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Is this a correct proof? |
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| Sep27-10, 12:45 AM | #1 |
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Is this a correct proof?
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! |
| Sep27-10, 07:38 PM | #2 |
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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. |
| Sep27-10, 11:08 PM | #3 |
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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. |
| Sep27-10, 11:16 PM | #4 |
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Is this a correct proof?
t
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. |
| Sep27-10, 11:27 PM | #5 |
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