Ant on a rubber rope

Ant on a rubber rope is a mathematical puzzle with a solution that appears counter-intuitive or paradoxical. It is sometimes given as a worm, or inchworm, on a rubber or elastic band, but the principles of the puzzle remain the same.

The details of the puzzle can vary,cite book | last = Gardner | first = Martin | authorlink = Martin Gardner | coauthors = | title = aha! Gotcha: paradoxes to puzzle and delight | publisher = W. H. Freeman and Company | date = 1982 | location = | pages = 145-146 | url = | doi = | id = | isbn = 0-7167-1361-6 ] cite web|url=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1042806422 |title=Walking On A Stretching Rubber Band |accessdate=2008-04-06 |last=Wu |first=William |date=2003-01-17 |work=Wu:riddles forum ] cite web|url=http://theproblemsite.com/problems/mathhs/2002/Oct_1.asp |title=The long walk |accessdate=2008-04-06 |first=Graeme |date=2002-10-01 |work=The Problem Site ] but a typical form is as follows.

: An ant starts to crawl along a taut rubber rope 1km long at a speed of 1cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1km per second (so that after 1 second it is 2km long, after 2 seconds it is 3km long, etc). Will the ant ever reach the end of the rope?

At first consideration it seems that the ant will never reach the end of the rope, but in fact it does (although in the form stated above the time taken is colossal). In fact, whatever the length of the rope and the relative speeds of the ant and the stretching, providing the ant's speed and the stretching remain steady the ant will always be able to reach the end given sufficient time.

A formal statement of the problem

The problem as stated above requires some assumptions to be made. The following fuller statement of the problem attempts to make most of those assumptions explicit.

: Consider a thin and infinitely stretchable rubber rope held taut along an $x$-axis with a starting-point marked at $x=0$ and a target-point marked at $x=c$, $c>0$.

: At time $t=0$ the rope starts to stretch uniformly and smoothly in such a way that the starting-point remains stationary at $x=0$ while the target-point moves away from the starting-point with constant speed $v>0$.

: A small ant leaves the starting-point at time $t=0$ and walks steadily and smoothly along the rope towards the target-point at a constant speed $alpha>0$ relative to the point on the rope where the ant is at each moment.

: Will the ant reach the target-point?

Solutions of the problem

An informal reasoned solution

If the speed at which the target-point is receding from the starting-point is less than the speed of the ant on the rope (i.e., if

Otherwise, we can still find a point on the rope that is receding at less than the speed of the ant on the rope by picking a suitable proportion of the distance from the starting-point to the target-point, e.g. $alpha/2v$ of the way along (any amount less than $alpha/v$ will work). Call this point $P\_1$. It seems clear that the ant will reach $P\_1$ (because it would eventually reach $P\_1$ by walking along the axis, and walking along the rope can only carry it further forward). Now, the point on the rope at twice that proportion (call it $P\_2$) is receding at exactly the same speed from $P\_1$ that $P\_1$ was receding from the starting-point (although it is by now rather further away). So the ant should be able to reach $P\_2$. And now, the point on the rope at three times the proportion (call it $P\_3$) is receding at exactly the same speed from $P\_2$ that $P\_2$ was receding from $P\_1$ (although it is much further away). So the ant should be able to reach $P\_3$. This continues, and because the proportion of the way from the starting-point to the target-point at which each point, $P\_1$, $P\_2$, $P\_3$, etc, is found is a fixed amount greater than the proportion for the previous point, the proportion will eventually reach and exceed 1, so the ant will eventually reach the target-point.

For the problem as originally stated, take $P\_1$ to be the point $1/200000th$ of the way along the rope. This point is travelling away from the starting-point at half the walking-speed of the ant, so the ant has no trouble reaching it. Point $P\_2$ is $1/100000th$ of the way along the rope and is travelling away from $P\_1$ at half the walking-speed of the ant, $P\_3$ is $3/200000ths$ of the way along, etc, so after repeating the achievement 200,000 times the ant reaches the end of the rope. However, as the distance gets longer each time, so the time to complete each $1/200000th$ of the way gets longer each time, it is clear that the time required for the ant to complete the journey will be very large. This solution does not provide any more precise indication of how long it will take.

A discrete mathematics solution

Although solving the problem appears to require analytical techniques, it can actually be answered by a combinatorial argument by considering a variation in which the rope stretches suddenly and instantaneously each second rather than stretching continuously. Indeed, the problem is sometimes stated in these terms, and the following argument is a generalisation of one set out by Martin Gardner, originally in Scientific American and later reprinted.

Consider a variation in which the rope stretches suddenly and instantaneously before each second, so that the target-point moves from $x=c,!$ to $x=c+v,!$ at time $t=0,!$, and from $x=c+v,!$ to $x=c+2v,!$ at time $t=1,!$, etc. Many versions of the problem have the rope stretch at the "end" of each second, but by having the rope stretch before each second we have disadvantaged the ant in its goal, so we can be sure that if the ant can reach the target-point in this variation then it certainly can in the original problem or indeed in variants where the rope stretches at the end of each second.

Let $heta(t),!$ be the proportion of the distance from the starting-point to the target-point which the ant has covered at time t. So $heta(0)=0,!$. In the first second the ant travels distance $alpha,!$, which is $frac\{alpha\}\{c+v\},!$ of the distance from the starting-point to the target-point (which is $c+v,!$ throughout the first second). When the rope stretches suddenly and instantaneously, $heta(t),!$ remains unchanged, because the ant moves along with the rubber where it is at that moment. So $heta(1)=frac\{alpha\}\{c+v\},!$. In the next second the ant travels distance $alpha,!$ again, which is $frac\{alpha\}\{c+2v\},!$ of the distance from the starting-point to the target-point (which is $c+2v,!$ throughout that second). So $heta(2)=frac\{alpha\}\{c+v\}+frac\{alpha\}\{c+2v\},!$. Similarly, for any $ninmathbb\{N\},!$, $heta(n)=frac\{alpha\}\{c+v\}+frac\{alpha\}\{c+2v\}+cdots+frac\{alpha\}\{c+nv\},!$.

Notice that for any $iinmathbb\{N\},!$, $frac\{alpha\}\{c+iv\}geqslantfrac\{alpha\}\{ic+iv\}=left(frac\{alpha\}\{c+v\}\; ight)left(frac\{1\}\{i\}\; ight),!$,so we can write $heta(n)geqslantleft(frac\{alpha\}\{c+v\}\; ight)left(1+frac\{1\}\{2\}+cdots+frac\{1\}\{n\}\; ight),!$. The term $left(1+frac\{1\}\{2\}+cdots+frac\{1\}\{n\}\; ight),!$ is a partial Harmonic series, which diverges, so we can find $Ninmathbb\{N\},!$ such that $1+frac\{1\}\{2\}+cdots+frac\{1\}\{N\}geqslantfrac\{c+v\}\{alpha\},!$, which means that $heta(N)geqslant1,!$.

Therefore, given sufficient time, the ant will complete the journey to the target-point. This solution could be used to obtain an upper-bound for the time required, but does not give an exact answer for the time it will take.

An analytical solution

A key observation is that the speed of the ant at a given time $t>0,!$ is its speed relative to the rope, i.e. $alpha,!$, plus the speed of the rope at the point where the ant is. The target-point moves with speed $v,!$, so at time $t,!$ it is at $x=c+vt,!$. Other points along the rope move with proportional speed, so at time $t,!$ the point on the rope at $x=X,!$ is moving with speed $frac\{vX\}\{c+vt\},!$. So if we write the position of the ant at time $t,!$ as $y(t),!$, and the speed of the ant at time $t,!$ as $y\text{'}(t),!$, we can write:

$y\text{'}(t)=alpha+frac\{v,y(t)\}\{c+vt\},!$

This is a first order linear differential equation, and it can be solved with standard methods. However, to do so requires some moderately advanced calculus. A much simpler approach considers the ant's position as a proportion of the distance from the starting-point to the target-point.

Consider coordinates $psi,!$ measured along the rope with the starting-point at $psi=0,!$ and the target-point at $psi=1,!$. In these coordinates, all points on the rope remain at a fixed position (in terms of $psi,!$) as the rope stretches. At time $tgeqslant0,!$, a point at $x=X,!$ is at $psi=frac\{X\}\{c+vt\},!$, and a speed $A,!$ relative to the rope in terms of $x,!$ is equivalent to a speed $frac\{A\}\{c+vt\},!$ in terms of $psi,!$. So if we write the position of the ant in terms of $psi,!$ at time $t,!$ as $phi(t),!$, and the speed of the ant in terms of $psi,!$ at time $t,!$ as $phi\text{'}(t),!$, we can write:

$phi\text{'}(t)=frac\{alpha\}\{c+vt\}$

$hereforephi(t)=int\{frac\{alpha\}\{c+vt\},dt\}=frac\{alpha\}\{v\}log(c+vt)+kappa$ where $kappa,!$ is a constant of integration.

Now, $phi(0)=0,!$ which gives $kappa=-frac\{alpha\}\{v\}log\{c\},!$,so $phi(t)=frac\{alpha\}\{v\}log\{left(frac\{c+vt\}\{c\}\; ight)\},!$.

If the ant reaches the target-point (which is at $psi=1,!$) at time $t=T,!$, we must have $phi(T)=1,!$ which gives us:

$frac\{alpha\}\{v\}log\{left(frac\{c+vT\}\{c\}\; ight)\}=1,!$

$herefore\; T=frac\{c\}\{v\}left(e^\{v/alpha\}-1\; ight),!$

As this gives a finite value $T,!$ for all finite $c,!$, $v,!$, $alpha,!$ ($v>0,!$, $alpha>0,!$), this means that, given sufficient time, the ant will complete the journey to the target-point. This formula can be used to find out how much time is required.

For the problem as originally stated, $c=1,mathrm\{km\},!$, $v=1,mathrm\{km\}/mathrm\{s\},!$ and $alpha=1,mathrm\{cm\}/mathrm\{s\},!$, which gives $T=(e^\{100,000\}-1),mathrm\{s\},!approx2.8\; imes10^\{43,429\},mathrm\{s\},!$. This is a truly vast timespan, vast even in comparison to the estimated age of the universe, and the length of the rope after such a time is similarly huge, so it is only in a mathematical sense that the ant can ever reach the end of this particular rope.

Applications of the problem

This problem has a bearing on the question of whether light from distant galaxies can ever reach us if the universe is expanding. If the universe is expanding uniformly, this means that galaxies that are far enough away from us will have an apparent relative motion greater than the speed of light. This does not violate the relativistic constraint of not travelling faster than the speed of light, because the galaxy is not "travelling" as such -- it is the space between us and the galaxy which is expanding and making new distance. The question is whether light leaving such a distant galaxy can ever reach us, given that the galaxy appears to be receding at a speed greater than the speed of light.

By thinking of light photons as ants crawling along the rubber rope of space between the galaxy and us, it can be seen that just as the ant will eventually reach the end of the rope, given sufficient time, so the light from the distant galaxy will eventually reach earth, given sufficient time.

References

External links

* [http://www.math.hmc.edu/funfacts Su, Francis E., et al. "Inchworm on a Rubber Rope." Mudd Math Fun Facts]
* [http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/664.shtml Waeber, Marie-Jo. "Puzzle involving exponential" on Cut the knot: Learn to enjoy!]