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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 ?

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  • $\begingroup$ 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 $\endgroup$ Commented Jan 14, 2019 at 9:24
  • $\begingroup$ @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. $\endgroup$ Commented Jan 14, 2019 at 9:26
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    $\begingroup$ Many time ? your $\hat {\phi}$ seems wrong $\endgroup$
    – seVenVo1d
    Commented Jan 14, 2019 at 10:51
  • $\begingroup$ @Reign ah crap gotta try again. $\endgroup$ Commented Jan 14, 2019 at 10:53
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    $\begingroup$ @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. $\endgroup$ Commented Jan 14, 2019 at 10:58

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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 $\phi$.

You got the first two right but the third should be

$$\hat\phi=-\sin\phi \hat i +\cos\phi \hat j$$

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;

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