Later On

A blog written for those whose interests more or less match mine.

A person-less variant of the Bernadete paradox

leave a comment »

From an intriguing post by Roy Cook in the Oxford University Press blog:

. . . Imagine that A, B, and C are points lying exactly one meter from the next, in a straight line (in that order). A particle p leaves point A, and begins travelling towards point B at exactly one second before midnight. The particle p is travelling at exactly one meter per second. The particle p will pass through B (at exactly midnight) and continue on towards C unless something prevents it from progressing further.

There is also an infinite series of force-field generators, which we shall call G1, G2, G3, and so on. Each force-field generator in the series will erect an impenetrable force field at a certain point between A and B, and at a certain time. In particular:

(1) G1 will generate a force-field at exactly ½ meter past B at ¼ second past midnight, and take the force-field down at exactly 1 second past midnight.

(2) G2 will generate a force-field at exactly ¼ meter past B at exactly 1/8 second past midnight, and take the force-field down at exactly 1/2 second past midnight.

(3) G3 will generate a force-field at exactly 1/8 meter past B at exactly 1/16 second past midnight, and take the force-field down at exactly 1/4 second past midnight.

And so on. In short, for each natural number n:

(n) Gn will generate a force-field at exactly 1/2n meter past B at exactly 1/2n+1 second past midnight, and take the force-field down at exactly 1/2n-1 second past midnight.

Now, what happens when p approaches B?

Particle p’s forward progress will be mysteriously halted at B, but p will not have impacted any of the barriers, and so there is no explanation for p’s inability to move forward. Proof: Imagine that particle pdid travel to some point x past B. Let n be the largest whole number such that 1/2n is less than x. Thenp would have travelled at a constant speed between the point ½n+2 meter past B and 1/2n meter past Bduring the period from ½n+2 second past midnight and 1/2n second past midnight. But there is a force-field at 1/2n+1 meter past B for this entire duration, so p cannot move uniformly from ½n+2 meter past Band 1/2n meter past B during this period. Thus, p is halted at B. But p does not make contact with any of the force-fields, since the distance between the mth force-field and p (when it stops at B) is 1/2m meters, and the mth force-field does not appear until 1/2m+1 second after the particle halts at B. . . .

Written by LeisureGuy

17 October 2016 at 3:35 pm

Posted in Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s