Dot product of a vector and del operator

$\begingroup$

How did this expansion come about.enter image description here

What is the physical significance of this expansion.

And what is the significance of $$\vec{A} . \nabla {\vec{A}}$$ And what does this mathematically represent.

$\endgroup$ 1

1 Answer

$\begingroup$

$\newcommand{\col}[3]{\left(\begin{matrix} #1 \\ #2 \\ #3 \end{matrix}\right)}$ $\newcommand{\vecc}[1]{\col{#1_x}{#1_y}{#1_z}}$ $$ \vec A \times (\nabla \times \vec A) = \vecc A \times \left( \vecc \partial \times \vecc A \right) = \vecc A \times \col{\partial_y A_z - \partial_z A_y}{\partial_z A_x - \partial_x A_z}{\partial_x A_y - \partial_y A_x} \\ = \col {A_y(\partial_x A_y - \partial_y A_x) - A_z(\partial_z A_x - \partial_x A_z)} {A_z(\partial_y A_z - \partial_z A_y) - A_x(\partial_x A_y - \partial_y A_x)} {A_x(\partial_z A_x - \partial_x A_z) - A_y(\partial_y A_z - \partial_z A_y)} = \col {(A_y \partial_x A_y + A_z \partial_x A_z) - (A_y \partial_y + A_z \partial_z) A_x} {(A_z \partial_y A_z + A_x \partial_y A_x) - (A_z \partial_z + A_x \partial_x) A_y} {(A_x \partial_z A_x + A_y \partial_z A_y) - (A_x \partial_x + A_y \partial_y) A_z} \\ = \col {(A_x \partial_x A_x + A_y \partial_x A_y + A_z \partial_x A_z) - (A_x \partial_x + A_y \partial_y + A_z \partial_z) A_x} {(A_x \partial_y A_x + A_y \partial_y A_y + A_z \partial_y A_z) - (A_x \partial_x + A_y \partial_y + A_z \partial_z) A_y} {(A_x \partial_z A_x + A_y \partial_z A_y + A_z \partial_z A_z) - (A_x \partial_x + A_y \partial_y + A_z \partial_z) A_z} = \col {\vec A \cdot \partial_x \vec A - (\vec A \cdot \nabla) A_x} {\vec A \cdot \partial_y \vec A - (\vec A \cdot \nabla) A_y} {\vec A \cdot \partial_z \vec A - (\vec A \cdot \nabla) A_z} = \col {\frac12 \partial_x (\vec A \cdot \vec A)} {\frac12 \partial_y (\vec A \cdot \vec A)} {\frac12 \partial_z (\vec A \cdot \vec A)} - \col {(\vec A \cdot \nabla) A_x} {(\vec A \cdot \nabla) A_y} {(\vec A \cdot \nabla) A_z} \\ = \frac12 \nabla (\vec A \cdot \vec A) - (\vec A \cdot \nabla) \vec A $$

Thus, $$(\vec A \cdot \nabla) \vec A = \frac12 \nabla (\vec A \cdot \vec A) - \vec A \times (\nabla \times \vec A)$$

$\endgroup$ 7

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like