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Calculating Curl With Index Notation

  1. Sep 30, 2008 #1
    Hi, does anyone know a link showing how to calculate curl with a Levi-Civita tensor. I can't figure it out but I am sure if I could see an actual example would be able to work out what is going on.
  2. jcsd
  3. Sep 30, 2008 #2


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    Isn't the curl of some vector, A, say [itex](\nabla \times \vec{A})_i[/tex] just [itex]\epsilon_{ijk}\partial_j A_k[/itex] ?
  4. Sep 30, 2008 #3
    It is, but I can't find out how to use it.
  5. Sep 30, 2008 #4


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    Have you tried just writing it out?

    If Ak= <f(x,y,z), g(x,y,z),h(x,y,z)> where x= x1, y= x2, z= x3[/sub], then [itex]\epsilon_{ijk}\partial_j A_k[/itex] is:

    We can simplify some of the "writing out" by noting that [itex]\epsilon_{ijk}= 0[/itex] if any of i, j, k are the same, [itex]\epsilon_{123}= \epsilon_{231}= \epsilon{312}= 1[/itex] and [itex]\epsilon_{132}= \epsilon{213}= \epsilon{321}= -1[/itex].

    So for B= curl A, we have
    [tex]B_x= B_1= \epsilon_{123}\partial_2\ A_3+ \epsilon{132}\partial_3 A_2[/tex]
    [tex]= \partial_2 A_3- \partial_3 A_2= \frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}[/tex]

    [tex]B_2= \epsilon_{213}\partial_1\ A_3+ \epsilon{231}\partial_3 A_1[/tex]
    [tex]= -\partial_1 A_3+ \partial_3 A_1= \frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}[/tex]

    [tex]B_3= \epsilon_{312}\partial_1\ A_2+ \epsilon{321}\partial_2 A_1[/tex]
    [tex]= \partial_1 A_2- \partial_2 A_1= \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}[/tex]
    which are, of course, the usual formulas for curl A.
  6. Sep 30, 2008 #5


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    So you have

    So to get the x component of the curl, for example, plug in x for k, and then there is an implicit sum for i and j over x,y,z (but all the terms with repeated indices in the Levi-Cevita symbol go to 0)

    [tex](\nabla\times\vec{A})_x = \partial_yA_z\epsilon_{yzx} + \partial_zA_y\epsilon_{zyx}=\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}[/tex]
  7. Sep 30, 2008 #6
    Thanks thats what I wanted, I thought once I had seen this I would be able to figure it out, unfortunately it's just not clicking for me.

    Can I check I have the right thinking....
    [itex]\epsilon_{ijk}\partial_j A_k[/itex]

    For B1 you set i to 1. then that leaves two combinations for [itex]partial_j A_k[/itex]
    [itex]partial_2 A_3[/itex] or [itex]partial_3 A_2[/itex]

    Then you can have any tensor as long as i is 1 i.e [itex]epsilon_{123}\ \epsilon_{112}\ \epsilon_{111}[/itex]
    but obviously any with two 1's in are zero.

    I think this gives the right answer but I should be thinking more in terms of implicit sums than what combination I have left.
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