# Curl of the position vector

## Homework Statement

compute the curl of:

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

## Homework Equations

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

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

## 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

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Dick
Homework Helper

## Homework Statement

compute the curl of:

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

## Homework Equations

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

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

## 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 :

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

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

The "(X)" is my cross product notation lol
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.

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.

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

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

Dick
Homework Helper
gradient of 1/r is $$-\frac{\vec{r}}{r^3}$$
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?

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

Dick
Homework Helper
Cool thanks for that connection i totally missed haha...does this look sufficient?
Well, no. You proved that the curl of any gradient vector is zero in the previous exercise. You then showed that the vector r over r^3 is the gradient of -1/r. So the curl of vector r over r^3 is...??? The curl of ANY gradient is zero. That's where the skipping of some calculation comes in. You don't have to repeat the previous proof.

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