Equivalence Between Polar Coordinates and Cartesian Coordinates?

Question

I am trying to better understand the relationship between Cartesian Coordinates and Polar Coordinates.

Here is what I understand:

When going from Cartesian Coordinates $(x, y)$ to Polar Coordinates $(r, \theta)$, the transformation is given by (I can draw a circle and use trigonometry to get these relationships):

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

The Jacobian matrix $J$ of this transformation contains all the first order partial derivatives:

$$ J = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{bmatrix} $$

The determinant of the Jacobian matrix, $\text{det}(J)$, is calculated as:

$$ \text{det}(J) = \frac{\partial x}{\partial r} \frac{\partial y}{\partial \theta} - \frac{\partial y}{\partial r} \frac{\partial x}{\partial \theta} = r\cos^2(\theta) + r\sin^2(\theta) = r $$

Here is where my confusion starts:

The above analysis supposedly gives us this classic relationship:

$$ dx \, dy = \text{det}(J) \, dr \, d\theta = r dr d\theta $$

I am confused as to how the above the analysis implies that $dx \, dy = \text{det}(J) \, dr \, d\theta = r dr d\theta$ is true.

I understand that the Jacobian matrix contains all the first order partial derivatives of the transformation. These partial derivatives describe small changes in the variable of interest.

But I don't understand:

• Why does the presence of all these derivatives in the same matrix relate the change/equivalence between the two coordinate systems?
• Why the determinant of the Jacobian matrix gives us the factor by which the transformation changes areas?

Can someone please help me out here?

Answer 1

If you want an explanation in the context of calculus, this is a mere consequence of the chain rule, which is applied many times due to the change of coordinates, with all the computations being somehow gathered together inside a matrix determinant for the sake of clarity. In a way, it can be deduced informally but straightforwardly with the help of a vectorial notation such as $$ \mathrm{d}x\,\mathrm{d}y = \mathrm{d}^2\vec{r}(x,y) = \frac{\partial(x,y)}{\partial(r,\theta)}\mathrm{d}^2\vec{r}(r,\theta) = \det J \,\mathrm{d}r\,\mathrm{d}\theta $$ But I guess this is not the explanation you are looking for, that is why here is a geometric approach.

Let's recall that your coordinate systems parametrize a surface $S$. This surface may be described conventionally by its normal vector field (cf. Gauss map). How is it constructed ? At a given point $p \in S$, the normal vector is by definition orthogonal to the surface $S$, and more precisely to its tangent plane at $p$. The latter is spanned by the tangential vectors $\frac{\partial\vec{r}}{\partial r}$ and $\frac{\partial\vec{r}}{\partial\theta}$, where $\vec{r} = (x,y)$. Note that they corresponds to the columns of the Jacobian matrix. In consequence, the normal vector is given by the cross product of the tangential vectors, i.e. $\vec{n} = \frac{\partial\vec{r}}{\partial\theta} \times \frac{\partial\vec{r}}{\partial r}$, while its norm represents the area of the parallelogram formed by the tangential vectors due to the properties of the cross product $-$ it has to be highlighted that this parallelogram doesn't belong to the surface $S$ but lies in the tangent plane. Yet, in $\Bbb{R}^2$, one has $\|\vec{n}\| = \left|\det\left(\frac{\partial\vec{r}}{\partial\theta},\frac{\partial\vec{r}}{\partial r}\right)\right|$, which is nothing else than the Jacobian $|\det J|$.

Now, this reasoning is also valid with an infinitesimal area. Moreover, since the surface element $\mathrm{d}S = \mathrm{d}x\,\mathrm{d}y$ is arbitrarily small, it can be identified with the associated parallelogram in the tangent plane, whose sides are given by $\mathrm{d}\vec{x}$ and $\mathrm{d}\vec{y}$, hence the normal vector $\mathrm{d}\vec{S} = \mathrm{d}\vec{x} \times \mathrm{d}\vec{y} = \vec{n}(x,y)\,\mathrm{d}x\,\mathrm{d}y = \vec{n}(r,\theta)\,\mathrm{d}r\,\mathrm{d}\theta$, where the last equality corresponds to the change of parametrization of the surface $S$ towards polar coordinates. Now, taking the norm of this expression in order to recover the area element, we get $\mathrm{d}S = \|\vec{n}\|\,\mathrm{d}r\,\mathrm{d}\theta = |\det J|\,\mathrm{d}r\,\mathrm{d}\theta$.

This treatment can be extended to higher-dimensional spaces; then, the cross product is replaced by the exterior product and the volume element $\mathrm{d}x_1 \cdots \mathrm{d}x_n$ is interpreted as a differential form.