Zeno's paradox

By James Thimont (P369) on Monday, December 13, 1999 - 12:43 pm :

Does anyone know how to resolve (mathematically) Zeno's paradox. It states that motion is impossible. To get from A to B, you must pass through a point P halfway between A and B, then pass through a point P1 halfway between P and B etc.
Thanks
James

By Dan Goodman (Dfmg2) on Monday, December 13, 1999 - 04:04 pm :

Mathematically there is no paradox, the essence of the problem is in the way it is stated. The paradox is (as you pointed out) that to get from A to B you need to pass through P, P1, P2, P3, etc. The reason it seems like a paradox is there is an implicit assumption that you cannot do an infinite number of things in a finite amount of time. It only takes half as long to get from

P_{n}

P_{n + 1}

as it does to get from

P_{n - 1}

P_{n}

. So, the total time it takes to get from A to B is the

\sum_{n = 0}^{\infty} (1 / 2)^{n}

which is equal to 2. It can be illuminating to draw the picture for this, have time on the x axis and distance on the y axis and plot the point corresponding to each of the

P_{n}

. In other words, if the rabbit (or whatever) is at point

P_{n}

at a distance

x_{n}

from A at time

t_{n}

, plot the point

(t_{n}, x_{n})

. You will be able to draw a line through all these points, and although there will be infinitely many points, they will all be on the line. The fact that you never start plotting points beyond B for instance, is due to the method of plotting points rather than a paradox. By saying he has to pass through all these points, you are merely looking in more and more detail at the points just to the left of B and not any to the right. Hope that helps.

By Michael Doré (P904) on Sunday, December 26, 1999 - 03:45 pm :

Perhaps one of Zeno's even more confusing paradoxes is the following. Consider a race between a tortoise and a hare. The hare can run much faster than the tortoise so the tortoise has a head start. In order to overtake the tortoise, the hare must first reach the initial position of the tortoise. But as the tortoise is also moving, by this time the tortoise will have moved on in front of its initial position and will still be ahead. So the hare must reach this new position -but by the time it does the tortoise will have moved further forward. And so on. Therefore it is impossible for the hare to catch up.

This is normally resolved in the same way as the above -while it is true that the hare must try to catch up with the tortoise an infinite number of times, the time it takes to do so forms a geometric progression with a convergent infinite sum. But I still find this confusing -the formula a/(1-r) for the sum of an infinite geometric progression is actually defined as the limit of the partial sum of the geometric progression as the number of terms tends to infinity. The formula a/(1-r) is proved using the limiting definition. But in the resolution of Xeno's paradox, we are not actually using the limit of a geometric progression -instead we need the actual sum of an infinite geometric progression in order to work out how long the hare takes to catch up. Can anyone explain why the limit happens to be equal to what we want here?

Thanks,

Michael

By Dan Goodman (Dfmg2) on Sunday, December 26, 1999 - 05:42 pm :

This is basically the same problem as the previous paradox, but looked at in a moving frame of reference. But anyway, let's ignore that for a moment.

I think that the problem you have with this resolution of the problem is saying that the limit of the partial sums is the actual infinite sum. I don't know if the concept of an actual infinite sum is meaningful, unless it is taken to mean the limit of the partial sums.

In slightly more physical terms, how about this procedure. Choose any distance bigger than 0, call it e say. However small this distance is, we can find a time less than 1 whereby the hare is closer than e to the tortoise. This is true for every e > 0, so it seems as though you can then say that the time it takes to reach the tortoise must be less than or equal to 1. Otherwise, if it took 2 seconds or more for instance, we could ask how far away the hare was at time t=1.5. This distance must be bigger than 0, let's call it e . But we know that we can find a time less than 1 where the distance is less than e , so in fact, it cannot have any position at this time, a contradiction. The only resolution to this contradiction is that he reaches it at time 1.

I hope that provided some physical intuition to why we say that the actual sum is the same as the limit of the partial sums, but if I've been unclear (which is certainly possible), just ask again.

By Michael Doré (P904) on Monday, December 27, 1999 - 04:53 pm :

Thanks for that. That all made sense and has improved my understanding. I guess the way to look at it is: as the number of steps tends to infinity, the time tends to T and the distance tends to S, where T is the time it takes for the hare to catch up and S is their displacement when it has done so. Therefore as time -> T then distance -> S. Now as distance is a continuous function of time, when time = T, distance = S as we wanted. It just seemed a bit strange that nowhere else in maths do you encounter the infinite geometric progression without it actually being a limit that you're calculating ...

Thanks for clearing that up,

Michael