May I assure you all that if anything travels faster than light, it does not travel in time, and nor does anything else. Causality is not violated, it just appears to be so to a particular observer. If I set off in my fast jet, break the sound barrier, and land (all in a straight line), I can then listen to myself arriving. I have not arrived before I set off. Likewise, if I fly my spaceship faster than light, and then come to a stop (in a straight line), I can see myself arriving. I have not arrived before I set off.
I think we can all agree that two clocks which are experiencing 1G run at the same rate, and therefore stay in sync. This is still true if one is experiencing 1G because it is on the surface of the Earth, and the other is undergoing an acceleration of 1G. Einstein’s equivalence principle and the experiment with the man in the chest :- “Relativity The Special And The General Theory”, Methuen and Co 1920, chapter XX .
I set off in my spaceship to Alpha Centauri with the rocket accelerating at 1G. At the half way point it turns round and decelerates at 1G. It comes to a stop at AC, and without cutting the motor, is immediately accelerated back towards Earth at 1G, again turning and decelerating at the half way point, coming to a halt at Earth. The total journey time was 7.8 years, and the total distance covered was 8.8 light years. The clocks on the spaceship were always subject to 1G as were those on Earth, therefore keeping in sync with each other. I certainly did not arrive at AC before I set off, and did not arrive back at Earth before I set off.
The above scenario is allowed. As the rocket and the rocket motor are in the same reference frame (FR), there is no mass increase between the two, and I can use the Lorentz transformations to show this.
Remember that the v in these equations refers to the velocity of the rocket relative to a different FR, which the observer is in (and by inference the propulsion force : the only experiments which have been done to show mass increase with velocity have been under those circumstances). The applicable equation is m = m0 / sqrt (1-(v/c) ^2 ). Because v = 0 (the motor is in the same FR as the rocket, so there is no relative velocity), v/c = 0, and (v/c) ^2 = 0. The square root of 1 - (v/c) ^2 = 1, and m = m0 / 1. Therefore m = m0. As the acceleration is held to 1G for a distance of 2.2 light years before turning round to decelerate, the space ship will reach a speed greater than that of light (the velocity in meters per second at turn around will be 63.87 X 10e7 [ c = 30 X 10e7] ), and time on this space ship will pass at exactly the same rate as back at home on earth.
I urge you to look up :- http://myweb.tiscali.co.uk/carmam/Hollings.html by Tom Hollings
And :- http://www.aquestionoftime.com/ by Hans Zweig