# Problem with Curl of Curl

• Jun 30th 2008, 11:42 AM
Schrodingerscat
Problem with Curl of Curl
I have been studying vector calculus, and came across what was to me an odd identity:

$\nabla\times\nabla\times\vec{F}=\nabla(\nabla\bull et\vec{F})-\nabla^2\vec{F}$

I noticed that the two terms on the right hand side are of different character.

The first term $\nabla(\nabla\bullet\vec{F})$, is the gradient of the divergence of the field, which is a vector quantity.

The second term however, $\nabla^2\vec{F}$ is a scalar quantity.

So how are we supposed to do addition between a vector and a scalar?

Something like a complex number?

Any help with understanding this would be great.

Thank you.
• Jun 30th 2008, 12:39 PM
Plato
Why do you say that $\nabla ^2 \overrightarrow F$ is a scalar?
By common understanding, $\nabla ^2 = \nabla \cdot \nabla = \frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }} + \frac{{\partial ^2 }}{{\partial z^2 }}$ which is a scalar.
Therefore $\nabla ^2 \overrightarrow F$ is a vector.
• Jun 30th 2008, 12:48 PM
Schrodingerscat
Isn't the Laplacian Operator the divergence of the gradient? Divergence is a scalar quantity. That was my point.

$\nabla^2\vec{F}=\nabla\bullet\nabla\vec{F}$

I stand corrected, you cannot take the Laplacian of a vector field. I am not entirely sure what the book is saying now.
• Jun 30th 2008, 01:02 PM
Plato
As I said above, by common understanding $\nabla ^2 = \nabla \cdot \nabla = \frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }} + \frac{{\partial ^2 }}{{\partial z^2 }}$ which is a scalar.
Therefore it is scalar operator applied to a vector field: $\left( {\nabla \cdot \nabla } \right)\vec F$

To quote Davis & Snider’s text “to interpret $(\nabla ^2) \overrightarrow F$ as div grad is rather strained”.
• Jun 30th 2008, 04:20 PM
Mathstud28
Quote:

Originally Posted by Schrodingerscat
I have been studying vector calculus, and came across what was to me an odd identity:

$\nabla\times\nabla\times\vec{F}=\nabla(\nabla\bull et\vec{F})-\nabla^2\vec{F}$

I noticed that the two terms on the right hand side are of different character.

The first term $\nabla(\nabla\bullet\vec{F})$, is the gradient of the divergence of the field, which is a vector quantity.

The second term however, $\nabla^2\vec{F}$ is a scalar quantity.

So how are we supposed to do addition between a vector and a scalar?

Something like a complex number?

Any help with understanding this would be great.

Thank you.

Maybe your misunderstanding was purely notational, but for those out there viewing this, I believe this is gotten anti-climactically by the triple product

$a\times(b\times{c})=b(a\cdot{c})-c(a\cdot{b})$
• Jul 1st 2008, 02:38 AM
topsquark
In case it helps the components of
$\nabla ^2 \overrightarrow F$
are
$\left ( \nabla ^2 \overrightarrow F \right ) _i = \nabla ^2 (F_i)$

(This is what I was taught. I believe that it is standard.)

-Dan

Edit: By the way, thank you Plato for a better way to indicate a vector. The \vec{ } command seems kind of "wimpy" to me. :)