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Differential Forms?

  1. Feb 25, 2007 #1
    Hi, I don't know if this is the right place to post, but can someone help me understand what differential forms are intuitively? And the wedge product intuitively? And finally, how can they help see the bigger picture of multiple integrals, curls, divergence, gradient, etc. I don't know that much, but we're learning it in our multivariable calculus class. Thanks.
     
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  3. Feb 25, 2007 #2

    mathwonk

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    a differential form on a single vector space V is a function of k variables which is alternating and linear in each variable separately. If k = the dimension of the vector space, then essentially the only such animal is the determinant. Thus they are as uintuitive as determinants, no more no less. The intuition for a determinant to me is as a measure of oriented volume.


    If k is less than the dimension of V, and the dimension is n > k, then there are exactly the binomial coefficient "n choose k" linearly independent such functions, formed as follows. Take k vectors from V and place them in order in a k by n matrix. Then choose k of the n columns, and take the determinant of the resulting square k by k subdeterminant.

    There are exactly "n choose k" ways to do this. Moreover one can form linear combinations of these basic functions and get thus a vector space of such functions of dimension "n choose k". Thus there is an "n choose k" dimensional vector space of such k dimensional oriented volume measures on an n dimensional vector space, dependeing on which set of k axes one projects ones k dimensional parallepiped onto before taking the determinant.

    Of course in the case of length, there is a formula known as pythgoras theorem that lest one calculate the usual length of a segment from the lengths of its projections noto the avrious axes, and this generalizes to the case of area of parallelograms in 3 space. So perhaps there is also a way of compouting k dimensional volume of a k block from the various projections onto different choices of sets of k axes in n space.

    Anyway, differential k forms on n space are a computations way of emasuring orienteds violume of k blocks in that space, generalizing determinants, and in fact are precisely computed as linear combinations of k fold subdeterminants of k by n matrices.


    now if we have a family of n dimensional vector spaces, say VxU where U is an open set in R^n, then we can have a linear combination of k forms with varying coefficients. i.e. the "n choose k" coefficients expressing our k form as a linear combination of the basic subdeterminants, can vary over the open set U, in a continuous or smooth way.

    This is called a differential k form on the tangent bundle VxU, of the opens et U. I.e. a smooth k form on a manifold, is locally a linear combination of k diomensional subdeterminants, with smoothly varying coefficients.


    so it is a way of measuring oriented k dimensional volume in the smoothly varying family of tangent spaces to the manifold.

    hows them apples?:grumpy:
     
  4. Feb 25, 2007 #3

    mathwonk

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    consequently, if we have a parametrized k dimensional surface in ouir manifold, by looking at the images in each tangent space, of the unit k block under the derivative of our parameter map, and evaluating our oriented volume measure on each image block, we get a numerical function back on the unit block which we can integrate. I.e. we can integrate k forms over parametrized k diml surfaces.

    i.e. you approximate a parametrized k surface at each point by a parametrized k block, and your k form measures its size. this approximates the stretching factor of your parameter map on the unit block, i.e. the size of the image surface.

    thus since determinants are the basic way of measuring volumes, differential forms are the basic way of expressing integrals over parametrized surfaces. hence they are the natural language for formulating theiorems about integrals, such as stokes, greens, gauss, divergence, etc...

    as to why those theorems are true, that relies on the FTC and fubinis theorem, i.e.repeated integration, which reduces them all to the fundamental theorem of calculus.:tongue:
     
    Last edited: Feb 25, 2007
  5. Feb 25, 2007 #4

    mathwonk

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    wedge product is just alternating multiplication, i.e. you want something that acts like multiplication in each variable separately, but that changes sign when you change the order of x and y, so instead of the dot product x1y1 + x2y2, you take the wedge product x1y2-y1x2. then when you change order you get y1x2-x1y2.

    this formula should look familiar from 2 by 2 determinants. this lets you make higher order determinants out of lower order ones, i.e. lets you get LaGranges inductive formula for expanding determinants.
     
    Last edited: Feb 25, 2007
  6. Feb 25, 2007 #5

    mathwonk

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    thus intuitively a differential k form is very simple. in each vector space it is an animal that waits for a k block and gobbles it up and spits out a number.

    and it behaves like a volume, i.e. if the k block has one side stretched by the factor c, then the number spit out is multiplied by c.

    and it is oriented, so if we interchange the ordering of two sides of the k block, the number spit out chjanges sign. thats all.

    but because we want to express them cimoputationally in terms of a bsis of them, it becoems operationally complicated. i.e. already the formula for the area of a parallelogram in terms of the 4 heights and bases is complicated ad-bc.

    the area of a parallelogram in space in terms of its porojections on the three pairs of axes is complicated, and the volume of an n block in terms of the various coordinates fo the spanning vectors is a determinant, hence coimp[licated.

    so k forms are intuitively easy, just oriented volume measures, but operationally complex. since the simplest ones are determinants, they cannot be made simpler than determinants.

    david bachmans little book on them studied here some time back is highly recommended for an intuitive grasp of them. the little technical flaws I argued about in the presentation at the time can safely be ignored.
     
  7. Feb 26, 2007 #6
    I'm been trying to answer this question for little while now.

    For what its worth, at least so far, I don't see much value in JUST differential forms/wedge products. As Mathwonk points out, it is basically just different notation for the volume invariant change of coordinates formula (i.e. the determinant of the jacobian).

    So it doesn't really give you much additional power compared to the usual vector calculus scale factors.

    The true power of comes when you add in all the other machinery that calculus on manifolds and differential topology brings to the table.

    I'm working thru:
    Bamberg, Sternberg: A course in mathemathics for students of physics
    Arnold: Mathematical methods of classical mechanics
    Lee: Introduction to smooth manifolds

    I was always unsatisfied with the apparently ad hoc manner of scale factors in usual vector calculus, and I wanted a better understanding.

    These 3 books have greatly increased my understanding. I still have a long way to go.

    brian
     
    Last edited: Feb 26, 2007
  8. Feb 26, 2007 #7
    Differwntial forms have no intuative interpretation. Seriously. Forms are operators on vectors. They exist in the dual space to vectors so their geometry is highly non-intuative. Anyway, here's something that may help.

    You may have seen the visual interpreatation of a one-form as a set of parallel gradient lines. The visual interpretation of a two form is two overlapping set of such lines creating a gird of parallelograms. Now, here's where forms get messy.

    Take a two form in two diemtsions. This form takes two vectors and then, careful now, returns the area of the parallelogram spanned by those vectors divided by the area of the parallelograms formed by the overlapping one form lines.

    Now, that second part about dividing by the area of the parallelogram may seem a little odd. That's because it is It's a big problem with forms, because what a form is giving you is not the area of the parallelogram spanned by two vectors, but the area of the parallelogram spanned by the vectors in the pullback space. Of course, areas in the pullback space aren't much use to you so nine times out of ten, the only two forms anyone talks about are dx^dy, i.e. the canonical "area" form in 2D, with the area of the parrallelograms spanned by the two form lines of dx and dy, equal to one.

    The situation gets even worse when you move to differentiation, with the omnipresent exterior derivative overshadowing all.

    dw. What is it? No idea. It's usually attempted to be passed off as the curl in 2d or divergence in 3d, but this is simply wrong. The curl is an operator on one vector field to produce another, and the divergence is an operator on a vector field to produce a scalar field. The exterior derivative of a one form produced a two form, and of a two form produces a three form. Of course, the coefficients of these produced two and three forms happen to coincide with the coeffients for the curl and divergence, but this isn't really very helpful.

    And the classic argument for forms is that they compress Maxwell's equations from four to three.

    dF=0
    d*F=J
    dJ=0

    Which means......? I have no idea. dw meant little to begin with and d*F=J means even less. At least in the vector form of Maxwells equations you could see what was going on. Differential forms tends to strip all semantic meaning from familiar equations.

    I've recently discovered Geometric Algebra. I'm not going to argue that it's some kind of miracle cure, as it has some serious, often glaring issues. But frankly, it's better. In Geometric Algebra notation, you set [tex]F=\mathbf{E}+i\mathbf{B}[/tex] and pretty much immediately get
    [tex]
    (\frac{1}{c}\partial_t +\nabla)F = \rho -\frac{1}{c}\mathbf{J}
    [/tex]
    Yeah. That's right. Maxwell's equations become a single line, first order wave equation. Semantic or what. I was pretty much sold on GA at this point, but YMMV.

    So basically, IMHO, forms work, but not very well. There are alternatives out there, and if forms are giving you pains in your back teeth, I would suggest looking them up.
     
  9. Feb 26, 2007 #8
    Perhaps I'm missing something subtle or just there's a matter of less than rigorous notation, but isn't that equation wrong on the grounds that the LHS is a matrix and the RHS is a vector, if that! Even [tex] \rho -\frac{1}{c}\mathbf{J}[/tex] seems to be questionable, you're adding a scalar and a 3-vector. Even if you meant to say [tex](\rho , -\frac{1}{c}\mathbf{J})[/tex] you're still mixing matrices and vectors.

    Or am I missing something?

    BTW, Mathwonk, excellent posts :)
     
  10. Feb 26, 2007 #9
    Elements in Geometric algebra exist in a 2^N dimensional "multivector" space, with the usual N dimensional vectors as a subset of that space. For example in 2d, you have a 4 dimensional extended space with 1, i, j and i^j as basis "vectors". As you can see, we have i and j as the usual vector basis, but now we append scalars and directed area elements, or "bivectors" (i^j) to them to form a more complete space.

    In this fashion, expressions like 7+4i-6j+3i^j are valid if one simply considers them as a "multivector" which componentwise would be (7,4,-6,3). Addition and subtraction work as expected. Multiplication however does not, with multiplication by another multivector potentially swapping elements about in a strange, but well defined way.

    Technically in GA, if I remember correctly, E is a regular vector as is J. Charge (rho) is a scalar and B is a "bivector", or directed area segement. This is analagous to the expression in differential forms notation of E as a one form, and B as a two form. In both cases the fact that E has units of V/m and B has units of Wb/m^2 is expressed clearly.

    GA takes a bit of getting used to, but you do get used to it. Something I can't readily say about differential forms.
     
  11. Feb 26, 2007 #10
    Is that not just another way of talking about [tex]\Omega^{\ast} = \Omega^{0} \oplus \Omega^{1} \oplus \Omega^{2}[/tex] ? Or is that kind of the point, it's an analogous way of describing the space of p-forms but without some of the complications?

    I'm only just getting into p-forms myself and I agree it can be somewhat perplexing at times. I didn't get the whole "It gives div, grad, curl and laplacian without coordinates" thing for quite a while but I don't agree that it's not geometrically intuitive. Once things begin to click into place, I think it does give a very useful way of describing geometric properties.

    Nakahara - Geometry, Topology and Physics gets onto differential forms about 1/3 the way through and it builds up to them by moving through a number of geometric systems, putting them in context. An excellent book for more mathematical physicists.
     
  12. Feb 26, 2007 #11
    No. Multivectors exists in the tangent space, by themselves. Forms exists in the dual space and you need to apply vectors to them to evaluate them to a number.

    But it doesn't give grad, curl and div. It just gives something that kinda sorta maybe but not really looks like grad, curl and div. dw isn't the curl. It's a two form. You have to give it vectors and integrate for it to have any solid meaning. As to geometric meaning, I've already discussed the major problem with forms, that is, the only R^3 geometry involved requires you to use the canonical volume forms. Which is a pretty big qualifier.

    I highly reccommend the following as an introduction to Geometric Algebra. Enjoy.
     
  13. Feb 26, 2007 #12
    If you're interested the Lee text I mention above gives a mapping from forms to the usual vector grad, curl, div.

    I.e. inverse(g)(df) where g is the metric maps the differential 1 form df to the usual gradient vector.

    Curl and div mappings are slightly more involved.
     
  14. Feb 26, 2007 #13

    Hurkyl

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    It is terribly arrogant of you to think that your intuition is maximal amongst human beings. Your inability (or, more likely, refusal) to intuit forms does not imply that nobody else can intuit them.



    Let's think physics in 3 dimensions for the moment. Would you agree that mass density is a pretty intuitive concept? Well, mass density is a differential 3-form. To wit, mass density is something you integrate over a volume to obtain the mass, and is rescaled if you make a coordinate transformation.

    Of course, if really didn't want to use differential forms (or didn't know they existed, so scalars and vectors are your only tools), you could instead work with the hodge dual. That is, instead of working with the mass-density 3-form G, you could work with the mass-density scalar p, which is defined so that G = p dV. (where dV is the canonical volume measure)

    Or worse, you could do what happened historically, and treat mass-density as a pseudoscalar, so as to maximize the technical awkwardness of the subject.
    In particular, this happens because that's the only way to get the notion of a pseudoscalar (or pseudovector) to work out properly -- if you strip dV off of the mass-density 3-form to turn it into a pseudoscalar, you're going to have to reattach it if you ever want to integrate anything.


    Now, let's move up to 4 dimensions. Here, we have the current density 3-form J: it's the thing you integrate over a 3-volume to obtain the charge lying on that 3-volume. (Compare with how awkward it is to define such a thing without using higher differential forms)

    Charge can be neither created nor destroyed: so the charge going into any 4-volume must equal the charge leaving the 4-volume. In other words:

    [tex]\int_{\delta R} J = 0[/tex]

    which, by Stoke's theorem, immediately gives us the charge continuity equation dJ = 0.

    Note that, if you were pretending current density is a vector, you would be using the (4-analogue of) divergence here.

    And once we have dJ = 0, that again immediately tells us that there is a 2-form M such that J = dM. With barely any thought at all, we're led directly to the (hodge dual of) electromagnetic field strength tensor!
     
  15. Feb 27, 2007 #14
    But if multivectors exist in the tangent space, do you not need to apply dual forms to them to get them to a number? A 0-form is a number or function, just like your '1', then I've got a 1-form and you've got i and j.

    It would seem to me you're algebra is very much like p-forms, the same 'total space', similar formulism.
    It does give the grad, curl and div. It depends what form w is. If w is a 0-form, then dw is grad. If w is a 1-form, then dw is curl. If w is a 2-form then dw is div.

    I was confused about it too, but once it falls into place it's very elegant.
    You can look at the coefficents of the various basis terms of your p-form.

    For instance, suppose [tex]\omega = \omega_{x}dx + \omega_{y} dy + \omega_{z} dz[/tex]. Here w is a 1-form.
    [tex]d\omega = \left( \frac{\partial \omega_{y}}{\partial x}- \frac{\partial \omega_{x}}{\partial y} \right) dx \wedge dy + \left( \frac{\partial \omega_{z}}{\partial y}- \frac{\partial \omega_{y}}{\partial z} \right) dy \wedge dz+ \left( \frac{\partial \omega_{x}}{\partial z}- \frac{\partial \omega_{z}}{\partial x} \right) dz \wedge dx[/tex]

    Immediately, you've got the coefficents for [tex]\nabla \times \omega[/tex].
    Are you sure? I've never seen such a requirement in a book? If anything, you can have spaces where the volume form doesn't exist due to lack of orientability.
     
  16. Feb 27, 2007 #15
    Actually, the existence of various, different 1-forms possible in 3 dimensions is what gives us three of the 3-dimensional geometries from Thurston's Geometrization Theorem (at least, according to the experts, it's a theorem now).

    This dovetails nicely into another interpretation of forms:

    Suppose in R^3 we have a smooth plane field of vectors. That is, at each point of space, we have a well-defined 2-dimensional subspace in the tangent space at that point. Smoothness essentially means that points nearby to each other will have plane fields that can smoothly transformed to each other.

    At each point x, the plane field can be defined at the zero set of some linear functional [itex]\omega: T_x R^3 \rightarrow R[/itex]. Thus, we have a 1-form [itex]\omega[/itex] defined on all of R^3.

    Properties of the plane field as a whole can be found by investigating the properties of this 1-form. For example, if it is the case that [itex]\omega \wedge d\omega=0[/itex] at all points on R^3, then the plane field is integrable, that is, at each point, one can embed a surface in R^3 that is tangent to the plane field at all of its points.

    If [itex]\omega \wedge d\omega\neq 0[/itex] at all points in R^3, then the plane field is maximally non-integrable. Which is to say, that there is no surface in R^3 that is tangent to the plane field everywhere -- the points on the surface where it is completely tangent to the plane field has measure zero. This is called a contact structure and has popped up repeatedly in both mathematics and physics (e.g. quantum physics and thermodynamics).

    Page 3 of this paper (http://www.math.gatech.edu/~etnyre/preprints/papers/legsur.pdf) shows a number of examples of these.

    Although clearly not appropriate for a high school level geometry course (or most undergrad courses for that matter), one finds that many geometrical structures lend themselves easily to a differential form convention.
     
    Last edited: Feb 27, 2007
  17. Feb 28, 2007 #16

    mathwonk

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    how silly of me, i thought i made this crystal clear to absolutely anyone.
     
  18. Mar 1, 2007 #17
    Face it, MW, this is The Question That Will Never Go Away. Every discipline has one (or several).

    In art history, it's "What does abstract art actually mean?"

    In physics, it's "So, what's a quick way to summarize general relativity?"

    Etc. etc.

    Sometimes this is asked in good faith. But I suspect that most of the time the inquirer is hoping to reveal that it's all just a complicated shell game invented by a group of pinheads who want to appear smart or to get rich or to be famous...

    What irks me, though, is that I personally spent 6 years of graduate school dedicated to learning about this subject, and yet some people seem to think that a few hours on the internet getting unpaid advice from complete strangers should be enough to understand the topic. And then get dismissive of it when they still "don't get it."
     
  19. Mar 1, 2007 #18

    mathwonk

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    yup, yup and (to satisfy length constraints) yup.
     
  20. May 4, 2009 #19
    Actually, you gave a very nice, some might say "intuitive" (at least "pictorial") description in the clipped material.

    If you'd like to see some interesting pictorial depictions of forms, you can have a look at Misner, Thorne & Wheeler's Gravitation. For example, pages 57, 100, 104, 106, 107, 109, 116 & 117.
     
  21. May 15, 2009 #20
    Also read Baez book "Gauge fields, knots and gravity". He gives elementary intuitive introductions to forms (and other topics).
     
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