yogi said:
Ok - if you go ahead and analyse it in the moving frame - what will you get - will you get the same result that the moving clock is behind the clock in the stationary frame - let's see what you come up with
OK, Einstein does not give any specific numbers in his problem, but let's say that in the original rest frame of the two clocks, they are 12 light-seconds apart, with the clock 1 at position x=0 and clock 2 at position x=12 l.s. Assume they are also synchronized in this frame. Then when clock 1 reads time t=0 seconds, it accelerates instantaneously to a velocity of 0.6c in the direction of clock 2, and travels at this velocity until it reaches clock 1 at t=12/0.6=20 seconds in the coordinates of this frame. At the moment they meet, clock 2 will of course read 20 seconds since this is its rest frame, but clock 1 will have been ticking at \sqrt{1 - 0.6^2} = 0.8 the normal rate, so it will only have ticked 0.8*20 = 16 seconds. Thus when they meet, clock 1 reads 16 seconds and clock 2 reads 20 seconds, according to this frame's prediction.
So, let's consider the coordinates of the following 4 events as seen in this frame:
EVENT A -- clock 1 reads 0 seconds and accelerates: x=0 l.s., t=0 s
EVENT B -- clock 2 reads 0 seconds: x=12 l.s., t=0 s
EVENT C -- clock 2 reads 7.2 seconds: x=12 l.s., t=7.2 seconds (the reason I picked this event will become apparent later)
EVENT D -- clock 1 and clock 2 meet, with clock 1 reading 16 s and clock 2 reading 20 seconds: x=12 l.s, t=20 s
In this frame, events A and B are simultaneous, because they both happen at the same coordinate time, while C is not simultaneous with either of them.
But now, let's transform events A, B, and C into an inertial frame moving at 0.6c relative to the first one, in the same direction that clock 1 moves afte accelerating. In this case the Lorentz transform for finding the coordinates of events in this frame would be:
x' = 1.25*(x - 0.6c*t)
t' = 1.25*(t - 0.6*x/c)
So, the coordinates of event A would be: x'=0 l.s., t'=0 s
The coordinates of event B would be: x'=15 l.s., t'=-9 s
And the coordinates of event C would be: x'=9.6 l.s., t'=0 s
So you can see that in
this frame, it is events A and C that are simultaneous, while event B happened prior to both of them--in other words, at the "same moment" that clock 1 read "0 seconds" and accelerated towards clock 2, clock 2 was reading 7.2 seconds (that was the definition of event C, remember) and was at a distance of 9.6 l.s. away in this frame. And once clock 1 accelerates, it is at rest in this frame, while clock 2 is moving towards it at 0.6c. So in this frame, we conclude that it will take 9.6/0.6 = 16 seconds for clock 2 to reach clock 1 (and if you map event D into this frame using the Lorentz transform, you find it does indeed have coordinates x'=0 l.s., t'=16 s in this frame).
Now,
if we assume the laws of physics work exactly the same in this frame as in all other frames, then since clock 1 is at rest in this frame it should be ticking at a normal rate, so it should have elapsed 16 seconds between the time it comes to rest and the time the two clocks meet; but since clock 2 is moving at 0.6c in this frame, it should be slowed down by a factor of 0.8, meaning it will have elapsed only 16*0.8 = 12.8 seconds between the moment clock 1 accelerates and the time they meet. But we already figured out that the time of clock 1 accelerating was simultaneous with the event of clock 2 reading 7.2 seconds in this frame (ie they were not synchronized to begin with), so it should read 7.2 + 12.8 seconds = 20 seconds when they meet.
So, to sum up: in the first frame, both clocks read a time of 0 when clock 1 accelerates, and it takes 20 seconds of coordinate time for them to meet, so that the slowed-down clock 1 only reads 16 seconds while the at-rest clock 2 reads 20 seconds when they meet. But in the second frame, clock 1 reads 0 but clock 2 reads 7.2 seconds at the moment clock 1 accelerates, and it takes 16 seconds of coordinate time for them to meet, so clock 1 elapses 16 seconds while the slowed-down clock 2 only elapses 12.8 seconds, meaning that it reads 7.2+12.8 = 20 seconds when they meet. So in both frames you predict that clock 1 reads 16 seconds and clock 2 reads 20 seconds when they meet, even though the two frames disagree about which was running slower as they approached each other, and they also disagree about simultaneity (ie what clock 2 read 'at the same moment' that clock 1 accelerated).
