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Einstein summation notation for magnetic dipole field

  1. Apr 30, 2013 #1
    I can do this derivation the old fashioned way, but am having trouble doing it with einstein summation notation.

    Since [itex]\vec{B}=\nabla \times \vec{a}[/itex]
    [itex]\vec{B}=\mu_{0}/4\pi (\nabla \times (m \times r)r^{-3}))[/itex]
    [itex]4\pi \vec{B}/\mu_{0}=\epsilon_{ijk} \nabla_{j}(\epsilon_{klm} m_{l} r_{m} r^{-3})[/itex]
    [itex]=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_{j}m_{l}r_{m}r^{-3}[/itex]

    here is where I am stumbling. My professor has for the next step

    [itex]=m_{l}(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})r^{-3} \delta_{jm}-3 r_{m}\hat{r}_{j}r^{-4})[/itex]

    but I don't really know how to get to that step
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2

    Fredrik

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    You may be interested in the The LaTeX guide for the forum. :smile: Link.
     
  4. Apr 30, 2013 #3
    You were too fast. Was trying to get it to work just needed to delete the spaces in brackets I guess.
     
  5. Apr 30, 2013 #4

    Fredrik

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    My first thought is that he's using the product rule for derivatives to evaluate ##\partial_j## acting on a product.
     
  6. Apr 30, 2013 #5

    WannabeNewton

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    First off, tell your professor that he is horribly butchering Einstein notation. Seriously, what was written down misses the entire point of the notation. Anyways, ##\frac{4\pi}{\mu_{0}}B^{i} = \frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = \epsilon^{ijk}\epsilon_{klm}m^{l}[r^{-3}\partial_{j}r^{m} + r^{m}\partial_{j}(r^{-3})]## hence ##\frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = 2\delta^{[i}_{l}\delta^{j]}_{m}m^{l}[r^{-3}\partial_{j}r^{m} + r^{m}\partial_{j}(r^{-3})]##. Now, ##\partial_{j}r^{m} = \delta^{m}_{j}## and ##\partial_{j}(r^{-3}) = -3(-r^i r_{i})^{-5/2}r_{k}\partial_{j}r^{k} = -3r^{-4}\hat{r}_{j}## giving us ##\frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = 2\delta^{[i}_{l}\delta^{j]}_{m}m^{l}[r^{-3}\delta^{m}_{j} -3r^{-4}\hat{r}_{j}r^{m}]## as desired.

    EDIT: By the way, in the above it should be ##(r^i r_{i})^{-5/2}## not ##(-r^i r_{i})^{-5/2}##; I've gotten too used to General Relativity xD.
     
    Last edited: Apr 30, 2013
  7. Apr 30, 2013 #6
    O wow, thanks. that makes much more sense now.
     
  8. Apr 30, 2013 #7
    Do you know of any materials online that would give more written out examples of such derivations with Einstein summation? I just need more practice
     
  9. Apr 30, 2013 #8

    WannabeNewton

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    I honestly can't think of any online resources off of the top of my head because I got used to the notation when learning special relativity (the text used was Schutz).
     
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