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Math Help - Question involving Levi-Civita symbol

  1. #1
    Junior Member CuriosityCabinet's Avatar
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    Question involving Levi-Civita symbol

    Can someone please explain to me why

    \epsilon_{ijk}\frac{\partial}{\partial x_i}\frac{\partial A_k}{\partial x_j} = 0

    where A is a constant vector field.
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  2. #2
    Member Ruun's Avatar
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    Re: Question involving Levi-Civita symbol

    \epsilon_{ijk} is a totally antisymmetric symbol and \frac{\partial^2 A_{k}}{\partial x_{i}\partial x_{j}} is symmetric under the exchange of i and j, because the partial derivatives commute.

    In your particular case, as \vec{A} is constant, the derivatives are always zero.
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  3. #3
    Junior Member CuriosityCabinet's Avatar
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    Re: Question involving Levi-Civita symbol

    OK thanks. So which part of the expression is zero? The \epsilon_{ijk} part?
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    Member Ruun's Avatar
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    Re: Question involving Levi-Civita symbol

    Your expression actually means

    \displaystyle\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk} \frac{\partial ^2 A_{k}}{\partial x_{i}\partial x_{j}}

    This is

    \epsilon_{111}\frac{\partial^2 A_{1}}{\partial x_{1}^2}+\epsilon_{112}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{2}}+\epsilon_{113}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{3}}+\epsilon_{121}\frac{\partial ^2 A_{1}}{\partial x_{2}\partial x_{1}}+\epsilon_{122}\frac{\partial ^2 A_{2}}{\partial x_{1}\partial x_{2}}+...

    The definition of \epsilon_{ijk} gives you a lot of zeros, in particular the only non zero are 1=\epsilon_{123}=\epsilon_{231}=\epsilon_{312} and -1=\epsilon_{321}=\epsilon_{132}=\epsilon_{213}. From this fact and that

    \frac{\partial ^2}{\partial x_{i} \partial x_{j}}=\frac{\partial ^2}{\partial x_{j} \partial x_{i}}

    you will have pairs of terms in your sum like

    \epsilon_{123}\frac{\partial A_{3}}{\partial x_{1}\partial x_{2}}+\epsilon_{213}\frac{\partial A_{3}}{\partial x_{2}\partial x_{1}}

    which cancel and gives you zero.
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  5. #5
    Junior Member CuriosityCabinet's Avatar
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    Re: Question involving Levi-Civita symbol

    Fantastic, problem solved.
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