Given a vector in the standard cartesian basis \begin{equation} v = \left( \begin{aligned} x\\ y \end{aligned}\right) \end{equation}

We may write its components in terms of the polar coordinates as follows: $$ v=\left( \begin{aligned} r \cos\theta\\ r \sin\theta \end{aligned}\right) $$

To put this in terms of the spherical basis, we can apply the jacobian transform: $$ \frac{\partial (x,y)}{\partial (r,\theta)} = \left(\begin{array}[cc] \cos\cos\theta & \sin\theta \\ -r \sin\theta & r \cos\theta \end{array}\right) $$

Which when applied to the vector $v$ gives

$$ \left( \begin{aligned} r \\ 0 \end{aligned}\right) $$

Clearly $\theta$ is not always $0$, what on earth am I doing wrong?


2 Answers 2


Your jacobian transform converts one set of Cartesian coordinates to another. Specifically, it converts coordinates of a tangent space at a particular point aligned with your original Cartesian basis vectors $\hat x, \hat y$ to coordinates of the same tangent space based on the polar basis vectors at that point, $\hat r, \hat \theta.$

The polar basis vector $\hat r$ of the tangent space at any given point is a unit vector in the same direction as that point's position vector. The position vector is therefore equal to $r \hat r$ where $r$ is the distance from the origin. That is what your calculations have confirmed.

Specifically, the $\hat \theta$ component of the position vector at any point is zero according to the way the polar basis vectors are defined at that point.

If you repeat this procedure at another point in a different direction from the origin, you will again find that its position vector is $r \hat r,$ which may seem paradoxical until you realize that you are working in a different tangent space there and the $\hat r$ of that tangent space is not the same as the $\hat r$ of the tangent space at the first point.

To put it another way, although you may be able to use a single Cartesian basis in exactly the same way at every point in the plane, there is no single "polar basis" that you can use in exactly the same way at every point in the plane. The polar basis at Cartesian coordinates $(1,0)$ is different from the polar basis at Cartesian coordinates $(0,1).$

There is no linear transformation between the Cartesian coordinates of a point and the polar coordinates of a point.

  • $\begingroup$ Thank you for this well thought out answer. Does this mean the answer in the following link is wrong? physics.stackexchange.com/questions/67066/… $\endgroup$
    – Craig
    Commented Oct 16, 2022 at 15:57
  • $\begingroup$ In the physics application, I think they are interested in vectors in the tangent space at a particular point. In the tangent space you certainly can have cylindrical basis vectors specific to that tangent space. I don't think physicists say "tangent space" but I believe it describes the way they are using those basis vectors. I'm fairly sure at least they are not trying to transform the spatial coordinates of that point. $\endgroup$
    – David K
    Commented Oct 16, 2022 at 16:09
  • $\begingroup$ I'm not sure how that translates to vector fields though. Can it be expanded to account for this? $\endgroup$
    – Craig
    Commented Oct 16, 2022 at 20:56
  • $\begingroup$ A vector field is a function from points in space to vectors. So at each point in space you have the vector that the vector field associates with that point. A lot of the things you would want to do with that vector involve things that also happen at exactly the same point, in which case it isn't a problem to use a local coordinate system that is defined for that particular point, as long as you consistently do everything in that coordinate system. It may not be convenient to take a vector sum of two field vectors at distant points this way, but why would you want to do that? $\endgroup$
    – David K
    Commented Oct 17, 2022 at 23:02

We have $v = (r\cos \theta, r\sin \theta) = f(r,\theta)$. To recover $(r,\theta)$, we simply apply the inverse map $f^{-1}(f(r,\theta))=(r,\theta)$. The inverse map is given by $f^{-1}(x,y) = (\sqrt{x^2+y^2},\tan^{-1}(y/x))$.


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