# Why should the position vector be noted as $R\hat{R}$ in spherical polar coordinates?

Why should the position vector be noted as $$R\hat{R}$$ in spherical polar coordinates? Now i did the calculation like this: $$\vec R = R \sin\theta \cos\phi \hat{i} + R \sin\theta \sin\phi \hat{j} + R \cos\theta \hat{k}$$ so now i am manipulating the unit vectors. As :- $$\hat{R}= \frac{\partial R}{\partial R}$$/$$| \frac{\partial R}{\partial R}|$$ = $$\sin\theta \cos\phi \hat{i} + \sin\theta \sin\phi \hat{j} + \cos\theta \hat{k}$$ by doing similiar calculations i found $$\hat{\theta}$$= $$\cos\theta \cos\phi \hat{i} + \cos\theta \sin\phi \hat{j} -\sin\theta\hat{k}$$. Similiarly i found $$\hat{\phi}$$ =$$\cos\phi \hat{j} - \sin\phi\hat{i}$$ now position vector can be written as $$\vec R= [\vec R. \hat{R}]\hat{R} + [\vec R. \hat{\theta}]\hat{\theta} + [ \hat{\phi}. \vec{R}] \hat{\phi}$$. Which gives me $$\vec{R} = R\hat{R} + R\sin\theta \hat{\phi}$$ not $$R\hat{R}$$ now where i am misunderstanding or miscalculating ?

• Maybe you made a mistake somewhere. But look at your expression of $\hat{R}$. Simply multiply it by $R$ and you get your first expression $\vec{R}$. The $\hat{\phi}$ vector is wrong. Look at en.wikipedia.org/wiki/Spherical_coordinate_system – E. Bellec Jan 14 '19 at 9:24
• @E.Bellec i am actually stuck on doing the derivation. Checked my results many time , so needed some help. Although i totally agree with you. – Nobody recognizeable Jan 14 '19 at 9:26
• Many time ? your $\hat {\phi}$ seems wrong – Layla Jan 14 '19 at 10:51
• @Reign ah crap gotta try again. – Nobody recognizeable Jan 14 '19 at 10:53
• @Reign well in cylindrical coordinates i found the radial vector that was $\rho \hat{\rho}$ so wanted to confirm for spherical coordinates. Made a crappy childish mistake and gotta try again. In other words i am reexamining my method of the derivation is right or not. I see where was my mistake. – Nobody recognizeable Jan 14 '19 at 10:58

The unit vectors for spherical coordinates are obtained by taking the derivatives of the coordinate transformation with respect to $$r$$, $$\theta$$, $$\phi$$, and normalizing to 1 if needed:
$$\begin{cases} x=R\sin\theta\cos\phi\\ y=R\sin\theta\sin\phi\\ z= R\cos\theta \end{cases}$$
This is because the vectors $$\frac{\partial \vec x}{\partial R}$$, $$\frac{\partial \vec x}{\partial \theta}$$, $$\frac{\partial \vec x}{\partial \phi}$$ tell you which are the directions you move when you “turn the knob” on one of your three coordinates $$r$$, $$\theta$$ or $$\phi$$.
$$\hat\phi=-\sin\phi \hat i +\cos\phi \hat j$$
The vector you wrote, if you check, has a direction which is radial in the $$x-y$$ plane, while it should be tangent to the circle described in $$x-y$$ plane by the $$\phi$$ angle;