Is there a better way to solve this proof using Levi Civita symbols?

[NexusN]

Homework Statement

$$\nabla \cdot(\vec E \times \vec H)$$=$$\vec H\cdot(\nabla \times \vec E) - \vec E\cdot(\nabla \times \vec H)$$

Homework Equations

The Attempt at a Solution

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

$$\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}$$
Thank you so much for your kind attention!

 

[Inferior89]

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

$$\nabla \cdot (\mathbf{u} \times \mathbf{v}) = \frac{\partial}{\partial x_i} (\epsilon_{ijk} u_j v_k )$$

$$= \epsilon_{ijk} \frac{\partial u_j}{\partial x_i}v_k + \epsilon_{ijk}u_j \frac{\partial v_k}{\partial x_i}$$

$$= \left( \epsilon_{kij} \frac{\partial u_j}{\partial x_i}\right)v_k - \left( \epsilon_{jik} \frac{\partial v_k}{\partial x_i}\right)u_j$$

$$= (\nabla \times \mathbf{u}) \cdot \mathbf{v} - (\nabla \times \mathbf{v}) \cdot \mathbf{u}$$

This is probably not the method you are 'supposed' to do but my point is that there are much nicer ways of doing it.

 

[NexusN]

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

$$\nabla \cdot (\mathbf{u} \times \mathbf{v}) = \frac{\partial}{\partial x_i} (\epsilon_{ijk} u_j v_k )$$

$$= \epsilon_{ijk} \frac{\partial u_j}{\partial x_i}v_k + \epsilon_{ijk}u_j \frac{\partial v_k}{\partial x_i}$$

$$= \left( \epsilon_{kij} \frac{\partial u_j}{\partial x_i}\right)v_k - \left( \epsilon_{jik} \frac{\partial v_k}{\partial x_i}\right)u_j$$

$$= (\nabla \times \mathbf{u}) \cdot \mathbf{v} - (\nabla \times \mathbf{v}) \cdot \mathbf{u}$$

This is probably not the method you are 'supposed' to do but my point is that there are much nicer ways of doing it.

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.

 

[╔(σ_σ)╝]

t

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.

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.

 

[NexusN]

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.

I am interested in that, thank you for your suggestion