
Interstellar TravelThe distances between stars are almost inconceivable. With a little practice, you can get a handle on it, but you have to build that perspective. If we were to fly to the nearest star at the speed of sound at sea level, then it would take about 3.8 million years to get there. Voyager 2 is the fastest "interstellar" spacecraft ever launched, and it is traveling at 15.9 km/s. It's not headed for the nearest star, but if it were it would take 81,000 years to get there. Even at the speed of light, it would take 4.3 years to get to the nearest star, and only eleven stars are within ten lightyears. But nature gives us a tiny opening. If you travel at near lightspeed, then you don't experience as much elapsed time, because the transformation between moving coordinate system is not the familiar Gallilean coordinate transformation:
This transformation applies very well to our everyday experience. Suppose we look at a train traveling down the track. Let the coordinate x describe the position along the track, and let there be a station at x = 0. If a train is traveling along the track with velocity v, then its position is x = vt. If, however you are inside the train, then you wouldn't keep track of your position according to the scenery flying by. You'd use a different coordinate system centered on some convenient spot, such as your seat. Let's call that position x′ = 0, and the seat in front of you is x′ = 1 m. (That's one meter. All of my variables are dimensioned.) The coordinate transformation describes how you would convert a position (x, t) in the reference frame of the station to a position (x′, t′) in the reference frame of the moving train. When Einstein worked out the special theory of relativity, it had been observed that the speed of light (in a vacuum) is apparently the same in any coordinate system. This just doesn't work out under a Gallilean transformation. If you send light from your seat to the seat in front of you, it travels at a velocity c. An observer in the station would have to see that the same signal traveled at a velocity c + v. The special theory of relativity simply states that the coordinate transformation between moving frames of reference is not the familiar Gallilean transformation, but is actually a Lorentz transformation:
where γ = √[1 – v^{2} / c^{2}]^{ –1}
With this coordinate transformation, moving frames of reference no longer share a common time coordinate, and they don't share a common scale for distance either. This takes some getting used to, and there are some common paradoxes that you should work through to understand it. One clear result, however, is that a passenger traveling at 0.9c doesn't take 4.8 years to travel the 4.3 lightyears to the nearest star. (Well, that's what we would see from Earth, but not what the passenger sees.) The passenger would find that it only took 2.1 years. (The passenger doesn't see him/herself traveling faster than the speed of light. Instead, she/he sees that the distance is only 1.9 lightyears.) The problem to play with is a simple integration exercise, coupled with some clear thinking. If we suppose that we eventually have the ability to harness enormous resources, but do not uncover new laws of physics, then it will always take individual humans years to travel between the stars. The problem is that we can't accelerate faster than our bodies can survive. So, if we assume that the passengers want to experience the journey at an acceleration of 1 g, then how much travel time do they experience on an interstellar journey? I'll use τ to represent the time coordinate as experienced by the passenger and X to represent the distance between the Earth and the destination (in Earth's frame of reference). Assume that the final velocity is zero—that is, the passenger wants to get off the ship at the other end, not just fly by at near lightspeed. Solving this problem doesn't actually involve any more physics. It does, however, require that you determine how the velocity v of the spaceship changes while the passengers experience an acceleration of g in their own frame of reference. 
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Thanks,
Steve