2
$\begingroup$

How find the divergence and Curl of the following:

$(\vec{a} \cdot \vec{r}) \vec{b}$,

where $\vec{a}$ and $\vec{b}$ are the constant vectors and $\vec{r}$ is the radius vector.

I have tried solving this by supposing $\vec{r} = (x,y,z)$ and got answer as

div($(\vec{a} \cdot \vec{r}) \vec{b}$) = $\vec{a} \cdot \vec{b}$

but I was wondering if anybody could help me to solve it by using the formulas involving Nabla Operator.

Thanks

$\endgroup$
2
$\begingroup$

If we use the well-known formula

$\nabla \cdot (f \vec X) = \nabla f \cdot \vec X + f \nabla \cdot \vec X, \tag{1}$

where $f$ is scalar function and $\vec X$ is a vector field (see http://en.m.wikipedia.org/wiki/Vector_calculus_identities), we find, since $\vec b$ is constant,

$\nabla \cdot ((\vec a \cdot \vec r) \vec b) = \nabla (\vec a \cdot \vec r) \cdot \vec b; \tag{2}$

at this point we pretty much have to use

$\vec a \cdot \vec r = a_x x + a_y y+ a_z z, \tag{3}$

where

$\vec a = (a_x, a_y, a_z)^T \tag{4}$

to obtain

$\nabla (\vec a \cdot r) = \vec a, \tag{5}$

whence

$\nabla \cdot ((\vec a \cdot \vec r) \vec b = \vec a \cdot \vec b. \tag{6}$

We may find $\nabla \times ((\vec a \cdot \vec r) \vec b$ in an analogous manner, from (see the same linked citing)

$\nabla \times (f \vec X) = \nabla f \times \vec X + f\nabla \times X; \tag{7}$

again since $\vec b$ is constant, and by (5)

$\nabla \times ((\vec a \cdot \vec r) \vec b) = \nabla (\vec a \cdot \vec r) \times \vec b = \vec a \times \vec b. \tag{8}$

We can use the vector identities to simplify the calculations somewhat, but at some point the actual coordinate expression for $\vec a \cdot \vec r$ must be invoked to finalize the work.

$\endgroup$
0
$\begingroup$

Hint:

$ \nabla \cdot (\phi \vec b)=\vec b \cdot \nabla \phi+\phi \nabla \cdot \vec b \qquad$ and $ \qquad \nabla \times (\phi \vec b)=\phi(\nabla \times\vec b) + (\nabla \phi) \times \vec b $

Where $\phi$ is the scalar function $(\vec a\cdot \vec r)$ (so you can easely find $\nabla \phi=\vec a$) and $\vec b$ is a constant.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.