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Two Spaceship Tachyon Paradox

Supplement to
Spacetime, Tachyons, Twins and Clocks

John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh

The main text describes a tachyon paradox that employs only one spaceship. That simplicity comes at the cost of assuming the possibility of a tachyon mirror that can reflect a tachyon back toward its source while maintaining the tachyon's backwards-in-time propagation.

An easy repair is to replace the mirror by a second spaceship that receives the tachyon and immediately emits a tachyon into the past along the requisite path. If pay attention to the details of how this second spaceship must perform, we arrive at the two spaceship version of the tachyon paradox.

As before, the two spaceship paradox employs spaceships run entirely by robots, programmed to behave in certain ways according to whether or not a tachyonic signal has been received by them. In the figure below, the robot controlled spaceship A is programmed simply to self destruct if it receives a tachyonic signal; and if it still exists later to emit a tachyonic signal into the past.

The robot controlled spaceship B is programmed to switch into an "activated" mode upon receipt of a tachyonic signal; and to transmit a tachyonic signal later only if it is in the activated mode. In addition the motions of the spaceships and timing of the emissions are all carefully pre-programed so that the spaceships are in just the right positions for the sending of a signal the other will receive; and the receiving of a signal the other will send.

tachyon paradox

Start the cycle with the sending of a tachyonic signal by spaceship A. Tracing through the effects of that signal soon leads to the outcome that spaceship A self-destructs prior to sending the tachyonic signal. So the signal is not sent. And if the signal is not sent, tracing through the effects leads to the conclusion that spaceship A does not self-destruct.

A sends a tachyon. arrow B receives a tachyon. arrow B sends a tachyon. arrow A receives a tachyon. arrow A does NOT send a tachyon.
A does NOT send a tachyon. arrow B does NOT receives a tachyon. arrow B doesNOT send a tachyon. arrow A does NOT receive a tachyon. arrow A does send a tachyon.


So A's tachyonic signal is sent. We have a contradiction:

A sends
a tachyon
if and
only if
A does not send
a tachyon.

Copyright John D. Norton. January 2001, September 2002; July 2006; February 3, 2007; January 23, September 24, 2008; January 21, 2010; February 1, 2012; September 24, 2015, August 25, 2018.