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Qestion: Vector field and (n-1)-form representation of current density

  1. Jan 4, 2006 #1

    mma

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    Question: Vector field and (n-1)-form representation of current density

    Electric current density can be represented by both a vector field and by a 2-form. Integrating them on a given surface must lead the same result. My question is, what is the relation between this vector field and the 2-form. More generally, in an n-dimensional space, between a vector field and an (n-1)-form, which results the same value when integrated on a (n-1) dimensional surface. Is the latter the Hodge-dual of the vector field regarded as an 1-form ? If yes, how can one see this?
     
    Last edited: Jan 4, 2006
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  3. Jan 4, 2006 #2

    robphy

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    In n-dim, Hodge-duality relates a k-form with an (n-k)-multivector.
    In 3-D, [tex]j^a=\epsilon^{abc}J_{bc}[/tex].
    You may enjoy the URLs in these earlier threads:
    https://www.physicsforums.com/showthread.php?t=92472
    https://www.physicsforums.com/showthread.php?t=89419
    https://www.physicsforums.com/showthread.php?t=18963
    http://www.ee.byu.edu/forms/
     
  4. Jan 4, 2006 #3

    mma

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    Dear Robphy, thank you very much.
    Are you sure that I will find the answer for this question in these references? And before I read all you showed, could you tell me the answer? Yes, or no? Maybe it's trivial for you but not for me, in spite that I know the definition of the Hodge dual.
    mma
     
  5. Jan 4, 2006 #4

    robphy

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    The tensorial relation I gave shows how the current-vector is associated with a current 2-form, in 3-D space. I literally took the Hodge-dual.

    As I said, "In n-dim, Hodge-duality relates a k-form with an (n-k)-multivector. " So, the "Hodge-dual of the vector field regarded as an 1-form" is incorrect... it is an (n-1)-form. I hope I understood your question.

    In the old threads I listed,
    the links to Burke's pages describes electromagnetism using differential forms... in fact, in a rather unique pictorial way... a sort of visual tensor algebra.
    the links to Bossavit's Applied differential geometry discusses forms in more detail than Burke (Bossavit's goal is electromagnetism).
    the BYU link describes a course in electrodynamics using differential forms.

    The subject is not trivial [to me]... I still struggle with developing my intuition and understanding of them. I have offered a set of references that have helped me get to where I am now.
     
  6. Jan 4, 2006 #5

    mma

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    Perhaps my English is too poor, sorry. I wanted to say that vector fields do not have Hodge dual, because only forms have it, so we have to regard this vector field as an 1-form (really, this 1-form is the dual of the vector field, but I wanted to avoid the usage of word "dual" in one sentence with two different meaning). The Hodge-dual of this 1-form is of course an (n-1)-form.
    Thank you again the references, I will study them.
     
  7. Jan 4, 2006 #6

    robphy

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    In my understanding, Hodge duality operates on totally-antisymmetric tensors, which include vectors va and 1-forms wa. Based on http://en.wikipedia.org/wiki/Hodge_star_operator , a metric is needed for the Hodge-dual to relate a k-form to a (n-k)-form... or a k-multivector to a (n-k)-multivector. [To distinguish the two duals, you could say metric-dual and Hodge-dual.]

    Your restriction may depend on your starting definitions and specification of the available structures [e.g. volume form, metric, etc...], which you may wish to present here.
     
  8. Jan 4, 2006 #7

    mma

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    Of course, va is an 1-form, but it acts on the dual space. That's why we must use for our purpose the metric-dual va of it (as long as I am right that the 2-form representing the current density is really a 2-form, and not a 2-multivector) . Certainly, this duality requires the same metric as the Hodge-dual requires.
     
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