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Curl in arbitrary coordinates

  1. Jun 27, 2007 #1

    nrqed

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

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

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

    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 [itex] d \phi = \partial_i \phi [/itex] 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 [itex]x^i [/itex] 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
     
    Last edited: Jun 27, 2007
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  3. Jun 28, 2007 #2

    mathwonk

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    learn differential forms.
     
  4. Jun 28, 2007 #3

    HallsofIvy

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    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.
     
  5. Jun 28, 2007 #4

    nrqed

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    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"!
     
  6. Jun 28, 2007 #5

    nrqed

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    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 [itex] d \phi [/itex] with components [itex] \partial_i \phi [/itex], 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 [itex] g^{ji} \partial_i \phi [/itex] but this does not give the components of the gradient, let's say in spherical coordinates, that we learn in elementary calculus.
     
  7. Jun 28, 2007 #6

    mathwonk

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    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.
     
  8. Jun 28, 2007 #7

    mathwonk

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    if i may make another suggestion, when asking for help, a comment like "in fact thats what Im trying to do" is more productive than "if you weren't so stupid youd realize that was what im 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.
     
  9. Jun 28, 2007 #8

    nrqed

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    I apologize sincerely.
    Since I started my post by mentioning the intrinsic expression [itex] \,^* dA[/itex] 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
     
  10. Jun 28, 2007 #9

    nrqed

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    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 [itex] \,^* dA [/itex] (which is obviously a one-form if I am in 3 dimensions) and not just [itex] dA [/itex]. 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.
     
  11. Jun 28, 2007 #10

    mathwonk

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    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.
     
  12. Jun 28, 2007 #11

    mathwonk

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    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 fucntion 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.
     
    Last edited: Jun 28, 2007
  13. Jun 28, 2007 #12

    mathwonk

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    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 lets 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 lets make a stab at it.
     
  14. Jun 28, 2007 #13

    mathwonk

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    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 wont let me complete this mess, if you believe that. but i assure you it works really, if done right....
     
    Last edited: Jun 28, 2007
  15. Jun 28, 2007 #14

    nrqed

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    :smile: 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 [itex] 1/(r sin \theta) [/itex] which looks like [itex] 1/\sqrt{g} [/itex]. In fact, the expression [itex] \sqrt{g} \epsilon_{ijk} \partial^j A^k [/itex] 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
     
  16. Jun 29, 2007 #15

    mathwonk

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    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.
     
  17. Jun 29, 2007 #16

    mathwonk

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    ok lets 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 lets 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 lets 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.
     
  18. Jun 29, 2007 #17

    mathwonk

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    of all the frustrating,.... i finished the calculation and the biorwser wont 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.
     
  19. Jun 29, 2007 #18

    mathwonk

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    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.)
     
  20. Jun 29, 2007 #19

    mathwonk

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    man this is frustrating.
     
  21. Jun 29, 2007 #20

    mathwonk

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    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,
     
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