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Originally Posted by Sam5
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Originally Posted by SeanF
So let's take a little quiz.
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I’m still waiting for you to answer my earlier questions: Please tell us how you came up with the 36 seconds to advance the A clock at the beginning of one of your earlier thought experiments. In the first half of the experiment you had A and C set at 00:00 at the beginning, and in the second half you had A set at 00:36 while you had C set at 00:00. They weren’t moving relative to each other in either half of the experiment. This advance for the A clock gave a false reading at the end of your experiment and made it appear that observer B saw A’s clock ticking rapidly, whereas, according to SR theory, he actually saw the A clock ticking more slowly. |
It takes all the fun out of it if you don't do it yourself, Sam5, but since you insist...
"Light is always propogated in empty space with a definite velocity
c which is independent of the state of motion of the emitting body." - Albert Einstein
We have an emitter/receiver with a mirror. A pulse of light is sent from the emitter, bounces off the mirror, and returns to the receiver. A clock on the emitter/receiver times the duration of the light's round-trip.
Since the velocity of light is
c regardless of the motion of the emitting body, the velocity of light relative to the emitter & mirror will be
c-v and
c+v, where v is the velocity of the emitter assembly relative to the observer.
So, t1 = d/(
c-v) and t2 = d/(
c+v), where t1 is the time from the emitter to the mirror, t2 is the time from the mirror back to the receiver, and d is the distance between the emitter/receiver and the mirror. t1+t2=t, of course, where t is the total round-trip time.
Now, if v=0, meaning the emitter assembly is motionless relative to the observer, then both
c-v and
c+v are equal to
c. This means t2=d/
c and t1=d/
c, so t1=t2. Therefore, t=t1+t2 becomes t=t1+t1, then t=t1*2, and finally t1=t*0.5. So, the light pulse spends half the total round-trip time on the way out and half on the way back.
Now, if v=0.6
c, then
c-v=0.4
c and
c+v=1.6
c. This means t2=d/1.6
c and t1=d/0.4
c. From that last equation, d=t1*0.4
c. Plugging that in for d in the next-to-last equation gives us t2=(t1*0.4
c)/1.6
c, which simplifies to t2=t1*0.25. Therefore, t=t1+t2 becomes t=t1+t1*0.25, or t=t1*1.25, or t1=t*0.8. So, the light pulse spends four-fifths (80%) the total round-trip time on the way out and only one-fifth (20%) on the way back.
If the round-trip time takes 120 seconds, then an observer stationary to the system says the light took 60 seconds out and 60 seconds back. An observer who sees the system moving at 0.6
c relative to himself says the light took 96 seconds out and 24 seconds back. 96-60 = 36 = 60-24. Therefore the "moving" observer concludes the clock is 36 seconds later when the pulse hits the mirror than the "stationary" observer does.
Specifically, if a clock at the emitter reads 11:59:00 when the pulse leaves and 12:01:00 when the pulse returns, the "stationary" observer would conclude the light hit the mirror when that clock read 12:00:00. However, an observer moving relatively at 0.6
c says the light would hit the mirror when that clock read 12:00:36.
If a clock located
at the mirror says 12:00:00 when the light hits the mirror, the "stationary" observer would conclude the two clocks are synchronized. The "moving" observer, however, would conclude that the emitter clock is 36 seconds ahead of the mirror clock.
So from this we can conclude that two clocks located one light-minute apart (which would be the required distance for the pulse to take two minutes round trip) which are synchronized in the "stationary" frame would be off from each other by 36 seconds in the relatively "moving" frame.
The fact that 36 seconds is exactly what I
needed to set the clocks off by to make the final numbers work out is not coincidence or something I "fudged." It is a result of the fact that Special Relativity is internally consistent and so it
will all work out in the end.
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Originally Posted by Sam5
And also, I asked you to tell us the start times on all your clocks, A, B, and C, as seen by their own observers, at the beginning of your last thought experiment. And I would like to know the speed of B relative to C.
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When adding relativistic velocities, v = ( v1 + v2 ) / ( 1 - ( v1 * v2) / (
c *
c ) ), where v is the total velocity, v1 and v2 are the separate original velocities, and
c, of course, is the speed of light. Plugging 0.6
c in for both v1 and v2 results in approximately 0.88
c (unless I did my math wrong), which should be the velocity of B relative to C and vice-versa.
You've got to know that equation (or know where to look it up
) if you're going to deal with SR...
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Originally Posted by Sam5
Some of the readers of this thread might not be as familiar with the SR numbers as we are, so when you artificially set your clocks to different times at the beginning of your thought experiments, so that you can get their times to work out the way you want them too later in the thought experiments, I think you need to tell everyone what those different beginning set-times are and why you used those particular numbers. Otherwise you might leave people thinking all the start times are 00:00 or 12:00:00, and I’m sure you wouldn’t want to give anyone a false impression or a misconception about any of your clock start times or your own thought experiments.
And when you have two different relative velocities in your thought experiments, please let us know what both of them are so we can run them through the Lorentz Transformation equation to find out what A, B, and C actually “see” on their own clocks and the other clocks.
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The explanation I gave above for 36 seconds is the basis for the simultaneity issue in SR. At the point when A & B pass each other in this experiment (which we can consider to be the "start of the experiment"), C is quite some distance away, and A, B, and C are all three in different inertial frames. Therefore, while it's easy and universal to say that A & B were both at 12:00:00 when they passed each other, the time displayed on C at that same moment (in other words, what C-clock "tick" event is simultaneous with the A-B-passing event) will depend on which of the three inertial frames you're looking from.
This will also be true for the question of what time is displayed on A when B & C pass each other, and also for the question of what time is displayed on B when A & C pass each other.