A Walk on the City Wall

On the Ja:Mathe! website the eminent importance of the Mathematical Institute in Göttingen has been emphasized on several occasions – and not only because it's the alma mater of the writer. On the one hand the Institute has a glorious history, on the other hand it stands out by its preferred environment, in proximity to the city wall. This old fortification is no longer a wall in the narrow sense, but rather an embankment covered with trees and grove and providing a walkway around the historic center. You get an impression how it looks like, if you watch this video on YouTube. if you speak Nepalese, you will even understand the commentary.

So, if you get stuck with a mathematical problem, it's a good idea to leave the Institute, traverse the Bürgerstraße, ascend to the city wall and take a walk around the historic center, which will take about a short hour. If you find a peer as company, listening patiently to your problem and asking an intelligent question from time to time, the probability to find a solution would increase dramatically.

At the beginning of the 20th century, this kind of doing mathematics was almost a trade mark of the Göttingen Institute. Emmy Noether and David Hilbert lectured while walking, not only while walking on the city wall, but also while hiking in the forests that surround Göttingen – obviously, the problems Noether and Hilbert worked upon were too difficult to be solved within a short hour.

But let's stay with the city wall and let's imagine two mathematicians, Emmy and David, who start a walk around the historic center counterclockwise, David keeping his position exactly one meter to the right of Emmy. Admittedly, this doesn't conform to etiquette, but most mathematicians are nerds anyway. After a short hour both come back to the staircase next to the Mathematical Institute, and they ask themselves by how many meters David's way exceeds Emmy's way. Well, walking around the city wall is almost like walking on a circle around the center. Let $r$ be the radius of this circle (in meters), then Emmy walked $2\pi r$ meters and David $2 \pi (r+1) = 2\pi r + 2\pi$ meters. Hence David walked exactly $2\pi$ meters more than Emmy, and this does not depend on the radius $r.$

Emmy and David were poor mathematicians, if they would not try to generalize the problem. Let's imagine Emmy walking $n$ kilometers straight ahead, then turning left on her heel by 180° and then walking back the same way, David always exactly one meter to the right of her. At the end point, which is the starting point, she has to turn left on her heel again, in order to be in the same position as she was, when the walk started. On the straight part of the way Emmy and David both walk the same distance. At the reversal point and at the end point Emmy turns on her heel, walking $0$ meters, and poor David has to circle around this point with a radius of one meter, in order to keep his distance to the right of Emmy. Only these two half circles of one meter radius each add to the length of David's way, so again he walks an additional distance of $2 \pi$ meters, and this doesn't depend on the length of the line segment $n.$

Every mathematician would be thrilled by these facts: Should it be possible that David's additional distance is always $2 \pi$ meters with every kind of circuit? One may consider another simple example: Emmy walks on a rectangular way counterclockwise, turning on her heel by 90° at the vertices. Now David has to run the quarter of a circle at every vertex, giving again an additional distance of $2 \pi$ meters. You may even assemble a complicated circuit out of many small straight lines not intersecting themselves. The computation may become a little cumbersome, but it will turn out that David walks an additional distance of $2 \pi$ meters, if he is always one meter to the right of Emmy and if they walk counterclockwise around the circuit.

Of course, this holds only for one lap around the circuit. If the two walk several times around the wall or around any other circuit, David has to walk the additional distance of $2 \pi$ for every lap. If the circuit has the shape of the number $8$ or of the symbol $\infty,$ David is on the outside of the circuit during the first loop and on the inside of the circuit during the second loop. Hence the distances of the two are the same. Generally it can be shown that the difference between the distances walked by the two is equal to the net rotation angle of the walkers in radian measure. If the two walk once around a circle or an rectangle, they look into the same direction when they come back to the starting point as they did when they started their walk. So the rotation angele is 360° or $2 \pi$ in radian measure. In case of the $8$, they rotate by 360° counterclockwise when walking the first loop, and then by 360° clockwise when walking the second loop, so the net rotation angle is $0$.

In his book [Nahin 2006] Paul Nahin proves a general theorem for (almost) arbitrarily shaped circuits: The difference of the distances the two walkers cover is always equal to the net rotation angle of the walkers in radian measure. The restriction almost refers to a circuit, where one walker turns that sharply towards the other one that the latter has to run backwards, to keep the distance of one meter. The problem goes back to a paper of R. Bruce Crofoot [Crofoot 2002] published in the Mathematics Magazine, and this article is written as a kind of advertising for it.

By the way, in its most simple form the problem is similar to a brainteaser, which baffles everyone, who hears about it for the first time: Google with the keywords rope and equator.