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Consider a particle at point, $\vec{r} = (r,\phi,\theta)$, where the azimuthal angle from $\hat{x}$ axis is $\phi$; polar angle from $\hat{z}$ axis is $\theta$. This particle has a velocity $\vec{v} = (\dot{r}, r \dot{\phi}, r \dot{\theta})$. How do I calculate the components of this velocity vector in a reference frame $(\hat{r}', \hat{\phi}', \hat{\theta}')$ that is rotated arbitrarily from the $(\hat{r}, \hat{\phi}, \hat{\theta})$ frame?

I have the rotation matrix that rotates the cartesian components, $\vec{r} = (x, y, z)$, into this new frame, lets call it $R(\alpha)$; i.e. $\vec{r}' = R(\alpha) \, \vec{r}$. So if this was just a position vector, I would transform to cartesian coordinates, rotate, and rotate back. But the velocity vector points from the position along the velocity direction. So the rotation must depend on the position, but that's not included in either the velocity vector itself, nor in the rotation matrix. Thus the same procedure doesn't work.

Does this mean that $(dr, d\phi, d\theta)$ transform differently from $(r, \phi, \theta)$? Or, perhaps I'm missing that there should be additional terms from the changes in the basis vectors (i.e. terms like $\frac{d\hat{\phi}}{dt}$)? Because this is an instantaneous velocity, and the tangent basis-vectors are orthogonal to eachother at the point $\vec{r}$, it almost seems like they could be treated as if they were cartesian?

Edit: perhaps the solution is to first translate the vector from the point $\vec{r}$ to the origin, then rotate, then translate to the rotated position? That seems reasonable, but how do I know (i.e. mathematically/rigorously) that translating, rotating, and translating back has the same effect (is this a question of commutation?)?

DilithiumMatrix
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  • In short, yes, they do transform differently because the basis vectors relative to which you have the coordinates $(\dot r, r\dot\phi, r\dot\theta)$ vary from point to point. In other words, the position is included tacitly in those basis vectors that you elided. You can’t really interpret those three numbers without knowing the position as well. – amd Dec 31 '19 at 00:37
  • Thanks @amd. Does that imply the following? 1) use cartesian velocity (and position), 2) rotate to the new frame (since cartesian-velocity rotates normally), then 3) only at the end convert to spherical coordinates. Edit: also, is there a simple way of "seeing" (e.g. from the metric of spherical coordinates, or something), that things won't transform nicely? I thought I remembered that as long as the bases were orthogonal, "things worked nicely". But perhaps the metric also needs to be spatially invariant? – DilithiumMatrix Dec 31 '19 at 00:42
  • Spherical basis vectors are always right-handed orthonormal, so the transformation between them is ”nice” in that it’s always going to be a rotation. I’ll have to think about it a bit to see if there’s a simple relationship between the global coordinate rotation and the local change of basis for the tangent space. – amd Dec 31 '19 at 01:18

1 Answers1

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A velocity vector has no position. It has only direction and magnitude. However, the coordinate frame in which the coordinates of your vector are measured is dependent on the position of the point at which the velocity was measured. So I think your concern is justified.

Assuming the coordinates of $\vec v$ were determined by the usual orthonormal basis for velocity at the spherical coordinates of your position vector $\vec r,$ you can express that basis in the (unrotated) Cartesian coordinates. You can express the coordinates of each basis vector either in terms of the spherical or Cartesian coordinates of $\vec r$ (using different formulas for spherical and Cartesian coordinates, of course).

Next you can find the coordinates of the position vector in the rotated frame. Using the same formulas you would use in the unrotated frame, you can find the orthonormal basis at that point, with each basis vector expressed in the rotated Cartesian coordinates.

Convert the rotated-frame Cartesian coordinates of each basis vector to Cartesian coordinates in the unrotated frame.

Now you have two sets of orthonormal basis vectors. The matrix that converts one to the other has the form of a rotation matrix. In general, however, it will be different from the rotation matrix that you already know for the rotation of the position coordinate frame.

As an aside, the velocity vector in an orthonormal basis should be something like $\vec{v} = (\dot{r}, r \dot{\phi}\sin\theta, r \dot{\theta})$. If you really mean what you wrote, you will have something other than a simple rotation matrix because you have to account for the foreshortening of the basis vector in the $\dot\phi$ direction. But if you just forgot to copy the factor $\sin\theta$ to the question then you’re probably OK.

David K
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  • Thank you for the thoughtful response (to a seemingly trivial question). But I'm still confused. I think the crux might come down to the idea that a "velocity vector has no position", which is something I 'know', but might not fully understand. For example, if we have a vector in a spherical basis, $\vec{a} = r \hat{r} + \theta \hat{\theta} + \phi \hat{\phi}$, this seems to imply that there is some position besides the origin, because if the base/tail of the vector is at the origin, then all vectors in spherical basis should have only a $\hat{r}$ component. No? – DilithiumMatrix Dec 31 '19 at 18:48
  • @DilithiumMatrix I might put it somewhat differently: a velocity vector at a point belongs to the tangent space at that point. In a Euclidean space, each tangent space is just another copy of it, which lets us blithely move these vectors from one tangent space to another. In that sense, velocity vectors have no position. I discuss these ideas briefly in this answer to a somewhat related question. – amd Dec 31 '19 at 20:08
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    I didn't mean to cause consternation over the issue of the "position" of a vector. It causes some trouble in various applications but isn't really the issue here. The issue rather is that according to the convention by which we write a vector basis $\hat r, \hat\phi, \hat\theta$, the direction of each of those vectors depends on a position. So it really doesn't matter whether you think a vector has a "position" or not; if you use this basis to assign coordinates to the vector, you need to pay attention to how the basis was defined. – David K Dec 31 '19 at 23:26
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    Note that there is no $\hat r, \hat\phi, \hat\theta$ defined by any point along the $z$ axis, including the origin. So any attempt to use such a basis to describe a velocity simply doesn't work. – David K Dec 31 '19 at 23:28
  • @amd I admit I hadn't been considering tangent spaces, but I had thought that the concept of tangent space is (or can be) defined so that $\mathbb R^3$ has only one tangent space. – David K Dec 31 '19 at 23:55