4
$\begingroup$

I know that it is a very old question but still I don't find any satisfactory solution for Achilles Paradox. Please explain me the fundamentals of Achilles paradox in terms of stage wise distance covered. Note that it is easily solvable in terms of time, but if you start analysing this event in terms of time, then there is not at all any paradox. So please explain in terms of stage wise distances only.

$\endgroup$
4
$\begingroup$

This is quite straightforward as long as you break it up into steps.

Let the tortoise travel with speed $v_t$ and Archilles with speed $v$. We split up the steps Archilles takes as \begin{equation} d_A = d_0 + d _1 + ... \end{equation} where $d_i $ is the distance travelled in a single step. The initial distance $d_0 $ is then just the initial separation between Archilles and the tortoise. We denote $\Delta t _i $ as the time of step $i $. The distance $d_1 $ is just how far the tortoise managed to step while Archilles was catching up to the tortoise's initial position. It is given by \begin{equation} d_1=v_t \Delta t_1 =v_t \frac{d_0}{v} \end{equation} Similarly, the distance $d_2 $ is given by \begin{equation} d_2=v_t \Delta t_2 = v_t \frac{d_1 }{v} = \left(\frac{v_t}{v} \right)^2 d_0 \end{equation} Its easy to see that this trend will continue: \begin{align} d_A &= d_0 + \frac{v _t }{v} d_0 +\left(\frac{v_t}{v} \right)^2 d_0 + ... \\ &= d _0 \left( 1 + \frac{ v _t }{ v} + \left( \frac{ v _t }{ v } \right) ^2 + ...\right) \end{align} In the brackets we have a geometric series. If $v_t\le v$ then, \begin{align} d _ A & = d_0\frac{1}{ 1 - v _t / v } \end{align} This is finite (and hence Archilles can catch up with the tortoise) as long as $v_t <v $. The apparent paradox is that Archilles needs to take an infinite number of steps. However, having to take an infinite number of steps doesn't mean it would take an infinite distance (or infinite time). An infinite series can either converge, as this one does, or diverge.


I think it's also interesting to solve this problem the simple way, avoiding splitting it up into steps. For Archilles and the tortoise representatively we have, \begin{equation} d_A=\frac{v}{t}\, , \quad d_t= \frac{v_t}{t}+d_0 \end{equation} Setting $d_A = d_t $ and solving these equations gives, \begin{equation} d_A=d_0 \frac{1}{1-v_t/v} \end{equation} as required.

$\endgroup$
  • 3
    $\begingroup$ This answer doesn't convince me. $1+\frac{1}{2}+\frac{1}{4}+...$ doesn't have to be equal to $2$, because the proof uses that if a number is infinitely near another number, they are the same (archimedian property). If you use other axioms, you will get different results. $\endgroup$ – jinawee Feb 9 '14 at 15:26
  • $\begingroup$ @jinawee, I'm confused. Are disputing the well-known result of a geometric series:en.wikipedia.org/wiki/Geometric_series ? $\endgroup$ – JeffDror Feb 9 '14 at 15:32
  • 2
    $\begingroup$ No, but solving Zeno's paradox is basically trying to prove the Archimedian property, which is a mathematical axiom. Since it's an axiom, you can't prove it. In p-adic numbers, $1+2+4+..=-1$, which of course, would give you incorrect predictions. $\endgroup$ – jinawee Feb 9 '14 at 15:46
  • 10
    $\begingroup$ Mathematicians can make arbitrary definitions (and seem to have a lot of fun doing it), but physicists are constrained to use the ones that generate results in a agreement with the way the world works. (I.e. the second panel of Every Major's Terrible.) $\endgroup$ – dmckee Feb 9 '14 at 16:51
  • 1
    $\begingroup$ With a little tweaking this argument can be used to prove that the series converges: you use your second expression: "speed difference gives time until they meet", then conclude that "ergo, this infinite series must converge to this number". $\endgroup$ – Floris May 27 '14 at 12:44
3
$\begingroup$

I don't know if I'll answer what you want, but is impossible not to talk about time; the point of the paradox is that both the Achilles will never reach the turtle, right? now, in the way the paradox is stated, it is said that at each "step" the turtle advances "x" space and Achilles will advance only a fraction of "x" space. The fallacy of the paradox lies in that the statement displaces the concept of time from real time to "step-time". So, Achilles in deed will reach the turtle only in infinity, but this doesn't happen in the real time, just in the mathematical steps, which are increasingly small. The place where they will meet can be solved as the result of a geometric series (see answer by JeffDror).

$\endgroup$
1
$\begingroup$

It's just a paradox, it's using weakness of human logic about infinite calculation in limited time. It convinces you think that you have to cover half of the distance between Achilles and turtle every passing instant of time. If you think deeper, you will notice that most of the concepts such as time, motion, distance, instant, speed lose meaning, their definitions are relative to each other indeed.

$\endgroup$
1
$\begingroup$

Not Newtonian, but still relevant:

Zeno's Paradox is based on the premise that the distance between Achilles and the tortoise is infinitely divisible. In theory, the Planck length is a lower limit on measurable distances. If this is true, eventually Achilles must move more than half the distance to an arbitrary point between himself and the tortoise if he moves at all.

$\endgroup$
  • 1
    $\begingroup$ This isn't the resolution to the paradox. There is no evidence that time or length is quantized. $\endgroup$ – Brandon Enright Feb 19 '14 at 23:39
  • 1
    $\begingroup$ True. I've edited my post to reflect this better. The other answers thus far have all been purely mathematical arguments, so I thought it would be helpful to the discussion to offer one based on physical principles. To my understanding (which is limited; I only have a B.Sc.) distances shorter than the Planck length have no physical meaning anyway. $\endgroup$ – PolskiPhysics Feb 20 '14 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.