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

[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).

 

[Chris Hillman]

Recommend some reading

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!

Glad you are interested in learning how physics employs exterior calculus! One good books for mathematicians who wish to learn a bit about physics might be:

Frankel, Geometry of Physics, Cambridge University Press. (Just skim the mathematical exposition; there is plenty of exposition of physics to be found.)

Going in the other direction:

Isham, Modern Differential Geometry for Physicists, World Scientific, 2005. (Covers basic manifold theory and exterior calculus too.)

 

[bhale]

The two vector expressions for the curl (with contravariant and covariant basis vectors, respectively, inserted into the sums) are not equal. Only the second form (with covariant basis vectors and contravariant components) can be derived rigorously from the curl, a vector defined in the Cartesian frame. (Note, you must equate vectors, not components.)

You cannot transform the second (correct) form into the first form because elements of the metric tensor do not commute with the differential operator. If the generalized frame metric tensor components are all constants the two forms give the same result for the curl.

 

[timur]

I did not read the pages 2-4, but think about what happens when there is no metric: if you can define something without using a metric, it is more fundamental than the dual.

 

[mmcf]

Sorry about this, but none of these formulas seem to work out for me so
could someone tell me where I'm going wrong? I
guess starting with the simplest one, the proposed form for the gradient of
a scalar field in terms of the exterior derivative is:

(d\phi)^t

where the transpose of a covariant vector is obtained by applying the
contravariant metric tensor. I'm going to write a one-form as df and a directional derivative along a coordinate x as \partial x. I'll only write \partial_x
when I'm taking a derivative of a function rather than talking about a vector field.
In a euclidean space the contravariant metric tensor can be expressed in cartesian
coordinates as:

g = \partial x \otimes \partial x + \partial y \otimes \partial y +<br /> \partial z \otimes \partial z

Let's pretend I'm interested in cylindrical coordinates
(r,\theta,z) because they require the least typing:

<br /> x = rcos(\theta)<br />
<br /> y = rsin(\theta)<br />
<br /> z = z<br />

I can take some derivatives to obtain:
<br /> \partial x = (x/r)\partial r + (-y/r^2)\partial \theta = cos(\theta)\partial<br /> r - (sin(\theta)/r) \partial \theta<br />
<br /> \partial y = (y/r) \partial r + (x/r^2) \partial \theta =<br /> sin(\theta)\partial r + (cos(\theta)/r) \partial \theta<br />

I then will get by substitution that:
<br /> \partial x \otimes \partial x = cos^2(\theta) \partial r \otimes \partial r<br /> + (sin^2(\theta)/r^2) \partial \theta \otimes \partial \theta +<br /> (-cos(\theta)sin(\theta)/r) \partial \theta \otimes \partial r +<br /> (-cos(\theta)sin(\theta)/r) \partial r \otimes \partial \theta<br />
<br /> \partial y \otimes \partial y = sin^2(\theta) \partial r \otimes \partial r<br /> + (cos^2(\theta)/r^2) \partial \theta \otimes \partial \theta +<br /> (cos(\theta)sin(\theta)/r) \partial \theta \otimes \partial r +<br /> (cos(\theta)sin(\theta)/r) \partial r \otimes \partial \theta<br />

Substituting those into the metric tensor in cartesian coordinates I get:
<br /> g = \partial r \otimes \partial r + (1/r^2) \partial \theta \otimes \partial \theta<br /> + \partial z \otimes \partial z<br />

I'm pretty sure that's the standard result you should get because it's the inverse of the covariant metric guy.

So to obtain the cylindrical coordinates gradient, I start by taking the
0-form \phi and applying an exterior derivative:
<br /> d\phi = (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz<br />

He's a one-form, and I'm interested in a vector, so I apply the metric tensor:
<br /> (d\phi)^t = \partial r &lt;\partial r,(\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz&gt; + (1/r^2) \partial \theta &lt;\partial \theta, (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz&gt; + \partial z &lt;\partial z, (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz&gt;<br />
<br /> = (\partial_r \phi)\partial r + ((\partial_\theta \phi)/r^2) \partial \theta<br /> + (\partial_z \phi) \partial z<br />
That's sort of messy but those &lt;x,y&gt; are supposed to be a natural pairing as Bishop and Goldberg say it.

If I'm supposed to assume that the classical gradient is this where you substitute
<br /> \partial r \rightarrow \hat{r}<br />
<br /> \partial \theta \rightarrow \hat{\theta}<br />
<br /> \partial z \rightarrow \hat{z}<br />

then this is off by a factor of 1/r along the \theta direction from the normal definition:
<br /> (\partial_r \phi)\hat{r} + ((\partial_\theta \phi)/r) \hat{\theta}<br /> + (\partial_z \phi) \hat{z}<br />

Any thoughts on where I went wrong?

 

[llarsen]

If I'm supposed to assume that the classical gradient is this where you substitute
<br /> \partial r \rightarrow \hat{r}<br />
<br /> \partial \theta \rightarrow \hat{\theta}<br />
<br /> \partial z \rightarrow \hat{z}<br />

then this is off by a factor of 1/r along the \theta direction from the normal definition:
<br /> (\partial_r \phi)\hat{r} + ((\partial_\theta \phi)/r) \hat{\theta}<br /> + (\partial_z \phi) \hat{z}<br />

Any thoughts on where I went wrong?

In your derivation above, you used coordinate vectors. This is very natural when you are using the modern differential geometry approach. However, the equation for the gradient is often presented using unit vectors rather than coordinate vectors. I think this is the source of your problem. It looks like \hat{\theta} represents a unit vector, whereas \partial \theta is a coordinate vector. The correct substitution is:

<br /> \partial \theta = r \hat \theta<br />

I believe that the overall derivation was correct. It seems to just be the last step substituting coordinate vectors with unit vectors that caused problem. The unit vector substitutions for r and z were OK since unit vectors and coordinate vectors are the same length for r and z in cylindrical coordinates.

 

[mmcf]

Alright, thanks. That seems to be my whole problem. It looks like I get the right expression for div A from
<br /> *d(*A^t)<br />
and curl A from
<br /> (*dA^t)^t<br />
just like it says in the thread as long as I consider my components to be in the form of the A^i in
<br /> \Sigma_i \frac{A^i \partial_i}{g(\partial_i,\partial_i)^{1/2}}<br />