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Problem with index notation

  1. Sep 27, 2012 #1
    now in index notation it is written as,
    but when I tried to prove it ,it just came out twice.can anyone tell how it is correct(given is the correct form).i really mean that i was getting four terms which gave twice of above after reshuffling so prove it.
  2. jcsd
  3. Sep 27, 2012 #2


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    All your formulae are correct.

    The left-hand side of your first equation reads in index notation
    [tex](\vec{r} \times \vec{\nabla}) \cdot (\vec{r} \times \vec{\nabla}) \phi= \epsilon_{jkl} r_k \partial_l (\epsilon_{jmn} r_m \partial_n \phi).[/tex]
    Now using
    [tex]\epsilon_{jkl} \epsilon_{jmn}=\delta_{km} \delta_{ln}-\delta_{kn} \delta_{lm},[/tex]
    you indeed get
    [tex](\vec{r} \times \vec{\nabla}) \cdot (\vec{r} \times \vec{\nabla}) \phi = r_{k} \partial_l(r_k \partial_l \phi)-r_k \partial_l(r_l \partial_k \phi).[/tex]

    The right-hand side of your first equation is
    [tex]\vec{r} \cdot [\vec{\nabla} \times (\vec{r} \times \vec{\nabla}) \phi] = r_j \epsilon_{jkl} \partial_k (\epsilon_{lmn} r_m \partial_n \phi).[/tex]
    Again we have
    [tex]\epsilon_{jkl} \epsilon_{lmn}=\epsilon_{ljk} \epsilon_{lmn}=\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}.[/tex]
    Thus we have
    [tex]\vec{r} \cdot [\vec{\nabla} \times (\vec{r} \times \vec{\nabla}) \phi] = r_j \partial_k (r_j \partial_k \phi)-r_j \partial_k (r_k \partial_j \phi).[/tex]
    This shows that indeed both expressions of your first equations are equal, because the only difference is the naming of the dummy-summation indices :-).
  4. Sep 28, 2012 #3
    [tex]\epsilon_{jkl} \epsilon_{jmn}=\delta_{km} \delta_{ln}-\delta_{kn} \delta_{lm},[/tex]
    I was aware of it which I have seen in butkov an year ago.but this is the first use of it.so thanks,van I think i am just becoming lazy.
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