Directional derivative $(\mathbf{A}\cdot\nabla)\mathbf{B}$ of the vector field $\mathbf{B}$

Question

While reading Introduction to Electrodynamics by David J. Griffiths, I have encountered some issue with the notation of the directional derivative of the vector field and I was wondering if there are any simple terms to put it.

The problem is the following notation: $$(\mathbf{A}\cdot\nabla)\mathbf{B}=\mathbf{A}\cdot\nabla\mathbf{B}$$ So both $\mathbf{A}$ and $\mathbf{B}$ are vectors, say in $\mathbb{R^3}$, then what is the meaning of $\nabla\mathbf{B}$? If it is just the jacobian matrix of the vector field $\mathbf{B}:\mathbb{R^3}\rightarrow\mathbb{R^3}$ then the dot between the terms is even more confusing because then I am not sure how the dot product of a vector with the matrix would be defined. Also what is the difference between $\nabla\cdot\mathbf{A}$ and $\mathbf{A}\cdot\nabla$?

The vector identity and the notation in question appear also in this article on Wikipedia without further treatment.

Answer 1

The easiest way to understand this is to see that $\mathbf{A}\cdot\nabla$ is the usual directional derivative operator. Just like how $\nabla = \hat{\mathbf{x}}\partial/\partial x + \hat{\mathbf{y}}\partial/\partial y + \hat{\mathbf{z}}\partial/\partial z$, we have $\mathbf{A}\cdot\nabla = A_x(\partial/\partial x) + A_y(\partial/\partial y) + A_z(\partial/\partial z)$. The only difference here is that it is being applied to a vector instead of a scalar.

The meaning of $\nabla\mathbf{B}$ is exactly what it means. Just like how the gradient of a scalar is a vector, the gradient of a vector is a second-rank tensor. So $\mathbf{A}\cdot\nabla\mathbf{B}$ is the contraction of the vector $\mathbf{A}$ with this tensor. So it is effectively the dot product of $\mathbf{A}$ with each row $\partial B_i/\partial x + \partial B_i/\partial y + \partial B_i/\partial z$ of this tensor, where $i = x, y, z$. That's three dot products in total, one for each component of $\mathbf{B}$, and they constitute the components of the result.

However, looking at it from this point of view is unsatisfying because it is not obvious from the dot notation which index of the tensor is being contracted. All of this can be made proper with tensor calculus and index notation, where it is written as $A^i \nabla_i B$ (or $A^i \partial_i B$ in Cartesian coordinates). From this, it is much clearer that (1) it is manifestly coordinate-independent and (2) it is the directional derivative of $\mathbf{B}$ along $\mathbf{A}$.

Answer 2

Here are a few examples to clarify some notations:

1. $\nabla$ notation: $\nabla$ is the $del$, which is a vector differential operator that is typically used in vector calculus and it can accept two different inputs:

   • $\nabla{\mathbf{B}}$: When the input is a vector field (assigns a vector to each point in space), the gradient becomes a second-rank tensor. It captures how each component of the vector field changes concerning each spatial direction.
   • $\nabla{f}$: When the input is a scalar field (assigns a single value to each point in space), the gradient is a vector. It describes the rate of change of the scalar field in each direction and points in the direction of the steepest increase.

2. Meaning of $\mathbf{A}\cdot\nabla\mathbf{B}$:

• $\mathbf{B}$ stands for another vector field.
• $(\mathbf{A} \cdot \nabla)\mathbf{B}$ is a useful operation that combines the two quantities $\mathbf{A}$ and the gradient of $\mathbf{B}$.
• It shows the directional rate of change of $\mathbf{B}$ vector with respect to $\mathbf{A}$ vector field in that direction.

3. Difference between $\nabla\cdot\mathbf{A}$ and $\mathbf{A}\cdot\nabla$:

• $\nabla\cdot\mathbf{A}$ is the divergence of the vector field $\mathbf{A}$, which gives a scalar field filed at each point that represents the rate of expansion of the vector field.
• $A\cdot\nabla$ is a function that is applied to a scalar or vector quantity. It stands for the rate of change of the scalar field (assuming $\mathbf{A}$ to be a scalar function) or gradient of the vector field (if $\mathbf{A}$ is a vector).

To sum it up, in linear algebra you can define a matrix as a combination of column vectors. Hence, the dot product between a matrix and a vector is basically the dot product between the matrix and each column vector within the matrix, which is the sum of the corresponding values of the matrix and the vector multiplied by each other. Thus when you perform a dot product between a vector and a matrix, it may be expressed as a sum of dot products between the vector and every column of a matrix. In the case $(\mathbf{A}\cdot\nabla)\mathbf{B}$, the dot product is performed by taking element by element in $\mathbf{A}$ and crossing the result with $\nabla\mathbf{B}$.