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

[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clumsy way of obscuring the quite simple differential calculus as now used in terms of differential forms, and it is so out of date, i have never even seen it!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.

 

[mathwonk]

well we just need a symbol for the duality between vector fiedls and covector fields, i.e.one forms, let's 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.

 

[mathwonk]

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.

 

[nrqed]

well we just need a symbol for the duality between vector fiedls and covector fields, i.e.one forms, let's 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

 

[mathwonk]

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!

 

[nrqed]

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!

You are very welcome. I am glad that this whole exercise did not end up being a complete waste of time for you. And I am glad that it got you interested in physics applications. If you have any thoughts or you realize anything about the physics applications of these concepts, please post them. But you are warned that in conventional physics or engineering books (or even in most undergraduate books of mathematical physics such as Arfken for example), vector calculus is developped without any mention of differenatial forms so it's a mess to sort out. (For example, Arfken obtains the expressions for the curl and divergence in arbitrary coordinates by taking a ratio of integrals over volumes of surfaces in a certain limit.) This brings up a mental block for physicists learning differential forms: almost everything that they seem to be useful form can be derived without their help, or so it seems.

Best regards

 

[nrqed]

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.

Very interesting!

But I am a bit confused by this trick. In other words, the units would give the answer, it would be given by how many factors of "meter" appears in the units.

So a velocity and an acceleration would be vectors. Ok. Butthen, a force would also be a vector. Ok.

Now, using

\vec{F} = q \vec{E} + q \vec{v} \times \vec{B} [/itex]<br /> <br /> this would seem to indicate that the E field is a vector but t hat the B field is a scalar!<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /><br /> <br /> So I am using this trick incorrectly?

 

[Hurkyl]

Well, the first thing is that cross products should always make you suspicious.


While the test I listed does distinguish amongst the things I listed, it can't tell the difference between a scalar and a rank-(1,1) tensor. I was assuming we weren't considering any of those!


If, by B, you mean the "thing that takes a velocity as input and gives you the value of magnetic force", then B really should be an (antisymmetric) rank-(1,1) tensor.

But I usually hear B specified as a pseudovector -- which usually means that you really should be using bivectors. In that case, the cross product becomes the dot product of velocity along one index of B -- in particular, the metric appears. (And the quadrupling of B exactly cancels the one-fourth that gets applied to the metric)

I'm not really sure about the "right" definitions of things if you stick to three-vectors. It's all behaves much more nicely in 3+1 dimensions.

 

[llarsen]

There is a group of electrical engineering professors at BYU that uses differential forms to teach electromagnetic theory. They have a website related to using differential forms in teaching EM theory. You might look around this site. The site does include stuff they hand out to students that give correct derivations of the laws of electromagnetism using differential forms and hodge star duality. (Note that if you do a google search on 'differential forms', this is one of the links on the first page at present.) Here is the page. Hope the materials are as useful to you as they have proved to me:

http://www.ee.byu.edu/forms/forms-home.html

 

[nrqed]

There is a group of electrical engineering professors at BYU that uses differential forms to teach electromagnetic theory. They have a website related to using differential forms in teaching EM theory. You might look around this site. The site does include stuff they hand out to students that give correct derivations of the laws of electromagnetism using differential forms and hodge star duality. (Note that if you do a google search on 'differential forms', this is one of the links on the first page at present.) Here is the page. Hope the materials are as useful to you as they have proved to me:

http://www.ee.byu.edu/forms/forms-home.html

Thank you. Very nice link. I appreciate you posting this.

 

[Chris Hillman]

Lest the answer the question get buried...

(Edit: Sorry all, just noticed that nrqed resurrected a thread from back in June rather than starting a new thread, which might have been a better idea.)

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?

Because the elementary vector calculus results depends upon a coincidence of small dimensions, that two-forms are dual to one-forms, and the second formula expresses that duality.

By the way, I'd urge you to use frames and to lose the \sqrt{g} factors, which aren't helping. For example, the classic book by Flanders, Differential Forms with Applications to the Physical Sciences, Dover reprint, uses frame fields. You'll learn other cool things, like how to do Riemannian geometry with frames and coframes (bases of orthonormal vectors and covectors, resp.).

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.

Right, no duality involved in this one, so curl of function = gradient in all dimensions.

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

The Hodge dual is your friend

Here's another question to make you think: what can you say about the variety of closed one-forms? (A form \sigma is closed if d \sigma=0).