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Operator del as div, grad, rot

  1. Nov 22, 2008 #1
    hi there!

    I have some questions concerning the del operator when you use it together with the epsilon tensor and kronecker delta:

    1. if you have:
    phi - scalar, B vector fields

    [tex]\partial_j(\phi B)_i[/tex]

    is it equal to: [tex]\partial_j(\phi B)_i=(\partial_j\phi)B_i+\phi(\partial_jB_i)[/tex]

    or I also have to index phi

    2. How do I know if del represents rot or div of a vector field in a mixed expression? (ok, for the rot we also need the epsilon tensor, but there are some mixed identities where I can`t figure it out)

    Here`s an example, maybe you could give me some advice on it:

    [tex]div(\vec E\times \vec B)=\vec B. rot \vec E- \vec E. rot \vec B[/tex]

    [tex]div(\vec E\times \vec B)=\partial_i(\vec E\times \vec B)_i=\partial_i\epsilon_{jki}E_j B_k=\epsilon_{ijk} \partial_i(E_j B_k)=\epsilon_{ijk} (\partial_i E_j)B_k + \epsilon_{ijk} E_j (\partial_i B_k)[/tex]

    and now I don`t know how to express [tex](\partial_i B_k)[/tex] again in vectors

    I would be glad I anyone could explain to me the entity of the operations and maybe give me some advice how to use them properly :)

    thanks in advance, marin
  2. jcsd
  3. Nov 22, 2008 #2


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    Hi Marin.

    Your question has nothing to do with the derivative. Note that the vector [itex]\phi \vec B[/itex] is defined as: multiply each component of [itex]\vec B[/itex] with the scalar [itex]\phi[/itex]. So in index notation, [itex](\phi B)_i = \phi B_i[/itex]. Both [itex]\phi[/itex] and [itex]B_i[/itex] are just numbers, so you do get what you thought it'd be.

    Look at the index placement. If the index on the del is contracted with one on a vector field, it's a div. If it's used with an epsilon, it's rot.

    Well, in the first term, you recognise
    [tex]\epsilon_{ijk} \partial_i E_j[/tex]
    which is just component-notation for [itex](\nabla \times E)_k[/itex]
    In general,
    [tex](A \times B)_k = \epsilon_{ijk} A_i B_j[/tex]

    For the second term, you can write
    [tex]\epsilon_{ijk} \partial_i B_k = - \epsilon_{ikj} \partial_i B_k[/tex]
    and you recognise the same structure (first two indices of epsilon symbol contracted with two vectors) so this is
    [itex]- (\nabla \times B)_j[/itex]
  4. Nov 22, 2008 #3
    now that was helpful! thanks, CompuChip!

    i tried this one:

    [tex]rot rot \vec E=\epsilon_{jki}\partial_j\epsilon_{lmk}\partial_l E_m=\epsilon_{kij}\epsilon_{klm}\partial_j\partial_l E_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j\partial_l E_k=\delta_{il}\delta_{jm}\partial_l\partial_j E_m-\delta_{im}\delta_{jl}\partial_j\partial_l E_k = \partial_i\partial_m E_k-\partial_j^2 E_i[/tex]

    i`ve just corrected it :) I think it should be correct now!
    Last edited: Nov 22, 2008
  5. Nov 23, 2008 #4


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    I don't think it's entirely right yet. On the left hand side, you should get a vector (rot E is a vector, and the rot of that is again a vector). On the right hand side, however, I see free indices i, m and k in one term (and i and j in the other one, although I suppose you mean [itex]\partial_j^2 = \sum_j \partial_j \partial_j[/itex]).

    Your calculation looks mostly correct though.
    Where you rewrite the product of epsilons, you suddenly have Ek, in the next step you have again Em, and finally it becomes Ek again :smile:

    Supposing you meant
    [tex]\partial_i \partial_m E_m - \partial_j \partial_j E_i[/tex]
    you have shown that
    [tex]\nabla \times (\nabla \times \vec E) = \nabla(\nabla \cdot \vec E) - \nabla^2 \vec E[/tex],
    is that right?
  6. Nov 23, 2008 #5
    yes, I wrote initially E_k then realised it must be E_m and tried to correct the latex version, but since there were lots of symbols I just forgot some :)

    so the Kronecker deltas contract the first possible vectors that come afterwards, del´s or not del´s, right?
  7. Nov 23, 2008 #6


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    Well, not the first possible vectors that come afterwards... just the vectors with the same indices. Of course you can write the delta's anywhere in your expression :smile:
    But otherwise, yes.

    Technically speaking, nabla is of course not a vector. But if you work in component notation, it doesn't really matter: don't think, just write :smile: - that's the beauty of it.
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