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Inversion of curl of A formula

  1. Jan 5, 2012 #1
    Hello! I'm reading up on Hamiltonian mechanics and i stumbled on the fact that the curl of the vector potential can be expressed as

    [tex]B_k = \sum_k \epsilon_{kij}\frac{\partial A_i}{\partial x_j}[/tex]

    Now the text that I'm reading says that this formula can be inverted as

    [tex] \sum_k \epsilon_{kij} B_k = \frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}[/tex]

    But I then wondered how this inversion would be accomplished?

    I suspect the formula [tex]\sum_k \epsilon_{kij} \epsilon_{klm}= \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}[/tex] to be involved.
     
  2. jcsd
  3. Jan 5, 2012 #2
    Just substitute the third formula into the first and you get the second.
    What is the problem here?
     
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