Divergence and curl identity

Question

I'm trying to prove

$div(F \times G) = G \cdot curl(F) - F \cdot curl(G)$

I tried expanding the left side and the right side but I'm getting $2(div(F \times G)) = G \cdot curl(F) - F \cdot curl(G)$. I'm not sure what to do with that "2". Thanks.

Answer 1

They are linear so that we suffice to consider the following two cases :

Case 1 : $F=(f,0,0),\ G=(g,0,0)$

$$ {\rm div}\ (F\times G)=0$$ And ${\rm curl}(F)=(0,\ast,\ast)$ so that $$ G\cdot {\rm curl} F - F\cdot {\rm curl} G=0$$

Case 2 : $F=(f,0,0),\ G=(0,g,0)$

$$ {\rm div}\ (F\times G)={\rm div} (0,0,fg)=f_zg+fg_z$$ And $${\rm curl}(F)=(0,f_z,-f_y),\ {\rm curl} (G)=(-g_z,0,g_x)$$ so that $$ G\cdot {\rm curl} F - F\cdot {\rm curl} G=f_zg+fg_z$$

Answer 2

If you're familiar with Einstein notation, $$\nabla \cdot(F\times G)=\nabla \cdot \epsilon_{ijk}F_i G_je_k = \delta_{mk}\epsilon_{ijk}\partial_{m}(F_iG_j) = \delta_{mk}\epsilon_{ijk}(\partial_m F_i)G_j+\delta_{mk}\epsilon_{ijk}F_i(\partial_m G_j)=\epsilon_{ijk}(\partial_k F_i)G_j+\epsilon_{ijk}F_i(\partial_k G_j)=(\nabla \times F)_{j}G_j+F_i(-\nabla \times G)_{i}$$ $$\Rightarrow \nabla \cdot(F\times G) = G\cdot \nabla \times F - F\cdot \nabla \times G$$

Now the factor 2 in your answer is not something that can be simplified away, so you may want to review your calculations