Differential geometry: If $\vec v = v^i \vec e_i$, then why is $\vec r = r \vec e_r$ in spherical coordinates?

Question

In differential geometry (and later carried over to GR) any abstract vector $\vec v$, exists on its own vector space.

We can then choose to represent this vector in a coordinate basis $\vec v = v^i \vec e_i$, and a fundamental statement is that this is independent of bases, so if we transform to a basis $\vec e^{'}_i$, it is $$\vec v = v^i \vec e_i = v^{'i} \vec e^{'}_i,$$ which later helps us find transformation laws etc.

So my silly question: As the former statement is true, why is the full position vector in 3D spherical coordinates $\vec r = r \vec e_r$, instead of $\vec r = r \vec e_r + \theta \vec e_{\theta} + \phi \vec e_{\phi}$?

P.S. I'm aware of this answer on math.se, where it is stated in a comment, that

In polar or spherical coordinates, the radial unit vector embeds the directional information through its dependence on the angular coordinate variables.

Knowing this, I'm still not sure, why $\theta = \phi=0$ in the above example.

I am mainly asking this, because the form of $\vec r$ determines the form of equations of motion, and using $\vec r = r \vec e_r + \theta \vec e_{\theta} + \phi \vec e_{\phi}$ would obviously give too many ficticious force terms in $\ddot{\vec r}$, but one could get the idea of using that form of $\vec r$.

Answer 1

You are confusing two spaces. A manifold, M, has a tangent space, Tp(M) at each point of the manifold. So, for example, on a sphere (not spherical coordinates, but a 2dim manifold) there is a tangent plane at each point and any vector in that space can be expressed in terms of the basis vectors of the space at that point. In some sense we can think of the vectors a being based at the point of the manifold but this isn't necessary. All the vector algebra we know from Euclidean space applied to Tp(M). Vectors based at different points in M cannot be added. They need to be parallel transported from one point to another to bring them into the same space.

Even Euclidean space has an infinite number of Tp(M) but they are all "parallel" copies of each other and provide no additional information for practical use.

In your example you are looking at the position vector of a point in 3-dim Euclidean space. This is a displacement vector from the point (0,0,0) to the point (x, y, z). Spherical coordinates are not the same thing as tangent planes on a sphere. Here we are using families of surfaces embedded in E3 (R3 with an identity matrix for a metric) to label points (r, theta, phi). The conversion of (x, y, z) to these coordinates is trivial (r*sin(theta)cos(phi), ...) = r(sin(theta)cos(phi), ...) = re_r. Coordinates in E3 are not comparable to tangents on M. Each coordinate surface may be a 2dim M in E3 and the local vectors coincide with those of Tp(M) for each family of surfaces. But the vector you are trying to describe does not live in any of those spaces.

Answer 2

Short answer: $\vec{r}$ is always along $\hat{e}_r$, therefore its decomposition cannot include other basis elements.

Position vector $\vec{r}$ is a specific kind of vector that is always along $\hat{e}_r$. The reason for this is $$ \hat{e}_r \triangleq \frac{\vec{r}}{|\vec{r}|} $$ for spherical coordinates for that point. Not every vector will carry this property.

One way to remove the confusion is to consider another vector, for example, a displacement vector $\Delta{\vec{r}} := \vec{r}_1 - \vec{r}_2$. Looking from the first point, the first term $\vec{r}_1$ is along $\hat{e}_r$. But, the second point does not (necessarily) lie on the line represented by $\hat{e}_r$ because the bases were defined with respect to the first point and $\vec{r}_2$ can be at any arbitrary point almost all of which not along $\hat{e}_r$.