Don't understand how to use jacobian for transformation of coordinates

Question

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.

Answer 1

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. $$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{pmatrix}\frac{\partial x}{\partial r}\;\;\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}\;\;\frac{\partial y}{\partial \theta}\end{pmatrix}=\begin{pmatrix}\cos(\theta)& -r\sin(\theta) \\ \sin(\theta)& r\cos(\theta)\end{pmatrix}$$

Answer 2

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.

Answer 3

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.