# Curl in arbitrary coordinates

by nrqed
Tags: arbitrary, coordinates, curl
HW Helper
P: 2,887
 Quote by 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)
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!
 Emeritus Sci Advisor PF Gold P: 16,092 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.
 Sci Advisor HW Helper P: 9,453 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.
HW Helper
P: 2,887
 Quote by 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)
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.
 Sci Advisor HW Helper P: 9,453 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.
 Sci Advisor HW Helper P: 9,453 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.
HW Helper
P: 2,887
 Quote by 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.
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.
Emeritus
PF Gold
P: 16,092
 Quote by nrqed 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)
 Sci Advisor HW Helper P: 9,453 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.
HW Helper
P: 2,887
 Quote by Hurkyl 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?
 Sci Advisor HW Helper P: 9,453 we seem to be converging on the same point of view.
 Emeritus Sci Advisor PF Gold P: 16,092 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.
HW Helper
P: 2,887
 Quote by mathwonk 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!
 Emeritus Sci Advisor PF Gold P: 16,092 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.
 Sci Advisor HW Helper P: 9,453 well we just need a symbol for the duality between vector fiedls and covector fields, i.e.one forms, lets call it #, so # of a one form is a vector field and vice versa. then curl of a vector field V, should be something like #(*d#V), i.e. #V is a oneform, then d#V is a 2 form, then *d#V is a oneform, then #*d#V is a vector field., but this looks too complicated. nonetheless i have seen some pretty complicated expressions in my life. the reason i thought this was simple, is that the only part of this that is purely differential forms, to me, is the d part. duality is always confusing. and using metrics to identify spaces that are dual, makes it harder to tell them apart, and harder to recognize which operations are natural for them.
 Sci Advisor HW Helper P: 9,453 metrics are useful in the study of differential forms however, in hodge theory as follows: when oine studies one forms say, it is interesting to examine which one forms are d of a function (i would say which ones are gradients). a necessary condition which is locally sufficient, but not globalkly so, is to have curl = 0, or d =0. then one wants to know how far from sufficient this condition is, so one studies the quotien space {w: dw=0}/{df: all f}. this space is often finite dimensional, and choosing a metric allows one to pick a natural space of one forms which is orthogonal to the denominator space, called the harmonic forms. this has some computational advantages, analogous to studying complex holomorphic functions by their real and imaginary (harmonic) parts.
HW Helper
P: 2,887
 Quote by mathwonk well we just need a symbol for the duality between vector fiedls and covector fields, i.e.one forms, lets call it #, so # of a one form is a vector field and vice versa. then curl of a vector field V, should be something like #(*d#V), i.e. #V is a oneform, then d#V is a 2 form, then *d#V is a oneform, then #*d#V is a vector field.,
That's indeed what Hurkyl wrote except that he used a small "t" to indicate that: $(\,^* d V^t)^t$
 but this looks too complicated. nonetheless i have seen some pretty complicated expressions in my life.
I hope this is not taking too much time away from your vacations!
 the reason i thought this was simple, is that the only part of this that is purely differential forms, to me, is the d part. duality is always confusing. and using metrics to identify spaces that are dual, makes it harder to tell them apart, and harder to recognize which operations are natural for them.
True. But this is the case in physics from the very start.
You see now why I have asked so many stupid questions in the past (and I am still doing that) as I am trying to "unconfuse" everything that was tangled up together in my physics background!

Regards
 Sci Advisor HW Helper P: 9,453 well you may have interested me in learning the physics of these matters, since i have realized it is also connected to hodge theory, so thank you!

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