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I fail to understand why the Jacobian matrix is used to transform Cartesian coordinates to polar coordinates.

If I'm not misunderstanding, it is assumed that the matrix $\begin{pmatrix}\cos(\theta)\;\;\;\; -r\sin(\theta) \\ \sin(\theta)\;\;\;\; r\cos(\theta)\end{pmatrix}$ is the matrix which I can multiply to the polar vector $(r, \theta)$ in order to obtain the values of the Cartesian vector $(x, y)$, but when i multiply the matrix and the vector, I get

$$x = r\cos(\theta) - r(\theta)\sin(\theta)$$

$$y = r\sin(\theta) + r(\theta)cos(\theta)$$

but what I expected was

$$x = r\cos(\theta)$$

$$y = r\sin(\theta) $$

Thank you very much.

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    $\begingroup$ A couple of comments: 1. The Jacobian matrix is not used to transform coordinates in that way. I think you misunderstood something you read. 2. Your question will be much easier to read if you use LaTex commands to write mathematics: for example, \alpha instead of @. 3. Your English is fine! $\endgroup$
    – user64687
    Commented Apr 25, 2013 at 20:04
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    $\begingroup$ The Jacobian tells you how to change differentials when moving between cartesian and polar coordinates. Notice that the determinant of your Jacobian is $r$ - let $J$ denote the Jacobian matrix, then you will have $dxdy = (\det J)drd\theta = rdrd\theta$. As Asal Beag Dubh points out, I think you have misread something. $\endgroup$ Commented Apr 25, 2013 at 20:15
  • $\begingroup$ Ok, thanks Asal. I read that the Jacobian matrix coincides with the matrix for linear transformations $A={a^i_j}$ , $x^i=a^i_j*z^j$ . Thank you $\endgroup$
    – Pedro
    Commented Apr 25, 2013 at 20:26
  • $\begingroup$ But from what you say, I understand that it must be only in some special conditions I have not read or understood correctly. $\endgroup$
    – Pedro
    Commented Apr 25, 2013 at 20:27
  • $\begingroup$ Yes, John. Thanks for your clarification. I definitely would have misunderstood. $\endgroup$
    – Pedro
    Commented Apr 25, 2013 at 20:30

3 Answers 3

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The Jacobian map is for transforming vectors expressed in terms of one set of coordinate basis vectors into another coordinate system's basis vectors. Positions like $(x,y)$ and $(r,\theta)$ are not expressed in terms of coordinate basis vectors, so it's inappropriate to use the Jacobian to try to convert between them.

Let $e_1, e_2$ be a pair of basis vectors. We can express positions on the 2d plane as $p = x e_1 + y e_2$.

Now, let $f(p) = p' = r e_1 + \theta e_2$. This looks like a change of coordinates, but it's really not--it's an active deformation of the plane into something where $r, \theta$ are "Cartesian" coordinates. This is just an active transformation, however, and fully equivalent to the passive change of coordinates that you're used to.

$f$ is appropriate to move positions to new positions, but it is not appropriate to move, for example, the tangent vector to a curve from one space to another (that is, to express such a tangent vector in terms of the polar coordinate basis vectors). For this, we need the Jacobian map $J_f$.

Example: let $\ell(t) = e_1 \cos t + e_2 \sin t$ be a curve that draws out the unit circle. It's clear that its derivative is the tangent vector $\dot \ell(t) = -e_1 \sin t + e_2 \cos t$. We can't transform this tangent vector using $f$; we must use $J_f$ instead.

(You'll note here I'm moving from Cartesian coordinates to polar, backwards from what you wanted* but the math is basically the same.)

Here's the Jacobian map:

$$\begin{align*} J_f(e_1) &= \frac{x e_1}{\sqrt{x^2 + y^2}} - y e_2 \\ J_f (e_2) &= \frac{y e_1}{\sqrt{x^2 + y^2}} + x e_2\end{align*}$$

Along the curve, $x = \cos t$ and $y = \sin t$, so we get

$$\begin{align*} J_f (\dot \ell(t)) &= -(\sin t )(e_1 \cos t - e_2 \sin t) + (\cos t)(e_1 \sin t + e_2 \cos t) \\ &= e_2\end{align*}$$

Remember that $e_2$ is associated with $\theta$--this says that, unsurprisingly, the velocity is entirely in the $\theta$ direction along this curve. We conclude that $\dot \ell(t) = J_f^{-1}(e_2) = e_\theta$.

In conclusion, we started with a tangent vector $\dot \ell(t)$ in our Cartesian coordinate system, and we moved it--using the Jacobian $J_f$--into a deformed plane where $(r,\theta)$ are "Cartesian" coordinates instead. The Jacobian is what moves tangent vectors from one space to another (or between coordinate systems), but positions are different and will always be handled by the full, nonlinear transformation.

One way you can remember this is that the Jacobian is like the derivative of the transformation, and so it's appropriate for moving things involving derivatives, like $\dot \ell(t)$, which is a velocity.

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  • $\begingroup$ Why is the transformation from cartesian to polar not a linear one. Could you pls say a bit about this. $\endgroup$
    – Shashaank
    Commented Oct 6, 2020 at 16:45
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First of all you have to understand that Jacobian is not a transformation of Cartesian Coordinates to Polar Coordinates. Infact, there exists no such linear transformation from Cartesian Coordinates to Polar Coordinates since the transformation is not linear. (lines cannot be mapped to circles).

Jacobian Metrix represents matrix with partial derivative of one vector with respect to another vector. i.e. $$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{pmatrix}\frac{\partial x}{\partial r}\;\;\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}\;\;\frac{\partial y}{\partial \theta}\end{pmatrix}=\begin{pmatrix}\cos(\theta)& -r\sin(\theta) \\ \sin(\theta)& r\cos(\theta)\end{pmatrix}$$

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  • $\begingroup$ Ok. I was wrong, now I have it all much clearer. Thank you very much. $\endgroup$
    – Pedro
    Commented Apr 25, 2013 at 20:33
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For transforming vector from a coordinate to another, you need to multiply the row vector $1\times n$ of the basis times the jacobian, and the column vector of the coefficient $n\times 1$ times the inverse jacobian. That's for maintain the invariance of the vector as a geometric object, so you use this equality:

$$ {\hat b}^T\cdot\ \hat v = b^T \cdot \Bbb J \cdot {\Bbb J}^{-1}\cdot v$$

And ${\hat b}^T = b^T \cdot \Bbb J$; $\hat v = {\Bbb J}^{-1}\cdot v$. Where $\hat b$ is the new coordinate basis, and $\hat v$ is the new coefficients.

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