Relativity Q

[losfomot]

Every once in a while I think of a scenario that I can't figure out in my head. I'm sure the solution is simply something I missed, but here it is:

 

2 ships in Earth's reference frame... we'll call it 'frame 1'. So Two ships leave Frame 1 and accelerate in a given direction: (we'll call it:) North. They accelerate together until they are traveling with a velocity of .99c relative to frame 1... So they have a relativistic factor of about 7... which means that (among other things) their clocks are now running 7 times slower than frame 1 clocks. The ships are at rest relative to each other and we'll call their current reference frame 'frame 2' In the next paragraph, we'll ignore frame 1 for a moment...

 

Ship 1 decides to accelerate away from frame 2 (leaving ship 2 in frame 2). Ship 1 accelerates away in a given direction: (we'll call it:) South. Ship 1 accelerates in that direction until it reaches a velocity of .99c relative to frame 2. According to relativity, because ship 1 accelerated away from frame 2 and is now coasting at .99c relative to frame 2, Ship 1's clock should be running 7 times slower than ship 2's clock.

 

Now we bring frame 1 back into it... if frame 2's (ship 2's) clock is running 7 times slower than frame 1's (Earth's) clock and Ship 1's clock is running 7 times slower than frame 2's (ship 2's) clock, then Ship 1's clock should be running 49 times slower than frame 1's (Earth's) clock...

 

But Ship 1's second acceleration was in the opposite direction of it's first acceleration, which means that Ship 1 should be back in Frame 1 (roughly the same frame of reference as Earth)

 

What am I missing here?

[timo]

You're probably not properly taking into account that velocity and acceleration are vectors, not just some positive scalar numbers. In case you are familiar with linear algebra, look up the matrix form of the Lorentz transformation and verify [math]\Lambda(-v) \Lambda(v) = 1 \neq \Lambda(2v) [/math].

[swansont]

Because the tranforms are not linear, you can't "daisy-chain' them in this fashion (which is another version of what Atheist said). You either have to compare two frames directly or use the correct velocity addition formula

 

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c1

[losfomot]

Yes, I understand what the actual relative velocities would be. My problem is with the no preferred frame part. I thought that, as the two ships traveled together with a velocity of .99c away from Earth, they could reset their speedometers (and directionometers) to zero, and assume that they were at rest, regardless of what Earth or the rest of the universe was doing. If this was the case, then it wouldn't matter what direction Ship 1 went in when it decided to leave ship 2. Ship 1's time would be dilated relative to ship 2, because ship 1 was the ship doing the accelerating. So ship 1's clock should run slower than ship 2's clock.

 

But, if the direction ship 1 chose happened to be back toward Earth, then the opposite would happen... ship 2's clock would be running slower.

[swansont]

The dilation is relative between two frames; if you want to know how clock 3 is dilated with respect to clock 1, you don't look at clock 2. You have to look at it from clock 1's frame.

[losfomot]

The dilation is relative between two frames; if you want to know how clock 3 is dilated with respect to clock 1, you don't look at clock 2. You have to look at it from clock 1's frame.

 

I understand that... and I don't feel like I'm mixing frames... (sigh)... my problem must be here:

 

I thought that, as the two ships traveled together with a velocity of .99c away from Earth, they could reset their speedometers (and directionometers) to zero, and assume that they were at rest, regardless of what Earth or the rest of the universe was doing. If this was the case, then it wouldn't matter what direction Ship 1 went in when it decided to leave ship 2 at .99c. Ship 1's time would be dilated relative to ship 2, because ship 1 was the ship doing the accelerating away from ship 2. So ship 1's clock should run slower than ship 2's clock.

 

we are only dealing with two clocks right now... ship 1's relative to ship 2's... so is the above paragraph wrong?

[swansont]

Ship 1 and ship 2 would each see the other's clock run slow while they are moving at constant v. You'll have a (separate) offset while clock 1 is undergoing an acceleration.

[losfomot]

Ship 1 and ship 2 would each see the other's clock run slow while they are moving at constant v.

 

Yes but this is a doppler effect. Who's clock really is running slower via a relativistic effect? Say ship 1 takes off in some direction at .99c (we'll pretend acceleration is nearly instantaneous just for simplicity) cruises for an hour (ship 1 time) stops (relative to ship 2), turns around and comes back toward ship 2 at .99c for another hour (ship 1 time), and then stops next to ship 2 and compares clocks. Since we are pretending the acceleration periods were instantaneous, a total of 2 hours have elapsed for ship 1... has more or less time elapsed for ship 2?

[Janus]

Yes but this is a doppler effect.

No, this is after the Doppler effect is factored out.

Who's clock really is running slower via a relativistic effect?

at any given point this depends on which ship you are making the determination from.

Say ship 1 takes off in some direction at .99c (we'll pretend acceleration is nearly instantaneous just for simplicity) cruises for an hour (ship 1 time) stops (relative to ship 2), turns around and comes back toward ship 2 at .99c for another hour (ship 1 time), and then stops next to ship 2 and compares clocks. Since we are pretending the acceleration periods were instantaneous, a total of 2 hours have elapsed for ship 1... has more or less time elapsed for ship 2?

 

More time has elasped for ship 2. But why this is so depends on which ship is made to make the detemination.

 

From ships 2's perspective, the following sequence of the events takes place.

 

1.Ship 1 speeds off at .99c and thus its clock runs slow by a factor of 7.

2.Ship 2 instantly turns around when it is .99 light hours distant and comes

back at .99c, its clock still running slow.

3. Ship 1 returns to Ship 2 after 2 hrs Ship 2 time, having aged only 2/7 of an hr by Ship 1's clock.

 

From Ship 1's perspective.

1. Ship 1 undergoes an instantaneous acceleration and the Relative velocity between ship 1 and 2 becomes .99c, at this point clock 2's clock runs slow by a factor of 7.

2. After 1/7 of an hour has passed on its own clock, Ship 2 undegoes a second instantanous acceleration, after which it has a .99c velocity towards ship 2. At the instant of acceleration ship 1 will determine that ship 2's clock has jumped forward in time by 1 47/49 hours, then runs slow by a factor of 7 thereafter.

3. Ship 1 and ship 2 meet up again in another 1/7 of an hour by Ship 1's clock. total time elasped on Ship 1's clock = 1/7hr + 1/7hr = 2/7hr.

Total time elasped on Ship 2's clock = 1/49hr + 1 27/49hr + 1/49hr = 2 hrs.

 

This jump forward that Ship1 sees in ship 2's clock is due to the fact that when Ship 1 goes from .99c away from ship 2 to .99c towards ship 2, it changes interial frames of reference.

[losfomot]

This jump forward that Ship1 sees in ship 2's clock is due to the fact that when Ship 1 goes from .99c away from ship 2 to .99c towards ship 2, it changes interial frames of reference.

 

Yeah, right... who are you selling this stuff to?

 

Just kidding, it all makes sense to me now. I was assuming that: when ship 1 accelerates away from ship 2 at .99c, ship 1 is now standing still relative to Earth (which is correct). But I was also assuming that, to cruise back to ship 2 at .99c, ship 1 would be again moving away from Earth at .99c. This is incorrect. Ship 1 would actually be moving at .99995c relative to Earth, in order to catch up to ship 2 at .99c (relative to ship 2). Man this stuff can get confusing!

[HannonRJ]

Have any of you done a rigorous analysis of any of Einstein's three (?) derivations of his SR tansformation equations?