Show that $\vec\nabla\times(\vec\nabla\times\vec A )=- \nabla^ 2\vec A + \vec\nabla (\vec\nabla\cdot\vec A )$

Question

I just started learning this and I don't understand much so how can I prove this? $$\vec\nabla\times(\vec\nabla\times\vec A )=-\vec\nabla^2\vec A +\vec\nabla (\vec\nabla\cdot\vec A )$$

Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. — Another User, Feb 07 '23 at 14:57
What have you tried? Have you attempted a brute force approach by carrying out the derivatives? — Mark Viola, Feb 07 '23 at 15:05
Possible duplicate: https://math.stackexchange.com/a/1108604/583883 — student91, Feb 07 '23 at 15:06

score 0 · Answer 1 · edited Jun 10 '23 at 08:30

0

$\def\n{\nabla}$The vector aspect can be addressed using the triple product (aka bac-cab) rule $a\times(b\times c) = b\,(a\cdot c)- c\,(a\cdot b)$, but the derivative aspect requires keeping the $\n$ operator on the LHS of each expression
$\eqalign{\n\times(\n\times c) &= \n\big(\n\cdot c\big)- \big(\n\cdot \n\big)\,c \\ &= \n\big(\n\cdot c\big)- \n^2c \\ }$

edited Jun 10 '23 at 08:30

user1942348

3,871

answered Feb 08 '23 at 17:37

greg

35,825

Show that $\vec\nabla\times(\vec\nabla\times\vec A )=- \nabla^ 2\vec A + \vec\nabla (\vec\nabla\cdot\vec A )$

1 Answers1