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In orthonormal curvilinear coordinate system, we define unit basis vector as $$\mathbf{\hat{e}}_u = \frac{1}{h_u} \frac{d\mathbf{r}}{du},$$ where $h_u = |\frac{d\mathbf{r}}{du}|$. Suppose we only know $\mathbf{e}_u$ in relation to Cartesian coordinates, are we able to construct coordinate system from there? Using polar coordinates $(s,\phi)$ as an example, given only the information $$\mathbf{\hat{s}}=\frac{x\mathbf{\hat{x}} + y \mathbf{\hat{y}}}{\sqrt{x^2+y^2}}, \qquad \mathbf{\hat{\phi}}=\frac{-y\mathbf{\hat{x}}+x\mathbf{\hat{y}}}{\sqrt{x^2+y^2}}, $$ arę we able to construct a coordinate system with $\mathbf{\hat{s}}$ and $\mathbf{\hat{\phi}}$ as basis unit vector? More generally, suppose I want to construct a new coordinate system $(u,v)$ but is only given $$ \mathbf{\hat{u}}=f(x,y)\mathbf{\hat{x}} + g(x,y)\mathbf{\hat{y}}, \qquad \mathbf{\hat{v}}=g(x,y)\mathbf{\hat{x}} - f(x,y)\mathbf{\hat{y}} $$ as the unit basis vector. Is there a way to find $$\mathbf{\hat{u}}=f'(u,v)\mathbf{\hat{x}} + g'(u,v)\mathbf{\hat{y}}, \qquad \mathbf{\hat{v}}=f'(u,v)\mathbf{\hat{x}} - g'(u,v)\mathbf{\hat{y}} ,$$ which gives a relationship between $x=f(u,v), y=g(u,v)$. This will enable the computation of norm $h_u$ of tangent vector $\frac{d\mathbf{r}}{du}$. Then, we are able to specify the differential in $(u,v)$ as $$ d\mathbf{r}= \mathbf{\hat{u}} h_u du + \mathbf{\hat{v}} h_v dv .$$

Sean
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1 Answers1

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A vector field is a coordinate basis (i.e. defines a coordinate system) if and only if their pairwise Lie brackets are zero. Intuitively, the vanishing of the Lie brackets means that the coordinate lines (i.e. the integral curves of the basis vector fields) are well-defined in the sense that they form closed grids locally. It ensures that traveling along a basis vector $\mathbf{e}_i$ and then along a different basis vector $\mathbf{e}_j$ is the same as first traveling along $\mathbf{e}_j$ and then along $\mathbf{e}_i$. Otherwise, the same coordinates will be assigned to different points and your coordinate system will be ill-defined.

Vincent Thacker
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  • Thanks for pointing it out. I have modified the general case such that $\mathbf{\hat{u}}$ and $\mathbf{\hat{v}}$ are orthogonal everywhere. For the polar coordinate case, is it possible to find $h_s, h_\phi$ given only the above information? – Sean Nov 16 '23 at 00:58
  • @Sean This has nothing to do with orthogonality. Also note that coordinate basis vectors are not in general of length 1 (the easiest example is polar/spherical coordinates). Once the Frobenius theorem is satisfied, the coordinate system will simply be the integral curves of your vector fields. You cannot assume that your basis vectors are normalized. – Vincent Thacker Nov 16 '23 at 01:26
  • In https://en.wikipedia.org/wiki/Curvilinear_coordinates, it mentioned that we can always define a normalized basis vector when they are orthogonal. Can you perhaps use polar coordinate as an example to demonstrate how the coordinate system will be the integral curves of the vector fields? – Sean Nov 16 '23 at 03:05
  • @Sean As I have already said, the coordinate basis is defined by $\mathbf{e}_i \equiv \partial \mathbf{r}/\partial x^i$, which you might recognize as the formula for the tangent vector to the i-th coordinate curve. If you are given $\mathbf{e}_i$, finding the integral curves amounts to solving the set of n differential equations. You can do that precisely when the given basis is a coordinate basis (i.e. satisfying the Frobenius theorem). As I have said, coordinate bases are not in general of unit length, but they are the most natural choice of basis. Don't obsess over normalization. – Vincent Thacker Nov 17 '23 at 08:25
  • (cont.) While nobody said you can't work with normalized basis, it hurts more than it helps. – Vincent Thacker Nov 17 '23 at 08:30
  • While I agree with everything @VincentThacker wrote I guess the following remark should be made: The Frobenius theorem deals with the integrability of distributions, that is, it gives a necessary and sufficient condition when for a set of vector fields there exists a tangent hyperplane. Thereby it is using the commutator. In contrast: Vanishing of the pairwise commutator of basis vector fields tell us if they are holonomic. In this answer ... – Kurt G. Nov 20 '23 at 10:26
  • ... I have pedagogically explained the case of polar coordinates. Holonomic bases are often called coordinate bases. In polar coordinates, ${\partial_r,\partial_\theta}$ is such a basis as it is using the "naked" partial derivatives w.r.t. the coordinates $r$ and $\theta,.$ The terminology might be a bit confusing since even a non holonomic basis gives rise to a coordinate system. In fact the non holonomic basis ${\partial_r,\frac1r\partial_\theta}$ has the same integral curves and equally gives rise to the polar coordinate system. This might be the reason why such systems are... – Kurt G. Nov 20 '23 at 10:30
  • better called frames. – Kurt G. Nov 20 '23 at 10:31
  • @KurtG. "the non holonomic basis has the same integral curves and equally gives rise to the polar coordinate system" How so? The integral curves are the same, but the $\theta$ coordinate won't be well-defined anymore, as only the unit circle will have $2\pi$, but other radii will have $\theta \in [0,2\pi r]$. This means that radial lines won't be lines of constant $\theta$. It will not work globally. – Vincent Thacker Nov 20 '23 at 10:41
  • @VincentThacker Viewed as geometric objects (neglecting their parametrization) the integral curves are the same. They are, as we know, the rays starting from the origin and the circles around it. I am not sure how you conclude that $\theta\in[0,2\pi\color{red}{r}],.$ In both bases ${\partial_r,\partial_\theta}$ and ${\partial_r,\frac1r\partial_\theta}$ the angle $\theta$ is in $[0,2\pi)$ and more than well defined. Only the length of the second basis vector moves with $r,.$ A "moving fame." – Kurt G. Nov 20 '23 at 10:46
  • @KurtG. Yes, the coordinate lines are geometrically the same. But the whole point is that the coordinates are canonically the parameters of the coordinate lines, and the coordinate basis vectors are the tangent vectors w.r.t. the parameters. If you use the normalized basis, your $\theta$ has to run from $0$ to $2\pi r$ in order to preserve this, as you shortened it by $1/r$. That's the whole point. It leads to a coordinate that is not well-defined. In other words, given only the basis, you won't be able to unambiguously reconstruct where the coordinates are along those lines. – Vincent Thacker Nov 20 '23 at 10:56
  • @VincentThacker Not sure if I agree with that. In formulas that display the $\nabla$ operator in spherical coordinates I see the normalized basis vector $\frac1r\partial_\theta$ all the time. This question might lead us too far astray. The point of my comment was that Frobenius' theorem is something else. This seems to have confused OP over here. Your comments on that are welcome, too. – Kurt G. Nov 20 '23 at 11:07
  • @VincentThacker If you mean that in the non holonomic basis the coordinate $\theta$ is not defined without $r$ I agree with that. – Kurt G. Nov 20 '23 at 11:18
  • @KurtG. The $\nabla$ operator can be expressed in any basis of the tangent space so it is irrelevant to the discussion. The whole question is that given basis vector fields, we want to reconstruct the coordinate system. To do that, you have to (1) construct the integral curves (i.e. coordinate lines), and (2) determine the values of each coordinate along each line. And for (2), the length of the vector matters a lot as it determines the spacing of the coordinates. Both bases agree on (1), but the normalized basis fails at (2). – Vincent Thacker Nov 20 '23 at 11:21
  • @VincentThacker Yes. I agree now, except that this is precisely what the vanishing of the commutator detects. I am going to try to dig out a version of Frobenius' theorem that should help to clarify this. – Kurt G. Nov 20 '23 at 11:26
  • @VincentThacker These notes seem a fairly digestible piece about it. It is quite easy to see that in $\mathbb R^2$ any two vector fields $X_1,X_2$ satisfy Frobenius' requirement that $[X_1,X_2]$ is a linear combination of the two. This is not equivalent to their commutator vanishing. The latter is called holonomy. The former total integrability (meaning: there exists a two-dimensional "sub" manifold of $\mathbb R^2$ such that $X_1,X_2$ are tangent to it). This is trivially the manifold $\mathbb R^2$ itself. – Kurt G. Nov 20 '23 at 13:47
  • @KurtG. Okay. Maybe I remembered Frobenius' theorem wrong. But what I do know is that a basis is a coordinate basis if and only if their Lie brackets are zero: https://math.stackexchange.com/a/3686403. So the normalized basis is involutive but does not define a consistent coordinate system. – Vincent Thacker Nov 20 '23 at 15:55
  • @VincentThacker I totally agree on that. After just cycling for 1 1/2 hours I also think the following: when we look at $\mathbb R^2$ in Cartesian vs polar coordinates we tend to mix up the manifold $M=\mathbb R^2$ with its tangent space $T_pM=\mathbb R^2$ at every $p\in M,.$ Imho: only $M$ needs a coordinate system to describe its points $p$ uniquely. I do not want to repeat the transformations from $(x,y)$ to $(r,\theta)$. These transformations don't have to change when we equip every tangent space with a holonomic or a non-holonomic basis. $M$ and $T_pM$ are two different things. – Kurt G. Nov 20 '23 at 17:30