# Does an accelerating spaceship move backwards due to length contraction?

Let's assume I have a spaceship in front of me let's say at 1000000km distance. Now let's assume I have also a stationary wall just behind the spaceship at 999999km. Initially the spaceship's speed is 0.

Now let's accelerate the spaceship rapidly to relativistic speeds.

Due to the length contraction the spaceship and it's distance from me contracts.

Now I see a paradoxical situation: I and the wall is in the same frame of reference. From the wall's point of view the spaceship come closer to the wall momentarily and zips away. From my point of view the spaceship can come closer to me than the wall (due to length contraction rate ), effectively smacking into the wall during the acceleration.

How to resolve this apparent paradox?

(or where is the origin of the contraction?)

The paradox of your situation lies in which frame of reference you're using. In reality, there is no paradox. It seems as though you have a decent grasp on relativistic effects, so I'll spare you the gory details. However, you problem is that in your and the wall's frame, the distance between you and the spaceship won't change.

Relativistic length contraction only affects things that are travelling at relativistic speeds in your frame. So in your frame, it is the spaceship that will contract, you will see it as being shorter, but the location of the ship won't change initially (assuming instantaneous acceleration). The same is true from the wall's perspective. What I believe you were thinking of is that in the spaceship's frame of reference, the distance between you and it will contract.

This doesn't present a problem though, because while the spaceship sees the distance between it and the wall shrink, the ratio of lengths between it to the wall and it to you remains the same. So no, it does not move backwards.

• So basically only the accelerating reference frames experience distance shrinking isn't it? Stationary reference frames seeing something accelerate won't. – Calmarius Apr 26 '13 at 7:14
• right, because in the non-accelerating frames, the distance between 2 objects should be the same as if there were no third accelerating object. – Jim Apr 26 '13 at 13:02

The distance between you and the rocket, measured in your rest frame as a function of your elapsed time, is given by the relativistic rocket equation:

$$d(t) = 10^9 + \frac{c^2}{a} \left(\sqrt{1 + \left(\frac{at}{c}\right)^2} - 1 \right)$$

I did a quick graph of this (setting the acceleration to a randomly selected value of 10$^6$ms$^{-2}$): For small times you get the usual quadratic dependance of distance on time, but relativistic effects quickly become significant and at large times the line becomes a straight line with gradient $c$ i.e. the speed we observe approaches $c$ asymptotically.

Anyhow, from time zero the distance increases smoothly from 1,000,000km i.e. it never decreases so the rocket never appears to move towards you. The length of the rocket will Lorentz contract, but the distance from you to the rocket does not.