# Length contraction alongside acceleration

Let's have a ship, a target and a ship traveler which we use as point of view. Assuming no other object are observed and we are so far from any other object that gravity distortion are negligible, I'm confused of the effect of length contraction during acceleration (usual length contraction seems quite clear).

If the ship is currently $5ly$ away from target and distance between them does not change (ship is at rest), then the ship begin to travel to target at speed $0.6c$, so the distance between the ship and the target by length contraction becomes: $$l = l_0 \sqrt{1-\frac{v^2}{c^2}}$$ Measuring the speed in "speed of light" unit where $c=1$: $$l = 5 \sqrt{1-\frac{(0.6)^2}{1}} = 5 \sqrt{0.64} = 5 * 0.8 = 4$$ So the distance is reduced by $1ly$ at that speed. However how fast can we reach that speed? If we reach it in a matter of minutes, let's say $1$ minute, then from the point of view of the ship traveler, the distance between his ship and the object was reduced by $1ly$ in $1$ minute. This exceeds the speed of light, which seems incorrect.

What is the expected behavior of the observed distance object from the point of view of the ship traveler, while he is accelerating from zero speed to near-light speed? How length contraction applies, but still keeping the speed of the change in distance less than the speed of light?

• "... the >>distance<< between his ship and the object was reduced by 1ly in 1 minute" - The observed distance to an object is also dependent on your motion relative to that object. It doesn't physically change the distance, and the object didn't physically become closer in any way (the time you need to reach it has been altered however) so no FTL violation has occurred. – Xeren Narcy Mar 3 '15 at 1:47

• As far as I know, I explain to myself both 5 effects of special theory on two reasons: ship traveler does not see the target, but light emitted from the target, therefore he must know that: 1. The light must travel to the ship traveler. 2. The light speed will always be $c$, no matter the speed change. However during acceleration it seems the traveler stay infinitesimal time in each inertial frame. – Plamen Dragiyski Mar 3 '15 at 8:59