The path described by a constantly accelerating particle is given by:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]x=c\sqrt{c^2/a'^2+t^2}[/itex]

where a prime denotes an observer traveling with the particle and a letter without a prime a resting observer.

If we leave the [itex]c^2/a'^2[/itex] out it reduces to x=ct, which makes sense. The distance traveled by a photon is simply it's speed times the travel time. If we think of a constantly accelerating particle, a particle which would first need to accelerate to a speed c (or at least close to it), we would expect that it would travel a shorter distance in the same amount of time, since it would first need to accelerate and hence travel at a smaller average speed. However, when you look at the equation above, you can see the travel time gets a little bonus, [itex]c^2/a'^2[/itex]:

[itex]x=c\left[t+f\left(c^2/a'^2\right)\right][/itex]

According to the equation, the slower one accelerates, the larger distance one will be able to travel. To me that does not make any sense at all, so I wondered whether I made any mistakes and decided to verify the answer on the internet. However, without blinking an eye, every single source I find online gives me the same answer.

So where do I go wrong in all this?

Sources:

http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])

http://www.physik.uni-leipzig.de/~schiller/ed10/Uniform relativistic acceleration.pdf

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# Hyperbolic path in Minkowski space

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