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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.

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I cannot really understand what you are trying to do. But I just wanted to mention that in your formula for $\vec r$ your $\theta$ and $\phi$ are reversed from usual usage. – Martin Argerami Jan 26 '14 at 12:37
There are two "usual" usages, unfortunately xD (this one blew my cover, I'm a physicist, we tend to use this one... I don't know why, maybe some frustrated fellow wanted to piss off mathematicians :D) – Schlomo Jan 26 '14 at 12:37
up vote 2 down vote accepted

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*}

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Ooooh, yes, yes, of course they are! A careless mistake, as I suspected, thank you! :) – Schlomo Jan 26 '14 at 14:11
You're very welcome. :) That's the sort of thing one ironically becomes less likely to notice the longer one stares. – Andrew D. Hwang Jan 26 '14 at 15:05
Hahaha, yes, exactly! – Schlomo Jan 26 '14 at 17:16

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