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Surface area vector

  1. Dec 18, 2014 #1
    Hello every one .
    What is the Surface Area vector form in exterior algebra ,I mean by that the Surface Area vector as an exterior form in 3D , just like the volume form .

  2. jcsd
  3. Dec 18, 2014 #2


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    Wouldn't that just be a 2-form? A n-form gives a definition of an n-dimensional volume, so an area should be associated with a 2-form.
  4. Dec 18, 2014 #3
    yes its a 2-forms , but 2-form is a co-variant second order tensor , but here the surface area is a vector ,
    this is why i want to know ,
    in mechanics the stress distribution formula is [itex] F^i = \sigma^\ij dA_j [/itex]
    where F is the force vector and (Sigma ) is the mechanical second order stree tensor and A is the Area vector
    while in exterior algebra it's written like this [itex] F^i = T^i_jk dx^j\wedgedx^k [/itex]
    where T is a third order tensor , when using calculus e.g. co-variant derivative , sigma with give 2 christoffel symbols while the T will give 3 christoffel symbols
  5. Dec 18, 2014 #4
    [itex] F^i =\sigma^ij dA_j [/itex]
    [itex] F^i= B^i_j_k dx^j \wedge dx^k [/itex]
  6. Dec 18, 2014 #5


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    The "surface area element" can be thought of as a "[3D-]vector" only in 3-D.
    In the cross-product, the oriented parallelogram formed from the factors is more fundamental than the vector perpendicular to that parallelogram.
    From a tensor algebra viewpoint, to get a vector from the oriented parallelogram,
    one has to use the Hodge-dual (often symbolized by *), which involves the [itex]\epsilon_{ijk}[/itex] symbol.
  7. Dec 18, 2014 #6
    you mean [itex] dA_i = \epsilon_ijk dx^j \wedge dx^k [/itex]
  8. Dec 18, 2014 #7
    [itex] dA^i = \epsilon _{ijk} dx^j \wedge dx^k [/itex]
  9. Dec 20, 2014 #8


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    Yes, it's basically this, but you might have some normalization factors in there, I'm not quite sure.

    EDIT: Oh, and in your formula the i has moved from lower index to upper index, so you have to raise the index in there somewhere. :)
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