# What is the divergence of a unit vector not in the r direction?

by SiggyYo
Tags: direction, divergence, unit, vector
 P: 5 Hi guys, I've run across a problem. In finding the potential energy between two electrical quadrupoles, I've come across the expression for the energy as follows: $U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\hat{k}\cdot \hat{r})^3-2(\hat{k}\cdot \hat{r})^2-(\hat{k}\cdot \hat{r}))\right],$ where $\hat{k}$ is the orientation of the quadrupoles, and $\hat{r}$ is the direction between the quadrupoles. If I let $\hat{r}$ be in the $\hat{z}$-direction, I get $U_{Q}=\frac{3Q_{0}}{4r^{4}}\left[(\hat{k}\cdot \nabla)(5(\cos{\theta})^3-2(\cos{\theta})^2-(\cos{\theta}))\right].$ My problem now is, that I don't know what to do about the divergence of the $\hat{k}$-vector. I would like to do the differentiation in cartesian coordinates, but have them translated into spherical polar coordinates. I know, that the result should probably involve a $\frac{1}{r}$-factor, but I can't seem to do it right. I've tried to rewrite $\hat{k}$ in polar coordinates and tried using the chain rule on the derivative, but I get 3 as an answer. So I don't know if the initial expression is wrong, or I just dont know how to take the derivative. Can anyone please help? Thanks,
 P: 5 Thank you chiro for the quick response. I am afraid I don't know what you mean. Wouldn't I just obtain the usual $x=r\sin{\theta}\cos{\phi}$ $y=r\sin{\theta}\sin{\phi}$ $z=r\cos{\theta}$? Also, I want $\hat{k}$ to be a unit vector, which gives me $r=1$. How do I take this into account, when trying to get a result with a factor of $\frac{1}{r}$? I am really lost on this one :P