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?

  • $\begingroup$ "... 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. $\endgroup$ Mar 3, 2015 at 1:47

1 Answer 1


This exceeds the speed of light, which seems incorrect.

During acceleration, the speed of light may seemingly be exceedet from the viewpoint of the accelerated observer. This is why one talks about "uniform relative velocity" when talking about inertial frames, in which the speed of light may not be exceedet. When you accelerate, you change your reference frame, so you are no longer in an inertial frame. Nevertheless, from the viewpoint of every inertial frame, the speed of light is of course never exceedet, since the rulers and clocks relative to which they measure the distance by time do not contract or dilate in their inertial frame. Also see Ruslans answear in this thread.

  • $\begingroup$ 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. $\endgroup$ Mar 3, 2015 at 8:59
  • $\begingroup$ What about looking at non-inertial frame (since infinitesimals are difficult to spot)? Is special theory of relativity is enough to calculate traveling parameters in non-inertial frame or we need to apply general theory of relativity then (because equivalence between acceleration and gravity)? $\endgroup$ Mar 3, 2015 at 9:21

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