1

Let us work in $\mathbb{R}^2$. If we are using Cartesian coordinates, then the position of a point $a$ can be given by its components, namely $a = a_x \hat{x} + a_y\hat{y}$. This applies globally as the unit vectors $\hat{x}, \hat{y}$ are constant functions. Now suppose we are working in polar coordinates, with unit vectors $\hat{r}$ and $\hat{\theta}$. In this case the direction of these unit vectors will change depending on the point $a$, so how would one interpret these unit vectors?

After thinking about it for a while, my interpretation is that suppose you have a point $a = (r, \theta)$. This specifies two unit vectors $\hat{r}(a)$ and $\hat{\theta}(a)$, which is just the normal Cartesian unit vectors but rotated by some angle $\theta$. Hence to express the position of $a$ in these unit vectors, we have to find its components in this rotated frame: $a = \tilde{a_x} \hat{r} + \tilde{a_y}\hat{\theta}$. This would imply there is a relation between $a_x$ and $\tilde{a_x}$ (and likewise for $a_y$), namely just a rotation. For example if I was to graph this, I would go $\tilde{a_x}$ units "horizontally" in the $\hat{r}$ direction and $\tilde{a_y}$ units "vertically" in $\hat{\theta}$ direction, so other than the rotation it is practically identical to a Cartesian coordinate system. Another consequence of this viewpoint is that all these "rotated coordinate systems" are all based at the origin $(0,0)$.

Is this the correct interpretation of polar unit vectors? Also, as someone who is self-studying differential geometry and smooth manifolds, how can all this be cast in that viewpoint?

CBBAM
  • 7,149
  • 1
    Possibly related. – Kurt G. Nov 24 '22 at 12:21
  • @KurtG. Thank you for that link, I took a look and it seems the unit vectors $\hat{r}(a)$ and $\hat{\theta}(a)$ are based at $a$ whereas I thought they should be based at the origin, is that correct? The picture I had in mind was that $\hat{r}(a)$ and $\hat{\theta}(a)$ simply rotate the Cartesian unit vectors $\hat{x}$ and $\hat{y}$, but does not translate them. – CBBAM Nov 24 '22 at 18:16

1 Answers1

1

First, understand that coordinates give you a way to move. If you are standing at a point on the plane, then Cartesian coordinates tells you that you can either move horizontally if you vary the $x$ coordinate, or vertically if you vary the $y$ coordinate. Polar coordinates now tell you at that same point, you can either move radially if you vary the $r$ coordinate, or ''angularly" if you move in the $\theta$ direction.

From the lens of differential geometry, your movement should not be thought of as constant speed. If you move in Cartesian coordinates, or in the radial direction in polar coordinates you will move at constant speed essentially because you are moving along straight lines. If you move in the theta direction, notice that if you take a theta slice, the further you are from the origin, the more distance you must transverse. Think of a slice of cake or pizza, with the tip of the pizza as the origin. Therefore, we must go with faster speed the further we are from the origin. This was essentially the upshot of the discussion we had in the linked post. Therefore, your $\partial_x(p),\partial_y(p),\partial_r(p),\partial_\theta(p)$ are just derivatives at $p$ with respect to these movements. All the above vectors live in your tangent space at $p$, namely $T_p\mathbb{R}^2$. We want to keep track of the point we're at, since this derivative information changes from point to point.

Mr. Brown
  • 1,867
  • Thank you for your answer, the differential geometry perspective is very helpful. I understand how polar coordinates work in the sense that $r$ moves you some distance radially outward from the origin and $\theta$ moves you angularly, my question was more in terms of the basis vectors. For example, consider the image in this answer, the polar unit vectors locally form a (rotated) Cartesian-like coordinate system. This is what I was trying to interpret. – CBBAM Nov 24 '22 at 19:13
  • 1
    The vectors in Kurt’s final picture aren’t normalized. These are velocity vectors corresponding to the movements in the r, $\theta$ directions through p as I’ve described above. – Mr. Brown Nov 24 '22 at 19:21
  • I think I understand now. If I may ask one more question, for a given vector $a = \tilde{a_x} \hat{r} + \tilde{a_y}\hat{\theta}$, how would one interpret this representation? Even though Kurt's vectors are not normalized, I thought practically the vector $a$ would be interpreted as the point $(\tilde{a_x}, \tilde{a_y})$ in the rotated (and possible translated?) coordinate system defined by $\hat{r}$ and $\hat{\theta}$ Kurt has in his image. – CBBAM Nov 24 '22 at 19:27
  • You notation is very odd. Your index on your coefficient should correspond to the index on the basis vector, but you are mixing the two. You should either write $a=a_x \partial_x+a_y \partial_y$ or $a=a_r \partial_r+a_\theta \partial_\theta$, depending on whether you choose Cartesian or polar coordinates. At a point $p$, this vector just lives in $T_p (\mathbb{R}^2-0)$ with the basis $\partial_r(p),\partial_\theta(p)$ if you choose polar coordinates. – Mr. Brown Nov 25 '22 at 04:18