Time Dilation and relative velocity, 2 girls, 3 atomic clocks?

Question

Right, Time dilation due to velocity has been experimentally demonstrated when a airline took an atomic clock up traveled at speed and then came back down again comparing the atomic clock with one on the ground it was measurably behind the control clock.

So... Girl 1 takes her atomic clock on one plane and girl 2 takes hers on another and the control clock stays on the ground. The two planes travel away from each other at the same velocity so to the control clock both plane-bound clocks should fall behind by the same amount but to girl 1 who accepts her position as reference zero girl 2 is traveling at a greater velocity and to girl 2 who accepts her position as reference zero girl 1 is traveling at a greater velocity. Both girls should expect that when they meet to compare clocks that the other girls is behind by an amount.

when they do meet how can they both see the others clock as being behind and will a ground observer see both as being equally behind the control.

Informed answers only please I will call you an idiot if you fail to understand the question.

Answer 1

That's a great question. The short answer is this: The concept that the "other ('moving') clock runs slow" applies only in the case where the observer is in an inertial reference frame (i.e. is not accelerating; or more generally, does not change their state of motion). There's also an effect due to gravity; but let's ignore that for now.



In the experiment you describe, presumably both girls motions are mirror images of each other (i.e. relative to the control clock they both leave at the same time, go the same speed and same distance, turn around at the same time, and arrive back at the same moment.)



Now let's say that each observer is sending out a radio "tick" each second. The other observers can count these ticks, and do a calculation (based on the tick frequency, relative speed, distance, etc.) to figure out the "current time" on the other clocks ("current time" in quotes because that is a relative quantity that not all observers will agree on).



Girl 1 observes the following (based on "tick" calculations):



1. On her (constant speed) outbound leg, when "T" represents the time on her own clock, she finds the control clock is lagging behind T, and Girl 2's clock is lagging still farther behind T. Note this is her conclusion even AFTER she accounts for the travel time of the "tick" signals.



2. At a certain point, she decellerates, stops, and accelerates in the direction back toward the clock. While this is going on, she observes that the control clock is suddenly ticking much FASTER than her own, and in fact it passes up her own clock's time (according to her calculations). She notes that Girl 2's clock does the same thing, though at a different rate. By the time Girl 1 has accelerated back up to speed, she finds that BOTH the control clock and Girl 2's clock are now AHEAD of her own.



3. As she continues at constant speed back toward the control clock, she observes that both the control clock and Girl 2's clock are ticking slower than her own. However, the readings on those clocks are still AHEAD of her own clock's reading.



4. By the time she gets back to the control clock, she finds that the control clock, though it had slowed down, is still reading ahead of her own. Meanwhile, Girl 2's clock has slowed down so much that it now reads the same as her own (Girl 1's) clock.



And Girl 2 records an exactly symmetrical chain of events. All three observers DISAGREE about whose clock was running fast or slow, and whose clock was ahead or behind at any particular instant. But they all agree that, once they were all back together again, the two girls' clocks both had the same reading, which was behing the control clock's reading.

Answer 2

"Time Dilation and relative velocity, 2 girls, 3 atomic clocks?"



Need a third girl... at rest.



"So... Girl 1 takes her atomic clock on one plane and girl 2 takes hers on another and the control clock stays on the ground. The two planes travel away from each other at the same velocity so to the control clock both plane-bound clocks should fall behind by the same amount but to girl 1 who accepts her position as reference zero girl 2 is traveling at a greater velocity and to girl 2 who accepts her position as reference zero girl 1 is traveling at a greater velocity."



Correct. Clock ticks from all clocks could be correctly anticipated with "relativistic Doppler shift".



"Both girls should expect that when they meet to compare clocks that the other girls is behind by an amount."



False. They can count the ticks, and know exactly what the clocks will say. So how are they going to meet, if they are going in opposite directions from a central clock? Around the world?



"when they do meet how can they both see the others clock as being behind and will a ground observer see both as being equally behind the control."



They agree their clocks will agree, because they could count the clock ticks. And the rest observer / clock agrees with both of them.



"Informed answers only please I will call you an idiot if you fail to understand the question."



Yes, that will really encourage participation.

Answer 3

It has to do with the person observing the other person. So if both person A and B are moving away from each other, then remember that person A is observing the "light coming from person B", not actually observing person B. If that makes sense. The light from the clock ticks takes longer to reach person A, because the plane has moved away a certain distance between ticks. Here is an article (below) which may help you.