# How to prove that $\nabla (a\cdot b)=(a\cdot \nabla )b+(b\cdot \nabla )a+a\times (\nabla \times b)+b\times (\nabla \times a)$?

I'm working hard to prove this.. $$\nabla (a\cdot b)=(a\cdot \nabla )b+(b\cdot \nabla )a+a\times (\nabla \times b)+b\times (\nabla \times a)$$ but I got $$\nabla (a\cdot b)=\nabla (a\cdot b)+\nabla (a\cdot b)$$ Is it the same or the answer will be $2\nabla (a\cdot b)$? I got confused on how to prove I let $a=a_1,a_2,a_3$ and $b=b_1,b_2,b_3$ I got $$a\times (\nabla \times b)+b\times (\nabla \times a)=\nabla (a.b)-b(a.\nabla )+\nabla (b.a)-a(b.\nabla )$$ then, \begin{align} \nabla (a\cdot b)&=(a\cdot \nabla )b+(b\cdot \nabla )a+\nabla (a.b)-b(a.\nabla )+\nabla (b.a)-a(b.\nabla ) \\ \nabla (a\cdot b)&=\nabla (a\cdot b)+\nabla (a\cdot b) \end{align} Is it the same? because there is no error when I'm checking it..

Use indices. \begin{align} LHS &= \left [ \nabla (a \cdot b) \right ]_i = (a\cdot b)_{,i} = (a_j b_j)_{,i} = a_{j,i} b_j + a_j b_{j,i} \\ RHS &= \left [(a \cdot \nabla) b + (b \cdot \nabla a) + a \times (\nabla \times b) + b \times (\nabla \times a) \right ]_i = \\ &= a_j b_{i,j} + b_j a_{i,j} + \color{red}{\epsilon_{ijk} a_j (\nabla \times b)_k} + \color{green}{\epsilon_{ijk} b_j (\nabla \times a)_k} = \\ &= \ldots +\ \color{red}{\epsilon_{kij} a_j \epsilon_{klm} b_{m,l}} = \ldots + \color{red}{(\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl})a_j b_{m,l}} = \ldots + \color{red}{(a_m b_{m,i} - a_l b_{i,l})} = \\ &= \ldots + \color{green}{\epsilon_{ijk} \epsilon_{klm} b_j a_{m,l}} = \ldots = \ldots + \color{green}{(b_m a_{m,i} - b_l a_{i,l})} = \\ &= \underline{a_j b_{i,j}} + \underline{\underline{b_j a_{i,j}}} + a_j b_{j,i} - \underline{a_j b_{i,j}} + b_j a_{j,i} - \underline{\underline{b_j a_{i,j}}} = a_j b_{j,i} + b_j a_{j,i} = LHS \end{align}

## PS

I used some tensorial notation rules.

1. $a \cdot b = a_j b_j$
2. $\left [ \nabla c \right ]_i = c_{,i}$
3. $a_j b_j = a_m b_m = \ldots = a_k b_k$ (dummy variable change)
4. $\epsilon_{ijk} = \epsilon_{kij}$ (Cyclic change of Levi-Civita symbol indices)
5. $[a \times b]_i = \epsilon_{ijk} a_j b_k$
6. $[\nabla \times a]_i = \epsilon_{ijk} a_{k,j}$
7. $\epsilon_{ijk} \epsilon_{imn} = (\delta_{jm}\delta_{kn} - \delta_{jn} \delta_{km})$ (Levi Civita contraction)
8. $\delta_{ij} a_i = a_j$
• I agree, however, somewhat mono-chromatically. Nice answer +1. Commented May 26, 2014 at 8:18
• I've always thought this was the least masochistic way to prove these identities because of the efficient notation, but this is notation the questioner likely hasn't encountered. Commented May 26, 2014 at 8:18
• @David H Yes, the other brute-force methods are not a good use of time in my estimation. With the index notation you can more clearly realize why these sort of identities are actually true! Commented May 26, 2014 at 8:20

Probably the best way to prove this is by using the $\delta_{ij}$ and $\epsilon_{ijk}$ identities: I'll use Einstein summation notation in what follows. For example, $$\delta_{ii} = \delta_{11}+\delta_{22}+\delta_{33}= 3$$ On the other hand, and with a bit more thinking, it can be shown: $$\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$ Furthermore, these symbols give us lucid formulations of the dot and cross products: $$A \cdot B = A_iB_j \delta_{ij} = A_iB_i \qquad \& \qquad A \times B = \epsilon_{ijk}A_iB_j\widehat{x_k}$$ where I denote $(\widehat{x_i})_j = \delta_{ij}$ so, we can write $A = A_1\widehat{x_1}+A_2\widehat{x_2}+A_3\widehat{x_3}$. The gradient and curl are naturally expressed in this formalism: $$\nabla f = (\partial_i f)\widehat{x_i}$$ and $$\nabla \times F = \epsilon_{ijk}(\partial_iF_j)\widehat{x_k}$$ Ok, now, I have shown you the toys. Let's play. \begin{align} A \times (\nabla \times B) &= \epsilon_{ijk} A_i (\nabla \times B)_j \widehat{x_k} \\ &= \epsilon_{ijk} A_i \epsilon_{lmj}(\partial_l B_m) \widehat{x_k} \\ &= -\epsilon_{ikj} \epsilon_{jlm} A_i (\partial_l B_m) \widehat{x_k} \\ &= -(\delta_{il}\delta_{km}-\delta_{im}\delta_{kl} )A_i (\partial_l B_m) \widehat{x_k} \\ &= \delta_{im}\delta_{kl}A_i (\partial_l B_m) \widehat{x_k} - \delta_{il}\delta_{km}A_i (\partial_l B_m) \widehat{x_k} \\ &= \delta_{im}A_i (\nabla B_m) - A_i (\partial_i B_k) \widehat{x_k} \\ &= A_i (\nabla B_i) - A_i \partial_i ( B_k \widehat{x_k}) \\ &= A_i (\nabla B_i) - (A \cdot \nabla)B \\ \end{align} By the same calculation with $A,B$ swapped, $$B \times (\nabla \times A) = B_i (\nabla A_i) - (B \cdot \nabla)A$$ However, we also know from the ordinary product rule, $$\partial_j [A_iB_i] = (\partial_j A_i)B_i+ A_i(\partial_jB_i)$$ hence, $$\nabla [A_iB_i] = (\nabla A_i)B_i+ A_i(\nabla B_i)$$ now, just assemble the pieces.

Lets see!

$$\nabla (a_1b_1+a_2b_2+a_3b_3) = \left\langle \frac {\partial(a_1b_1+a_2b_2+a_3b_3)}{\partial x},....\right\rangle$$

The first term can be expanded as $$\frac {\partial(a_1b_1+a_2b_2+a_3b_3)}{\partial x} = a_1\frac {\partial b_1}{\partial x}+b_1\frac {\partial a_1}{\partial x}+a_2\frac {\partial b_2}{\partial x}+b_2\frac {\partial a_2}{\partial x}+a_3\frac {\partial b_3}{\partial x}+b_3\frac {\partial a_3}{\partial x}$$

Now look carefully what you have to prove. It is easy from here! Though the algebra can be little tedious!