Physics against mathematics. Zeno’s Paradox

[Vadim Matveev]

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.
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 spacecraft s 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 spacecraft s:
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 .

 

[HallsofIvy]

“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 spacecraft s:
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

Pretty trivial problem.
a) Setting up a coordinate system fixed on "Totoise", with the 0 point on the stern of the "Tortoise", the bow of "Achilles" is initially at -1000 m and is moving at 0,1 m/s and so will take 1000/0,1= 10000 seconds to dock.

b) Setting up a coordinate system fixed on "Achilles", with the 0 point on the bow of the "Achilles", the stern of the "Tortoise" is initially at 1000 m and is moving at -0,1 m/s and so will take
1000/0,1= 10000 seconds to dock.

c) You haven't given enough information. What is the direction of "Tortoise" and "Achilles" relative to M? If M is on the line of motion of "Tortoise" and "Achilles", then we would multiply each distance and speed by the appropriate contraction factor. If the line of motion of the two ships relative to each other is perpendicular to motion relative to M, the factor is 1. Any other situation, the factor depends upon the angle between the two. In any case, the time would be 10000 seconds times the appropriate contraction factor.

 

[Vadim Matveev]

a) According to the condition of the task the speed of "Achilles" relative to "Tortoise" is equal not to 0,1 m/s, but to 1 m/s. Therefore "Achilles" will take 1000 seconds to dock.

b) Certainly, the speed of "Tortoise" relative to "Achilles" is equal to 1 m/s too and Tortoise" will take in the rest frame of reference of "Achilles" the same 1000 seconds to dock.

c) This question is more complex.
The direction of the motions of "Achilles" and "Tortoise" is so, that "Achilles" and "Tortoise" in the frames of references L and M are moving along the line them connecting. If in the frame of reference L "Tortoise" is moving to the right at a rate of 0,1 m/s, then "Achilles" it moving to the right at a rate of 1,1 m/s (difference of the speeds is 1 m/s, the time of rendezvous transfer is 1000 s).
If in the frame of reference M "Tortoise" is moving with the speed of light to the right, then A is moving with the speed of light to the right too. The contraction of the distance between "Achilles" and "Tortoise" has not mutch significance. This is easy to understand, if we remember that the jointing of "Achilles" and "Tortoise" will occur in 1000 s after their start according to the clocks of "Achilles" and "Tortoise". If at the initial moment the clock of "Achilles" shows the proper time 3 hours, 00 min, 00 seconds, then at the moment of jointing of "Achilles" and "Tortoise" the clock will show the proper time: 3 hours, 16 min. 40 seconds.
If the observers of the frame of reference M begin to observe the spacecraft s, when the clock of "Achilles" is showing 3 hours, 00 min, 00 seconds, then they must longly (infinitely) wait till this clock will show 3 hours, 16 min, 40 seconds.
This confirms that the proximity of spacecraft s to each other cannot be considered as the condition, which accelerates the docking of the spacecraft s.

 

[drag]

Privet Vadim !

Hmm... What's the problem ? A real spacecraft can't
move at the speed of light. And if, for a moment,
we assume it could - you wouldn't be able to see it
unless it rammed you and it would "experience" no time.

Live long and prosper.

 

[Vadim Matveev]

Privet, drag!
It is a logical problem, which can bee solved only by physical (but not mathematical) methods (mathematics describe the unconditional in real life and well-known fact, which contradicts the logic of the ancient Greek Zeno).
As far as moving of real spacecraft s in real frames of references at the speed of light, seeing the Invisible and ramming the observer, you are right. I should forego observing of such objects.

 

[Jonas Ciurlionis]

I think in a way this Zeno problem can be reduced to very basical logical problem.
Two basic steps to solve the paradox would be:
1. To understand that "if p then q" basic implication is false. Achilles doesn't need to reach the exact point where the tortoise is (T) in order to win the race. as there's no physical possibility (from classical point of view) for two bodies to occupy the same place at the same time.
2. to make the same reference system for both racers. In the paradox the Tortoise reference system is the racing track and Achilles' is the trajectory of the Tortoise. This is another basic mistake.
Best.
See also the next issue of "Problemos"