# Problem to do with curls

1. Jan 20, 2009

### Mashell

If r is the position vector in three dimensions, a is a constant vector and $$\nabla$$ is the gradient operator, evaluate

$$\nabla$$ x (a x r).

I don't understand what a constant vector is, it can't be the same as a scalar function. I've searched for a while and couldn't find anything that made sense, so any help would be vey much appreciated.

2. Jan 20, 2009

### gabbagabbahey

A constant vector is a vector whose length and direction do not depend on position.

For example; v=i+2j+687k qualifies as a constant vector. Its components are not functions of position.

In contrast, the vector v=3x2j is not a constant vector since its lentgh depends on position (specifically where you are along the x-axis)

Since constant vectors do not vary with position, their curl, divergence and gradient are all zero.

3. Jan 20, 2009

### Mashell

but if the constant vector is first "crossed" with the position vector... shouldn't it give me some values...

4. Jan 20, 2009

### Mashell

ohhhh but they cancel out... ahhh i see. Your right. Thank you gabbagabbahey :)

5. Jan 20, 2009

### gabbagabbahey

Careful...

$$\vec{\nabla}\times\vec{a}=\vec{\nabla}\cdot\vec{a}=\vec{\nabla}(\vec{a})=0$$

But that doesn't necessarily mean

$$\vec{\nabla}\times(\vec{a}\times\vec{r})=0$$

You should find the following vector identity useful:

$$\vec{\nabla}\times(\vec{A}\times\vec{B})=(\vec{B}\cdot\vec{\nabla})\vec{A}-(\vec{A}\cdot\vec{\nabla})\vec{B}+\vec{A}(\vec{\nabla}\cdot\vec{B})-\vec{B}(\vec{\nabla}\cdot\vec{A})$$

6. Jan 20, 2009

### Mashell

hmmm okay, if A is a constant vector and div A is equal to 0 then the above formula reduces to

$$\vec{\nabla}\times(\vec{A}\times\vec{B})=(2(\vec{B}\ \cdot \vec{\nabla})\vec{A})$$

right?

is

$$(\vec{B}\ \cdot \vec{\nabla}) = (\vec{\nabla}\ \cdot\vec{B})$$

7. Jan 20, 2009

### gabbagabbahey

No,

$$(\vec{B}\ \cdot \vec{\nabla}) \neq (\vec{\nabla}\ \cdot\vec{B})$$

In Cartesian coordinates,

$$(\vec{B}\ \cdot \vec{\nabla})= B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_z\frac{\partial}{\partial z}$$

which is a differential operator that operates on whatever vector is on the right of the operator.

Of course, any differential operator that operates on a constant vector will be taking some derivative of some constant and therefor gives a result of zero.

Therefor, $$(\vec{r}\cdot \vec{\nabla})\vec{a}=0$$

And

$$\vec{\nabla}\times(\vec{a}\times\vec{r})=-(\vec{a}\cdot \vec{\nabla})\vec{r}+\vec{a}(\vec{\nabla}\cdot\vec{r})$$

8. Jan 20, 2009

### gabbagabbahey

You can then use the fact that $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$ to simplify further.