How to define vectors in spherical coordinate system?

[f3sicA_A]

Homework Statement

Let us take an example of circular ring of radius $r$ on the $x$-$y$ plane centered at the origin, a vector begins at this ring and terminates on the +$z$ axis, what are components of the vector in spherical coordinate system?

Relevant Equations

$$\pmb{v}=v_r\pmb{\hat{r}}+v_\theta\pmb{\hat{\theta}}+v_\phi\pmb{\hat{\phi}}$$

I am extremely confused with how to represent vectors that do not start at the origin in spherical coordinate system. If they did start at the origin, the vector to any point is simply $r\pmb{\hat{r}}$; however, what if it does not start at the origin as in the question above? One thing I can think of is representing the final vector as a sum of two separate vectors: first is the vector from the terminal point to the origin, and second is from the origin to the end point:

$$\pmb{v}=r_1\pmb{\hat{r}}_1+r_2\pmb{\hat{r}}_2$$

Where $\pmb{\hat{r}}_1$ points towards the origin and $\pmb{\hat{r}}_2$ points towards the end point Would this be correct? If so, as far as I can think, we can represent all vectors using this convention, so then why would we even need $\pmb{\hat{\theta}}$ and $\pmb{\hat{\phi}}$?

 

[PeroK]

In spherical coordinates, each point in space is specified by three coordinates $(r, \phi, \theta)$ (or $(r, \theta, \phi)$). That gives you a unique set of coordinates for each point, except along the z-axis, where the azimuthal angle is undefined. It also gives you a prescription for getting from the origin to any point: it's a distance $r$ along a line set at the given azimuthal and polar angles.

The first thing to notice about spherical coordinates is that the basis vectors vary with position. For each point, the radial unit vector $\mathbf{\hat r}$ points away from the origin. Technically, this vector is a local vector, which applies to vectors at the point in space. However, because Euclidean space is uniform, you can translate the vector $r\mathbf{\hat r}$ to the origin to get the position vector for the given point. Note that $\mathbf{\hat r}$ itself is not well-defined at the origin. This is something of a trick, using the uniformity of Euclidean space. If and when you come to study curved manifolds, there is no unique way to translate a local vector to the origin, and so the concept of a position vector is lost!

Technically, at each point the three basis vectors define a local tangent space. If a particle has a velocity at a point, you can define this velocity as $\mathbf v = v_r\mathbf{\hat r} + v_\phi\mathbf{\hat \phi} + v_\theta\mathbf{\hat \theta}$. This again is a local vector and technically lives in the local tangent space. But, again, due to the uniformity of Euclidean space, we can move this local vector around and compare it with the velocity of another particle at another point. However, as we move this physical vector around, its components change, as the basis vectors change. In Cartesian coordinates, two particle (at two different points) have the same velocity if the components of their velocity vector are equal. Whereas, in spherical coordinates, the components will be different and you'll have work to do to check that two spatially separated vectors are parallel.

In summary, the best way to think about spherical basis vectors are as objects defined locally at each and every point. The position vector $r \mathbf{\hat r}$ is something of any exception.

 

[Orodruin]

The position vector rr^ is something of any exception.

I disagree, I would say that the position vector is local as well. It is just defined in terms of the affine structure of the base manifold (and as such requires an affine base space to be defined). That you can draw vectors between points is also an artefact of an affine base space.