# How to simplify expressions with del (or nabla) in them?

I always find it difficult to simplify expressions or open brackets in expressions that have a 'Del' (or 'Nabla') in them.

For example, how would one go about simplifying this expression?:

$$\nabla\boldsymbol{\cdot}(\phi\nabla\psi)$$

($$\phi$$ and $$\psi$$ are both scalar fields)

I need it to become: $$[\phi\nabla^2\psi + (\nabla\phi)\boldsymbol{\cdot}(\nabla\psi)]$$

I would also love to know how to simplify those standard equations mentioned in Griffiths (for example - the expansion of the 'curl of the curl' of a vector field)

The only method I know is to

1. find out every single term in the expression (in terms of $$a_x$$, $$a_y$$ etc.)
2. and then cancel out the terms
3. and then find patterns and regroup the terms in the remaining expression

Is there a faster way to approach these 'simplify' (or 'expand') problems? Maybe there are some tricks or formulas that I am unaware of (maybe something analogous to the uv-rule for differentiating the product of two functions in simple calculus) $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$

I understand that the uv-rule seems to work on my original expression. But I would still love some sort of formalization. The problem I have is that, in simple calculus, multiplying two functions does not have two meanings.

With Nabla however, I have two choices - Dot product and Cross product.

And I also have three choices for differentiation - Gradient, Divergence and Curl

To explain my concern better, try answering what would have been the simplification if the original expression was - $$\nabla \times (\phi\nabla\psi)$$

or maybe $$\nabla(v\boldsymbol{\cdot}\nabla\psi)$$ where $$v$$ is a vector-field

For the analogy, these three questions become the same question -

"Differentiation of something multiplied by the differential of something else"

• It is exactly analogous, try the product rule with $u\cdot v’$ instead of $uv$ and you’ll see Commented Jul 23, 2020 at 12:39
• Yeah, I noticed exactly that. But I would still love a proper formula or at least some sort or formalization. The problem I have is that, in simple calculus, multiplying two functions does not have two meanings. With Nabla however, I have two choices - Dot product and Cross product. To explain my concern better, try answering what would have been the simplification if the original expression was - $\nabla \times (\phi\nabla\psi)$ Commented Jul 23, 2020 at 12:44
• en.wikipedia.org/wiki/Vector_calculus_identities may be what you're looking for. In particular, the example you mention appears under "Summary of important identities : Second derivatives". Commented Jul 23, 2020 at 12:48
• Suffix notation and Einstein summation convention helps. Also remember $\require{color}{\color{red}\dots}\partial_i f{\color{blue}\dots}={\color{red}\dots}(\partial_i f){\color{blue}\dots}+{\color{red}\dots}f\partial_i{\color{blue}\dots}$ Commented Jul 23, 2020 at 12:48
• @user10354138 - your comment looks interesting. Can you explain it further. I am aware of the convention, but can you help me understand how it helps here? Commented Jul 23, 2020 at 13:11

You are computing the divergence of the vector field $$\left(\phi \frac{\partial \psi}{\partial x_i}\right)_{i=1,\cdots,n}$$, so you just get $$\sum_{i=1}^n \frac{\partial}{\partial x_i} \left(\phi \frac{\partial \psi}{\partial x_i}\right)$$

using the product rule you simply get $$\sum_{i=1}^n \left(\frac{\partial \phi}{\partial x_i} \frac{\partial \psi}{\partial x_i} + \phi \frac{\partial^2 \psi}{\partial x_i^2}\right) = \nabla \phi \cdot \nabla \psi + \phi \nabla^2\psi$$

once you know the result, you can "build" some mnemonic related to the product rule, but you still need to know what first and second order operators you must use.

• I like this answer as a proof, but don't you think that it is based on the tedious process that I mentioned in my question? Sure, the summation signs make it compact, but it's still the same thing. I was really hoping I'd learn some new trick from this question. Anyways, thanks for the answer! Commented Jul 23, 2020 at 13:19
• @TANMAYJOHRI It is not that tedious... and there is really no other process that does not involve memorizing some multivariable flavor of the derivative of a product. Commented Jul 23, 2020 at 13:44

There are many identities in vector calculus that can be referred to simplify such expressions.

Using $$\nabla\cdot(\phi\mathbf A)=\phi\nabla\cdot\mathbf A+(\nabla\phi)\cdot\mathbf A$$, which looks analogous to the product rule of differentiation, you get $$\nabla.(\phi\vec\nabla\psi)=\phi\nabla^2\psi + (\vec\nabla\phi).(\vec\nabla\psi)$$

EDIT:

Consider $${\color{red}{\mathbf C}}\times(\mathbf A\times\mathbf B)=\mathbf A(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf B)-\mathbf B(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf A)$$ and $${\color{red}{\mathbf\nabla}} \times (\mathbf{A} \times \mathbf{B}) \ =\ \mathbf{A}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{B}) \,-\, \mathbf{B}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{A}) \,+\, (\mathbf{B} {\cdot} \nabla) \mathbf{A} \,-\, (\mathbf{A} {\cdot} \nabla) \mathbf{B}$$ .

Where is the analogy? I think after deriving a few formulae listed in the link attached, you could tell apart where the analogy works and where it doesn't.

• Yeah, this identity works here perfectly! But I was hoping for some trick that would help me correctly apply the analogy everytime. Do you happen to know any such trick? Thanks for the answer by the way! Commented Jul 23, 2020 at 13:21
• @TANMAYJOHRI Have a look at the edit. Commented Jul 23, 2020 at 13:31
• @TANMAYJOHRI I think you should specifically ask, "Where does the analogy work...?" in your title. Commented Jul 23, 2020 at 13:37

I'll expand what I've commented above here.

Using suffix notation and summation convention (since we are working with $$\mathbb{R}^n$$ there is no need to distinguish upstairs and downstairs indices, so just write everything downstairs), you can get, for example, \begin{align*}\require{color} [\nabla\times(\mathbf{A}\times\mathbf{B})]_i &=\epsilon_{ijk}\partial_j(\mathbf{A}\times\mathbf{B})_k\\ &=\epsilon_{ijk}\epsilon_{k\ell m}\partial_j A_\ell B_m\\ &=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})\partial_j A_\ell B_m\\ &={\color{red}\partial_j A_i B_j}-{\color{blue}\partial_j A_j B_i}\\ \end{align*} Note that, by convention, $$\partial$$ acts on everything to its right. Whereas in the case of $$\mathbf{C}\times(\mathbf{A}\times\mathbf{B})$$ we would just "take out" the $$A_i$$ and $$B_i$$ from the red and blue term respectively, here we can't move them in front of the differential operator without paying for the noncommutativity: \begin{align*} {\color{brown}\partial_j A_i} B_j&={\color{brown}A_i\partial_j}B_j+({\color{brown}\partial_jA_i})B_j\\ &=A_i\partial_jB_j+B_j\partial_jA_i\\ \partial_j A_j B_i&=B_i\partial_jA_j+A_j\partial_jB_i. \end{align*} So $$\nabla\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}$$ and you can see the additional terms are precisely what we get from moving something behind a $$\nabla$$ to in front of that $$\nabla$$.

Now we have done the calculation, you might reasonably ask the question: Can one immediately get from $$\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-\mathbf{B}(\mathbf{C}\cdot\mathbf{A})$$ to a formula for $$\nabla\times(\mathbf{A}\times\mathbf{B})$$? To start with, we check the naive substitution still gives terms that make sense (i.e., it doesn't leave dangling $$\nabla$$). Then we see the formula involves pushing $$\mathbf{A}$$ (or $$\mathbf{B}$$ in the second term) in front of $$\mathbf{C}$$, so we need to compensate that by having something from $$\nabla\mathbf{A}$$ (or $$\nabla\mathbf{B}$$). So the formula has to read something like $$\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\nabla\mathbf{A})\ast\mathbf{B}-(\nabla\mathbf{B})\ast\mathbf{A}$$ where $$\ast$$ does some contraction between the rank-2 tensor and the vector. Now it is not hard to see in $$(\nabla\mathbf{A})\ast\mathbf{B}$$ the $$\mathbf{B}$$ has to contract with the $$\nabla$$ rather than $$\mathbf{A}$$ (because the term we are correcting has that), hence we obtain $$\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}.$$

Curl of a curl: Similarly, $$\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-(\mathbf{C}\cdot\mathbf{A})\mathbf{B}$$ The right hand side still make sense when $$\mathbf{A}=\mathbf{C}=\nabla$$. $$\nabla\times(\nabla\times\mathbf{B})=\nabla(\nabla\cdot\mathbf{B})-(\nabla\cdot\nabla)\mathbf{B}+\color{red}\text{correction}$$ We note that we didn't push any vector field pass a $$\nabla$$, so there are no correction terms. (We did change the order of $$\mathbf{C}$$ and $$\mathbf{A}$$ in $$\mathbf{A}(\mathbf{C}\cdot\mathbf{B})$$ but they are both the differential operator $$\nabla$$ so the symmetry of partial derivatives means there are no correction term).

However, I'd seriously advice against doing this eyeballing for anything more complicated. To see why, think about $$(\mathbf{A}\times\nabla)\times\mathbf{B}=\mathbf{A}\cdot\nabla\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})=\mathbf{A}\times(\nabla\times\mathbf{B})+(\mathbf{A}\cdot\nabla)\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})$$.