Swapping modulus and argument in polar coordinates

Question

Let $y=f(x)$ in Cartesian coordinates. Swapping the $x$ coordinate with the $y$ coordinate has the effect of reflection about the axis $y=x$.

But if $r=f(\theta )$ in polar coordinates (where $r$ is the modulus and $\theta$ is the argument), what happens when we swap $r$ with $\theta$?

For example, what's the difference between the polar graphs of $r=\theta ^2$ and $\theta =r^2$? Do they look the same? Or is there some kind of reflection as well?

Answer 1

The polar plots of $r= \theta^2, \sqrt \theta, \theta$ are shown in blue,red and grey respectively.

The gray line does not reflect red/blue profiles in a way presently imaginable by me.

This is perhaps because $(r,\theta)$ do not have the same physical dimension and belong to a common type or genre.

$Z=r e^{i \theta}$ is a complex number. I do not know what $U = \theta e^{ir}$ really is.

A very good question, but my answer may be is no good. That is indirectly suggested geometric reciprocal requirement is not answered.

Answer 2

Polar coordinates we call well known mapping $\mathbb{R}^2 \to \mathbb{R}^2$, from $(x,y)$ to $(\theta, r)$ using formulas $x = r\cos \theta$, $y = r\sin \theta$, $r \geqslant 0,\theta \in [0, 2\pi) $.

As to plane $(\theta, r)$, then it is usual cartesian coordinates, usual $\mathbb{R}^2$, and you can think about it exactly as you think about $(x,y)$. $r=\theta^2$ is exactly parabola. $\theta=r^2$ is both branches of square root.

We use polar coordinate, when some function/ curve looks "difficult" for $(x,y)$ and by mapping it to $(\theta, r)$ plane we obtain more "easy" case. Most known example is circle $x^2+y^2=r^2$, which by polar coordinate moves to interval $[0, 2\pi) \times \{1\}$. Disk $x^2+y^2\leqslant r^2$ is mapped to rectangle $[0, 2\pi) \times [0,1]$.

Addition. Now about swapping variables. By definition axial symmetry is not identical Orthogonal Transformation which have line of fixed points. This line is called symmetry axis. To obtain for point $M$ symmetrical point $M'$ with respect to symmetry axis one need to draw perpendicular line to symmetry axis from $M$ and take point $M'$ on this perpendicular on other side of symmetry axis on same distanse as $M$.

For example, if we consider $y=x$ as symmetry axis, then for point $(a,b)$ symmetrical point is $(b,a)$.

So, on $\mathbb{R}^2$ swapping coordinates i.e. having graph $y=f(x)$ and considering $x=f(y)$ is exactly creating symmetry with respect to line $y=x$. Same is, of course if we speak about $r=f(\theta)$ and considering $\theta=f(r)$ - they are symmetric with respect to line $r=\theta$.

Another question is what gives swapping variables for $(x,y)$ in $(\theta, r)$ and reverse. Let's consider firstly "polar plane". As is stated above, swapping variables there means symmetry with respect to line $r=\theta$. Last is well known Archimedean spiral on "cartesian plane". So swapping coordinates $\theta$ and $r$ gives on plane $(x,y)$ graphs "symmetric" with respect to spiral $r=\theta$ which is same as $\sqrt{x^2+y^2}=\arctan \frac{y}{x}$. For example parabola $r=\theta^2$, which is some type of spiral on $(x,y)$, after swapping gives $\theta=r^2$, or taking its one branch, $r=\sqrt{\theta}$ is again some spiral on $(x,y)$.

Summing up:

1. parabola $y=x^2$ is axial symmetric with respect to square root $x=y^2$ using symmetry axis line $y=x$.

2. In "polar" language spiral $r=\theta^2$ is "spirally" symmetric with respect to spiral $\theta=r^2$ using symmetry "axis" spiral $r=\theta$

Second example. Let's take in polar plane $r=\tan\theta$ i.e. points $(\theta,\tan\theta)$. Swapping variables give $\theta=\tan r$ i.e. points $(\tan r,r)$. Obviously $(\theta,\tan\theta)$ is axially symmetrical to $(\tan r,r)$ with respect to symmetry axis $\theta=r$. Now if we consider corresponding points on $(x,y)$ plane, then symmetry axis $\theta=r$ creates spiral, while $r=\tan\theta$ and $\theta=\tan r$ create some corresponding curves on $(x,y)$: $\sqrt{x^2+y^2}=\frac{y}{x}$ and $\arctan \frac{y}{x}=\tan \sqrt{x^2+y^2}$. Obviously $(x,y)$ curves are not axially symmetrical.

If it sounds acceptable, we can call "spirally" symmetrical on plane $(x,y)$ such points, which preimages are axially symmetrical on plane $(\theta, r)$ with respect to symmetry axis $\theta=r$.

Using this term we can call $\sqrt{x^2+y^2}=\frac{y}{x}$ and $\arctan \frac{y}{x}=\tan \sqrt{x^2+y^2}$ "spirally" symmetrical on plane $(x,y)$.