Show $\nabla\cdot\left(\mathbf{F}\times\mathbf{G}\right)=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G})$

Question

Question as follows.

Suppose that $\mathbf{F}$,$\mathbf{G}:\mathbb{R^3}\rightarrow\mathbb{R^3}$ and $\phi:\mathbb{R^3}\rightarrow\mathbb{R}$ are smooth. Show using the summation convention that $$\nabla\cdot\left(\mathbf{F}\times\mathbf{G}\right)=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G}).$$

So far I have $$\mathrm{LHS}=\partial_i\mathbf{e}_i\cdot\left(F_j\mathbf{e}_j\times G_k\mathbf{e}_k\right)=\partial_i\mathbf{e}_i\cdot(F_j G_k \epsilon_{jki}\mathbf{e}_i)=\partial_iF_jG_k\epsilon_{ijk}.$$

I'm under the impression that this should be $0$ as expanding the $\mathrm{RHS}$ gives $2\partial_iF_jG_k\epsilon_{ijk}$. Is this true and is it true because for $\partial_iF_jG_k\ne0$ iff $i=j$ or $i=k$ but if $i=j$ or $i=k$ then $\epsilon_{ijk}=0$?

Remember that the left-hand side needs the product rule - the partial derivative $\partial_i$ does not commute with $F_j$ and $G_k$ — Empy2, Jun 23 '13 at 06:48
So I don't even need the condition $i=k$? Saying that $i=j\Rightarrow \epsilon_{ijk}=0$ is enough? — jamesh625, Jun 23 '13 at 07:04
No, $\partial_i(F_jG_k)=F_{j,i}G_k+F_jG_{k,i}$, where the comma means the partial derivative. — Empy2, Jun 23 '13 at 07:11

score 4 · Answer 1 · answered Jun 23 '13 at 07:11

$$ \nabla \cdot \left ( \mathbf F \times \mathbf G\right ) = \left( \mathbf F \times \mathbf G\right )_{i,i} = \left ( e_{ijk} F_j G_k\right )_{,i} = e_{ijk}F_{j,i}G_k + e_{ijk}F_j G_{k,i} = \\ = G_k e_{kij}F_{j,i} + F_j e_{jki} G_{k,i} = G_k e_{kij}F_{j,i} - F_j e_{jik} G_{k,i} = \\ = G_k \left (\nabla \times \mathbf F \right )_k - F_j \left( \nabla \times \mathbf G\right )_j = \mathbf G \cdot \left (\nabla \times \mathbf F \right ) - \mathbf F \cdot \left( \nabla \times \mathbf G\right ) $$

score 3 · Answer 2 · answered Oct 11 '16 at 07:48

The same calculation that Kaster, fully expanded: \begin{align*} \nabla\cdot(\mathbf{F}\times\mathbf{G}) &= \partial_1(F_2\,G_3 - F_3\,G_2) + \partial_2(F_3\,G_1 - F_1\,G_3) + \partial_ 3(F_1\,G_2 - F_2\,G_1)\\ &= (\partial_1 F_2)\,G_3 - (\partial_1 F_3)\,G_2 + (\partial_2 F_3)\,G_1 - (\partial_2 F_1)\,G_3 + (\partial_3 F_1)\,G_2 - (\partial_3 F_2)\,G_1\\ &\qquad+ F_2\,(\partial_1 G_3) - F_3\,(\partial_1 G_2) + F_3\,(\partial_2 G_1) - F_1\,(\partial_2 G_3) + F_1\,(\partial_3 G_2) - F_2\,(\partial_3 G_1)\\ &= G_1\,(\partial_2 F_3 - \partial_3 F_2) + G_2\,(\partial_3 F_1 - \partial_1 F_ 3) + G_3\,(\partial_1 F_2 - \partial_2 F_1)\\ &\qquad- F_1\,(\partial_2 G_3 - \partial_3 G_2) - F_2\,(\partial_3 G_1 - \partial_1 G_3) - F_3\,(\partial_1 G_2 - \partial_2 G_1)\\ &= \mathbf{G}\cdot(\nabla\times\mathbf{F}) - \mathbf{F}\cdot(\nabla\times\mathbf{G}). \end{align*}

Show $\nabla\cdot\left(\mathbf{F}\times\mathbf{G}\right)=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G})$

2 Answers2

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