Bell's Spaceship Paradox Refuting Special Relativity
Two spaceships with a distance of one light year (1 LY) between each other are so positioned that both are at rest at a distance of 20 LY from the sun. Because both spaceships are in the same inertial frame as the sun, the spaceship clocks can by synchronized with a clock on the sun.
The paradox arises if both spaceships start accelerating in direction of their tie line (tangentially to the sun) in exactly the same way. At an acceleration of 40 m/s2 relative to the sun a spaceship reaches 60% of the speed of light and covers a distance of half a light month in around 52 days. From the earth the movement of the two spaceships can be observed with a delay of 20 years.
Inertial frame S
¦ 1991-1-1
V Spaceship 2
1991-1-1 ¦ v = 0
o ¦
our solar ¦ 1991-1-1
system V Spaceship 1
¦ v = 0
¦
Situation before acceleration: both spaceships are at rest relative to our solar system; date and time can be synchronized.
Acceleration phase: Both spaceships accelerate at the same time during 52 days at 40 m/s2 (all values with reference to frame S).
Inertial frame S Inertial frame S'
¦
¦ 1991-2-18
1991-2-22 V Spaceship 2
o ¦ v = 0.6 c
our solar ¦
system ¦ 1991-2-18
V Spaceship 1
¦ v = 0.6 c
Situation after acceleration: Both spaceships show same date and time and move at 60% of c relative to our solar system.
Question concerning relativistic length contraction: Has the distance between the spaceships decreased in frame S or increased in S'?
Does this so-called Bell's spaceship paradox refute Special Relativity? We can clear up this question by examining why this theory predicts that a 1 LY long body moving at a distance of 20 LY at 0.6 c appears shorter to observers on the earth.
If Dx, Dt, etc. denote coordinate differences, the Lorentz transformation takes in the case of a relative speed of 0.6 c this form:
Dx' = 1.25 (Dx - 0.6 c Dt) [ 1a ]
Dt' = 1.25 (Dt - O.6/c Dx) [ 1b ]
Dx = 1.25 (Dx' + 0.6 c Dt') [ 2a ]
Dt = 1.25 (Dt' + O.6/c Dx') [ 2b ]
The proper length Dx' of the body in the moving frame S' is 1 LY. A superficial examination could suggest that equation [2a] entails an elongation by the factor 1.25 (resulting in 1.25 LY) for the frame S. But correct is exactly the contrary: a contraction to 0.8 LY (1 LY/1.25).
The elongation is correct if simultaneity in frame S' is assumed (i.e. Dt' = 0). However, what is simultaneous in this frame is not simultaneous in S. Equation [2b] entails that time displacement in S, assuming simultaneity in S', is a function of the spatial coordinate Dx':
Dt(Dx') = 1.25 [O.6/c Dx'] [ 3 ]
For the 1 LY long body (i.e. Dx' = 1 LY) the formula gives a displacement of 0.75 years or nine months. Observing both ends of the body simultaneously in frame S' corresponds in frame S to a length of 1.25 LY, but e.g. 1991-2-22 at the back and 1991-11-22 at the front.
For the actual length in frame S, simultaneity in this frame is required. Because the front end (1991-11-22) has moved at 0.6 c nine months longer than the back end (1991-2-22), the distance covered only by the front end must be subtracted from the 1.25 LY.
length = 1.25 LY - [0.75 years * 0,6 c] = 0.8 LY [ 4 ]
After having understood this principle of length contraction the reasons of the above paradox become obvious. Before the acceleration phase, simultaneity in the frame of the sun entailed simultaneity in the frame of the two spaceships. The identical acceleration phases of both spaceships guarantee (in frame S) that in both spaceships the same time passes. The result is a situation inconsistent with the Lorentz transformation: what is simultaneous in frame S must also be simultaneous for the spaceships in spite of being located at two different points of frame S'.