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.

  • 2
    $\begingroup$ $\mathbf\nabla$ 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. $\endgroup$ Commented Feb 15, 2020 at 9:09
  • 5
    $\begingroup$ @ThomasFritsch It does make sense as it is follows from the chain rule of differentiation. You can see the whole expression as an operator. $\endgroup$
    – my2cts
    Commented Feb 15, 2020 at 9:47
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Feb 15, 2020 at 10:34
  • 3
    $\begingroup$ 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 $\endgroup$ Commented Feb 15, 2020 at 13:13
  • 1
    $\begingroup$ @Knight On the contrary, it is completely clear and consistent. $\endgroup$
    – my2cts
    Commented Feb 15, 2020 at 15:01

2 Answers 2


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.


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

  • $\begingroup$ My apologies but my mathematical knowledge has reached just the introduction of Tensor, so I’m quite unable to understand that Levi-Civita Tensor. $\endgroup$
    – user240696
    Commented Feb 15, 2020 at 13:24
  • $\begingroup$ It is just an extension of the concept of the matrix. $\endgroup$
    – my2cts
    Commented Feb 15, 2020 at 15:04
  • $\begingroup$ I learned that vectors are just one rank Tensor and we can represent tensors in a matrix array, how should I begin my study of Tensors ? $\endgroup$
    – user240696
    Commented Feb 15, 2020 at 15:07
  • $\begingroup$ You should start with linear algebra. $\endgroup$
    – my2cts
    Commented Feb 15, 2020 at 18:53
  • $\begingroup$ In this case, $\nabla_i\epsilon_{i j k}A_j B_k$ means $$\frac{d}{d x}(A_2 B_3-A_3 B_2)+\frac{d}{d y}(A_3 B_1-A_1 B_3)+\frac{d}{d z}(A_1 B_2-A_2 B_1).$$ If you are unfamiliar with tensors, my personal opinion is that this is correct but not the best approach for you right now. $\endgroup$ Commented Sep 14, 2020 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.