Inverse Functions & Curvilinear Coordinates
When coordinate transformations can be reversed - and the Jacobian test - Kaplan §2.12
Prereq: §2.10 Implicit Functions
We transform coordinates all the time - polar, cylindrical, spherical. But when can we go backwards? And what happens at the points where we can't?
Mappings and Their Inverses
Picture a GPS that converts latitude and longitude into a flat map position. It's a mapping from one coordinate system to another. Can you always go back from the map to the globe? Not at the poles - every longitude maps to the same point. Let's see exactly when mappings can be reversed.
Start simple: a linear map
Consider $x = 2u + v$ and $y = u - v$. Can we recover $(u, v)$ from $(x, y)$? Let's try. Add the equations: $x + y = 3u$, so $u = (x+y)/3$. Subtract twice the second from the first: $x - 2y = 3v$, so $v = (x - 2y)/3$. Done - the inverse exists everywhere.
Now look at the Jacobian:
The Jacobian is nonzero everywhere. Coincidence? No. It's exactly the condition that makes inversion possible.
Now nonlinear: the squaring map
Take $x = u^2 - v^2$ and $y = 2uv$. If you think of a complex number $z = u + \mathrm{i}v$, this is just its square: $z^2 = (u+iv)^2 = u^2 - v^2 + 2uv \mathrm{i}$. The Jacobian is:
Near $(u,v) = (1, 0)$: $J = 4 \neq 0$. We can locally invert. But at the origin: $J = 4(0 + 0) = 0$. And indeed, every nonzero point $w = x + iy$ has two square roots ($z^2 = w$ has two solutions). Near the origin, those two preimages collide and the mapping folds space onto itself. The vanishing Jacobian predicts exactly this collapse.
Connection to the Implicit Function Theorem
Why does $J \neq 0$ guarantee a local inverse? Because the inverse satisfies $f(u,v) - x = 0$ and $g(u,v) - y = 0$ - two implicit equations for $(u,v)$ in terms of $(x,y)$. The IFT from §2.10 says we can solve for $(u,v)$ precisely when the Jacobian $\partial(f,g)/\partial(u,v) \neq 0$.
See it in action
The visualization below shows a grid in $(u,v)$-space and its image in $(x,y)$-space. Switch between mappings to see how different Jacobians create different distortions.
The Jacobian Reciprocal Rule
Suppose we have a mapping $\mathbf{x} = \mathbf{f}(\mathbf{u})$ and its inverse $\mathbf{u} = \mathbf{g}(\mathbf{x})$. Composing them gives the identity: $\mathbf{g}(\mathbf{f}(\mathbf{u})) = \mathbf{u}$. What does this tell us about the Jacobians?
The derivation
Start from the identity:
Differentiate both sides using the chain rule for Jacobian matrices:
On the left, we have the product of two Jacobian matrices. On the right, the identity matrix. Now take the determinant of both sides. The determinant of a product is the product of determinants, and $\det(I) = 1$:
The Jacobian of the inverse is the reciprocal of the Jacobian of the forward map.
Verify with polar coordinates
Forward: $x = r\cos\theta$, $y = r\sin\theta$, so $J_{\text{forward}} = r$. The reciprocal rule predicts $J_{\text{inverse}} = 1/r$. Let's check directly.
The inverse is $r = \sqrt{x^2 + y^2}$, $\theta = \arctan(y/x)$. Computing the partials:
The inverse Jacobian determinant:
Curvilinear Coordinates
Every coordinate system comes with a natural grid. In Cartesian coordinates, the grid lines are horizontal and vertical. But what does the "grid" look like in polar coordinates? In any arbitrary coordinate system?
Coordinate curves
Here's the idea. Take a mapping $(u,v) \mapsto (x,y)$. Fix $v = c$ (some constant) and vary $u$ - you trace out a $u$-curve. Fix $u = c$ and vary $v$ - you trace out a $v$-curve. These two families of curves form a grid in the $(x,y)$-plane.
Polar example: Fix $\theta = c$ (vary $r$) - you get a ray from the origin. Fix $r = c$ (vary $\theta$) - you get a circle. Rays and circles: that's the polar grid. These two families are orthogonal everywhere, except at the origin where every ray converges to a single point.
The Jacobian as area distortion
A small rectangle $du \times dv$ in parameter space maps to a small (curved) parallelogram in $(x,y)$-space. The area of that parallelogram is $|J|\,du\,dv$. So the Jacobian is the local area magnification factor.
For polar coordinates, $J = r$: a patch $dr \times d\theta$ has area $r\,dr\,d\theta$. Near the origin ($r \approx 0$), the area shrinks to zero - the coordinate grid gets infinitely compressed, and the Jacobian reflects this.
Explore the grid
The dual-panel below highlights a single grid cell and its image. The area ratio equals $|J|$. Use the slider to change grid density and the presets to switch coordinate systems.
The Big Three Coordinate Systems
Three coordinate systems show up again and again in physics and mathematics. Each one has a Jacobian that encodes the geometry of its grid, and a singularity where that grid degenerates.
Polar coordinates ($\mathbb{R}^2$)
$x = r\cos\theta$, $y = r\sin\theta$. The Jacobian is $J = r$, the area element is $dA = r\,dr\,d\theta$, and the mapping breaks at $r = 0$ (all $\theta$ values map to the origin).
Cylindrical coordinates ($\mathbb{R}^3$)
Just polar coordinates plus a vertical axis: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. The Jacobian is again $J = r$ (the extra $z$-row and $z$-column contribute a factor of 1), the volume element is $dV = r\,dr\,d\theta\,dz$, and the mapping breaks along the entire $z$-axis ($r = 0$).
Spherical coordinates ($\mathbb{R}^3$)
$x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$, where $\rho$ is the distance from the origin, $\phi$ is the polar angle from the positive $z$-axis, and $\theta$ is the azimuthal angle in the $xy$-plane.
The $3\times 3$ Jacobian matrix is:
We expand along the third row (it has the fewest terms, and one zero). The third row has entries $\cos\phi$, $-\rho\sin\phi$, and $0$:
The first $2 \times 2$ determinant: $\rho^2\cos\phi\sin\phi(\cos^2\theta + \sin^2\theta) = \rho^2\cos\phi\sin\phi$. The second: $\rho\sin^2\phi(\cos^2\theta + \sin^2\theta) = \rho\sin^2\phi$. So:
Geometric meaning: $\rho^2\sin\phi$ is the product of three lengths: $d\rho$ (radial thickness), $\rho\,d\phi$ (polar arc), and $\rho\sin\phi\,d\theta$ (azimuthal arc). These three edges of the volume element are mutually perpendicular.
Singularities: $J = 0$ when $\rho = 0$ (origin) or $\sin\phi = 0$ (i.e., $\phi = 0$ or $\pi$, the north and south poles of the $z$-axis). At these points the azimuthal circle degenerates to a point.
| System | Mapping | Jacobian | Volume/Area | Singularity |
|---|---|---|---|---|
| Polar | $(r\cos\theta,\; r\sin\theta)$ | $r$ | $r\,dr\,d\theta$ | $r = 0$ |
| Cylindrical | $(r\cos\theta,\; r\sin\theta,\; z)$ | $r$ | $r\,dr\,d\theta\,dz$ | $r = 0$ ($z$-axis) |
| Spherical | $(\rho\sin\phi\cos\theta,\;\rho\sin\phi\sin\theta,\;\rho\cos\phi)$ | $\rho^2\sin\phi$ | $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ | $\rho=0$, $\phi=0,\pi$ |
Explore spherical coordinates
The 3D scene below shows the three families of coordinate surfaces: a sphere ($\rho = c$), a cone ($\phi = c$), and a half-plane ($\theta = c$). Drag the sliders to move the intersection point. The readout shows the Cartesian coordinates and the Jacobian value.
Practice Problems - §2.12
From Kaplan, problems after §2.12
For polar coordinates $x = r\cos\theta$, $y = r\sin\theta$:
(a) Verify that $dx = \cos\theta\,dr - r\sin\theta\,d\theta$ and $dy = \sin\theta\,dr + r\cos\theta\,d\theta$.
(d) Show that $\partial(x,y)/\partial(r,\theta) = r$ and $\partial(r,\theta)/\partial(x,y) = 1/r$.
Step 1 - Part (a): Compute the total differentials by taking partials:
$dx = \dfrac{\partial x}{\partial r}\,dr + \dfrac{\partial x}{\partial \theta}\,d\theta = \cos\theta\,dr + (-r\sin\theta)\,d\theta = \cos\theta\,dr - r\sin\theta\,d\theta$ ✓
$dy = \dfrac{\partial y}{\partial r}\,dr + \dfrac{\partial y}{\partial \theta}\,d\theta = \sin\theta\,dr + r\cos\theta\,d\theta$ ✓
Step 2 - Part (d), forward Jacobian:
$$\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r$$Step 3 - Part (d), inverse Jacobian via reciprocal rule:
By the reciprocal rule, $\partial(r,\theta)/\partial(x,y) = 1/J = 1/r$.
Direct check: From $r = \sqrt{x^2+y^2}$, $\theta = \arctan(y/x)$:
$$\frac{\partial(r,\theta)}{\partial(x,y)} = \begin{vmatrix} x/r & y/r \\ -y/r^2 & x/r^2 \end{vmatrix} = \frac{x^2}{r^3} + \frac{y^2}{r^3} = \frac{r^2}{r^3} = \frac{1}{r} \;\;\checkmark$$For the mapping $x = u^2 - v^2$, $y = 2uv$:
(a) Compute the Jacobian $\partial(x,y)/\partial(u,v)$.
(b) Find $(\partial u/\partial x)_y$ and $(\partial v/\partial x)_y$.
Step 1 - Part (a): The Jacobian matrix is:
$$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 2u & -2v \\ 2v & 2u \end{vmatrix} = 4u^2 + 4v^2 = 4(u^2 + v^2)$$Step 2 - Part (b): The inverse Jacobian matrix is:
$$\frac{\partial(u,v)}{\partial(x,y)} = \frac{1}{4(u^2+v^2)} \begin{pmatrix} 2u & 2v \\ -2v & 2u \end{pmatrix}$$Reading off the entries:
$\left(\dfrac{\partial u}{\partial x}\right)_y = \dfrac{2u}{4(u^2+v^2)} = \dfrac{u}{2(u^2+v^2)}$
$\left(\dfrac{\partial v}{\partial x}\right)_y = \dfrac{-2v}{4(u^2+v^2)} = \dfrac{-v}{2(u^2+v^2)}$
Step 3 - Singularity check: At the origin $(u,v) = (0,0)$, both denominators $2(u^2 + v^2)$ vanish - the inverse partial derivatives blow up. This is consistent with $J = 4(u^2+v^2) = 0$ at the origin. The squaring map is not locally invertible there.
For spherical coordinates $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$, compute the Jacobian $J = \partial(x,y,z)/\partial(\rho,\phi,\theta)$.
Step 1: Write the $3\times 3$ Jacobian matrix:
$$\begin{pmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \\ \cos\phi & -\rho\sin\phi & 0 \end{pmatrix}$$Step 2: Expand along the third row (entry $(3,3) = 0$ kills one term). We need cofactors $M_{31}$ and $M_{32}$:
$M_{31} = \begin{vmatrix} \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \end{vmatrix}$
$= \rho^2\cos\phi\sin\phi\cos^2\theta + \rho^2\cos\phi\sin\phi\sin^2\theta = \rho^2\cos\phi\sin\phi$
Step 3:
$M_{32} = \begin{vmatrix} \sin\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \sin\phi\sin\theta & \rho\sin\phi\cos\theta \end{vmatrix}$
$= \rho\sin^2\phi\cos^2\theta + \rho\sin^2\phi\sin^2\theta = \rho\sin^2\phi$
Step 4: Combine with signs from cofactor expansion along row 3:
$J = \cos\phi \cdot (+M_{31}) + (-\rho\sin\phi)\cdot(-M_{32}) + 0$
$= \cos\phi \cdot \rho^2\cos\phi\sin\phi + \rho\sin\phi \cdot \rho\sin^2\phi$
$= \rho^2\sin\phi\cos^2\phi + \rho^2\sin\phi\sin^2\phi = \rho^2\sin\phi$
Geometric interpretation: $\rho^2\sin\phi = (\rho) \cdot (\rho\sin\phi) \cdot 1 \times d\rho\,d\phi\,d\theta$, the product of three mutually perpendicular edge lengths of the volume element.