# 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. –  James S. Cook May 26 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. –  David H May 26 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! –  James S. Cook May 26 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!

-