## Salmon On Austins

In Chapter 8 one of the short problems, posed by A. K. Austin of the University of Sheffield, England, aroused considerable controversy among readers. Indeed, the problem proved to be an amusing new variant of Zeno's famous paradox of Achilles and tire Tortoise, and one that, so far as I know, had never been Formulated before. Here is how I phrased the problem and its answer:

"A boy, a girl and a dog are at the same spot on a straight road. The boy and the girl walk forward—the boy at four miles per hour, the girl at three miles per hour. As they proceed, the dog trots back and forth between them at 10 miles per hour. Assume that each reversal of its direction is instantaneous. An hour later, where is the dog and which way is it facing?"

Answer: "The dog can be at any point between the boy and the girl, facing either way. Proof: At the end of one hour, place the dog anywhere between the boy and the girl, facing in either direction. Time-reverse all motions and the three will return at the same instant to the starting point."

Even before this answer appeared I began receiving letters from readers protesting that the problem is meaningless because its initial conditions are logically contradictory. No matter how small we make the starting interval, many wrote, the dog will have to make an infinity of reversals that would drive it crazy. Others contended that the three "points" (as in all such problems, the boy, the girl and the dog symbolize ideal points) could never get started because the "instant" they did so the dog would either leap ahead of both boy and girl or run the opposite way, thereby ceasing to be between the boy and girl.

As Wesley C. Salmon, a noted philosopher of science, immediately recognized, Austin's paradox has innumerable other forms, one of the simplest of which is a time reversal of the familiar puzzle about two locomotives and a bird. The locomotives, starting at A and B, 30 miles apart, move toward each other on the same track at, say, 15 miles per hour until they collide at C. A bird, starting at A, flies back and forth at 60 miles per hour between the locomotives until they collide. How long is the bird's path? There is no need to sum an infinite series. Since the bird flies for one hour, the path must be 60 miles. If we time-reverse the event, specifying that the bird end at A, a unique zigzag path is defined that the bird can travel in either direction.

Suppose, however, we do not state where the bird must be after the locomotives have moved backward to points A and B. Without this information a unique path for the bird cannot be defined. Because the bird can now take an infinity of possible paths, the most we can say is that the backward-flying bird must end somewhere between A and B.

But is it really permissible to say this? No, many mathematicians contend, because a singularity arises in the time-reversed version that creates contradictory initial conditions. "There is no general justification in analysis," one mathematician put it, "for inverting the limit operator." When the locomotives move toward each other, it is only the bird's position that converges. "The velocity vector diverges, so that there is the same difficulty (as in Austin's problem) in finding a unique inverse to the limit process. The accepted rules of differential calculus have evolved because if followed properly they avoid contradictions."

It is helpful to plot a space-time graph of the bird's path from A to C' [see Figure 82]. Of course, we cannot finish drawing the bird's path to C' because the zigzags are infinite, but we certainly can assume that the ideal line exists. Surely if this line can go down from A to C', there is no logical objection to saying that it can go up from C' to A. If the final destination of the bird is not specified, an uncountable infinity of such graphs can start at C' and end anywhere on the track between A and B. It is true that calculus cannot solve Austin's similar problem if "solve" means to pinpoint the dog's final position, but Austin's "solution" is precisely one that shows this to be impossible. Since the dog is not told how to start, it can start in any way it pleases provided it always stays between the boy and the girl. Consequently its path can end anywhere between boy and girl.

Salmon has commented on Austin's problem as follows:

"Almost everyone has heard the old chestnut about the bird that flies back and forth between two approaching locomotives . . . [as given above]. Or, to achieve historical perspective, suppose Achilles is pursuing the tortoise and a Trojan fly buzzes back and forth between them. Given a set of velocities and distances, and our latter-day assurance that Achilles will overtake the tortoise at a determinate time and place (see my book Zeno's Paradoxes), we can easily figure out how far the fly will travel. Up to this point we have no new Zenonian paradoxes. . . . We see that Austin's problem is just the time-reversal of the bird-and-train problem.

"In order to retain historical perspective, let us go back to Achilles and the tortoise. In spite of the initial handicap traditionally imposed on Achilles, he catches the tortoise, and to redress the grievance he has long held against Zeno he keeps on running, steadily increasing his lead over the fortunate tortoise. [I consider the tortoise fortunate in this version of the tale, at least in comparison with Lewis Carroll's account "What the Tortoise Said to Achilles," in which Achilles stops and seats himself on the back of the tortoise, much to the tortoise's discomfort.] Now consider the Trojan fly, which attempts to continue flying back and forth between the two runners even after the faster overtakes the slower. When Achilles and the tortoise are just even, the fly finds itself precisely in the position of Austin's dog.

"For the sake of definiteness, say that the tortoise travels at one mile per hour, Achilles at five miles per hour (he has been running since 500 b.c., so that he is not as fleet as he once was) and the fly at 10 miles per hour. They all arrive at the common meeting point without difficulty. How can they go on? If the three start simultaneously from the common point, the fly immediately either advances ahead of both or moves behind both, each of which violates the condition that the fly be always in the interval between the two (end points included). It would seem we could argue that in any time interval e > 0, however small, the tortoise travels a distance of le, Achilles runs a distance of 5e and the fly goes lOe. Hence in an arbitrarily small time after the meeting the fly leaves the interval between the tortoise and Achilles. Even if we have shown how Achilles can perform the 'supertask' of catching the tortoise, and how the tortoise can perform the 'supertask' of initiating its motion, it appears that the fly now faces the new 'supertask' of continuing to fly back and forth between Achilles and the tortoise after the tortoise has been overtaken. In other words, the fly now faces the supertask of not passing Achilles!

"The apparent difficulty seems to me analogous to the problem pointed out by Zeno in his regressive dichotomy paradox. There is no doubt that the fly will outdistance both Achilles and the tortoise if it flies steadily in one direction without turning around, even in the arbitrarily small period of duration e. This fact does not render the fly's motion impossible, however, since no matter how small a time interval we choose the fly has already reversed its direction during that interval (infinitely many times, so that it is really quite dizzy). This simply means that there is no initial nonzero interval during which it flies straight without reversing its direction; thus it does not follow that the fly immediately leaves the interval between the tortoise and Achilles. In fact, we can see precisely how the fly's rapid reversals enable it to stay between Achilles and the tortoise after the meeting by examining the time reversal of this motion in the fly's approach to the point of meeting from the earlier side. The fact that the fly does not traverse an initial nonzero straight path is analogous to the fact that the tortoise, in leav ing its starting point, does not traverse any initial nonzero segment of its path. The lack of a suitable initial segment is not a serious obstacle to either of them.