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 $\vec r$.

  • 1
    $\begingroup$ You should try drawing the vector field you propose. You'll see that it is not $\vec r$. $\endgroup$ – Danu May 15 '19 at 19:49

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 $\hat{e}_r$.

| cite | improve this answer | |
  • $\begingroup$ Right, I didn't think of this example with $\Delta\vec r$. In this example $\vec r_2$ would then have some angular components right? Also sorry, but I don't know what that symbol with the equals-triangle means. $\endgroup$ – AtmosphericPrisonEscape May 15 '19 at 20:04
  • $\begingroup$ @AtmosphericPrisonEscape Yes, then you can express $\vec{r}_2$ in terms of $\hat{e}_{\theta}$ and $\hat{e}_{\phi}$ as well. And, my bad, it's definition symbol, one of those, anyways. $\endgroup$ – acarturk May 15 '19 at 20:07
  • $\begingroup$ Just for clarification: In spherical/cylindric coordinates, every position vector, carries its own basis, so that $\vec r_1 = r_1 \vec e_r(\theta_1, \phi_1)$ and $\vec r_2 = r_2 \vec e_r(\theta_2, \phi_2)$. But If I wanted to compute $\Delta \vec r$, I need to choose one of those bases, for example $\vec e_r(\theta_1, \phi_1)$, and then I get $\vec r_2 = c_1 \vec e_r(\theta_1, \phi_1) + c_2 \vec e_{\theta}(\theta_1, \phi_1) + c_3 \vec e_{\\phi}(\theta_1, \phi_1)$, with some constants $c_i$, right? $\endgroup$ – AtmosphericPrisonEscape May 15 '19 at 20:15
  • $\begingroup$ Yes, but to not loose generality (for other kinds of bases), I would advise you to include $r_1$ in arguments for all basis vectors. $\endgroup$ – acarturk May 15 '19 at 20:22

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.

| cite | improve this answer | |
  • $\begingroup$ So you're saying $r\vec e_r + \theta \vec e_{\theta} ..$ doesn't exist? I can surely make a picture of it, attached to some Tp(M). Also it has a coordinate representation in Euclidean space, so surely a representation in spherical ones must exist? What about the example in the other answer $\Delta \vec v = \vec r_2-\vec r_1$, that must have a representation in all coordinates of arbitrary Tp(M) (simply because $v'^i e'_i = v^i e_i)$? $\endgroup$ – AtmosphericPrisonEscape May 15 '19 at 21:22
  • $\begingroup$ I am not saying it does not exist. I am saying that they do not do what you think they do. You are trying to write a displacement in R3 by moving out way from (0,0,0) and then attributing more displacement to some arc in S2. These are living in different places. Also, I think you are confusing a coordinate function for an index. In some cases there is a one-to-one mapping but in general they represent different things altogether. So it is not valid to say xiei --> re_r + theta*e_th + ... etc. $\endgroup$ – ggcg May 15 '19 at 21:36
  • $\begingroup$ This is a common misunderstanding when first learning diff geom. Another is why the n-tuple (x, y, z) does not require a covariant derivative. These are scalar functions defined on M the n-tuple does not line in Tp(M). $\endgroup$ – ggcg May 15 '19 at 21:38
  • $\begingroup$ Fair point, the $v^i$ are functions and not coordinates, including for $\vec r$ in the spherical case. Only in euclidean space they happen to coincide with the coordinates. So for $\vec r$ it is $v_{\theta}(r,\theta,\phi)=v_{\phi}(r,\theta,\phi)=0$, but not for general vectors $\vec v$, where I would need to do the parallel transport to get the $v^i$, (just paraphrasing what you just explained) right? $\endgroup$ – AtmosphericPrisonEscape May 15 '19 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.