# Special relativity particle observation puzzle

1. Jul 7, 2007

### StatusX

Here's a little special relativity puzzle that I found interesting.

Say we have two particles initially at rest in a frame O and separated by a distance L. They begin to uniformly accelerate at t=0 in a direction along the line separating them, until they reach a velocity v at some time t, and then return to inertial motion. Say the first particle follows a path x(t) in O. Then since both particles undergo the same motion, the second particle follows the path x(t)+L. Thus at any time, the distance measured in O between the two particles is L. But this means that in the final rest frame of the particles, the separation has expanded to $\gamma$ L. What's going on? What happens to the separation as observed by the two particles during this process?

2. Jul 7, 2007

### JesseM

I think the answer would have to do with the relativity of simultaneity--if they both start and stop accelerating simultaneously in O, then in any other inertial frame, they started accelerating at different moments and stopped accelerating at different moments, so the distance between them would be constantly changing.

3. Jul 7, 2007

### StatusX

That's true, but a little more analysis shows an interesting effect. Consider what the separation the particles observe during their acceleration.

4. Jul 7, 2007

### pmb_phy

As observed in the reference frame of the trailing particle, assumed for simplicity to be accelerating uniformly, will reckon the leading particle to be higher up in a gravitational field, i.e. the one produced by acceleration. The leading particle in this accelerating frame will be reckoned, due to gravitational time dilation, to having his clock to be running at a higher rate as reckoned by the trailing particle. Thus in the rest frame of the trailing particle the leading particle will be measured to be accelerating in the trailing particle's rest frame and thus the distance measured in that frame will be an increasing function of time.

Pete

5. Jul 7, 2007

### pervect

Staff Emeritus
Assuming that t=0 is defined in frame O, this is known as Bell's spaceship paradox.

see for example:
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken]

The two particles do in fact separate as measured in their own frame. If one desires the particles to maintain a constant distance ('Born rigid motion'), the lead particle must have a lower proper acceleration than the trailing particle.

In the modified version, the lead particle accelerates for a longer proper time at a lower acceleration, winding up with the same velocity as the tailing particle.

In this modified version, the lead particle maintains a constant proper distance from the trailing particle, and they both maintain a constant proper distance from some particular "pivot point".

In the unmodified problem, the particles separate.