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Vector cross product

  1. Nov 18, 2007 #1
    Is there such a thing as a cross product for R4 vectors? Can you use the permutation symbol to express it in the same way that it can be expressed in R3?
    Would the correct way to write it be: e _{i,j,k,l} u _{j} v _{k}?
  2. jcsd
  3. Nov 18, 2007 #2

    Ben Niehoff

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    There is such a thing, but in R4 the cross product will involve three vectors u, v, and w given by the determinant

    [tex]\left| \begin{array}{cccc} \hat \imath & \hat \jmath & \hat k & \hat \ell \\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{array} \right|[/tex]

    The resultant vector (call it z) has a direction perpendicular to each of u, v, w, and has a magnitude equal to the signed volume of the parallelopiped spanned by u, v, w.

    In terms of the alternating symbol, you would have

    [tex]z_i = \epsilon_{ijkl}u_jv_kw_l[/tex]

    Look up "exterior algebra" for more fun related to concepts like this in higher dimensions. :D
    Last edited: Nov 18, 2007
  4. Nov 18, 2007 #3
    Your construction gives an order two tensor instead of a vector.
    Recall that the scalar triple product gives the signed volume of the parallelepiped spanned by 3 vectors in R3. This is the inner product of uxv, denoting the orientation and area of the base, with the third vector w. In Rn, we can construct a "cross product" that generalizes this property of n-volume, but now the product is only defined for n-1 vectors specifying the (n-1)-dimensional "base", not just 2 vectors as in R3.
    An equivalent method of generalizing comes from generalizing the Hodge star operator's form that in R3, the cross product is just [itex]\ast(u\vee v)[/itex] where * is the operator and [itex]\vee[/itex] is the exterior product.
    Last edited: Nov 18, 2007
  5. Nov 19, 2007 #4
    Thank you, one and all. You are most kind and informative. I'm studying Tensor notation presently and was fiddling with it. I was aware off Scalar & Cross products from undergraduate studies but never thought about them in higher dimensions - until now. That Hodge star and wedge product thingy is "out there" for me at this time. Can anyone suggest a website I can use that has a glossary and definition of such mathematical formalism. I know it's a little like explaining something in French to an Englishman - or something in English to an American for that matter - hold off on the brickbat yooose guyz, I'm from Brooklyn, NY - or something in English to a guy from Brooklyn, but just the same I'd like something definitive to refer to. Thanks.
    John Bowers
  6. Nov 22, 2007 #5

    Chris Hillman

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    For a clear explanation of wedge product and the Hodge star, try the classic book by Flanders, Differential Forms and their Physical Applications, Dover reprint. For a physics-oriented introduction to tensors, try Frankel, Geometry of Physics.

    There is no cross product in the sense of an alternating binary product of two four-vectors; this is addressed in many textbooks and also an expository paper by (IIRC) Massey which appeared long ago in (IIRC) the A. M. Monthly. The exterior product has the properties jbowers desires, but as already noted, in higher than three dimensions, bivectors are no longer in natural correspondence with vectors via the Hodge star, which is the special feature of three dimensions which makes the usual cross product work.
    Last edited: Nov 22, 2007
  7. Jan 13, 2010 #6
    It is a well known fact that vector product of two vectors exists only in 1,3 and 7 dimensions. There is some relation between 7-dimensional vector product and electronmagnetics as studied by Feynmen. One can search for the calssical paper of Alfred Gray or other sources. :smile:
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