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In polar coordinates, the origin has $r = 0$, but $\theta$ is not unique.

what sort of problems does this create, and how can I resolve them? For example, suppose an ant is wandering around a plane. Its speed is

$$s = \sqrt{\dot{r}^2 + r^2 \dot{\theta}^2}$$

but if the ant wanders through the origin a quantity like $\dot{\theta}$ is undefined. In this particular case I can deal with it because the limit

$$\lim_{r\to 0}\ r^2\dot{\theta}^2$$

is defined. Similarly, if I want to find an area with integration, I'd need to look at the Jacobian

$$\left|\begin{array}{cc}\partial r / \partial x & \partial r / \partial y \\ \partial \theta / \partial x & \partial \theta / \partial y\end{array}\right|$$

which is not defined at the origin. Again I can get around it. If I want the area of the unit circle, for example, I can take

$$\lim_{\epsilon \to 0} \int_{\theta = 0}^{2\pi}\int_{r=\epsilon}^1 r\ \textrm{d}r\textrm{d}\theta$$

How do I know I can always work around things like this? If I receive some other coordinate system, how can I tell if the points with no unique coordinates are going to give me trouble?

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Yet another good argument for $\arg(0)$ being indeterminate... –  J. M. Apr 14 '11 at 1:01
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4 Answers

up vote 10 down vote accepted

For integrals you can generally ignore stuff like this. This is because integrals ignore what's happening on a set of measure zero, so as long as the part of the thing you're integrating over where $r = 0$ has measure zero, you can just ignore that part. For example, if you're trying to figure out something about a particle involving an integral, and the set of times that particle is at the origin has measure zero, you're fine.

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@Qiochu Thanks. How about derivatives? –  Mark Eichenlaub Apr 14 '11 at 0:35
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You are correct that the Jacobian from Cartesian to polar coordinates is singular at the origin. In practice what this means is that the transformation looses one degree of freedom at the origin. Put in simple terms, there are directions in which you cannot infinitesimally move at the origin. For example, if your state is $r=0$ and $\theta = 0$, then a move in the $x$ direction of $dx$ is achieved by $dr$ in polar coordinates, however, a move in the $y$ direction of $dy$ requires a change of $dr$ in the $r$ direction but requires a change of $\pi/2$ (finite) in the $\theta$ direction. Thus the degree of freedom is lost in the $y$ direction and only moves in the $x$ direction are possible. The same problem occurs in the lat-long parameterization of the sphere at the North and South poles.

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Here's a problem that my students constantly run into. Suppose you have the following two variable functions and want to compute their limits at $(x,y)$ $\rightarrow (0,0)$

$f(x,y)$=${x^2-y^2}\over{{x^2-y^2}}$ and $g(x,y)$=$6x^3\over{\sqrt{x^2-y^2}}$

Since the limit is at the origin it's a good idea to use polar coordinates since $r$ approaches 0 along any path to the origin so $\theta$ can seemingly be ignored. However, that is where the problem lies. For the second function, $g$, the limit exists since sending $r$ to 0 produces a limit of 0 regardless of the value of $\theta$. On the other hand, if you try to use polar coordinates to evaluate the limit of $f$ you may incorrectly prove that this limit exists since sending $r$ to 0 produces different limits depending on the value of $\theta$ (and many students ignore $\theta$ as they simply focus on the fact that $r\rightarrow 0$).

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Sorry but how is this relevant to the question? –  Eric O. Korman Apr 13 '11 at 23:32
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Yes, origin is not well defined.

It's customary to say origin is (0,0) in analogy to cartesian coordinates.

HOWEVER you should be aware this is a convention and you shouldn't use it in a "limits" problem (as in : substitute 0,0 and get the limit) since functions are likely to be discontinuous there.

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I don't completely understand your answer. Can you give an example of a function discontinuous at the origin as you were mentioning, please? –  Mark Eichenlaub Apr 14 '11 at 0:38
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