# Special relativity puzzle

Tags:
1. Aug 10, 2014

### JanEnClaesen

Two spaceships approach an observer from an equal distance and from an opposite direction with an equal speed in the observer's intertial reference frame O.
The speed of a spaceship in the intertial reference frame of the other spaceship S is 0.8c , what is the speed of one of the spaceships in O?

I proceeded as follows:

Let 2l' be the distance between the two spaceships in S.
In S the two spaceships will collide after a time t' = 2.5l'/c

Let F be the squareroot of 1 - v²/c² with v the sought speed
In O the two spaceships will collide when vt = l or vFt' = l'/F (O has to correct for what he perceives as the time dilations and space contractions of the measurements made in S)

Substituting we get the equation x(1-x²)=0.4 with x = v/c, if you solve the equation you conclude that this line of reasoning was wrong (but when we replace 0.4 by 0.375 we do get the right solution, which is 0.5c).

What happened?

Last edited: Aug 10, 2014
2. Aug 10, 2014

### Staff: Mentor

3. Aug 10, 2014

### Staff: Mentor

You forgot to allow for relativity of simultaneity. It's not sufficient to allow for time dilation and length contraction, you also have to remember that if the two spaceships are the same distance from O at the same time according to O, they won't be according to a spaceship observer.

Thus, if the two ships are approaching the observer from opposite directions at the same speed from the same distance in the observer's frame... Then in the spaceship frame they won't be equidistant from the observer and if the distance between the ships is 2l' the distance to the observer won't be l'.