What’s with the spherical basis vectors?

Question

My short calculation Regarding the representation of the point (1,1,0) in spherical coordinates led me to an error. Have I miscalculated or is my view on the point being specified by $r=\sqrt{2}$, $\phi= \pi/4$ and $\theta =\pi /2$ Somehow wrong in the first place? My heads stuck for the moment :(

Answer 1

If you want to convert to spherical coordinates, you need to do the following

1. Calculate distance from origin - $r$

$$r = ||p|| = \sqrt 2$$

2. Calculate $\theta$ - angle made by position vector with z-axis: Since the vector is in $xy$ plane, this angle will be $\frac{\pi}{2}$

3. Calculate $\phi$ - angle made by projection of position vector onto $xy$ plane with x-axis: Since the vector is already in the x-y plane, the projection is itself, and it makes an angle of $\frac{\pi}{4}$ with it

Hence, point in $(r,\theta, \phi)$ system is $(\sqrt 2, \frac{\pi}{2}, \frac{\pi}{4})$, which is what you have got in the calculations. What you did after that I do not understand

Answer 2

There are two mistakes. The first is that $r$ should be $\sqrt{2}$. The second, and perhaps more confusing thing is that in spherical coordinates, the radius vector $\mathbf{r}$ is simply $\mathbf{r} = r \mathbf{\hat{r}}$, it is NOT $ r \mathbf{\hat{r}} + \theta \mathbf{\hat{\theta}} + \phi \mathbf{\hat{\phi}}$. This is a very common misconception.

So, yes, although the spherical coordinate representation of the point $p = (1,1,0)\in \Bbb{R}^3$ is $r=\sqrt{2}, \theta = \pi/2, \phi = \pi/4$, the radius vector is still \begin{align} \mathbf{r} &= \sqrt{2}\,\, \mathbf{\hat{r}} \end{align}