Examining the paradox of “Achilles and tortoise” one should remember that this is only one of several paradoxes, showing (in the opinion of the Eleatics) the illusoriness of motion. The final conclusion of paradox consists not in the fact that Achilles never will overtake tortoise in the finishing stage, but in the fact that Achilles never will move from the place. Indeed, if between Achilles and tortoise is located a second tortoise, that runs away from Achilles, then Achilles will not overtake it too, and, therefore, he will never run half-way to the first tortoise. If between the second tortoise and Achilles there is a third tortoise, then, etc.... To put it briefly, Achilles will be not at all able to move from the place and to begin motion.(adsbygoogle = window.adsbygoogle || []).push({});

The mathematical negative solution of Zeno’s paradox, showing the inaccuracy of the conclusion of Zeno is well-known. The positive physical solution of this paradox showing the correctness of the conclusion of Zeno under uncommon conditions indicated in the solution is proposed in the decision below).

Imagine two spacecraft ”Achilles” and ”Tortoise”, the length of each of them is equal to 10 m, performing prolonged interstellar overflight, being located at a certain distance from each other.

Let us assume that there was a need in a docking of spacecrafts in the process of flight, for realization of which “Achilles” at a certain moment of time begins uniform and rectilinear motion to “Tortoise” and in a certain time performs rendezvous contact with it.

It asks itself, how long will take the process of the rendezvous transfer of “Achilles” and”Tortoise”, if at the start moment the nose of “Achilles” is located at a distance of 1000 meters from the stern of “Tortoise”, and the speed of its movement to ”Tortoise” is equal to 1 m/s (relative to “Tortoise”)?

Answer the questions considering the spacecrafts:

a) in the frame of reference K, motionlessly connected with the the “Tortoise”;

b) in the frame of reference L, in which “Tortoise” at a rate of 0,1 m/s moves away from it overtaking “Achilles” (“Achilles” moves in this frame of reference with a velocity of 1,1 m/s);

c) in the frame of reference M, in which “Tortoise” moves with the speed so close as desired to the speed of light but not exceeding it

According to the Zenon's logic, answer, how many completed transfer phases will be required to “Achilles” in order to overtake to “Tortoise”, if by completed transfer phase we understand the motion of “Achilles” from the starting-place ( or from the final place of previous completed transfer phase) to that place of space, where for a while back “Tortoise” was located (“Achilles” at the moment of the completion of completed transfer phase must completely, but not partially, occupy the place, where “Tortoise” was located)?

Answers:

a) in the frame of reference K “Achilles” will overtake tortoise for one uncompleted transfer phase (“Achilles” does not at the moment of rendezvous contact completely occupy the place, where “Tortoise” was and is located) in such a way:

1) Starting position (A is “Achilles”, T is “Tortoise”):

A--------------------------T

2) First uncompleted transfer phase (“Achilles” does not occupy the place, where “Tortoise” was and stays put). Rendezvous contact is completed. The situation is the following (docking):

--------------------------AT

Rendezvous transfer time in case a) is equal to 1000 seconds;

b) in the frame of reference L “Achilles” will overtake “Tortoise” in the second uncompleted transfer phase, after previous first completed transfer phase lasted 918,1818 seconds on the way of 1010 meters in such a manner:

1) Starting position:

A---------------------------T

2) First completed phase (“Achilles” does occupy the place, where “Tortoise” was). Rendezvous transfer is uncompleted.

----------------------------A----T

3) Second uncompleted transfer phase (“Achilles” does not occupy the place, where “Tortoise” was).

Rendezvous contact is completed.

-------------------------------------AT

Rendezvous transfer time of “Achilles” and “Tortoise” in case b) is equal to 1000 seconds;

c) in the frame of reference M as large as desired quantity of approach completed phases will be required to “Achilles”, in order to overtake “Tortoise”. Rendezvous transfer time as great as desired, if the speed of “Tortoise” is so close as desired to the speed of light (“Achilles” will “never” overtake “Tortoise”).

1) Starting position:

A---------------------------T

2) First completed phase (“Achilles” does occupy the place, where “Tortoise” was).

-----------------------------A---------------------------T

3) Second completed transfer phase (“Achilles” does occupy the place, where “Tortoise” was).

----------------------------------------------------------A---------------------------T

All subsequent phases are analogous.

The clock in the window of “Achilles” stops as a result of the relativistic time dilation and for ever shows the time of starting.

More about the relativistic effect of Zenon’ paradox you can (if you can) read in Russian in http://www.sciteclibrary.ru/rus/catalog/pages/5588.html .

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# Physics against mathematics. Zeno’s Paradox

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