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Monday, January 08, 2007
Interstellar travel

One big topic of perennial interest is interstellar travel and all its sundry issues. I've fiddled with a few equations for relativistic rockets and produced some interesting results - at least to me. For example, the regular equation for motion under continuous, unvarying acceleration is usually written like so...

s = Vo.t + 1/2.a.t^2

...where s is displacement (distance), t is acceleration time, Vo the initial velocity (can be 0 or even negative.) For a complete journey where you accelerate and deccelerate (which is negative acceleration, really) from one location to another, the equation simplifies to...

s = 1/4.a.t^2

and the amount of time you travel is...

t = [4.s/a]^0.5

Now the interesting relativistic thing is that time becomes a dimension as well, and your time displacement is calculated like so...

t^2 = 4.s/a + (s/c)^2

...c being the speed of light. Notice how it's just like the Newtonian equation, plus a component for light's journey as well. In this case all the units are standard SI, metres, seconds, their combinations and c = 299,792,458 m/s exactly. If we use years for time and lightyears for distance a constant, k, comes into play like so...

t^2 = 4.s.k/a + s^2

...k is c/yr, where 'yr' is the standard tropical year of 1900, what's used to calculate an official light-year, some 31,556,925.9747 seconds. But the difference between a tropical year from 1900 (365.2421987 days) and a standard Julian year (365.25 days) is negligible for most purposes, so in either case k is 9.5.

Notice, however, that the time displacement we calculated is for observers at rest relative to the destination. On-board ship the time observed is more complicated and requires hyperbolic equations, oddly enough...

t = 2.s/a.arcosh[1 + a.s/2.c^2]

...which I'll go into some more next time.

Posted at 7:57 am by Adam

 

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