Achilles without tortoise

Question

I am Achilles, and, with no tortoise in front of me, I start running on a straight line.

The $1$st metre, I cover in $1/2$ second.
The $2$nd metre in $1/4$ second.
The $3$rd metre in $1/8$ second.
The $n$-th metre in $1/2^n$ second, $n = 1,2,3,… $

If I can do this, then as time $t = 1/2 + 1/4 + 1/8 + ··· + 1/2^n$ tends to $1$, the distance from my initial position tends to infinity, because $d = 1 + 1 + 1 + ··· + 1 = n$. Also my velocity tends to infinity as well.

The question is: where could I be at times after 1 second? Can you suggest possible positions?

Or does it follow that I cannot run this way, that my velocity cannot tend to infinity? But I move in abstract space! So, have I proved that in abstract space my velocity is bounded? But, if instead of $1/2^n$ I use $1/n$, then time also tends to infinity, so there is no problem!

So, does anybody know a clear answer to Zeno-type paradoxa concerning the subdivision of time?

Answer 1

It is worth noting that there are some really quite deep issues about classical physics lurking in the region of this question.

Newtonian gravitational theory, for example, in fact doesn't rule out particles "accelerating out of the universe" in finite time. For example, there is a paper by J. N Mather and R. McGehee on "Solutions of the collinear four body problem which become unbounded in a finite time" (in J. Moser, ed., Dynamical Systems, Theory and Applications (Springer, 1975)). This involves four point-mass particles moving under mutual gravitational attraction, and in effect the potential energy given up by two of the particles as they approach each other while they accelerate away together is given up to the other particles (by an infinite number of bounces in finite time) to kick all the particles to an infinite distance ("out of the universe") in finite time! Very, very cute.

Now for a fun implication: Newtonian gravitational theory time reverses. So the equations allow the reversed solution which gives an empty universe up to time $t$, and then four collinear "space invaders" appear from infinity, one accelerating in from one direction, two from the other, with the fourth particle bouncing madly between them ...

Which nicely shows that, without additional side constraints, Newtonian gravitational theory isn't in itself a deterministic theory (given an empty universe up to now, it could continue empty, or you could next get the four collinear point-masses coming in at infinity, both consistently with the theory).

Fun, eh?

[There's more about this sort of thing in a very famous book, John Earman's A Primer of Determinism (Reidel 1986).]

Answer 2

If your distance travelled is $d$ after time $t$, then your description suggests (at least for positive integer $d$) that $$t = 1-2^{-d}.$$ This is equivalent to $$d = - \dfrac{\log(1-t)}{\log(2)}.$$

Now we see the problem. For $t \gt 1$, we have $1-t \lt 0$, and taking the logarithm of a negative number does not give a real number.

Answer 3

The only mathematical difficulty I see is the following: There exist "nice" functions $f: (0,1) \to \mathbb{R}$ such that there is no "reasonable" way to prolong $f$ to a function $\tilde{f}: (0,1] \to \mathbb{R}$. As Henry rightly points out, $f(t) := - \log(1-t)$ is one such function. There are of course a lot of others.

In the presented setting, there is an intuitive reason to wonder about $f(1)$ (and $f(t)$ for $t>1$), but it doesn't seem to me that there should be any mathematical reason for this to be justified. From physics perspective, I think you are cheating a little bit: On one hand, you are assuming that everything in your model behaves nicely enough for things like velocity to be defined (corresponging roughly to position being a differentiable function of time), and for limits to work nicely. On the other hand, you are considering a particle that reaches some sort of infinity in finite time, which is problematic because "nice" functions don't reach infinity when time is finite (what would being differentiability mean?, for example). Physics just does not work well with actual infinities, that's all, I think.

Answer 4

The question is: where could I be at times after 1 second? Can you suggest possible positions?

Achilles, after this second you’ll exhaust your space. Indeed, each inch of it will be already behind you. So then, because your world line will be unextendable, you’ll have to return to Elysium. But you'll have a lot of fresh impressions.

But, if instead of $1/2^n$ I use $1/n$, then time also tends to infinity, so there is no problem!

Indeed, here is no problem, but it is quite a boring way of spending time. You can ask Sisyphus or Danaides about such things.

Answer 5

As I got a little lost in your edited post, I will stick to the old version instead. I hope you do not mind.

We have Achilles running to infinity. He will cover n^th meter of his journey in $\frac1{2^n}$ of a second. What is his distance after one second?

Let us think of a second as of a constant amount of time. Let us say that infinity means "regardless of how much quantity you have, you can have more". Then, I assume, we have got infinite amount of "time" and "space".

(Before we go on, we need to distinguish between "time" and "space", variable quantities without physical meaning, and the time and space, objects of physical reality. "Time" and "space" (or distance) are numbers - nothing less, nothing more.)

Because "space" is infinite, regardless how far we are from the start, we can go further. The same applies to time; "however late it is, it can get later".

And Achilles - you, my friend - runs. Let us say you started running when "time" and "distance" were "zero" or "no quantity". Because both "time" and "space" are infinite, you can "go further" and it can "get later". This is true whenever, wherever you are.

And now, you introduced a function that assigns "how long it took you to cover n^th meter": $f(n) = \frac1{2^n}$. This function takes the ordinal number of the meter (1^st, 2^nd etc.) and gives you how much of a "second", that is how much of some constant quantity we have picked above, it took you to cover it.

I will introduce a little different function: $$t(n) = \sum_{k=1}^\infty f(k).$$ This function takes the ordinal number of the meter again and returns how much time it took you to get there.

You would like to know where you are after one second. Well, this question makes no sense. You run so fast (and more importantly, still faster) that you will cover the whole infinite space (any distance from the start) in less than one second. That is not surprising given the setup you chose. The function $t(n)$ I defined is a "map" between natural numbers $(1, 2, 3 \dots)$ and the interval $[0, 1)$ of real numbers. It will never reach one, because at the time one, you would "be in infinity" and that is not possible, because you can go further. And it can be later, too: if it "is time $\frac12 = 0.5$ seconds", it can be time $\frac34 = 0.75$ seconds, then $\frac78=0.875$ and so on and so on.

You see? The time had not stopped, it keeps running. But you chose to run so fast, that the question "what happens at time one" no longer makes sense. Nothing can happen at time one, as at the time one, you would reach infinity.

This is different from the original idea of Zeno. Neither Achilles, nor the poor tortoise were accelerating infinitely.

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Mathematically, this means that the range of the function $g(t)$ is $[0, 1)$. The inverse function that tells your distance in meters from the start given the time therefore has the domain of $[0, 1)$. You cannot ask what $g^{-1}(1)$ equals to, as the function is not defined for $x \ge 1$.

Let us define Zeno as the function \begin{align*} z(t)= \begin{cases} 0 & \text{$t \ne 1$,} \\ 1 & \text{$t = 1$,} \end{cases} \end{align*} that is Zeno shouts "Time!" at the time one. Your position at the time one is not defined - that is the whole problem - but Zeno still can shout after one second.

Neither "time" nor "space" are limited. But they are variables bound by the rules you chose for them.

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Note that you do not need Achilles, Zeno and tortoise to have fun. Look at the function plot of the tangent function. Let the $x$ axis stand for time and the $y$ axis stand for the distance traveled. Where are you at time $\frac\pi2$ seconds? (Tangent is really a funny example: it suggest you will get into negative infinity.)

One final note. By finding function $f$ that maps between natural numbers and the real interval $[0, 1)$, you have actually proved there are not less numbers inside the interval $[0, 1)$ than all the natural numbers (thanks, Alex). Isn't it nice?