of something less natural, i.e. components gij using a metric.

perhaps you have a need for this, but i do not see it.

I think that it's because we are talking about different curls.

My goal (and it may be that the goal in itself seems useless but that's a different discusssion) is to recover, from a differential form approach, the following expression that we learned in elementary calculus:


\nabla \times \vec{A} = \frac{1}{r sin \theta} \bigl( \frac{\partial ( sin \theta A_{\phi})}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi} \bigr) \hat{r} + \ldots

This is what I am trying to obtain starting from differential forms.


regards

 

[mathwonk]

what is del ? or nabla, or whatever?

what is A? presumably A is the vector of coordinates of the differential one form Ar dr + Atheta dtheta, but maybe not?

what coordinates are yoiu using? spherical? are del and A both in spherical coords?this very old fashioned notation is just to me a clumsy way of obscuring the quite simple differential calculus as now used in terms of differential forms, and it is so out of date, i have never even seen it! (believe it or not.)

(i took elementary calculus out of loomis and sternberg and was never exposed to this old maxwellian version of vector calc notation, that originally arose using quatern ions, and has been outmoded in mathematics for over 50 years.)

but i agree it would be fun to see then turn out the same. but i am confident there is no difficuklty about this if we just make clear what the symbols mean.

 

[Hurkyl]

Is \nabla \times A not equal to ({}^\star d(A^t))^t?

(Where the transpose of a vector is the one-form you get by contracting with the metric, and conversely)

 

[mathwonk]

come to think of it, we were being taught this clunky old stuff in physics class. i just never learned it.

are in three dimensions? anyway i promise you these are just primitive versions of d(fdx) = df^dx.

 

[mathwonk]

why bring in a metric to compute an exterior derivative, when it does not depend on one?

you guys seem bound to make something simple look complicated.

 

[nrqed]

what is del ? or nabla, or whatever?

what is A? presumably A is the vector of coordinates of the differential one form Ar dr + Atheta dtheta, but maybe not?

what coordinates are yoiu using? spherical? are del and A both in spherical coords?


this very old fashioned notation is just to me a clumsy way of obscuring the quite simple differential calculus as now used in terms of differential forms, and it is so out of date, i have never even seen it! (believe it or not.)

(i took elementary calculus out of loomis and sternberg and was never exposed to this old maxwellian version of vector calc notation, that originally arose using quatern ions, and has been outmoded in mathematics for over 50 years.)

but i agree it would be fun to see then turn out the same. but i am confident there is no difficuklty about this if we just make clear what the symbols mean.

yes, this is the expression in spherical coordinates.

This is the expression given for example in undergraduate electricity and magnetism formula when applying the curl of electric or magnetic fields (or of the vector potential) in spherical coordinates.

 

[nrqed]

Is \nabla \times A not equal to ({}^\star d(A^t))^t?

(Where the transpose of a vector is the one-form you get by contracting with the metric, and conversely)

Yes, that sounds right! This is what I am getting from reading Frankel. And I *think* this *does* correspond to the formula \sqrt{g} \epsilon^{ijk} \partial_j A_k I mentioned early on. I have to check this.

What about the gradient? It would be d \phi, not (d \phi)^t, right? (i.e. we would not go back to a vector by contracting with the metric)

Thanks!

 

[Hurkyl]

It depends on what you mean by "gradient". I always thought it meant the covector that yields directional derivatives, but I've seen people insist that it means the vector pointing in the direction of greatest ascent.

If you mean the vector, then you'll have to transpose.

 

[mathwonk]

this may seem simple minded but think what i am asking is whether del means (d/dx, d/dy)

or whether it means (d/dr, d/dtheta).

i suspect the reason this curl expression looks funny, is that it stands for the polar transfrom of d of the cartesian transform of a one form given in polar coords.i.e. (d/dx,d/dy) X (Ar, Atheta) is going to look different from

(d/dr, d/dtheta) X (Ar, Atheta).there is to me still no reason at all to bring in any stars or duals or metrics into this calculation.

 

[nrqed]

Is \nabla \times A not equal to ({}^\star d(A^t))^t?

(Where the transpose of a vector is the one-form you get by contracting with the metric, and conversely)

It seems almost right except for one very nagging detail.

To make it work, I would instead need to use ({}^\star d(A))^t, i.e. I would have to treat the components of the "vector field \vec{A}" as if they were already the components of a one-form. This is strange.

 

[mathwonk]

this seems to me to be an object lesson in the unnecessary confusion introduced by trying to pretend that a space is isomorphic to its dual, by means of a metric, when it is clearer to keep dual spaces distinguished.

 

[mathwonk]

ok, i think i see you are saying "curl" as an operation on a vector field, which of course makes no sense unless you have a metric, since properly it means the exterior derivative of a one form.

so you have to change your vector field into a one form, then take d, then change it back. uggh. all this compounded by changing coordinates.

same confusion for "gradient" which to me is just a one form associated to a function, but to some people is a vector field (artificially) dual to that one form.

this may be the underlying mystery you are grappling with, i.e. i think curl of a vector field is not a natural operation, undefined without a metric, but d of a one form is.

 

[nrqed]

this seems to me to be an object lesson in the unnecessary confusion introduced by trying to pretend that a space is isomorphic to its dual, by means of a metric, when it is clearer to keep dual spaces distinguished.

agreed. Unfortunately, in elementary physics the distinction is not made. So as physicists, we learn to calculate the curl of vector fields (well, they are called vector fields but in restrospect it's not clear if they were vector fields or components of differential forms), to take divergence and gradient, to integrate, to use Stokes and Green's theorems, and so on without ever using differential forms.

Now I am trying to connect everything I have learned to the more powerful language of differential forms but this requires that I disentangle everything that was patch together in physics. Unfortunately, this is very difficult because most physicists don't know well (or at all) the language of differential forms and mathematicians are not necessarily used to physics applications. And in addition it's ahrd to find references explaining clearly the connection between the two languages.

 

[Hurkyl]

It seems almost right except for one very nagging detail.

To make it work, I would instead need to use ({}^\star d(A))^t, i.e. I would have to treat the components of the "vector field \vec{A}" as if they were already the components of a one-form. This is strange.

That doesn't work, because dA is gibberish. d acts on forms, A is not a form. What's wrong with what I wrote? When I worked it out, I thought I got the right answer, except possibly with the wrong overall sign. (Which is easy to fix)

 

[mathwonk]

thank you. i think iam beginning to grasp the situation. do you want the curkl of a vector field to be VECTOR FIELD?

so poerhaps do you first turn the vector field into a one form using a metric? then take d of the one form, getting a 2 form, then take the hodge dual of the 2 form getting a one form?

then turn that back into a vector field using the metric?

No wonder it is complicated. then there are the changes of coordinates for all these operations!

so do we have two dual operations, between one forms and 2 forms, and also between vectors and one forms, and we are also changing coordinates?

I am feeling too much on vacation to deal with this mess. but hurkyl seems up for it.

you might look in loomis and sternberg, or nickerson, spencer and steenrod, or spivak, or marsden and tromba, or maybe wendell fleming.

 

[nrqed]

That doesn't work, because dA is gibberish. d acts on forms, A is not a form. What's wrong with what I wrote? When I worked it out, I thought I got the right answer, except possibly with the wrong overall sign. (Which is easy to fix)

I understand that it makes only sense to apply the exterior derivative on a one-form.

What I mean is that it seems to me that in order to get the usual expression (that I learned in undergraduate school), it looks as if I must assume that the three components of the "vector field" A_r, A_{\theta} and A_{\phi} must be treated as the three components of a one -form. In undergrad E&M, say, everything is called a vector field so when three components of something is given, it's not clear if it's really the components of a vector field or the components of a one-form (or who knows, maybe even the three independent components of a two form in 3 dimensions!). It seems to me that I need to treat the three components of the supposed vector field as really being the components of a one-form.
I will post the details a bit later (maybe tomorrow).

Do you see what I am trying to say?

 

[mathwonk]

we seem to be converging on the same point of view.

 

[Hurkyl]

If you intend to define a vector field valued curl of a covector field, then ({}^\star dA)^t does appear to be a reasonable definition.

 

[nrqed]

thank you. i think iam beginning to grasp the situation. do you want the curkl of a vector field to be VECTOR FIELD?

You see, this is part of the difficulty I am having. Because in undergraduate physics, everything si called a vector field. Let's say we are working with three functions A_r, A_{\theta} and A_{\phi} (where the position of the indices has no special meaning). Then it's not clear at all if these are meant to be the components of a vector field, or the components of a differential one-form (or, who knows, the three independent components of a two form in 3 dimensions!). This is part of what makes my job so hard. I need to figure out if three functions are really the components of a vector field or of a one-form. In the end, I wlaso want to figure out what the different physical quantities (electric field, magnetic field, current density etc etc ) that we use in undergraduate physics are vector fields, one-forms, etc.

So I was trying to {\bf use} the expressions given in undergraduate physics textbooks for the curl, gradient and divergences to figure out if those functions that are differentiated are truly components of vector field or something else.

so poerhaps do you first turn the vector field into a one form? then take d of the one form, getting a 2 form, then take the hodge dual of the 2 form getting a one form?

then turn that back into a vector field?

No wonder it is complicated. then there are the changes of coordinates for all these operations!

yes, but the point was that if I could get an expression in terms of exterior derivatives and hodge dual and so on, the result would be completely coordinate independent. This is what I was trying to get at!

so we have two dual operations, between one forms and 2 forms, and also between vectors and one forms, and we are also changing coordinates?

I am feeling too much on vacation to deal with this mess. but hurkyl seems up for it.

Believe it or not, I am on vacation too and I am going insane with that stuff!

 

[Hurkyl]

Double the size of everything: that will tell you what you need to know.

3-forms will be divided by 8.
2-forms will be divided by 4.
1-forms will be halved.
scalars will be unchanged.
vectors will be doubled.
bivectors will be quadrupled
trivectors will be octupled.


(Note that if you chance coordinates (x', y', z') = (x/2, y/2, z/2), then the rescaled thing in (x', y', z') coordinates looks the same as the original in (x, y, z) coordinates -- except, of course, that the metrics are different)


For example, consider mass density. A 1 kg cube 1m on a side has density 1 kg / m^3. A 1 kg cube 2m on a side has density (1/8) kg / m^3. Thus, mass density is best represented by a 3-form.

You could see this directly too: density directly tells you how much mass there is in a volume, which is precisely what 3-forms do.



Rescaling by -1 gives a simpler test, but can't distinguish between everything. But it helps for some things -- e.g. it tells you that a vector-valued cross product of vectors is not a very good idea.