I'm quite comfortable with vector calculus in all sorts of coordinate systems, but for the love of me, I can't seem to figure out where did I go wrong in this simple derivation of the position vector in spherical coordinates.
Maybe my morning coffee didn't kick in yet, but I'd still appreciate your help!
So, using spherical coordinates $(r, \theta, \phi)$, we can write the Cartesian position vector $$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$$ as $$\vec{r} = (r \sin{\theta} \cos{\phi}) \hat{i} + (r \sin{\theta} \sin{\phi}) \hat{j} + (r \cos{\theta}) \hat{k}$$
In the next step, I substitute for Cartesian unit vectors expressed in spherical unit vectors, i.e. $$\hat{i} = (\sin{\theta} \cos{\phi} )\hat{r} + (\cos{\theta} \cos{\phi} )\hat{\theta} – (\sin{\theta})\hat{\phi}$$
$$\hat{j} = (\sin{\theta} \sin{\phi} )\hat{r} + (\cos{\theta} \sin{\phi} )\hat{\theta} + (\cos{\theta})\hat{\phi}$$
$$\hat{k} = (\cos{\theta}) \hat{r} – (\sin{\theta})\hat{\theta}$$
So, when I do that, you can see that my $\hat{\phi}$ terms don't cancel out and I don't get $$ \vec{r} = r\hat{r},$$
which is what I should get.
Help me spot the mistake, my brain refuses to cooperate today.
Best Answer
According to Wikipedia, the $\hat{\phi}$ components of $\hat{\imath}$ and $\hat{\jmath}$ are, respectively, $-\sin\phi$ and $\cos\phi$; that is, \begin{align*} \hat{\imath} &= (\sin{\theta} \cos{\phi} )\hat{r} + (\cos{\theta} \cos{\phi})\hat{\theta} - (\sin{\phi})\hat{\phi}, \\ \hat{\jmath} &= (\sin{\theta} \sin{\phi} )\hat{r} + (\cos{\theta} \sin{\phi})\hat{\theta} + (\cos{\phi})\hat{\phi}, \\ \hat{k} &= (\cos{\theta}) \hat{r} - (\sin{\theta})\hat{\theta}. \end{align*}