# The strange character of operator $\nabla$

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.

• $\mathbf\nabla$ is an operator, i.e. it is meant to operate on the the thing written on it's right side. Therefore, writing $\mathbf A\cdot \mathbf\nabla$ doesn't make sense. You are just overstretching the notation. – Thomas Fritsch Feb 15 at 9:09
• @ThomasFritsch It does make sense as it is follows from the chain rule of differentiation. You can see the whole expression as an operator. – my2cts Feb 15 at 9:47
• I think overstretching the notation (as a computer scientist I might say we are overloading it) is the key problem - the standard treatment ignores the operator character of $\nabla$, pretends it is a vector and just avoids doing formulas where things turn strange because students are not yet used to operators. – Anders Sandberg Feb 15 at 10:34
• Re As you all know learning Maths from a Physics teacher always leads to some gigantic misconceptions. Learning mathematics from a mathematics teacher oftentimes leads to even greater misconceptions. The problem is that you haven't developed a tolerance for abuse of notation. Mathematics, like physics, is chock full of abuses of notation. Another way to put it: "The student of mathematics has to develop a tolerance for ambiguity. Pedantry can be the enemy of insight." Gila Hanna – David Hammen Feb 15 at 13:13
• @Knight On the contrary, it is completely clear and consistent. – my2cts Feb 15 at 15:01

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 $$\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$$.