The strange character of operator $\nabla$

Question

I was first introduced to the mathematical operation gradient, divergence and curl not in Mathematics but during my studies of Electromagnetism. As you all know learning Maths from a Physics teacher always leads to some gigantic misconceptions.

I studied that divergence of a vector field $\mathbf A$ is $$ div~\mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} $$

And similarly divergence and curl were defined (by writing the div and curl before the vector valued function on LHS). After this the symbol $\nabla$ was introduced and it was said in my book (Feynman Lectures on Physics Vol 2, Griffiths Introduction to Electrodynamics) that $\nabla$ was a vector $$ \nabla =\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle$$So, divergence is our normal dot product, divergence of any vector field $\mathbf{A}$ can be written as $$ div~\mathbf{A} = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle ~\cdot~ \langle A_x, A_y , A_z \rangle$$ $$ div~\mathbf{A} = \nabla \cdot \mathbf{A}$$ So, the divergence is just the dot product of $\nabla$ with the field whose divergence we want. My first doubt is that in vector algebra we can write $$ \mathbf A \cdot \mathbf B = \mathbf B \cdot \mathbf A $$ but when it comes to our $\nabla$ we find $$ \nabla \cdot \mathbf A \neq \mathbf A \cdot \nabla$$ the RHS in the above relation is something else.

Second problem comes when we define the product $\nabla$ with some other vector, we know from vector algebra $$ \mathbf A \cdot \left( \mathbf B \times \mathbf C \right) = \mathbf B \cdot \left ( \mathbf C \times \mathbf A \right ) = \mathbf C \cdot \left ( \mathbf A \times \mathbf B \right )$$ Now, if we replace $\mathbf A$ by $\nabla$ then $$ \nabla \cdot \left ( \mathbf B \times \mathbf C \right) \neq \mathbf B \cdot \left ( \mathbf C \times \nabla \right) \neq \mathbf C \cdot \left ( \nabla \times \mathbf B \right)$$

Some people say $\nabla \cdot \left (\mathbf B \times \mathbf C\right) $ should be seen as the derivative of a product, even if we accept it that way then also we have few problems, we know $$ \frac{d}{d\vec r} \left( \mathbf B (\vec r) \times \mathbf C (\vec r) \right) = \mathbf B'(\vec r) \times \mathbf C (\vec r) + \mathbf B(\vec r) \times \mathbf C '(\vec r) $$ but replacing $\frac{d}{d\vec r}$ by $\nabla$ and writing the RHS as it is is not that indisputable, you see we got many choices $$ \nabla \cdot \left (\mathbf B \times \mathbf C \right) = \left ( \nabla \cdot \mathbf B \right) \mathbf C + \mathbf B \left ( \nabla \cdot \mathbf C\right) $$

$$\nabla \cdot \left (\mathbf B \times \mathbf C \right) = \left (\nabla \times \mathbf B \right) \mathbf C + \mathbf B \left ( \nabla \times \mathbf C\right)$$

$$ \nabla \cdot \left (\mathbf B \times \mathbf C \right) = \left ( \mathbf B \times \nabla \right) \mathbf C + \mathbf B \left ( \nabla \times \mathbf C\right)$$
There are three more but I'm not writing it as you all have got an idea about what I'm saying. I want to know why we chose this one $$\nabla \cdot \left( \mathbf A \times \mathbf B \right) = (\nabla \times \mathbf A) \cdot \mathbf B + \mathbf A \cdot ( \mathbf B \times \nabla)$$ from the others.

I request you all to please describe the actual character of operator $\nabla$ and clarify my doubts that I have described above. I need an explanation of why $\nabla$ was defined in such a strange way.

Answer 1

It’s not $\nabla$ which is behaving strangely, it’s $\frac{d}{d x}$. You note that $A\cdot\nabla\ne\nabla\cdot A$, by which you mean $A\cdot\nabla f\ne\nabla\cdot(A f)$ in general for a test function $f$. But it’s also true that $g\frac{d}{d x}\ne\frac{d}{d x}g$, in the same sense that $g\frac{d f}{d x}\ne\frac{d}{d x}(g f)$ in general. That second statement is about functions $f, g:{\mathbb R}\to{\mathbb R}$ and has nothing to do with vectors. You wouldn’t think to treat $\frac{d}{d x}$ as a real number, so you shouldn’t treat $\nabla$ as a vector and expect everything to work.

For better or for worse, it turns out that grad, div and curl can conveniently be written as $\nabla f$, $\nabla\cdot A$ and $\nabla\wedge A$ respectively. It’s traditional to write $\nabla^2$ for the operator $\nabla^2 f=\nabla\cdot(\nabla f)$, and sometimes also for the operator $\nabla^2 A=\nabla(\nabla\cdot A)$. Finally, $\nabla\wedge\nabla=0$ in the sense that $\nabla\wedge(\nabla f)=0$. But for more complicated relations between these operators, you should not expect any such neat coincidence to occur.

Answer 2

All properties follow directly from the definition equation 2, and the definition of dot and vector products. By the way, if the vectors A, B, C are constant the same rules apply as for ordinary vectors.

The best way to deal with such quantities is to drop the vector and vector product notation and work with the 3D fully antisymmetric Levi-Civita tensor $\epsilon_{ijk}$, which is 1 if ijk is an even permutation of 123, -1 if it is an odd permutation and otherwise 0. With this $$\nabla \cdot \left( \mathbf A \times \mathbf B \right) = \nabla_i \epsilon_{ijk} A_j B_k \,.$$ Summation over i,j,k is understood. A useful relation is $$\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$$.