Again, sorry for the naff title.

Find $\displaystyle \nabla\times(r\mathbf{r})$, where $\displaystyle r=|\mathbf{r}|$ and $\displaystyle \mathbf{r}=(x,y,z)$.

I ended up with an answer of $\displaystyle (0,0,0)$, which, again, I doubt is right.

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- Nov 29th 2009, 11:49 AMchella182Gradient, divergence and curl - nabla crossed with |r|r
Again, sorry for the naff title.

*Find $\displaystyle \nabla\times(r\mathbf{r})$, where $\displaystyle r=|\mathbf{r}|$ and $\displaystyle \mathbf{r}=(x,y,z)$.*

I ended up with an answer of $\displaystyle (0,0,0)$, which, again, I doubt is right. - Dec 1st 2009, 09:54 PMqmech
You're right. A radial field has no curl.

- Dec 2nd 2009, 02:48 AMchella182
Cheers. Yeah, like I said in my other thread, turns out I was doing all of this shizz right when I thought I was doing it wrong :p