# Curl of the position vector

1. Jan 9, 2013

### blackhole007

1. The problem statement, all variables and given/known data

compute the curl of:

$$\vec{r}$$
and
$$\frac{\vec{r}}{r^3}$$

2. Relevant equations

$$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$

$$r^3=(x^2+y^2+z^2)^\frac{3}{2}$$

3. The attempt at a solution

I figured out that the curl of $$\vec{r} = 0$$ as my book says it should be....

however...I also need to prove the second vector field up there also is equal to 0...Is it as simple as saying this?:

since $$∇\times\vec{r}=0$$... therefore, $$∇\times\vec{r}(\frac{1}{r^3})=0$$ as well. ?????

My book states that we should be able to see that these equal zero before doing the calculations to save us time

Last edited: Jan 9, 2013
2. Jan 9, 2013

### Dick

It's simple to say that. To prove it you need a vector calculus identity about the curl of a vector times a function. Do you know one? And yes, if you have that, there is no need do the whole calculation.

3. Jan 9, 2013

### blackhole007

We have not learned one yet, the last problem I did was a proof of $$∇\times∇ψ(\vec{r})=0$$ and that was the problem directly before this, so no identities like you mentioned thus far.

4. Jan 9, 2013

### Dick

Ok, that works too. What's the gradient of 1/r?

Last edited: Jan 9, 2013
5. Jan 9, 2013

### blackhole007

gradient of 1/r is $$-\frac{\vec{r}}{r^3}$$

6. Jan 9, 2013

### Dick

Ok, so? The curl of that must then be zero. That's what you proved in the previous exercise, right? How does that relate to your problem?

Last edited: Jan 9, 2013
7. Jan 9, 2013

### blackhole007

Cool thanks for that connection i totally missed haha...does this look sufficient?