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

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:

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

• Also, why do you think you should arrive at the expression at the end? Where does that expression come from? And why do you think you "should" arrive at anything clean? There is a point at which the clean expressions of vector calculus give out, and the notation necessarily becomes messier (and, say, you'd just stop at the explicit coordinate expression you've already found); just where this point sits is a subjective, personal thing, but I for one tend to think that the coordinate expression you've found is clearer and more transparent than the opaque monster at the bottom. – E.P. Oct 30 at 12:21
• The notation $\vec{\nabla} \cdot \left (\vec{a} \cdot \vec{b} \right )$ makes no sense, as the object inside the parentheses is a scalar, and the only way it can interact with $\nabla$ is via the gradient, $\vec{\nabla}\left (\vec{a} \cdot \vec{b} \right )$. You've tried to write its divergence, which is only defined for vector objects. – E.P. Oct 30 at 12:30
• (Also, I'd argue that $\nabla$ is already a vector object, and that $\vec\nabla$ is redundant notation which only serves to visually complicate formulae without adding anything at all, but that's just my opinion.) – E.P. Oct 30 at 12:31
• The second dot is wrong too. The correct way to denote directional derivatives is $(\vec a\cdot\nabla)\vec b$. The directional derivative $\vec a\cdot\nabla$ is a scalar operator and cannot be dotted with anything. – E.P. Oct 30 at 13:32

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.

• Identity $(1)$ is a little more palatable when $\vec{a}=\vec{b}$ which then reads $\nabla (\frac{1}{2}a^2) = \vec{a} \times \text{curl}(\vec{a}) + (\vec{a}\cdot \nabla )\vec{a}$. If $\vec{a}=\vec{u}$ is the velocity this has some relevance to fluid mechanics in that you can use it to prove Bernoulli's theorem. – jacob1729 Oct 30 at 15:55
• "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"." --------Griffith used this identity in differentiating the Lienard-Wiechert potentials. – verdelite Oct 30 at 19:38
• Sometime one just needs weird things. I once needed that for any three vector fields $a \times (\nabla\times(b\times c)) + b \times (\nabla\times(c\times a)) + c \times (\nabla\times(a\times b))\\ - (a\times b)(\nabla \cdot c) - (b\times c)(\nabla \cdot a) - (c\times a)(\nabla \cdot b) \\ = \nabla(a\cdot(b\times c))$ – mike stone Oct 30 at 19:45
• @all Yes, I agree, sometimes one just needs weird things. But that doesn't mean that the form $(1)$ has any right to call itself "right" by any stretch. – E.P. Oct 30 at 20:35