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Simplify the proof of different vector calculus identities

  1. Jul 19, 2015 #1
    Is there a way to simplify the proof of different vecot calculus identities, such as grad of f*g, which is expandable. And also curl of the curl of a field. Is there a more convenient way to go about proving these relations than to go through the long calculations of actually performing the curl and grad etc? I know nothing about geometric algebra, but is that a good way?
     
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  3. Jul 19, 2015 #2
  4. Jul 19, 2015 #3
    vecid.gif
    Im talking about all these identities, is there a branch of mathematics that simplifies the proofs of these, and lets me avoid expending the vectors and del operators?
     
  5. Jul 19, 2015 #4

    robphy

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  6. Jul 24, 2015 #5

    Fredrik

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    You can use the Levi-Civita symbol without knowing anything about tensors other than the convention to not bother to write a summation sigma for an index that appears exactly twice. For example, ##\varepsilon_{ijk}A_j(B\times C)_k## really means ##\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}A_j(B\times C)_k##. The proof of the first identity in post #3 goes like this:
    \begin{align*}
    &(A\times(B\times C))_i =\varepsilon_{ijk}A_j(B\times C)_k =\varepsilon_{ijk}A_j\varepsilon_{klm}B_l C_m =(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})A_jB_lC_m
    =A_jB_iC_j -A_jB_jC_i\\
    &=B_i(A\cdot C)-C_i(A\cdot B) =(B(A\cdot C)-C(A\cdot B))_i.
    \end{align*} The Levi-Civita symbol ##\varepsilon_{ijk}## is defined by saying that ##\varepsilon_{123}=1## and that an exchange of any two indices changes the sign of ##\varepsilon_{ijk}##. (For example ##\varepsilon_{132}=-\varepsilon_{123}=-1##). Note that this implies that ##\varepsilon_{ijk}=0## when two of the indices have the same value. (If ##i=j##, then ##\varepsilon_{ijk}=\varepsilon_{jik}=-\varepsilon_{ijk}##). The third equality in the calculation above involves one of a small number of identities that you have to prove before you can really start working with this notation.
     
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