How can a continuous 1-G thrust get you to Alpha Centauri without exceeding the speed of light half way there?
A simple but incorrect calculation would predict the following velocity at the turn-around point:
distance = 1/2 acceleration x time^2 and since a = 980 cm/sec/sec and 4.2 light years/2 = 2 x 10^18 centimeters, you get time = 64 million seconds. Then, from v = acceleration x time = 980 x 64 million = 6.3 x 10^10 cm/sec which is 2.1 times the speed of light. So, if you accelerated at 1-G for 64 million seconds, and then decelerated at 1-G for 64 million seconds, you would arrive at Alpha Centauri in 128 million seconds or 4.1 years. The problem is that at your half-way point, you would have reached twice the speed of light which is an impossibility.
What has not been included is the correct dynamics of the process according to special relativity. We have neglected the fact that 1-G of thrust acting on a body whose mass is steadily increasing by
M =m(rest)/( 1 - (V/c)^2 )^2 1/2
results in less and less velocity change as the acceleration continues. By the time the acceleration has gotten you to 0.8c, the effective mass your motors are pushing against has grown to 1.7 times the rocket's rest mass, and the gain in velocity is only about 1/2 as much per unit time. The best you can hope for is to get to the turn around point at about 0.5 c which means that the trip would take a bit more than 4.2 x /0.5 = 8.4 years when the time to accelerate to 0.5c is included properly.
Glad that's figured out...