How fast can we get to Alpha Centauri?

Suppose in the near future we send an Antimatter rocket that is capable of constant $$1g$$ acceleration to our nearest star, Alpha Centauri, $$4.3$$ light years away and suppose we want the spacecraft to reach the destination in the shortest time possible. This means the spacecraft will accelerate at $$1g$$ to the half way point and then decelerate at $$1g$$ for the rest of the journey.

Using Newtonian mechanics I have found that the entire journey will take $$4$$ years as measured from Earth with a maximum velocity of $$v_{max}=6.31*10^8 ms^{-1}$$ ($$2.1c$$ - faster than speed of light) at the midpoint. However, nothing can travel faster than the speed of light so applying the formula $$v(t)=\frac{at}{\sqrt{1+\frac{a^2t^2}{c^2}}},$$ where $$v(t)$$ is the velocity at time $$t$$, $$a$$ is the acceleration of $$1g$$ and $$t$$ is the time as measured from Earth to the model, I get a maximum velocity of $$v_{max}=2.7*10^8$$ ($$0.9c$$).

What if we are in the rocket? Our destination will appear closer to us due to length contraction, not only that, but because of time dilation, our journey will take even less time.

1) How do I apply length contraction and time dilation to the model?

2) Are there any other special relativity effects that I should implement into the model?

• Jun 15 '19 at 20:30
• You need to integrate your $v(t)$ to get $x(t)$ so that you can find what $t$ makes $x$ equal 4.3 ly. Jun 15 '19 at 21:19
• @PM2Ring I have looked at the link. How do they derive the rocket equation $T = \frac{c}{a}\sinh{\frac{at}{c}}$? Also why does the time dilation equation $\Delta T = \gamma \Delta t$ not give the same answers? Jun 21 '19 at 18:22
• @G.Smith I have done that. I believe all my results are still in the rest frame. How do I work out the results in the rocket's frame of reference? Jun 21 '19 at 18:25

You swapped $$T$$ & $$t$$ in your equations in the comment; $$T$$ is the proper time in the ship frame.
To get the equation for $$t$$ in terms of $$T$$ we can't simply multiply, because $$\gamma$$ isn't constant. We need to integrate $$dt=\gamma dT$$. I show how to find $$v$$ in terms of $$T$$ from 1st principles towards the end of this answer. The result is
$$v=c\tanh\left(\frac{aT}{c}\right)$$
We can now use $$\gamma=\frac1{\sqrt{1-(v/c)^2}}$$ to obtain $$\gamma=\cosh(aT/c)$$
We now integrate $$t=\int\gamma\,dT$$, which gives us $$t=\frac ca\sinh\left(\frac{aT}{c}\right)$$