Polar coordinates with arc-length instead of angle Is it possible to define coordinates on the 2d cartesian plane with arc length and radius instead of angle and radius. For example I could have
$$
\begin{split}
s(x,y) &= \sqrt{x^2 + y^2 } \arctan(y/x)\\
r(x,y) &= \sqrt{x^2 + y^2 }
\end{split}
$$
are there any subtle points about coordinate systems that prevent this from meeting the definition of  a proper coordinate system?
 A: I am guessing that the coordinates you want actually are
\begin{split}
s(\rho,\theta) &= \rho\theta,\\
r(\rho,\theta) &= \rho,
\end{split}
where $(\rho,\theta)$ are the standard polar coordinates.
(I used $\rho$ for the polar-coordinates radius in order to be able to make sense of the expression of your radius in terms of the polar coordinates.)
As pointed out in the comments, the arc tangent function is not good for such a definition because its range is only the angles $-\frac\pi2 < \theta < \frac\pi2.$
There is no value for $\arctan(y/x)$ when $x = 0$,
and the value of $\arctan(y/x)$ duplicates the value of $\arctan(-y/-x),$
hence coordinates in the first quadrant would be duplicated in the third.
You can improve your form of the definition a little by means of
the two-parameter arc tangent function,
$\DeclareMathOperator{\atan}{atan2}\atan(y,x)$,
whose range is $(-\pi,\pi]$, so at least it covers all points of the plane uniquely:
\begin{split}
s(x,y) &= \left(\sqrt{x^2 + y^2}\right) \atan(y,x),\\
r(x,y) &= \sqrt{x^2 + y^2},
\end{split}
But there is still a discontinuity along the negative $x$ axis in this definition.
For a point $(x,y)$ where $x < 0$ and $y\approx 0,$ if $y$ is positive the $s$ coordinate is approximately $\pi x,$ but if $y$ is negative the $s$ coordinate is approximately $-\pi x.$
Note that standard polar coordinates finesse the discontinuity by allowing multiple values of the polar coordinates at any given point.
For example, the point with Cartesian coordinates $(-1,0)$ has polar coordinates
$(1,\pi)$, $(1,-\pi)$, $(1,3\pi)$, $(1,-3\pi)$, $(-1,0)$, $(-1,2\pi)$, and many other choices. We don't define $\rho$ and $\theta$ as functions of $x$ and $y$, and this is what allows us to move along a differentiable path anywhere in the plane while varying $r$ and $\theta$ continuously.
However, there is another feature polar coordinates have, which is that we can define local coordinate vectors $\hat \rho$ and $\hat \theta$ which are orthogonal anywhere except at the origin.
That is, if you are anywhere except at the origin, a small change in $\rho$ (while holding $\theta$ constant) will take you in a direction perpendicular to the direction you would go with a small change in $\theta$ (while holding $\rho$ constant). Your coordinates do not have that property. (Note that this is not a fatal flaw; there are useful coordinate systems, such as oblique coordinates, that do not have orthogonal coordinate vectors.)
