# Kinematics in polar coordinates

1. Feb 23, 2015

It it bad practice to consider values of $r$ that are greater than or equal to zero, while ignoring negative values? Do I lose any information in my analysis of motion? I understand what values of $r < 0$ represent, and I'm willing to use them in a pure mathematics context. In classical mechanics, however, the reason I try to avoid using negative values is because, well, it's more intuitive to consider positive values. So, does the convention $r ≥ 0$ have any shortcomings when it comes to PHYSICS?
Also, I am also curious as to which interval $θ$ is usually taken to be in. $-π$ to $π$? $0$ to $2π$? It doesn't matter as long as the interval spans an entire circle once, right? The thing is, I usually see the relationship between cartesian and polar coordinates outlined by the equation $θ = arctan (\frac{y}{x})$, although the arctangent function has a range from $-\frac{π}{2}$ to $\frac{π}{2}$ (only the first and fourth quadrants in the xy-plane). Should this be avoided as well?

2. Feb 23, 2015

### A.T.

In programming you should use the atan2 function:

http://en.wikipedia.org/wiki/Atan2

I guess that is also what is implicitly meant in the coordinate conversion.

3. Feb 23, 2015

Interesting. I had no idea such a function exists; I have never studied programming.

4. Feb 23, 2015

### jbriggs444

As long as you have coordinate values for all of the positions that you want to consider, it is of no particular importance to the physics whether you choose to cover the left half of the coordinate plane by considering angles greater than pi/2 (or less than -pi/2) or by considering negative values of r.

However... What does matter to the physics are boundaries where you have two points that are near one another in a physical sense but far from one another in terms of the coordinate system. [The folks over in the relativity forum would probably use terms like "manifold", "coordinate patch", "continuous" and "open set"] If you use negative values of r to model the left half of the coordinate plane then you have introduced a "seam" in the coordinate system. When the angle goes from just under pi/2 to just over pi/2, the r value has to go from positive to negative. You have a coordinate discontinuity without a corresponding physical discontinuity. This is in addition to the coordinate singularity you have at (0,0) and the seam that you have where the angle goes from pi to -pi. As long as you stay away from the seams, the physics formulas work fine.

Note that this is not an area where I have had any formal training. The terminology above may not be proper.

5. Feb 24, 2015

So, introducing seams should be avoided; which means using negative values of $r$ to model the left quadrants should be avoided, right? I don't get why negative values of $r$ are used in the first place. Couldn't we just as well represent all points in the left quadrants with positive values of $r$ (provided we use angles between 0 and 2π)? Isn't this a lot easier? I can visualize radial velocity and acceleration in terms of positive values of $r$, but I don't have the same intuition for negative values. Is it a matter of preference? I find motion in polar coordinates complicated enough already (my first encounter); it took me a while (and a lot of ink) to try to grasp the derivation of acceleration in polar coordinates geometrically, let alone doing so with consideration of negative $r$.
Also, does physics require a unique representation of each point in a polar coordinate system?

Last edited: Feb 24, 2015
6. Feb 24, 2015

### jbriggs444

It seems that I misread your initial post. You were arguing in favor of restricting attention to positive values of r rather than the opposite. Yes, I agree that is easier.
Based on what I just finished reading in http://en.wikipedia.org/wiki/Manifold, the answer is:

Locally, yes. The mapping between physical points and coordinate tuple within a coordinate chart must be continuous and invertible. The polar coordinate system amounts to a "coordinate chart" that meets this requirement everywhere but at the seams -- where continuity breaks down. If you use a -pi to +pi system, the seam is between the second and third quadrants. If you use a 0 to 2pi system, the seam is between the first and fourth quadrants. You can stitch the two schemes together with transition maps into a global "atlas" that avoids the seams. This leaves the center point to be handled somehow.

Last edited: Feb 24, 2015
7. Feb 24, 2015

### HomogenousCow

You want to go with r>0,r=0 for simple transformation laws.

8. Feb 24, 2015