How to further expand $\text{grad} \left( \vec{a} \cdot\vec{b} \right ) = \vec{\nabla} \left (\vec{a} \cdot\vec{b} \right )$?

Question

With $\vec{a}, \vec{b}: \mathbb{R}^3 \to \mathbb{R}^3$ vector fields:

I want to expand $\text{grad} \left( \vec{a} \cdot \vec{b} \right ) = \vec{\nabla} \left (\vec{a} \cdot \vec{b} \right )$.

So I started with:

$\left [\vec{\nabla}\left (\vec{a} \cdot \vec{b} \right ) \right ]_i = \partial_i\left (a_j b_j \right ) \overset{\text{Product rule}}{=} \left ( \partial_i a_j \right ) b_j + a_j \left( \partial_i b_j \right )$

But where to go from there? In the end I'm supposed to arrive at:

$\text{grad} \left( \vec{a} \cdot \vec{b} \right ) =\vec{a} \times \left(\vec{\nabla} \times \vec{b}\right) + \vec{b} \times \left(\vec{\nabla} \times \vec{a}\right) + \left(\vec{b} \cdot\vec{\nabla}\right) \vec{a} + \left(\vec{a} \cdot\vec{\nabla} \right) \vec{b}$

Answer 1

So I started with:

$\left [\vec{\nabla}\left (\vec{a} \cdot \vec{b} \right ) \right ]_i = \partial_i\left (a_j b_j \right ) \overset{\text{Product rule}}{=} \left ( \partial_i a_j \right ) b_j + a_j \left( \partial_i b_j \right )$

But where to go from there?

Why do you feel the need to "go" anywhere from there? That's a perfectly acceptable answer.

I think, in the end I should arrive at:

$\vec{a} \times \left(\vec{\nabla} \times \vec{b}\right) + \vec{b} \times \left(\vec{\nabla} \times \vec{a}\right) + \left(\vec{b} \cdot\vec{\nabla}\right) \cdot\vec{a} + \left(\vec{a} \cdot\vec{\nabla} \right) \cdot\vec{b}$

If you have some external reason to conclude that this form is, because of your particular circumstances, more useful than anything else, then to prove that $$ \vec{\nabla}\left (\vec{a} \cdot \vec{b} \right ) = \vec{a} \times \left(\vec{\nabla} \times \vec{b}\right) + \vec{b} \times \left(\vec{\nabla} \times \vec{a}\right) + \left(\vec{b} \cdot\vec{\nabla}\right) \cdot\vec{a} + \left(\vec{a} \cdot\vec{\nabla} \right) \cdot\vec{b} \tag 1 $$ the simplest way is to start with the right-hand side and show that it reduces to the component identity that you've already found, $$ \vec{\nabla}\left (\vec{a} \cdot \vec{b} \right ) = \hat e_i ( \partial_i a_j ) b_j + \hat e_i a_j ( \partial_i b_j ). \tag 2 $$ To do that, you work each term individually, so e.g. \begin{align} \vec{a} \times \left(\vec{\nabla} \times \vec{b}\right) & = \hat e_i \epsilon_{ijk} a_j \epsilon_{klm}\partial_l b_m \\& = \epsilon_{kij}\epsilon_{klm} \hat e_i a_j \partial_l b_m \\& = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \hat e_i a_j \partial_l b_m \\& = \hat e_i a_j \partial_i b_j - \hat e_i a_j \partial_j b_i, \tag 3 \end{align} and then you work from there.

However, to be honest, I'm not sure there are any particular circumstances in which showing the identity $(1)$ is useful, other than "a textbook problem asked me to".

---

What do you do in the 'real' world? Well, there's a ton of different notations for $(2)$, depending on what you want to do (say, you might introduce $\nabla\vec a$ as a matrix, and then use notation like $(\nabla\vec a)\cdot \vec b $, being careful to specify how the left and right dot products of that matrix are defined in terms of the matrix indices, or you might use notation like $\nabla(\vec a\cdot \vec b) = \nabla_\vec{a}(\vec a\cdot \vec b) + \nabla_\vec b(\vec a\cdot \vec b)$ where the subscripts indicate what function the differential operator acts on, among other alternatives) but there's no one-size-fits all "best" notation for the object that you've already found. It comes down to personal preference and what fits best the requirements of the calculation you're working in.