# Thought Experiment on Relativity

1. Jan 23, 2010

### MonroeIns

Hey, I am 6 hours into a 12 hour audio course on Relativity. I love the course, but the bad thing is that you obviously can't ask questions. So here is a thought experiment I was hoping people could help me with.

I am an observer standing between two rocket ships, R1 and R2. Each rocket starts flying at .75C in the opposite direction. R2 has a flash light on the rear of the ship shining back toward R1.

1. From the observer's frame of reference how quickly will the rockets be moving away from each other?
2. Will R1 ever see the light from the flash light shining back at it? If so, how quickly will R2 be moving from R1's perspective.

Now let's take the same rocket ships and attach a string to the back of R2 that is wound on a spool which is attached to the back of R1. As the string is pulled off the spool, R1 should be able to calculate how quickly the ships are moving apart based on how quickly the spool turns.

1. What would R1 calculate the speed at which it is moving away from R2?

Thanks a lot for your help!

2. Jan 23, 2010

### jambaugh

From the central observer's frame of reference the distance between the ships is increasing at twice 0.75C No paradox there.
Yes. This is the classic "addition of velocities" problem in SR. Here's my "quick and dirty" answer. In Galilean (i.e. pre-Einsteinian) relativity we parametrize boosts with change in velocities and the velocities add. In SR however boosts are a pseudo-rotation (using hyperbolic trigonometry) of the 4-velocity and thus the correct parameter is a hyperbolic angle. The 3-velocity (sticking to one spatial direction) is now a slope and thus the hyperbolic tangent of a boost parameter (let's use beta). To convert to typical units we must multiply by C. So let's consider what $$R_1$$ sees.

From $$R_1$$'s perspective the central observer is boosted relative to him with boost parameter:
$$\beta : C\tanh(\beta) = 0.75C$$
or
$$\beta = \tanh^{-1}(0.75)\approx 0.973$$

The other ship, $$R_2$$ has been boosted twice as much so from $$R_1$$'s perspective it has a velocity of:
$$C\tanh(2\beta) = C\tanh(1.946)=0.96C$$

Let me see... the string is fixed to $$R_2$$ so it is moving at his same velocity. As above $$R_1$$ will see the string unwinding and feeding out at 0.96C. Its no different than if $$R_1$$ were just flying along the string already laid out behind $$R_2$$. From the central observer's perspective that string is flying by at speed $$0.75C$$, the same speed he see's $$R_2$$ traveling. The resolution of these is a matter of the interplay of time dilation and length contraction.

OK?

3. Jan 23, 2010

### MonroeIns

Thanks a lot! So do we see in the observable universe objects, from our frame of reference, moving away from each other at faster than the speed of light?

4. Jan 23, 2010

### jambaugh

In the additive sense yes. They however are still able from our perspective able to signal each other because we see each emitting signals which travel at the speed of light in our frame and thus (additively) traveling away from the signaler faster than is the recipient.

Nothing weird here. Two light pulses moving away from each other are moving away at speed 2c. If you like that's the relativistic speed limit for two objects relative motion as seen by a third observer.

5. Jan 24, 2010

### GRDixon

My calculator says that jambaugh got it right. Here's another take on 2. Let R1 be at rest in IRF K, and ME (midway between the 2 ships) be at rest in K'. u2', the speed of R2 relative to K', is .75c. And v, the speed of K' relative to K, is also .75c. Thus use the addition of velocities to get the speed of R2 relative to K: u2 = (.75c + .75c)/)(1 + .752) = .96C.

The string will indicate that, relative to R2's rest frame, R1 is moving away at a speed of .96c. If R1 reverse engineers this, he'll find that the other ship is moving away from himself at the same speed. Judicious application of the Lorentz transformation will reveal that either ship is moving away from ME at a speed of .75c.

6. Jan 26, 2010

### DanRay

This is a slightly altered version of an old problem that is said to be the type of problem that led both Poincare and Einstein to adapt the Lorentz Transformations with a slight change from what Lorentz designed them for. Namely to calculate the contraction of "each body of light" as it moved against the ether current. Poincare never did give up believing in the ether but Einstein did. Both Einstein and Poincare believed that somthing like a paradox occured with this type of problem because obviously either rocket moving at .75 c can send light signals to you and you can send them to the other but both declared that the two rockets could never comunicate with each other if their velocities are simply added because that would mean they are moving at 1.5c relative to each other and the light could never catch up. But as long ago as 1959 I realized and proved (not that anybody listened) this is not the case!

When either Rocket sends a light signal that signal propagates from the point in space occupied by the rocket at that moment and proceeds at c from that stationry point toward the other rocket which is only moving .75c. If the rockets syncronize clocks when they pass your position and R1 sends a signal 1 second later it will catch R2 in just 7 seconds and if R2 reflects the signal back to R1 it will arrive at 49 seconds on their clocks. No matter what the two rocket's speeds are if they are less than c the light can always catch up and no transformations are needed. The problem with the Lorentz Tramsformations in this type of problem is that they keep the light tied to the source as it moves not to the point or points in spece at which the light is propagated.

Incidentally if they keep reflecting the light back and forth; by the time it reaches R2 the 5th time it will take 8 years and 39+ days. That's with two leap years.

Last edited: Jan 26, 2010
7. Jan 28, 2010

### MonroeIns

So I have thought about this some more and have one more question. Lets say R2 does not turn on the flash light on the back the rocket ship until a few seconds after it takes off. We have stated that in the frame of reference of R1, R2 is moving at less than the speed of light. So that light should "catch up" to R1. But in the frame of reference of the observer, R1 and R2 are moving away from each other at 1.5C. If this is the case, when the light passes the observer on its way "catching up" to R1, wouldn't he observe the light moving at faster than 1.5C? Otherwise, how would the oberver observe the light "catching up" at all?

Thanks. I appreciate your help.

8. Jan 28, 2010

### GRDixon

The speed of light is independent of the speed of its source. Thus from the observer's point of view, the speed of the light from R2 is c, which is greater than R1's speed of .75c. The light from R2 overtakes R1.

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