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Generalizing: Curl of a Function

  1. Nov 29, 2009 #1
    Hello. I have tried to search all day for the answer to my question but could not find anything.

    Is there a way to generalize the curl of a vector field to dimensions greater than three? It seems pretty straightforward to find the gradient and divergence of higher dimensional vector fields, but my calculus book only offers the definition: curl F := ∇×F, which appears to only work if F has three components. Or am getting something wrong?

    I am also only in my first real analysis course, so I have not come across this kind of topic in that class yet.
  2. jcsd
  3. Nov 29, 2009 #2


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    Not easily. The problem is that the best way to generalize the cross product itself is to use the "alternating tensor". That is the operator represented by [itex]\epsilon_{i_1i_2\cdot\cdot\cdot i_n}[/itex] equal to 1 if [itex]i_1i_2\cdot\cdot\cdot i_n[/itex] is an even permutation of [itex]1 2 3 \cdot\cdot\cdot\ n[/itex], -1 if it s an odd permutation, and 0 if neither of those is true.
    For example, when n= 3, [itex]\epsilon_{ijk}[/itex] has 27 entries. [itex]\epsilon_{123}= \epsilon_{231}= \epsilon_{312}= 1[/itex], [itex]\epsilon_{132}= \epsilon_{213}= \epsilon_{321}= -1[/itex], and the other 21 entries are all 0.

    With that [itex]w_i= \sum_{j=1}^3\sum_{k=1}^3\epsilon_{ijk}u_jv_k[/itex] gives [itex]w_1= \epsilon_{123}u_{2}v_{3}+ \epsilon_{132}u_3v_2= u_2v_3- u_3v_2[/itex]
    [itex]w_2= \epsilon_{213}u_1v_3+ \epsilon_{231}u_3v_2= -u_2v_3+ u_3v_2[/itex]
    [itex]w_3= \epsilon_{312}u_1v_2+ \epsilon_{321}u_2v_1= u_1v_2- u_2v_3[/itex]
    precisely the cross product.

    But, in n dimensions, [itex]\epsilon[/itex] has n indices. In order to get a vector result, we would have to multiply n-1 vectors, not just two.
  4. Nov 29, 2009 #3
    In a general space whose dimension is not necessarily three, the generalization of curl is a Hodge dual of an exterior derivative of the vector field. The vector field is a 1-form, its exterior derivative is a 2-form, and its Hodge dual is a (n-2)-form. Which is a scalar (a number) for n=2, a vector for n=3, a higher-rank tensor for n>=4.
  5. Nov 30, 2009 #4
    Thank you for the help! Looks like I have a few things to learn about the topic.
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