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Homework Help: Is this a correct proof?

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\nabla \cdot(\vec E \times \vec H)[/tex]=[tex]\vec H\cdot(\nabla \times \vec E) - \vec E\cdot(\nabla \times \vec H)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]\begin{array}{l}
    \nabla \cdot(\vec E \times \vec H)\\
    = \left[ {\frac{1}{{{h_1}{h_2}{h_3}}}\left( {{{\hat a}_1}\frac{\partial }{{\partial {l_1}}}{h_2}{h_3} + {{\hat a}_2}\frac{\partial }{{\partial {l_2}}}{h_1}{h_3} + {{\hat a}_3}\frac{\partial }{{\partial {l_3}}}{h_1}{h_2}} \right)} \right]\cdot[({{\hat a}_1}{E_1} + {{\hat a}_2}{E_2} + {{\hat a}_3}{E_3}) \times ({{\hat a}_1}{H_1} + {{\hat a}_2}{H_2} + {{\hat a}_3}{H_3})]\\
    = \left[ {\frac{1}{{{h_1}{h_2}{h_3}}}\left( {{{\hat a}_1}\frac{\partial }{{\partial {l_1}}}{h_2}{h_3} + {{\hat a}_2}\frac{\partial }{{\partial {l_2}}}{h_1}{h_3} + {{\hat a}_3}\frac{\partial }{{\partial {l_3}}}{h_1}{h_2}} \right)} \right]\cdot\left| {\begin{array}{*{20}{c}}
    {{{\hat a}_1}}&{{{\hat a}_2}}&{{{\hat a}_3}}\\
    {{E_1}}&{{E_2}}&{{E_3}}\\
    {{H_1}}&{{H_2}}&{{H_3}}
    \end{array}} \right|\\
    = \left[ {\frac{1}{{{h_1}{h_2}{h_3}}}\left( {{{\hat a}_1}\frac{\partial }{{\partial {l_1}}}{h_2}{h_3} + {{\hat a}_2}\frac{\partial }{{\partial {l_2}}}{h_1}{h_3} + {{\hat a}_3}\frac{\partial }{{\partial {l_3}}}{h_1}{h_2}} \right)} \right]\cdot[{{\hat a}_1}({E_2}{H_3} - {H_2}{E_3}) + {{\hat a}_2}({E_3}{H_1} - {H_3}{E_1}) + {{\hat a}_3}({E_1}{H_2} - {H_1}{E_2})]\\
    = \frac{1}{{{h_1}{h_2}{h_3}}}\left\{ {\frac{\partial }{{\partial {l_1}}}[{h_2}{h_3}({E_2}{H_3} - {H_2}{E_3})] + \frac{\partial }{{\partial {l_2}}}[{h_1}{h_3}({E_3}{H_1} - {H_3}{E_1})] + \frac{\partial }{{\partial {l_3}}}[{h_1}{h_2}({E_1}{H_2} - {H_1}{E_2})]} \right\}
    \end{array}[/tex]

    [tex]\begin{array}{l}
    \vec H\cdot(\nabla \times \vec E) - \vec E\cdot(\nabla \times \vec H)\\
    = ({{\hat a}_1}{H_1} + {{\hat a}_2}{H_2} + {{\hat a}_3}{H_3})\cdot\frac{1}{{{h_1}{h_2}{h_3}}}\left| {\begin{array}{*{20}{c}}
    {{{\hat a}_1}{h_1}}&{{{\hat a}_2}{h_2}}&{{{\hat a}_3}{h_3}}\\
    {\frac{\partial }{{\partial {l_1}}}}&{\frac{\partial }{{\partial {l_2}}}}&{\frac{\partial }{{\partial {l_3}}}}\\
    {{h_1}{E_1}}&{{h_2}{E_2}}&{{h_3}{E_3}}
    \end{array}} \right| - ({{\hat a}_1}{E_1} + {{\hat a}_2}{E_2} + {{\hat a}_3}{E_3})\cdot\frac{1}{{{h_1}{h_2}{h_3}}}\left| {\begin{array}{*{20}{c}}
    {{{\hat a}_1}{h_1}}&{{{\hat a}_2}{h_2}}&{{{\hat a}_3}{h_3}}\\
    {\frac{\partial }{{\partial {l_1}}}}&{\frac{\partial }{{\partial {l_2}}}}&{\frac{\partial }{{\partial {l_3}}}}\\
    {{h_1}{H_1}}&{{h_2}{H_2}}&{{h_3}{H_3}}
    \end{array}} \right|\\
    = ({{\hat a}_1}{H_1} + {{\hat a}_2}{H_2} + {{\hat a}_3}{H_3})\cdot\frac{1}{{{h_1}{h_2}{h_3}}}[{{\hat a}_1}({h_1}\frac{\partial }{{\partial {l_2}}}{h_3}{E_3} - {h_1}\frac{\partial }{{\partial {l_3}}}{h_2}{E_2}) + {{\hat a}_2}({h_2}\frac{\partial }{{\partial {l_3}}}{h_1}{E_1} - {h_2}\frac{\partial }{{\partial {l_1}}}{h_3}{E_3}) + {{\hat a}_3}({h_3}\frac{\partial }{{\partial {l_1}}}{h_2}{E_2} - {h_3}\frac{\partial }{{\partial {l_2}}}{h_1}{E_1})]\\
    - ({{\hat a}_1}{E_1} + {{\hat a}_2}{E_2} + {{\hat a}_3}{E_3})\cdot\frac{1}{{{h_1}{h_2}{h_3}}}[{{\hat a}_1}({h_1}\frac{\partial }{{\partial {l_2}}}{h_3}{H_3} - {h_1}\frac{\partial }{{\partial {l_3}}}{h_2}{H_2}) + {{\hat a}_2}({h_2}\frac{\partial }{{\partial {l_3}}}{h_1}{H_1} - {h_2}\frac{\partial }{{\partial {l_1}}}{h_3}{H_3}) + {{\hat a}_3}({h_3}\frac{\partial }{{\partial {l_1}}}{h_2}{H_2} - {h_3}\frac{\partial }{{\partial {l_2}}}{h_1}{H_1})]\\
    = \frac{1}{{{h_1}{h_2}{h_3}}}({h_1}{H_1}\frac{\partial }{{\partial {l_2}}}{h_3}{E_3} - {h_1}{H_1}\frac{\partial }{{\partial {l_3}}}{h_2}{E_2} + {h_2}{H_2}\frac{\partial }{{\partial {l_3}}}{h_1}{E_1} - {h_2}{H_2}\frac{\partial }{{\partial {l_1}}}{h_3}{E_3} + {h_3}{H_3}\frac{\partial }{{\partial {l_1}}}{h_2}{E_2} - {h_3}{H_3}\frac{\partial }{{\partial {l_2}}}{h_1}{E_1})\\
    - \frac{1}{{{h_1}{h_2}{h_3}}}({h_1}{E_1}\frac{\partial }{{\partial {l_2}}}{h_3}{H_3} - {h_1}{E_1}\frac{\partial }{{\partial {l_3}}}{h_2}{H_2} + {h_2}{E_2}\frac{\partial }{{\partial {l_3}}}{h_1}{H_1} - {h_2}{E_2}\frac{\partial }{{\partial {l_1}}}{h_3}{H_3} + {h_3}{E_3}\frac{\partial }{{\partial {l_1}}}{h_2}{H_2} - {h_3}{E_3}\frac{\partial }{{\partial {l_2}}}{h_1}{H_1})\\
    = \frac{1}{{{h_1}{h_2}{h_3}}}[({h_1}{H_1}\frac{\partial }{{\partial {l_2}}}{h_3}{E_3} + {h_3}{E_3}\frac{\partial }{{\partial {l_2}}}{h_1}{H_1}) - ({h_1}{H_1}\frac{\partial }{{\partial {l_3}}}{h_2}{E_2} + {h_2}{E_2}\frac{\partial }{{\partial {l_3}}}{h_1}{H_1}) + ({h_2}{H_2}\frac{\partial }{{\partial {l_3}}}{h_1}{E_1} + {h_1}{E_1}\frac{\partial }{{\partial {l_3}}}{h_2}{H_2})\\
    - ({h_2}{H_2}\frac{\partial }{{\partial {l_1}}}{h_3}{E_3} + {h_3}{E_3}\frac{\partial }{{\partial {l_1}}}{h_2}{H_2}) + ({h_3}{H_3}\frac{\partial }{{\partial {l_1}}}{h_2}{E_2} + {h_2}{E_2}\frac{\partial }{{\partial {l_1}}}{h_3}{H_3}) - ({h_3}{H_3}\frac{\partial }{{\partial {l_2}}}{h_1}{E_1} + {h_1}{E_1}\frac{\partial }{{\partial {l_2}}}{h_3}{H_3})]\\
    = \frac{1}{{{h_1}{h_2}{h_3}}}(\frac{\partial }{{\partial {l_2}}}{h_1}{h_3}{E_3}{H_1} - \frac{\partial }{{\partial {l_3}}}{h_1}{h_2}{E_2}{H_1} + \frac{\partial }{{\partial {l_3}}}{h_1}{h_2}{E_1}{H_2} - \frac{\partial }{{\partial {l_1}}}{h_2}{h_3}{E_3}{H_2} + \frac{\partial }{{\partial {l_1}}}{h_2}{h_3}{E_2}{H_3} - \frac{\partial }{{\partial {l_2}}}{h_1}{h_3}{E_1}{H_3})\\
    = \frac{1}{{{h_1}{h_2}{h_3}}}\{ \frac{\partial }{{\partial {l_1}}}[{h_2}{h_3}({E_2}{H_3} - {E_3}{H_2})] + \frac{\partial }{{\partial {l_2}}}[{h_1}{h_3}({E_3}{H_1} - {E_1}{H_3})] + \frac{\partial }{{\partial {l_3}}}[{h_1}{h_2}({E_1}{H_2} - {E_2}{H_1})]\} \\
    = \nabla \cdot(\vec E \times \vec H)
    \end{array}[/tex]



    Thank you so much for your kind attention!
     
  2. jcsd
  3. Sep 27, 2010 #2
    Well the proof is probably correct but are you sure you aren't supposed to be doing a more 'refined' method than just writing out the whole definition and 'brute forcing' it.

    As an example, in index notation you can do something like

    [tex] \nabla \cdot (\mathbf{u} \times \mathbf{v}) = \frac{\partial}{\partial x_i} (\epsilon_{ijk} u_j v_k ) [/tex]

    [tex]= \epsilon_{ijk} \frac{\partial u_j}{\partial x_i}v_k + \epsilon_{ijk}u_j \frac{\partial v_k}{\partial x_i} [/tex]

    [tex]= \left( \epsilon_{kij} \frac{\partial u_j}{\partial x_i}\right)v_k - \left( \epsilon_{jik} \frac{\partial v_k}{\partial x_i}\right)u_j [/tex]

    [tex] = (\nabla \times \mathbf{u}) \cdot \mathbf{v} - (\nabla \times \mathbf{v}) \cdot \mathbf{u} [/tex]

    This is probably not the method you are 'supposed' to do but my point is that there are much nicer ways of doing it.
     
  4. Sep 27, 2010 #3
    Thank you so much for your reply.
    At the moment I did not have other ideas so I just brute-forced it by expanding.
    I am now learning some vectors in the year 2 of electronic engineering.

    I am new to your method and I think I will need to take some time to understand it thoroughly, but it is really nice and short!
    Thanks again for your help.
     
  5. Sep 27, 2010 #4
    t
    This is levi civita symbols. Been an engineering student myself, this is not something they teach use in engineering mathematics.

    I think your method would suffice considering you were probably not taught a better way of doing it. Alternatively, you could look up levi civita symbols and the kronecker delta function.
     
  6. Sep 27, 2010 #5
    I am interested in that, thank you for your suggestion:smile:
     
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