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Electromagnetic fields derivation

  1. Jul 21, 2015 #1
    1. The problem statement, all variables and given/known data
    In a book I find the following derivation:
    [tex] \int (J \cdot \nabla ) \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'}= -\sum_{i=1}^3 \int J_i \frac{\partial}{\partial r_i'} \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'} \\
    = -\sum_{i=1}^3 \int J_i \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^2r_{j\neq i}'\bigg|_{r_i' = -\infty}^\infty + \int \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} \nabla' \cdot \mathbf{J} d^3\mathbf{r'} [/tex]



    2. Relevant equations
    None

    3. The attempt at a solution
    I understand the first line, but in the second line I don't understand how the second term comes about. I dont see how the divergence of J is taken as the nabla operator doesnt operate on it. If I calculate for example one of the three terms:
    [tex] \int J_x \frac{\partial}{\partial x'} \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'} [/tex]
    I don't see how a term with [itex] \frac{\partial J_x}{\partial x'}[/itex] comes out of this?
     
  2. jcsd
  3. Jul 21, 2015 #2

    jasonRF

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    Science Advisor
    Gold Member

    Here is a hint: integration by parts
     
  4. Jul 21, 2015 #3
    the x component of j*grad (r/r3) is j*grad(rx/r3), not jxgrad(rx/r3) you then use the vector calculus identity
    div(fV)=fdivV+V*grad(f)
     
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